【数学题笔记本】——The-math-problems-notebook

Valentin Boju Louis Funar The Math Problems Notebook Birkha¨user Boston • Basel • Berlin Valentin Boju MontrealTech Institut de Technologie de Montreal P.O. Box 78575, Station Wilderton Montreal, Quebec, H3S 2W9 Canada valentinboju@montrealtech.org Louis Funar Institut Fourier BP 74 URF Mathematiques Universite´ de Grenoble I 38402 Saint Martin d’Heres cedex France funar@fourier.ujf-grenoble.fr Cover design by Alex Gerasev. Mathematics Subject Classification (2000): 00A07 (Primary); 05-01, 11-01, 26-01, 51-01, 52-01 Library of Congress Control Number: 2007929628 ISBN-13: 978-0-8176-4546-5 e-ISBN-13: 978-0-8176-4547-2 Printed on acid-free paper. c©2007 Birkha¨user Boston All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkha¨user Boston, c/o Springer Science+Business Media LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbid- den. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 9 8 7 6 5 4 3 2 1 www.birkhauser.com (TXQ/SB) The Authors Valentin Boju was professor of mathematics at the University of Craiova, Romania, until his retirement in 2000. His research work was primarily in the field of geome- try. He further promoted biostatistics and biomathematics as a discipline within the Department of Medicine of the University of Craiova. He left Romania for Canada, where he has taught at MontrealTech since 2001. He was actively involved in coach- ing college students in problem-solving and intuitive mathematics. In 2004, he was honored with the title of Officer of the Order “Cultural Merit,” Category “Scientific Research.” MontrealTech Institut de Technologie de Montréal P.O. Box 78575 Station Wilderton Montréal, Quebec, H3S 2W9 Canada e-mail: valentinboju@montrealtech.org Louis Funar has been a researcher at CNRS, at the Fourier Institute, University of Grenoble, France, since 1994. During his college years, he participated three times in the International Mathematics Olympiads with the Romanian team in the years 1983–1985, winning bronze, gold, and silver medals. His main research interests are in geometric topology and mathematical physics. Institut Fourier, BP 74 Mathématiques UMR 5582 Université de Grenoble I 38402 Saint-Martin-d’Hères cedex France e-mail: funar@fourier.ujf-grenoble.fr To all schoolteachers, particularly to my wonderful parents, Evdochia and Nicolae Boju, who genuinely believed in the education of the younger generation. To my parents, Maria and Ioan Funar, who laid the foundation for our collaboration, as a late reward for their hard work and help, for their enthusiasm, determinedness, and strength during all these years. Preface The authors met on a Sunday morning about 25 years ago in Room 113. One of us was a college student and the other was leading the Sunday Math Circle. This circle, within the math department of the University of Craiova, gathered college students who possessed a common passion for mathematics. Most of them were participating in the mathematical competitions in vogue at that time, namely the Olympiads. Others just wanted to have good time. There were similar math circles everywhere in the country, in which high-school students were committed to active training for math competitions. The highest stan- dard was achieved by the selective training camps organized at the national level, which were led by professional coaches and mathematicians who were able to stimu- late and elicit high performance. To mention a few among the leaders, we recall Dorin Andrica, TomaAlbu, TituAndreescu, Mircea Becheanu, Ioan Cuculescu, Dorel Mihet, and Ioan Tomescu. The Sunday Math Circle shifted partially from its purpose of Olympiad training by following freely the leader’s ideas and thus becoming a place primarily intended for disseminating beautiful mathematics at an elementary level. The fundamental texts were the celebrated book of Richard Courant and Herbert Robbins with the mysterious title What is mathematics?, and the book of David Hilbert and Stefan Cohn–Vossen entitled Geometry and the Imagination. The participants soon realized the differences in both scale and novelty between the competition-type math problems that they encountered every day and the mathematics built up over hundreds of years and engaged in by professional mathematicians. Competition problems have to be solved in a short amount of time, and people compete against each other. These might be highly nontrivial, unconventional prob- lems requiring deep insights and a lot of imagination, but still they have the advantage that one knows in advance that they have a solution. In the real mathematical world, problems often had to wait not merely years but sometimes centuries to be solved. A working mathematician could go for months or years trying to solve a particular prob- lem that had not been solved before and take the risk of bitterly failing. Moreover, the sustained effort needed for accomplishing such a task would have to take into account a delicate balance between the accumulation of knowledge, methods, and tools and viii Preface the creation from scratch of a path leading to a solution. This intellectual adventure is filled with suspense and frustration, since one does not know for sure what one is trying to prove or whether it is indeed true. Real mathematics seemed to many of us to be too far away from competition problems. The philosophy of the Sunday Math Circle was that, in contrast to what we might think, the border between the two kinds of mathematics could be so vague and flexible that at times one could cross it at an early age. The history of mathematics abounds in examples in which a fresh mind was able to find an unexpected solution that specialists had been unable to find. One needs, however, the right training and the unconventional sense of finding the inspiring problem. Eventually, once a solution has been found, mathematicians are then willing to try to understand it even better, and other solutions follow in time, each one simpler and clearer than the previous one. In some sense, once solved, even the hardest problems start losing slowly their aura of difficulty and eventually become just problems. Problems that are today part of the curriculum of the average high-school student were difficult research problems three hundred years ago, solved only by brilliant mathematicians. This phenomenon demonstrates the evolution that language and science have undergone since then. The authors conceived the present book with the nostalgia of the “good old times" of the Sunday Math Circles. We wanted something that carries the mark of that philos- ophy, namely, a number of challenging math problems for Olympiads with a glimpse of related problems of interest for the mathematician. The present text is a collection of problems that we think will be useful in training students for mathematical com- petitions. On the other hand, we hope that it might fulfill our second goal, namely, that of awakening interest in advanced mathematics. Thus its audience might range from college students and teachers to advanced math students and mathematicians. The problems in each section are in increasing order of difficulty, so that the reader give some of the problems a try. We wanted to have 25% easy problems concerning basic tools and methods and consisting mainly of instructional exercises. The beginner might jump directly to the solutions, where a concentrate of the general theory and basic tricks can be found. The largest chunk contains about 50% problems of medium difficulty, which could be useful in training for mathematical competitions from local to international levels. The remaining 25% might be considered difficult problems even for the experienced problem-solver. These problems are often accompanied by comments that put the results in a broader perspective and might incite the reader to pursue the research further. The problems serve as an excuse for introducing all sorts of generalizations and closely related open problems, which are spread among the solutions. Some of these are truly outstanding problems that resisted the efforts of mathematicians over the centuries, such as the congruent numbers conjecture and the Riemann hypothesis, which are among seven Millennium Prize Problems that the Clay Mathematics Insti- tute recorded as some of the most difficult issues with which mathematicians were struggling at the turn of the second millennium and offered a reward of one million dollars for a solution to each one. In mathematics the frontier of our knowledge is still open, and it abounds in important unsolved problems, many of which can be Preface ix understood at the undergraduate level. The reader might be soon driven to the edge of that part of mathematics where he or she could undertake original research. We drew inspiration from the spirit of the famous books by R. K. Guy and his collaborators. Nevertheless, our aim was not to build up a collection of open questions in elementary mathematics but rather to offer a journey among the basic methods in problem-solving. Developing intuition and strengthening the most popular techniques in mathematical competitions are equally part of the goal. In the meantime, we offer the enthusiastic reader a brief glimpse of mathematical research, a place where problems yet unsolved long for deliverance. The present collection of problems evolved from a notebook in which the second- named author collected the most interesting and unconventional problems that he encountered during his training for mathematical competitions in the 1980s. In the tradition of the Romanian school of mathematical training, he encountered problems inspired by both Russian Olympiads and American Competitions. Some (if not most) of the problems have already appeared elsewhere, especially in Kvant, Matematika v Shkole, American Mathematical Monthly, Elemente der Mathematik, Matematikai Lapok, Gazeta Matematica, and so, in a sense, the collection gained a certain cos- mopolitan flavor. We have given the source in the problem section, when known, and more detailed information in the comments within the solutions part. The original set of problems is complemented by more basic exercises, which aim at introducing many of the tricks and methods of which the competitor should be aware. Years later, we followed the destiny of some of the most intriguing questions from the notebook. Some of them led to developments that are well beyond the scope of this book, and we decided to outline a few of them. We have supplied a short glossary containing some less usual definitions and identities in the geometry of triangles and the solution of Pell’s equation. In order to help the reader find his or her way through the book, we have provided both an index concerning all mathematical results needed in the proofs, which are usually stated at the place where they are used first, and an index of mathematical terms at the end of the book. The authors have benefited from discussions, corrections, help, and feedback from several people, whom we want to thank warmly: Dorin Andrica, Barbu Berceanu, Roland Bacher, Maxime Wolff, Mugurel Barcaˇu, Ioan Filip, and Simon György Szatmari. We also thank Ann Kostant, Editorial Director, Springer, and Avanti Paranjpye, Associate Editor, Birkhäuser Boston, for their productive suggestions. One of them led to the “Index of Topics and Methods,” which might be useful for a better reception of the book by readers. We are very grateful to Elizabeth Loew, our Production Editor, for her patience and dedication to accuracy and excellence. Finally, we are thankful to David Kramer for his thorough copyediting corrections. Valentin Boju and Louis Funar Montreal, Canada and Saint-Martin-d’Hères, France July 2006 Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Part I Problems 1 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2 Algebra and Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.2 Algebraic Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.3 Geometric Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.1 Synthetic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.2 Combinatorial Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.3 Geometric Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Part II Solutions and Comments to the Problems 5 Number Theory Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 6 Algebra and Combinatorics Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 6.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 6.2 Algebraic Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 6.3 Geometric Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 7 Geometry Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 7.1 Synthetic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 7.2 Combinatorial Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 xii Contents 7.3 Geometric Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 8 Analysis Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 9 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 9.1 Compendium of Triangle Basic Terminology and Formulas . . . . . . . 217 9.1.1 Lengths in a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 9.1.2 Important Points and Lines in a Triangle . . . . . . . . . . . . . . . . . 218 9.1.3 Coordinates in a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 9.2 Appendix: Pell’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 Index of Mathematical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Index of Mathematical Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Index of Topics and Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 Part I Problems 1 Number Theory Problem 1.1. Show that we have Ckn ≡ 0 (mod 2) for all k satisfying 1 ≤ k ≤ n − 1 if and only if n = 2β , where β ∈ Z∗+. Here Ckn denotes the number of combinations, i.e., the number of ways of picking up a subset of k elements from a set of n elements. Known also as the binomial coefficient or choice number and sometimes denoted as( n k ) it is given by the formula Ckn = ( n k ) = n! k!(n − k)! where the factorial n! represents n! = 1 × 2 × 3 × · · · × n. Problem 1.2. Let P = anxn + · · · + a0 be a polynomial with integer coefficients. Suppose that there exists a number p such that: 1. p does not divide an; 2. p divides ai , for all i ≤ n − 1; 3. p2 does not divide a0. Then P is an irreducible polynomial in Z[x]. Problem 1.3. Given mi, bi ∈ Z+, i ∈ {1, 2, . . . , n}, such that gcd(mi,mj ) = 1 for all i �= j , there exist integers x satisfying x ≡ bi (mod mi) for all i. This result is usually known as the Chinese remainder theorem. Problem 1.4. If gcd(a,m) = 1, then it is a classical result of Euler that we have the following congruence: aϕ(m)+1 ≡ 0 (mod m), where ϕ(m) is the Euler totient function, which counts how many positive integers smaller than m are relatively prime to m. Prove that this equality holds precisely for those numbers a,m such that for any prime number p that divides a, if pk divides m then pk also divides a. 4 1 Number Theory Problem 1.5. (Kvant) Let p > 2 be a prime number and ak ∈ {0, 1, . . . , p2 − 1} denote the value of kp modulo p2. Prove that p−1∑ k=1 ak = p 3 − p2 2 . Problem 1.6. If k ∈ Z+, then show that [√ k2 + 1 + · · · + √ k2 + 2k ] = 2k2 + 2k, where [x] denotes the integer part, i.e., the largest integer smaller than x. Problem 1.7. (Amer. Math. Monthly) If n is not a multiple of 5, then P = x4n + x3n + x2n + xn + 1 is divisible by Q = x4 + x3 + x2 + x + 1. Problem 1.8. (Amer. Math. Monthly) Let p be an odd prime and k ∈ Z+. Show that there exists a perfect square the last k digits of whose expansion in base p are 1. Problem 1.9. Any natural number greater than 6 can be written as a sum of two numbers that are relatively prime. Problem 1.10. (International Math. Olympiad) Prove that it is impossible to extract an infinite arithmetic progression from the sequence S = {1, 2k, 3k, . . . , nk, . . . }, where k ≥ 2. Problem 1.11. (Amer. Math. Monthly) Prove that ba−j+1 dividesCjba if a, b ≥ 2, j ≤ a + 1. Problem 1.12. Solve in integers the following equations: (1) x2 = y2 + y3; (2) x2 + y2 = z2. Problem 1.13. Let a, b, c, d ∈ Z+ be such that at least one of a and c is not a perfect square and gcd(a, c) = 1. Show that there exist infinitely many natural numbers n such that an + b, cn + d are simultaneously perfect squares if one of the following conditions is satisfied: 1. b and d are perfect squares; 2. a + b, c + d are perfect squares; 3. a(d − 1) = c(b − 1). Moreover, there do not exist such numbers if a = 1, b = 0, c = 4k2 − 1, d = 1. Problem 1.14. If N = 2 + 2√28n2 + 1 ∈ Z for a natural number n, then N is a perfect square. Problem 1.15. Let n ≥ 5, 2 ≤ b ≤ n. Prove that [ (n − 1)! b ] ≡ 0 (mod b − 1). 1 Number Theory 5 Problem 1.16. (Amer. Math. Monthly) Prove that for every natural number n, there exists a natural number k such that k appears in exactly n nontrivial Pythagorean triples. Problem 1.17. (Amer. Math. Monthly) Let n, q ∈ Z+ be such that all prime divisors of q are greater than n. Show that (q − 1)(q2 − 1) · · · (qn−1 − 1) ≡ 0 (mod n!). Problem 1.18. Every natural number n ≥ 6 can be written as a sum of distinct primes. Problem 1.19. Every number n ≥ 6 can be written as a sum of three numbers that are pairwise relatively prime. Problem 1.20. (Romanian training camp, Sinaia 1984) Find all pairs of integers (m, n) such that Cnm = 1984, where Cnm = m!n!(m−n)! denotes the usual binomial coefficient. Problem 1.21. (Romanian training camp, Sinaia 1984) Find the set A consisting of natural numbers n that are divisible by all odd natural numbers a with a2 < n. Problem 1.22. Prove that 21092 − 1 is divisible by 10932. Problem 1.23. (Amer. Math. Monthly) Let n ≥ 0, r > 1, and 0 < a ≤ r be three integers. Prove that the number n, when written in base r , has precisely ∞∑ k=1 [ nr−k + ar−1 ] − [ nr−k ] digits that are greater than or equal to r − a. Problem 1.24. Find a pair (a, b) of natural numbers satisfying the following proper- ties: 1. ab(a + b) is not divisible by 7. 2. (a + b)7 − a7 − b7 is divisible by 77. Problem 1.25. (International Math. Olympiad 1984) Let 0 < a < b < c < d be odd integers such that 1. ad = bc 2. a + d = 2k, b + c = 2m, for some integers k and m. Prove that a = 1. Problem 1.26. (Amer. Math. Monthly) Find those subsets S ⊂ Z+ such that all but finitely many sums of elements from S (possibly with repetitions) are composite numbers. 6 1 Number Theory Problem 1.27. (Amer. Math. Monthly) Prove that for any natural n > 1, the number 2n − 1 does not divide 3n − 1. Problem 1.28. (Putnam Competition 1964) Let un be the least common multiple of the first n terms of a strictly increasing sequence of positive integers. Prove that ∞∑ n=1 1 un ≤ 2. Find a sequence for which equality holds above. Problem 1.29. (Amer. Math. Monthly) Let ϕn(m) = ϕ(ϕn−1(m)), where ϕ1(m) = ϕ(m) is the Euler totient function, and set ω(m) the smallest number n such that ϕn(m) = 1. If m < 2α , then prove that ω(m) ≤ α. Problem 1.30. Let f be a polynomial with integer coefficients and N(f ) = card{k ∈ Z; f (k) = ±1}. Prove that N(f ) ≤ 2 + deg f , where deg f denotes the degree of f . Problem 1.31. Prove that every integer can be written as a sum of 5 perfect cubes. Problem 1.32. If n ∈ Z, then the binomial coefficient C2n = (2n)!(n!)2 has an even number of divisors. Problem 1.33. (Amer. Math. Monthly) Prove that every n ∈ Z+ can be written in precisely k(n) different ways as a sum of consecutive integers, where k(n) is the number of odd divisors of n greater than 1. Problem 1.34. (Amer. Math. Monthly) Let π2(x) denote the number of twin primes p with p ≤ x. Recall that p is a twin prime if both p and p + 2 are prime. Show that π2(x) = 2 + ∑ 7≤n≤x sin ( π 2 (n + 2) [ n! n + 2 ]) sin ( π 2 n [ (n − 2)! n ]) for x > 7. Problem 1.35. (Kvant) 1. Find all solutions of the equation 3x+1 + 100 = 7x−1. 2. Find two solutions of the equation 3x + 3x2 = 2x + 4x2 , and prove that there are no others. Problem 1.36. (Kvant) Let σ(n) denote the sum of the divisors of n. Prove that there exist infinitely many integers n such that σ(n) > 2n, or even stronger, such that σ(n) > 3n. Prove also that σ(n) < n(1 + log n). Problem 1.37. (American Competition) Let ai be natural numbers such that gcd(ai, aj ) = 1 and the ai are not prime numbers. Show that 1 a1 + · · · + 1 an < 2. 1 Number Theory 7 Problem 1.38. Let σ(n) denote the sum of divisors of n. Show that σ(n) = 2k if and only if n is a product of Mersenne primes, i.e., primes of the form 2k − 1, for k ∈ Z. Problem 1.39. (Putnam Competition) Find all integer solutions of the equation |pr − qs | = 1, where p, q are primes and r, s ∈ Z \ {0, 1}. Problem 1.40. Consider an arithmetic progression with ratio between 1 and 2000. Show that the progression does not contain more than 10 consecutive primes. Problem 1.41. (Amer. Math. Monthly) Let a1 = 1, an+1 = an + [√ an ] . Show that an is a perfect square iff n is of the form 2k + k − 2. Problem 1.42. Recall that ϕ(n) denotes the Euler totient function (i.e., the number of natural numbers less than n and prime to n), and that σ(n) is the sum of divisors of n. Show that n is prime iff ϕ(n) divides n − 1 and n + 1 divides σ(n). Problem 1.43. (Amer. Math. Monthly) A number is called ϕ-subadditive if ϕ(n) ≤ ϕ(k) + ϕ(n − k) for all k such that 1 ≤ k ≤ n − 1, and ϕ-superadditive if the reverse inequality holds. Prove that there are infinitely many ϕ-subadditive numbers and infinitely many ϕ-superadditive numbers. Problem 1.44. (Amer. Math. Monthly) Find the positive integers N such that for all n ≥ N , we have ϕ(n) ≤ ϕ(N). Problem 1.45. A number n is perfect if σ(n) = 2n, where σ(n) denotes the sum of all divisors of n. Prove that the even number n is perfect if and only if n = 2p−1(2p −1), where p is a prime number with the property that 2p − 1 is prime. Problem 1.46. (Elemente der Mathematik) A number n is superperfect if σ(σ(n)) = 2n, where σ(k) is the sum of all divisors of k. Prove that the even number n is superperfect if and only if n = 2r , where r is an integer such that 2r+1 − 1 is prime. Problem 1.47. Ifa, b are rational numbers satisfying tan aπ = b, thenb ∈ {−1, 0, 1}. Problem 1.48. (Amer. Math. Monthly) Let An = run + svn, n ∈ Z+, where r, s, u, v are integers,u �= ±v, and letPn be the set of prime divisors ofAn. ThenP =⋃∞n=0 Pn is infinite. Problem 1.49. Solve in natural numbers the equation x2 + y2 + z2 = 2xyz. Problem 1.50. (Amer. Math. Monthly) Find the greatest common divisor of the fol- lowing numbers: C12n, C32n, C52n, . . . , C 2n−1 2n . Problem 1.51. (Amer. Math. Monthly) Prove that if Sk = ∑ni=1 kgcd(i,n), then Sk ≡ 0 (mod n). 8 1 Number Theory Problem 1.52. Prove that for any integer n such that 4 ≤ n ≤ 1000, the equation 1 x + 1 y + 1 z = 4 n has solutions in natural numbers. Problem 1.53. (Nieuw Archief v. Wiskunde) For a natural number n we set S(n) for the set of integers that can be written in the form 1+g+· · ·+gn−1 for some g ∈ Z+, g ≥ 2. 1. Prove that S(3) ∩ S(4) = ∅. 2. Find S(3) ∩ S(5). Problem 1.54. If k ≥ 202 and n ≥ 2k, then prove that Ckn > nπ(k), where π(k) denotes the number of prime numbers smaller than k. Problem 1.55. (Amer. Math. Monthly) Let Sm(n) be the sum of the inverses of the integers smaller than m and relatively prime to n. If m > n ≥ 2, then show that Sm(n) is not an integer. Problem 1.56. Solve in integers the following equations: 1. x4 + y4 = z2; 2. y3 + 4y = z2; 3. x4 = y4 + z2. Problem 1.57. Show that the equation x3 + y3 = z3 + w3 has infinitely many integer solutions. Prove that 1 can be written in infinitely many ways as a sum of three cubes. Problem 1.58. Prove that the equation y2 = Dx4 + 1 has no integer solution except x = 0 if D �≡ 0,−1, 3, 8 (mod 16) and there is no factorization D = pq, where p > 1 is odd, gcd(p, q) = 1, and either p ≡ ±1 (mod 16), p ≡ q ± 1 (mod 16), or p ≡ 4q ± 1 (mod 16). Problem 1.59. (Amer. Math. Monthly) Find all numbers that are simultaneously tri- angular, perfect squares, and pentagonal numbers. Problem 1.60. (Amer. Math. Monthly) Find all inscribable integer-sided quadrilater- als whose areas equal their perimeters. Problem 1.61. (Amer. Math. Monthly) Every rational number a can be written as a sum of the cubes of three rational numbers. Moreover, if a > 0, then the three cubes can be chosen to be positive. Problem 1.62. 1. Every prime number of the form 4m + 1 can be written as the sum of two perfect squares. 1 Number Theory 9 2. Every natural number is the sum of four perfect squares. Problem 1.63. Every natural number can be written as a sum of at most 53 integers to the fourth power. Problem 1.64. Let G(k) denote the minimal integer n such that any positive integer can be written as the sum of n positive perfect kth powers. Prove that G(k) ≥ 2k +[ 3k 2k ] − 2. Problem 1.65. (Amer. Math. Monthly) LetR(n, k) be the remainder whenn is divided by k and S(n, k) = k∑ i=1 R(n, i). 1. Prove that limn→∞ S(n, n)n2 = 1 − π 2 12 . 2. Consider a sequence of natural numbers (ak) growing to infinity and such that limk→∞ ak log kk = 0. Prove that limk→∞ S(kak, k)k2 = 14 . Problem 1.66. (Nieuw Archief v. Wiskunde) Let a1 < a2 < a3 < · · · < an < · · · be a sequence of positive integers such that the series ∞∑ i=1 1 ai converges. Prove that for any i, there exist infinitely many sets of ai consecutive integers that are not divisible by aj for all j > i. Problem 1.67. (Amer. Math. Monthly) Denote by Cn the claim that there exists a set of n consecutive integers such that no two of them are relatively prime. Prove that Cn is true for every n such that 17 ≤ n ≤ 10000. Problem 1.68. (Schweitzer Competition) Prove that a natural number has more di- visors that can be written in the form 3k + 1, for k ∈ Z, than divisors of the form 3m − 1, for m ∈ Z. Problem 1.69. (Amer. Math. Monthly) A number N is called deficient if σ(N) < 2N and abundant if σ(N) > 2N . 1. Let k be fixed. Are there any sequences of k consecutive abundant numbers? 2. Show that there are infinitely many 5-tuples of consecutive deficient numbers. Problem 1.70. (Amer. Math. Monthly) Does there exist a nonconstant polynomial an2+bn+c with integer coefficients such that for any natural numbers m all its prime factors pi are congruent to 3 modulo 4? Prove that for any nonconstant polynomial f with integer coefficients and any m ∈ Z there exist a prime number p and a natural number n such that p divides f (n) and p ≡ 1 (mod m). 2 Algebra and Combinatorics 2.1 Algebra Problem 2.1. Set Sk,p = ∑p−1i=1 ik , for natural numbers p and k. If p ≥ 3 is prime and 1 < k ≤ p − 2, show that Sk,p ≡ 0 (mod p). Problem 2.2. Let P = a0+· · ·+anxn and Q = b0+· · ·+bmxm be two polynomials with m ≤ n. Then, deg gcd(P,Q) ≥ 1 if and only if there exist two polynomials K and L, such that deg K ≤ m− 1, deg L ≤ n− 1, and K ·P = L ·Q. Prove that this is equivalent to the vanishing of the following (n + m) × (n + m) determinant: det ⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜ ⎝ a0 a1 a2 · · · am−1 · · · an−1 an 0 0 · · · 0 0 a0 a1 · · · am−2 · · · an−2 an−1 an 0 · · · 0 ... ... ... ... ... ... ... ... ... 0 0 0 · · · a0 · · · an−m−1 an−m an−m+1 · · · an 0 0 0 · · · 0 · · · an−m−2 an−m−1 an−m · · · an−1 ... ... ... ... ... ... ... ... 0 0 0 · · · 0 0 a0 a1 a2 · · · am b0 b1 · · · bm−1 bm 0 0 0 0 · · · 0 0 b0 · · · bm−2 bm−1 bm 0 0 0 · · · 0 ... ... ... ... ... ... ... ... ... 0 0 · · · b0 b1 · · · · · · · · · 0 · · · 0 ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = 0. Problem 2.3. Prove that if P,Q ∈ R[x, y] are relatively prime polynomials, then the system of equations P(x, y) = 0, Q(x, y) = 0, has only finitely many real solutions. 12 2 Algebra and Combinatorics Problem 2.4. Let a, b, c, d ∈ R[x] be polynomials with real coefficients. Set p = ∫ x 1 ac dt, q = ∫ x 1 ad dt, r = ∫ x 1 bc dt, s = ∫ x 1 bd dt. Prove that ps − qr is divisible by (x − 1)4. Problem 2.5. 1. Find the minimum number of elements that must be deleted from the set {1, . . . , 2005} such that the set of the remaining elements does not contain two elements together with their product. 2. Does there exist, for any k, an arithmetic progression with k terms in the infinite sequence 1, 1 2 , . . . , 1 2005 , . . . , 1 n , . . .? Problem 2.6. (Amer. Math. Monthly) Consider a set S of n elements and n+1 subsets M1, . . . ,Mn+1 ⊂ S. Show that there exist r, s ≥ 1 and disjoint sets of indices {i1, . . . , ir} ∩ {j1, . . . , js} = ∅ such that r⋃ k=1 Mik = s⋃ k=1 Mjk . Problem 2.7. Let p be a prime number, and A = {a1, . . . , ap−1} ⊂ Z∗+ a set of integers that are not divisible by p. Define the map f : P(A) → {0, 1, . . . , p − 1} by f ({ai1 , . . . , aik }) = k∑ p−1 aip (mod p), and f (∅) = 0. Prove that f is surjective. Problem 2.8. Consider the function Fr = xr sin rA + yr sin rB + zr sin rC, where x, y, z ∈ R, A + B + C = kπ , and r ∈ Z+. Prove that if F1(x0, y0, z0) = F2(x0, y0, z0) = 0, then Fr(x0, y0, z0) = 0, for all r ∈ Z+. Problem 2.9. Let T (z) ∈ Z[z] be a nonzero polynomial with the property that |T (ui)| ≤ 1 for all values ui that are roots of P(z) = zn − 1. Prove that either T (z) is divisible by P(z), or else there exists some k ∈ Z+, k ≤ n − 1, such that T (z) ± zk is divisible by P(z). The same result holds when instead of P(z), we consider zn + 1. Problem 2.10. 1. If the map x �→ x3 from a group G to itself is an injective group homomorphism, then G is an abelian. 2. If the map x �→ x3 from a group G to itself is a surjective group homomorphism, then G is an abelian. 3. Find an abelian group with the property that x �→ x4 is an automorphism. 4. What can be said for exponents greater than 4? 2.2 Algebraic Combinatorics 13 Problem 2.11. (Amer. Math. Monthly) Let V be a vector space of dimension n > 0 over a field of characteristic p �= 0 and let A be an affine map A : V → V . Prove that there exist u ∈ V and 0 ≤ k ≤ np such that Aku = u. Problem 2.12. (Putnam Competition 1959) Find the cubic equation the zeros of which are the cubes of the roots of the equation x3 + ax2 + bx + c = 0. Problem 2.13. (Putnam Competition 1956) Assume that the polynomials P,Q ∈ C[x] have the same roots, possibly with different multiplicities. Suppose, moreover, that the same holds true for the pair P + 1 and Q + 1. Prove that P = Q. Problem 2.14. (Amer. Math. Monthly) Determine r ∈ Q, for which 1, cos 2πr, sin 2πr are linearly dependent over Q. Problem 2.15. (Amer. Math. Monthly) 1. Prove that there exist a, b, c ∈ Z, not all zero, such that |a|, |b|, |c| < 106∣∣ ∣a + b√2 + c√3 ∣ ∣∣ < 10−11. 2. Prove that if 0 ≤ |a|, |b|, |c| < 106, a, b, c ∈ Z, and at least one of them is nonzero, then ∣∣∣a + b√2 + c√3 ∣∣∣ > 10−21. Problem 2.16. (Amer. Math. Monthly) Prove that if n > 2, then we do not have any nontrivial solutions for the equation xn + yn = zn, where x, y, z are rational functions. Solutions of the form x = af, y = bf, z = cf , wheref is a rational function anda, b, c are complex numbers satisfyingan+bn = cn, are called trivial. Problem 2.17. (Kvant) A table is an n× k rectangular grid drawn on the torus, every box being assigned an element from Z/2Z. We define a transformation acting on tables as follows. We replace all elements of the grid simultaneously, each element being changed into the sum of the numbers previously assigned to its neighboring boxes. Prove that iterating this transformation sufficiently many times, we always obtain the trivial table filled with zeros, no matter what the initial table was, if and only if n = 2p and k = 2q for some integers p, q. In this case we say that the respective n × k grid is nilpotent. 2.2 Algebraic Combinatorics Problem 2.18. Let us consider a four-digit number N whose digits are not all equal. We first arrange its digits in increasing order, then in decreasing order, and finally, we subtract the two obtained numbers. Let T (N) denote the positive difference thus obtained. Show that after finitely many iterations of the transformation T , we obtain 6174. 14 2 Algebra and Combinatorics Problem 2.19. Find an example of a sequence of natural numbers 1 ≤ a1 < a2 < · · · < an < an+1 < · · · with the property that every m ∈ Z+ can be uniquely written as m = ai − aj , for i, j ∈ Z+. Problem 2.20. (Amer. Math. Monthly) Consider the set of 2n integers {±a1,±a2, . . . , ±an} and m < 2n. Show that we can choose a subset S such that 1. The two numbers ±ai are not both in S; 2. The sum of all elements of S is divisible by m. Problem 2.21. (Proposed for the International Math. Olympiad) Show that for ev- ery natural number n there exist prime numbers p and q such that n divides their difference. Problem 2.22. An even number, 2n, of knights arrive at King Arthur’s court, each one of them having at most n − 1 enemies. Prove that Merlin the wizard can assign places for them at a round table in such a way that every knight is sitting only next to friends. Problem 2.23. (Putnam Competition) Let r, s ∈ Z+. Find the number of 4-tuples of positive integers (a, b, c, d) that satisfy 3r7s = lcm(a, b, c) = lcm(a, b, d) = lcm(a, c, d) = lcm(b, c, d). Problem 2.24. (Putnam Competition) 1. Let n ∈ Z+ and p be a prime number. Denote by N(n, p) the number of binomial coefficients Csn that are not divisible by p. Assume that n is written in base p as n = n0 +n1p+· · ·+nmpm, where 0 ≤ nj < p for all j ∈ {0, 1, . . . , m}. Prove that N(n, p) = (n0 + 1)(n1 + 1) · · · (nj + 1). 2. Write k in base p as k = k0 + k1p + · · · + kjpj , with 0 ≤ kj ≤ p − 1, for all j ∈ {0, 1, . . . , s}. Prove that Ckn ≡ Ck0n0Ck1n1 · · ·C kj nj (mod p). Problem 2.25. (Putnam Competition) Define the sequence Tn by T1 = 2, Tn+1 = T 2n − Tn + 1, for n ≥ 1. Prove that if m �= n, then Tm and Tn are relatively prime, and further, that ∞∑ i=1 1 Ti = 1. Problem 2.26. Let α, β > 0 and consider the sequences [α], [2α], . . . , [kα], . . . ; [β], [2β], . . . , [kβ], . . . , where the brackets denote the integer part. Prove that these two sequences taken together enumerate Z+ in an injective manner if and only if α, β ∈ R \ Q and 1 α + 1 β = 1. 2.2 Algebraic Combinatorics 15 Problem 2.27. We say that the sets S1, S2, . . . , Sm form a complementary system if they make a partition of Z+, i.e., every positive integer belongs to a unique set Si . Let m > 1 and α1, . . . , αm ∈ R+. Then the sets Si = {[nαi], where n ∈ Z+} form a complementary system only if m = 2, α−11 + α−12 = 1, and α1 ∈ R \ Q. Problem 2.28. Let f : Z+ → Z+ be an increasing function and set F(n) = f (n) + n, G(n) = f ∗(n) + n, where f ∗(n) = card({x ∈ Z+; 0 ≤ f (x) < n}). Then {F(n); n ∈ Z+} and {G(n); n ∈ Z+} are complementary sequences. Conversely, any two complemen- tary sequences can be obtained this way using some nondecreasing function f . Problem 2.29. Let M denote the set of bijective functions f : Z+ → Z+. Prove that there is no bijective function between M and Z. Problem 2.30. (Amer. Math. Monthly) Let F ⊂ Z be a finite set of integers satisfying the following properties: 1. For any x ∈ F there exist y, z ∈ F such that x = y + z. 2. There exists n such that for any natural number 1 ≤ k ≤ n, and any choice of x1, . . . , xk ∈ F , their sum x1 + · · · + xk is nonzero. Prove that card(F ) ≥ 2n + 2. Problem 2.31. (Amer. Math. Monthly) For a finite graph G we denote by Z(G) the minimal number of colors needed to color all its vertices such that adjacent vertices have different colors. This is also called the chromatic number of G. Prove that the inequality Z(G) ≥ p 2 p2 − 2q holds if G has p vertices and q edges. Problem 2.32. LetDk be a collection of subsets of the set {1, . . . , n} with the property that whenever A �= B ∈ Dk , then card(A ∩ B) ≤ k, where 0 ≤ k ≤ n − 1. Prove that card(Dk) ≤ C0n + C1n + C2n + · · · + Ck+1n . Problem 2.33. Prove that 1 p! n∑ k=0 (−1)n−kCknkp = ⎧ ⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎩ 0, if 0 ≤ p < n, 1, if p = n, n/2, if p = n + 1, n(3n+1) 24 , if p = n + 2, n2(n+1) 48 , if p = n + 3, n(15n3+30n2+5n+1) 1152 , if p = n + 4. 16 2 Algebra and Combinatorics Problem 2.34. Write lcm(a1, . . . , an) in terms of the various gcd(ai, . . . , aj ) for subsets of {a1, . . . , an}. Problem 2.35. Let f (n) be the number of ways in which a convex polygon with n+1 sides can be divided into regions delimited by several diagonals that do not intersect (except possibly at their endpoints). We consider as distinct the dissection in which we first cut the diagonal a and next the diagonal b from the dissection in which we first cut the diagonal b and next the diagonal a. It is easy to compute the first values of f (n), as follows: f (1) = 1, f (2) = 1, f (3) = 3, f (4) = 11, f (5) = 45. Find the generating function F(x) =∑ f (n)xn and an asymptotic formula for f (n). Problem 2.36. Find the permutation σ : (1, . . . , n) → (1, . . . , n) such that S(σ) = n∑ i=1 |σ(i) − i| is maximal. Problem 2.37. (Amer. Math. Monthly) On the set Sn of permutations of {1, . . . , n} we define an invariant distance function by means of the formula d(σ, τ ) = n∑ i=1 |σ(i) − τ(i)|. What are the values that d could possibly take? Problem 2.38. (Preliminaries for the International Math. Olympiad, Romania) The set M = {1, 2, . . . , 2n} is partitioned into k sets M1, . . . ,Mk , where n ≥ k3 + k. Show that there exist i, j ∈ {1, . . . , k} for which we can find k + 1 distinct even numbers 2r1, . . . , 2rk+1 ∈ Mi with the property that 2r1 − 1, . . . , 2rk+1 − 1 ∈ Mj . Problem 2.39. (Preliminaries for the International Math. Olympiad, Great Britain) Let S be the set of odd integers not divisible by 5 and smaller than 30m, where m ∈ Z∗+. Find the smallest k such that every subset A ⊂ S of k elements contains two distinct integers, one of which divides the other. Problem 2.40. Prove that ∏ 1≤j nn−1. Show that there exist distinct prime numbers p1, p2, . . . , pn such that C + j is divisible by pj . Problem 2.47. (Nieuw Archief v. Wiskunde) Let p be a prime number and f (p) the smallest integer for which there exists a partition of the set {2, 3, . . . , p} into f (p) classes such that whenever a1, . . . , ak belong to the same class of the partition, the equation k∑ i=1 xiai = p does not have solutions in nonnegative integers. Estimate f (p). Problem 2.48. (Schweitzer Competition) Consider m distinct natural numbers ai smaller than N such that lcm(ai, aj ) ≤ N for all i, j . Prove that m ≤ 2 [√ N ] . Problem 2.49. The set M ⊂ Z+ is called A-sum-free, where A = (a1, a2, . . . , ak) ∈ Z k+, if for any choice of x1, x2, . . . , xk ∈ M we have a1x1 + a2x2 + · · · + akxk �∈ M . If A,B are two vectors, we define f (n;A,B) as the greatest number h such that there exists a partition of the set of consecutive integers {n, n + 1, . . . , h} into S1 and S2 such that S1 is A-sum-free and S2 is B-sum-free. Assume that B = (b1, b2, . . . , bm) and that the conditions below are satisfied: a1 + a2 + · · · + ak = b1 + b2 + · · · + bm = s, and min 1≤j≤k aj = min1≤j≤m bj = 1, k,m ≥ 2. Prove that f (n;A,B) = ns2 + n(s − 1) − 1. Problem 2.50. (Amer. Math. Monthly) Let 1 ≤ a1 < a2 < · · · < an < 2n be a sequence of natural numbers for n ≥ 6. Prove that min i,j lcm(ai, aj ) ≤ 6 ([n 2 ] + 1 ) . Moreover, the constant 6 is sharp. 18 2 Algebra and Combinatorics Problem 2.51. (Amer. Math. Monthly) Let 1 ≤ a1 < a2 < · · · < ak < n be such that gcd(ai, aj ) �= 1 for all 1 ≤ i < j ≤ k. Determine the maximum value of k. Problem 2.52. (International Math. Olympiad 1978) Consider the increasing se- quence f (n) ∈ Z+, 0 < f (1) < f (2) < · · · < f (n) < · · · . It is known that the nth element in increasing order among the positive integers that are not terms of this sequence is f (f (n)) + 1. Find the value of f (240). Problem 2.53. (Amer. Math. Monthly) We define inductively three sequences of in- tegers (an), (bn), (cn) as follows: 1. a1 = 1, b1 = 2, c1 = 4; 2. an is the smallest integer that does not belong to the set {a1, . . . , an−1, b1, . . . , bn−1, c1, . . . , cn−1}; 3. bn is the smallest integer that does not belong to the set {a1, . . . , an−1, an, b1, . . . , bn−1, c1, . . . , cn−1}; 4. cn = 2bn + n − an. Prove that 0 < n ( 1 + √3 ) − bn < 2 for all n ∈ Z+. 2.3 Geometric Combinatorics Problem 2.54. We consider n points in the plane that determine C2n segments, and to each segment one associates either +1 or −1. A triangle whose vertices are among these points will be called negative if the product of numbers associated to its sides is negative. Show that if n is even, then the number of negative triangles is even. Moreover, for odd n, the number of negative triangles has the same parity as the number p of segments labeled −1. Problem 2.55. (Amer. Math. Monthly) Given n find a finite set S consisting of natural numbers larger than n with the property that for any k ≥ n the k × k square can be tiled by a family of si × si squares, where si ∈ S. Problem 2.56. We consider 3n points A1, . . . , A3n in the plane whose positions are defined recursively by means of the following rule: first, the triangle A1A2A3 is equilateral; further, the points A3k+1, A3k+2, and A3k+3 are the midpoints of the sides of the triangle A3kA3k−1A3k−2. Let us assume that the 3n points are colored with two colors. Show that for n ≥ 7 there exists at least one isosceles trapezoid having vertices of the same color. Problem 2.57. Is there a coloring of all lattice points in the plane using only two colors such that there are no rectangles with all vertices of the same color whose side ratio belongs to { 1, 12 , 1 3 , 2 3 } ? 2.3 Geometric Combinatorics 19 Problem 2.58. (Amer. Math. Monthly) Let G be a planar graph and let P be a path in G. We say that P has a (transversal) self-intersection in the vertex v if the path has a (transversal) self-intersection from the curve-theoretic viewpoint. Let us give an example: Take the point 0 in the plane and the segments 01, 02, 03, 04 going counter- clockwise around 0. Then a path traversing first 103 and then 204 has a (transversal) self-intersection at 0, while a path going first along 102 and further on 304 does not have a (transversal) self-intersection. Prove that any connected planar graph G with only even-degree vertices admits an Eulerian circuit without self-intersections. Recall that an Eulerian circuit is a path along the edges of the graph that passes precisely once along each edge of the graph. Problem 2.59. (Hungarian Competition) Let us consider finitely many points in the plane that are not all collinear. Assume that one associates to each point a number from the set {−1, 0, 1} such that the following property holds: for any line determined by two points from the set, the sum of numbers associated to all points lying on that line equals zero. Show that, if the number of points is at least three, then to each point one is associated with 0. Problem 2.60. (Amer. Math. Monthly) If one has a set of squares with total area smaller than 1, then one can arrange them inside a square of side length √ 2, without any overlaps. Problem 2.61. Prove that for each k there exist k points in the plane, no three collinear and having integral distance from each other. If we have an infinite set of points with integral distances from each other, then all points are collinear. Problem 2.62. (International Math. Olympiad 1984) Let O,A be distinct points in the plane. For each point x in the plane, we write α(x) = x̂OA (counterclockwise). Let C(x) be the circle of center O and radius |Ox| + α(x)|Ox| . If the points in the plane are colored with finitely many colors, then there exists a point y with α(y) > 0 such that the color of y also belongs to the circle C(y). Problem 2.63. (Amer. Math. Monthly) Let k, n ∈ Z+. 1. Assume that n−1 ≤ k ≤ n(n−1)2 . Show that there exist n distinct points x1, ..., xn on a line that determine exactly k distinct distances |xi − xj |. 2. Suppose that [n2 ] ≤ k ≤ n(n−1)2 . Then there exist n points in the plane that determine exactly k distinct distances. 3. Prove that for any ε > 0, there exists some constant n0 = n0(ε) such that for any n > n0 and εn < k < n(n−1)2 , there exist n points in the plane that determine exactly k distinct distances. Problem 2.64. (International Math. Olympiad 1978) Show that it is possible to pack 2n(2n+ 1) nonoverlapping pieces having the form of a parallelepiped of dimensions 1 × 2 × (n + 1) in a cubic box of side 2n + 1 if and only if n is even or n = 1. 20 2 Algebra and Combinatorics Problem 2.65. (Putnam Competition 1964) Let F be a finite subset of R with the property that any value of the distance between two points from F (except for the largest one) is attained at least twice, i.e., for two distinct pairs of points. Prove that the ratio of any two distances between points of F is a rational number. 3 Geometry 3.1 Synthetic Geometry Problem 3.1. Let I be the center of the circle inscribed in the triangle ABC and consider the points α, β, γ situated on the perpendiculars from I on the sides of the triangle ABC such that |Iα| = |Iβ| = |Iγ |. Prove that the lines Aα,Bβ,Cγ are concurrent. Problem 3.2. We consider the angle xOy and a pointA ∈ Ox. Let (C) be an arbitrary circle that is tangent to Ox and Oy at the points H and D, respectively. Set AE for the tangent line drawn from A to the circle (C) that is different from AH . Show that the line DE passes through a fixed point that is independent of the circle (C) chosen above. Problem 3.3. Let C be a circle of center O and A a fixed point in the plane. For any point P ∈ C, let M denote the intersection of the bisector of the angle ÂOP with the circle circumscribed about the triangle AOP . Find the geometric locus of M as P runs over the circle C. Problem 3.4. Let ABC be an isosceles triangle having |AB| = |AC|. If AS is an interior Cevian that intersects the circle circumscribed about ABC at S, then describe the geometric locus of the center of the circle circumscribed about the triangle BST , where {T } = AS ∩ BC. Problem 3.5. Let AB,CD,EF be three chords of length one on the unit circle. Then the midpoints of the segments |BC|, |DE|, and |AF | form an equilateral triangle. Problem 3.6. (Amer. Math. Monthly) Denote by P the set of points of the plane. Let � : P ×P → P be the following binary operation: A�B = C, where C is the unique point in the plane such that ABC is an oriented equilateral triangle whose orientation is counterclockwise. Show that � is a nonassociative and noncommutative operation satisfying the following “medial property”: 22 3 Geometry (A � B) � (C � D) = (A � C) � (B � D). Problem 3.7. Consider two distinct circles C1 and C2 with nonempty intersection and let A be a point of intersection. Let P,R ∈ C1 and Q,S ∈ C2 be such that PQ and RS are the two common tangents. Let U and V denote the midpoints of the chords PR and QS. Prove that the triangle AUV is isosceles. Problem 3.8. (Amer. Math. Monthly) If the planar triangles AUV, VBU , and UVC are directly similar to a given triangle, then so is ABC. Recall that two triangles are directly similar if one can obtain one from the other using a homothety with positive ratio, rotations and translations. Problem 3.9. (Amer. Math. Monthly) Find, using a straightedge and a compass, the directrix and the focus of a parabola. Recall that the parabola is the geometric locus of those points P in the plane that are at equal distance from a point O (called the focus) and a line d called the directrix. Problem 3.10. Prove that if M is a point in the interior of a circle and AB ⊥ CD are two chords perpendicular at M , then it is possible to construct an inscribable quadrilateral with the following lengths: ∣∣|AM| − |MB|∣∣, |AM| + |MB|, ∣∣|DM| − |MC|∣∣, |DM| + |MC|. Problem 3.11. (Amer. Math. Monthly) If the Euler line of a triangle passes through the Fermat point, then the triangle is isosceles. Problem 3.12. Consider a point M in the interior of the triangle ABC, and choose A′ ∈ AM,B ′ ∈ BM , and C′ ∈ CM . Let P,Q,R, S, T , and U be the intersections of the sides of ABC and A′B ′C′. Show that PS, TQ, and RU meet at M . Problem 3.13. Show that if an altitude in a tetrahedron crosses two other altitudes, then all four altitudes are concurrent. Problem 3.14. Three concurrent Cevians in the interior of the triangle ABC meet the corresponding opposite sides at A1, B1, C1. Show that their common intersection point is uniquely determined if |BA1|, |CB1|, and |AC1| are equal. Problem 3.15. (International Math. Olympiad 1983) Let ABCD be a convex quadri- lateral with the property that the circle of diameterAB is tangent to the lineCD. Prove that the circle of diameter CD is tangent to the line AB if and only if AD is parallel to BC. Problem 3.16. Let A′, B ′, C′ be points on the sides BC,CA,AB of the triangle ABC. Let M1,M2 be the intersections of the circle A′B ′C′ with the circle ABA′ and let N1, N2 be the analogous intersections of the circle A′B ′C′ with the circle ABB ′. 1. Prove that M1M2, N1N2, AB are either parallel or concurrent, in a point that we denote by A′1; 3.2 Combinatorial Geometry 23 2. Prove that the analogously defined points A′1, B ′1, C′1 are collinear. Problem 3.17. (Putnam Competition) A circumscribable quadrilateral of area S =√ abcd is inscribable. Problem 3.18. (Nieuw Archief v. Wiskunde) Let O be the center of the circumcircle, Ge the Gergonne point, Na the Nagel point, and G1, N1, the isogonal conjugates of G and N , respectively. Prove that G1, N1, and O are collinear (see also the Glossary for definitions of the important points in a triangle). 3.2 Combinatorial Geometry Problem 3.19. Consider a rectangular sheet of paper. Prove that given any ε > 0, one can use finitely many foldings of the paper along its sides in either 2 equal parts or 3 equal parts to obtain a rectangle whose sides are in ratio r for some r satisfying 1 − � ≤ r ≤ 1 + �. Problem 3.20. (Amer. Math. Monthly) Show that there exist at most three points on the unit circle with the distance between any two being greater than √ 2. Problem 3.21. (Komal) A convex polygon with 2n sides has at least n diagonals not parallel to any of its sides. Problem 3.22. Let d be the sum of the lengths of the diagonals of a convex polygon P1 . . . Pn and p its perimeter. Prove that for n ≥ 4, we have n − 3 < 2 d p < [n 2 ] [n + 1 2 ] − 2. Problem 3.23. (Amer. Math. Monthly) Find the convex polygons with the property that the function D(p), which is the sum of the distances from an interior point p to the sides of the polygon, does not depend on p. Problem 3.24. Prove that a sphere of diameter 1 cannot be covered by n strips of width li if ∑n i=1 li < 1. Prove that a circle of diameter 1 cannot be covered by n strips of width li if ∑n i=1 li < 1. Problem 3.25. (Kvant) Consider n points lying on the unit sphere. Prove that the sum of the squares of the lengths of all segments determined by the n points is less than n2. Problem 3.26. (Kvant) The sum of the vectors −−→OA1, . . . ,−−→OAn is zero, and the sum of their lengths is d . Prove that the perimeter of the polygon A1 . . . An is greater than 4d/n. Problem 3.27. (Amer. Math. Monthly) Find the largest numbers ak , for 1 ≤ k ≤ 7, with the property that for any point P lying in the unit cube with vertices A1 . . . A8, at least k among the distances |PAj | to the vertices are greater or equal than ak . 24 3 Geometry Problem 3.28. (Amer. Math. Monthly) The line determined by two points is said to be admissible if its slope is equal to 0, 1,−1, or ∞. What is the maximum number of admissible lines determined by n points in the plane? Problem 3.29. If A = {z1, . . . , zn} ⊂ C, then there exists a subset B ⊂ A such that ∣ ∣ ∑ z∈B z ∣∣ ≥ π−1 n∑ i=1 |zi |. Problem 3.30. (Amer. Math. Monthly) Let A1, . . . , An be the vertices of a regular n-gon inscribed in a circle of center O. Let B be a point on the arc of circle A1An and set θ = ÂnOB. If we set ak = |BAk|, then find the sum n∑ k=1 (−1)kak in terms of θ . Problem 3.31. (Amer. Math. Monthly) Consider n distinct complex numbers zi ∈ C such that min i �=j |zi − zj | ≥ maxi≤n |zi |. What is the greatest possible value of n? Problem 3.32. The interior of a triangle can be tiled by n ≥ 9 pentagonal convex surfaces.What is the minimal value ofn such that a triangle can be tiled byn hexagonal strictly convex surfaces? Problem 3.33. (Putnam Competition) We say that a transformation of the plane is a congruence if it preserves the length of segments. Two subsets are congruent if there exists a congruence sending one subset onto the other. Show that the unit disk cannot be partitioned into two congruent subsets. Problem 3.34. Prove that the unit disk cannot be partitioned into two subsets of diameter strictly smaller than 1, where the diameter of a set is the supremum distance between two of its points. Problem 3.35. A continuous planar curve L has extremities A and B at distance |AB| = 1. Show that for any natural number n there exists a chord determined by two points C,D ∈ L that is parallel to AB and whose length |CD| equals 1 n . Problem 3.36. The diameter of a set is the supremum of the distance between two of its points. Prove that any planar set of unit diameter can pe partitioned into three parts of diameter no more than √ 3 2 . Problem 3.37. 1. Prove that a finite set of n points in R3 of unit diameter can be covered by a cube of side length 1 − 23n(n−1) . 3.2 Combinatorial Geometry 25 2. Prove that any planar set of n points having unit diameter can pe partitioned into three parts of diameter less than √ 3 2 cos 2π 3n(n−1) . Problem 3.38. Prove that any convex set in Rn of unit diameter having a smooth boundary can be partitioned into n + 1 parts of diameter d < 1. Problem 3.39. Let D be a convex body in R3 and let σ(D) = supπ area(π ∩ D), where the supremum is taken over all positions of the variable plane π . Prove that D can be divided into two parts D1 and D2 such that σ(Di) < σ(D). Problem 3.40. If we have k vectors v1, v2, . . . , vk in Rn and k ≤ n + 1, then there exist two vectors making an angle θ with cos θ ≥ − 1 k−1 . Equality holds only when the endpoints of the vectors form a regular (k − 1)-simplex. Problem 3.41. 1. Consider a finite family of bounded closed convex sets in the plane such that any three members of the family have nonempty intersection. Prove that the intersection of all members of the family is nonempty. 2. A set of unit diameter in R2 can be covered by a ball of radius √ 1 3 . Problem 3.42. (Amer. Math. Monthly) Let T be a right isosceles triangle. Find the disk D such that the difference between the areas of T ∪ D and T ∩ D is minimal. Problem 3.43. (Amer. Math. Monthly) Let r be the radius of the incircle of an arbitrary triangle lying in the closed unit square. Prove that r ≤ √ 5−1 4 . Problem 3.44. Let P be a point in the interior of the tetrahedron ABCD, with the property that |PA| + |PB| + |PC| + |PD| is minimal. Prove that ÂPB = ĈPD and that these angles have a common bisector. Problem 3.45. (Putnam Competition 1948) Let OA1, . . . , OAn be n linearly inde- pendent vectors of lengths a1, . . . , an. We construct the parallelepipedH having these vectors as sides. Then consider the n altitudes in H as a new set of vectors and further, construct the parallelepiped E associated with the altitudes. If h is the volume of H and e the volume of E, then prove that he = (a1 . . . an)2. Problem 3.46. Let F be a symmetric convex body in R3 and let AF,λ denote the family of all sets homothetic to F in the ratio λ that have only boundary points in common with F . Set hF (λ) for the greatest integer k such that AF,λ contains k sets with pairwise disjoint interiors. Prove that hF (λ) ≤ (1 + 2λ) 3 − 1 λ3 . 26 3 Geometry Problem 3.47. Let � denote the square of equations |xi | ≤ 1, i = 1, 2, in the plane, and let A = (a1|a2) be an arbitrary nonsingular 2 × 2 matrix partitioned into two columns. We identify each column with a vector in R2. Prove that the following inequality holds: min A max x∈� ∣∣∣ ∣ 〈a1, x〉〈a2, x〉 det A ∣∣ ∣∣ = 1 2 , where 〈x, y〉 = x1x2 + y1y2 is the usual scalar product. Problem 3.48. We denote by δ(r) the minimal distance between a lattice point and the circle C(O, r) of radius r centered at the origin O of the coordinate system in the plane. Prove that lim r→∞ δ(r) = 0. Problem 3.49. (Putnam Competition) Consider a curve C of length l that divides the surface of the unit sphere into two parts of equal area. Show that l ≥ 2π . Problem 3.50. (Amer. Math. Monthly) Let K be a planar closed curve of length 2π . Prove that K can be inscribed in a rectangle of area 4. Problem 3.51. 1. Consider a family of plane convex sets with area a, perimeter p, and diameter d. If the family covers area A, then there exists a subfamily with pairwise disjoint interiors that covers at least area λA, where λ = a a+pd+πd2 . 2. Assume that any two members of the family have nonempty intersection. Prove that there exists then a subfamily with pairwise disjoint interiors that covers area at least μA, where μ = a πd2 . Problem 3.52. (Putnam Compettion 1957) Let C be a regular polygon with k sides. Prove that for every n there exists a planar set S(n) ⊂ R2 such that any subset consisting of n points of S(n) can be covered by C, but S(n) itself cannot be included in C. Problem 3.53. Let M be a convex polygon and let S1, . . . , Sn be pairwise disjoint disks situated in the interior of M . Does there exist a partition M = D1 ∪ · · · ∪ Dn such that Di are convex disjoint polygons, each of which contains precisely one disk? Problem 3.54. Consider an inscribable n-gon partitioned by means of n − 2 nonin- tersecting diagonals into n− 2 triangles. Prove that the sum of the radii of the circles inscribed in these triangles does not depend on the particular partition. Problem 3.55. (Elemente der Mathematik) Prove that in an ellipse having axes of lengths a and b and total length L, we have L > π(a + b). Problem 3.56. (Kvant) Let F be a convex planar domain and F ′ denote its image by a homothety of ratio − 12 . Is it true that one can translate F ′ in order for it to be contained in F ? Can the constant 12 be improved? Generalize to n dimensions. 3.3 Geometric Inequalities 27 Problem 3.57. (Nieuw Archief v.Wiskunde)A classical theorem, due to Cauchy, states that a strictly convex polyhedron in R3 whose faces are rigid must be globally rigid. Here, rigidity means continuous rigidity in the sense that any continuous deformation of the polyhedron in R3 that keeps the lengths of edges fixed is the restriction of a deformation of rigid Euclidean motions of three-space. Prove that a 3-dimensional cube immersed in Rn remains rigid for all n > 3. Problem 3.58. Consider finitely many great circles on a sphere such that not all of them pass through the same point. Show that there exists a point situated on exactly two circles. Deduce that if we have a set of n points in the plane, not all of them lying on the same line, then there must exist one line passing through precisely two points of the given set. Problem 3.59. Given a finite set of points in the plane labeled with +1 or −1, and not all of them collinear, show that there exists a line determined by two points in the set such that all points of the set lying on that line are of the same sign. Problem 3.60. (Amer. Math. Monthly) If Q is a given rectangle and ε > 0, then Q can be covered by the union of a finite collection S of rectangles with sides parallel to those of Q in such a way that the union of every nonoverlapping subcollection of S has area less than ε. Problem 3.61. Prove that the 3-dimensional ball cannot be partitioned into three sets of strictly smaller diameter. 3.3 Geometric Inequalities Problem 3.62. If a, b, c, r, R are the usual notations in the triangle, show that 1 2rR ≤ 1 3 (∑ 1 a )2 ≤ ∑ 1 a2 ≤ 1 4r2 . Problem 3.63. (Amer. Math. Monthly) If a, b, c are the sides of a triangle, then prove that (b+c) 2 4bc ≤ mawa and b 2+c2 2bc ≤ maka , where ma,wa, ka denote respectively the lengths of the median, bisector, and altitude issued from A. Problem 3.64. (Putnam Competition) If S(x, y, z) is the area of a triangle with sides x, y, z, prove that √ S(a, b, c) +√S(a′, b′, c′) ≤ √S(a + a′, b + b′, c + c′). Problem 3.65. (Nieuw Archief v. Wiskunde) It is known that in any triangle we have the inequality 3 √ 3r ≤ p ≤ 2R + (3√3 − 4)r, where p denotes the semiperimeter. Prove that in an obtuse triangle we have (3 + 2√2)r < p < 2R + r. 28 3 Geometry Problem 3.66. Prove the Euler inequality R ≥ 2r. Problem 3.67. Prove that in a triangle we have the inequalities 36r2 ≤ a2 + b2 + c2 ≤ 9R2. Problem 3.68. (Amer. Math. Monthly) 1. Let ABC and A′B ′C′ be two triangles. Prove that a2 a′ + b 2 b′ + c 2 c′ ≤ R2 (a ′ + b′ + c′)2 a′b′c′ . 2. Derive that a2 + b2 + c2 ≤ 9R2, cosA cosB cosC ≤ 1 8 . Problem 3.69. (Elemente der Mathematik) Prove that the following inequalities hold in a triangle: 4 ∑ cyclic hAhB ≤ 12S √ 3 ≤ 54Rr ≤ 3 ∑ cyclic ab ≤ 4 ∑ cyclic rArB. Problem 3.70. (Amer. Math. Monthly) Prove that in an any triangle ABC, we have √ 1 + 8 cos2 B sin A + √ 1 + 8 cos2 C sin B + √ 1 + 8 cos2 A sin C ≥ 6. Problem 3.71. Let P be a point in the interior of the triangle ABC. We denote by Ra,Rb, Rc the distances from P to A,B,C and by ra, rb, rc the distances to the sides BC,CA,AB. Prove that ∑ cyclic R2a sin2 A ≤ 3 ∑ cyclic r2a , with equality if and only if P is the Lemoine point (i.e., the symmedian point). Problem 3.72. Prove the inequalities 16Rr − 5r2 ≤ p2 ≤ 4R2 + 4Rr + 3r2. Problem 3.73. Prove the following inequalities, due to Roché: 2R2 + 10Rr − r2 − 2(R − 2r) √ R2 − 2Rr ≤ p2 ≤ 2R2 + 10Rr − r2 + 2(R − 2r) √ R2 − 2Rr. 4 Analysis Problem 4.1. (Amer. Math. Monthly) Prove that z ∈ C satisfies |z| − �z ≤ 12 if and only if z = ac, where |c − a| ≤ 1. We denote by �z the real part of the complex number z. Problem 4.2. Let a, b, c ∈ R be such that a + 2b + 3c ≥ 14. Prove that a2 + b2 + c2 ≥ 14. Problem 4.3. Let fn(x) denote the Fibonacci polynomial, which is defined by f1 = 1, f2 = x, fn = xfn−1 + fn−2. Prove that the inequality f 2n ≤ (x2 + 1)2(x2 + 2)n−3 holds for every real x and n ≥ 3. Problem 4.4. (Amer. Math. Monthly) Prove the inequality min ( (b − c)2, (c − a)2, (a − b)2 ) ≤ 1 2 ( a2 + b2 + c2 ) . Generalize to min1≤k 1 for r �= s. Prove that ∞∑ n=1 1 a3n converges. Problem 4.12. (Amer. Math. Monthly) Consider the sequence Sn given by Sn = n + 12n+1 n∑ i=1 2i i . Find limn→∞ Sn. Problem 4.13. (Putnam Competition 1951) Prove that whenever a, b > 0 we have ∫ 1 0 ta−1 1 + tb dt = 1 a − 1 a + b + 1 a + 2b − 1 a + 3b + · · · . Problem 4.14. (Putnam Competition 1951) Let a, b, c, d ∈ Z∗+, and r = 1 − ab − cd . If r > 0 and a + c ≤ 1982, then r > 119833 . Problem 4.15. (Amer. Math. Monthly) Let ai ∈ R. Prove that nmin(ai) ≤ n∑ i=1 ai − S ≤ n∑ i=1 ai + S ≤ nmax(ai), where (n − 1)S2 =∑1≤i 2005? Also, prove that lim n→ an n = 1 3 . Problem 4.30. Compute the integral f (a) = ∫ 1 0 log(x2 − 2x cos a + 1) x dx. Problem 4.31. (Amer. Math. Monthly) Let −1 < a0 < 1, and define an =( 1 2 (1 + an−1) )1/2 for n ≥ 1. Find the limits A,B, and C of the sequences An = 4n(1 − an), Bn = a1 · · · an, Cn = 4n(B − a1a2 · · · an). Problem 4.32. (Amer. Math. Monthly) Let consider the sequence given by the recur- rence a1 = a, an = a2n−1 − 2. Determine those a ∈ R for which (an) is convergent. 4 Analysis 33 Problem 4.33. Let 0 < a < 1 and I = (0, a). Find all functions f : I → R satisfying at least one of the conditions below: 1. f is continuous and f (xy) = xf (y) + yf (x). 2. f (xy) = xf (x) + yf (y). Problem 4.34. If a, b, c, d ∈ C, ac �= 0, prove that max(|ac|, |ad + bc|, |bd|) max(|a|, |b|)max(|c|, |d|) ≥ −1 + √5 2 . Problem 4.35. (Putnam Competition) Let ∑∞i=1 xi be a convergent series with de- creasing terms x1 ≥ x2 ≥ · · · ≥ xn ≥ · · · > 0 and let P be the set of numbers which can be written in the form ∑ i∈J xi for some subset J ∈ Z+. Prove that P is an interval if and only if xn ≤ ∞∑ i=n+1 xi for every n ∈ Z+. Problem 4.36. (Amer. Math. Monthly) Does there exist a continuous function f : (0,∞) → R such that f (x) = 0 if and only if f (2x) �= 0? What if we require only that f be continuous at infinitely many points? Problem 4.37. (Kvant) Find the smallest number a such that for every real polynomial f (x) of degree two with the property that |f (x)| ≤ 1 for all x ∈ [0, 1], we have |f ′(1)| ≤ a. Find the analogous number b such that |f ′(0)| ≤ b. Problem 4.38. Let f : R → R be a function for which there exists some constant M > 0 satisfying |f (x + y) − f (x) − f (y)| ≤ M, for all x, y ∈ R. Prove that there exists a unique additive function g : R → R such that |f (x) − g(x)| ≤ M, for all x ∈ R. Moreover, if f is continuous, then g is linear. Problem 4.39. (Amer. Math. Monthly) Show that if f is differentiable and if lim t→∞ ( f (t) + f ′(t)) = 1, then lim t→∞ f (t) = 1. Problem 4.40. (Putnam Competition) Let c be a real number, and let f : R → R be a smooth function of class C3 such that limx→∞ f (x) = c and limx→∞ f ′′′(x) = 0. Show that limn→∞ f ′(x) = limx→∞ f ′′(x) = 0. 34 4 Analysis Problem 4.41. (Putnam Competition) Prove that the following integral equation has at most a continuous solution on [0, 1] × [0, 1]: f (x, y) = 1 + ∫ x 0 ∫ y 0 f (u, v) du dv. Problem 4.42. (Putnam Competition) Find those λ ∈ R for which the functional equation ∫ 1 0 min(x, y)f (y) dy = λf (x) has a solution f that is nonzero and continuous on the interval [0, 1]. Find these solutions. Problem 4.43. (Amer. Math. Monthly) Let X be an unbounded subset of the real numbers R. Prove that the set AX = {t ∈ R; tX is dense modulo 1} is dense in R. Problem 4.44. (Putnam Competition) Consider P(z) = zn + a1zn−1 + · · · + an, where ai ∈ C. If |P(z)| = 1 for all z satisfying |z| = 1, then a1 = · · · = an = 0. Problem 4.45. Let I ⊂ R be an interval and u, v : I → R smooth functions satisfy- ing the equations u′′(x) + A(x)u(x) = 0, v′′(x) + B(x)v(x) = 0, where A,B are continuous on I and A(x) ≥ B(x) for all x ∈ I . Assume that v is not identically zero. If α < β are roots of v, then there exists a root of u that lies within the interval (α, β), unless A(x) = B(x), in which case u and v are proportional for α ≤ x ≤ β. Problem 4.46. (Amer. Math. Monthly) LetV be a finite-dimensional real vector space and f : V → R a continuous mapping. For any basis B = {b1, b2, . . . , bn} of V , consider the set EB = {z1b1 + · · · + znbn, where zi ∈ Z}. Show that if f is bounded on EB for any choice of the basis B, then f is bounded on V . Problem 4.47. (Amer. Math. Monthly) It is known that if f, g : C → C are entire functions without common zeros then there exist entire functions a, b : C → C such that a(z)f (z) + b(z)g(z) = 1 for all z ∈ C. 1. Prove that we can choose a(z) without any zeros. 2. Is it possible to choose both a and b without zeros. Problem 4.48. (Amer. Math. Monthly) Consider a compact set X ⊂ R. Show that a necessary and sufficient condition for the existence of a monic nonconstant polyno- mial with real coefficients h ∈ R[x] such that |h(x)| < 1 for all x ∈ X is the existence of monic nonconstant polynomial g(x) ∈ R[x] such that |g(x)| < 2 for all x ∈ X. Prove that 2 is the maximal number with this property. Part II Solutions and Comments to the Problems 5 Number Theory Solutions Problem 1.1. Show that we have Ckn ≡ 0 (mod 2) for all k satisfying 1 ≤ k ≤ n − 1 if and only if n = 2β , where β ∈ Z∗+. Here Ckn denotes the number of combinations, i.e., the number of ways of picking up a subset of k elements from a set of n elements. Known also as the binomial coefficient or choice number and sometimes denoted as( n k ) it is given by the formula Ckn = ( n k ) = n! k!(n − k)! where the factorial n! represents n! = 1 × 2 × 3 × · · · × n. Solution 1.1. Denote by expk(m) the maximal exponent of m in k, i.e., the maximal r such that mr divides k. Since Ckn = n!k!(n−k)! , the exponent of 2 in Ckn has the value expCkn (2) = expn!(2) − expk!(2) − exp(n−k)!(2), which is, by hypothesis, strictly positive for all considered k. Now, the exponent of 2 in a factorial is given by the following formula: expm!(2) = ∞∑ i=1 [m 2i ] , where the brackets denote the integer part. This implies that our claim is equivalent to the inequality ∑ i=1 [ n 2i ] − ∑ i=1 [ k 2i ] − ∑ i=1 [ n − k 2i ] > 0. Let us now suppose that n = 2β + s, with 1 ≤ s < 2β . If we take k = s, then the exponent above can be calculated as 38 5 Number Theory Solutions ([ 2β + s 2 ] + · · · ) − ([ s 2 ] + · · · ) − ([ 2β 2 ] + · · · ) = 0, contradicting our assumptions. Thus, it is necessary that n = 2β . The sufficiency is established as follows. We have the identity (a+b)2 = a2 +b2 in Z/2Z. Using induction on β, one proves that (x + y)2β = x2β + y2β ∈ Z/2Z. In particular, (x + 1)2α = x2α + 1, and therefore, by identifying the coefficients of the binomial expansion on the left- and right-hand sides, we obtain the claim. Comments 1 Using the formula Ckn = Ckn−1 + Ck−1n−1 , we find that Ckn ≡ 1 (mod 2)for all k ≤ n if and only if n = 2α − 1. Problem 1.2. Let P = anxn + · · · + a0 be a polynomial with integer coefficients. Suppose that there exists a number p such that: 1. p does not divide an; 2. p divides ai , for all i ≤ n − 1; 3. p2 does not divide a0. Then P is an irreducible polynomial in Z[x]. Solution 1.2. Let ϕ : Z → Z/pZ be the reduction modulo p and let ϕ : Z[x] → Z/pZ[x] be the map consisting of taking the reduction modulo p of all coefficients. This map is actually a homomorphism of rings, meaning that it respects addition and multiplication of polynomials. We have ϕ(P (x)) = ϕ(an)xn �= 0. Assume that P is not irreducible, and so P = P1 · P2. Then ϕ(P ) = ϕ(P1) · ϕ(P2). Moreover, the only way to write ϕ(an)xn ∈ Z/pZ[x] as a product of two polynomials is βxk · γ xl , where k + l = n and βγ = ϕ(an). In particular, if we write P1(x) = β0 + · · · + βkxk , P2(x) = γ0 + · · · + γlxl , then ϕ(βk) = β, ϕ(βi) = 0 if i < k, and ϕ(γi) = γ, ϕ(γi) = 0 if i < l. Eventually, this yields a0 = β0γ0 ≡ 0 (mod p2), which contradicts our assumptions. Comments 2 This result is known as the Eisenstein criterion. Using this result we can give an alternative solution to Problem 1.1 above. Assume that Ckn ≡ 0 (mod 2) for k ∈ {1, . . . , n − 1}. Then the Eisenstein criterion shows that the polynomial P(x) = (x + 1)n + 1 is irreducible. This means that Q(x) = P(x − 1) = xn + 1 is equally irreducible. On the other hand, if we can write n = 2a(2b + 1), with a, b ∈ Z+, then xn + 1 is divisible by x2a + 1; thus irreducibility of Q(x) implies that b = 0. Therefore, n = 2a , as claimed. Problem 1.3. Given mi, bi ∈ Z+, i ∈ {1, 2, . . . , n} such that gcd(mi,mj ) = 1 for all i �= j , there exist integers x satisfying x ≡ bi (mod mi) for all i. This result is usually known as the Chinese remainder theorem. 5 Number Theory Solutions 39 Solution 1.3. For n = 2, we have m1Z + m2Z = Z. Therefore there exist ai ∈ miZ such that a1 +a2 = 1. Then x = a1b2 +a2b1 satisfies the conditions of the statement. Since mi are relatively prime, for each j ∈ {1, 2, . . . , n} there exists yj ∈ Z sat- isfying the conditions yj ≡ 1 (mod mj) and yj ≡ 0 (mod m1 · · ·mj−1mj+1 · · ·mn). Then take x = b1y1 + · · · + bnyn, which satisfies the system of congruences. Problem 1.4. If gcd(a,m) = 1, then it is a classical result of Euler that we have the following congruence: aϕ(m)+1 ≡ 0 (mod m), where ϕ(m) is the Euler totient function, which counts how many positive integers smaller than m are relatively prime to m. Prove that this equality holds precisely for those numbers a,m such that for any prime number p that divides a, if pk divides m then pk also divides a. Solution 1.4. Let p be a prime number and e the maximal exponent such that pe divides m. If p divides a, then pe divides a. If p does not divide a, then aϕ(pe) ≡ 1 (mod pe), because ϕ(m) is a multiple of ϕ(pe). Therefore aϕ(m) ≡ 1 (mod pe). This implies that pe divides a(aϕ(m) − 1). Hence m divides a(aϕ(m) − 1). In order to establish the converse, let p be a divisor of a such that pk divides m but pk does not divide a. Then pk does not divide a(aϕ(m) −1), because obviously, p and aϕ(m) − 1 are relatively prime. Consequently, m is not a divisor of a(aϕ(m) − 1). Problem 1.5. Let p > 2 be a prime number and let ak ∈ {0, 1, . . . , p2 − 1} denote the value of kp modulo p2. Prove that p−1∑ k=1 ak = p 3 − p2 2 . Solution 1.5. We have (p − k)p = (−k)p + C1pp(−k)p−1 + · · · + pp ≡ −kp (mod p2). Therefore, ap−k = p2 − ak . In particular, summing up these equalities, we obtain ∑k−1 p−1 ap−k = p2(p − 1) − ∑p−1 k=1 ak , which yields ∑p−1 k=1 ak = p 2(p−1) 2 . Problem 1.6. If k ∈ Z+, then show that [√ k2 + 1 + · · · + √ k2 + 2k ] = 2k2 + 2k, where [x] denotes the integer part, i.e., the largest integer smaller than x. Solution 1.6. We have √ k2 + m − k = m√ k2+m+k , and if 1 ≤ m ≤ 2k, then m 2k+1 ≤ m√ k2+m+k ≤ m 2k . Summing up all terms, we obtain k = 2k∑ m=1 m 2k + 1 ≤ ( 2k∑ m=1 √ k2 + m ) − 2k2 ≤ 2k∑ m=1 m 2k = k + 1 2 , whence the claim. 40 5 Number Theory Solutions Problem 1.7. If n is not a multiple of 5, then P = x4n + x3n + x2n + xn + 1 is divisible by Q = x4 + x3 + x2 + x + 1. Solution 1.7. If ξ is a fifth primitive root of unity, then P(ξ) = ξ4n + ξ3n + ξ2n + ξn + 1 = ξ 5n − 1 ξn − 1 = 0. Hence P = (x − ξ)(x − ξ2)(x − ξ3)(x − ξ4)R = Q · R, for some polynomial R. Comments 3 More generally, x(m−1)n + · · · + 1 is divisible by xm−1 + · · · + 1 if gcd(m, n) = 1. Problem 1.8. Let p be an odd prime and k ∈ Z+. Show that there exists a perfect square the last k digits of whose expansion in base p are 1. Solution 1.8. Use induction on k. It is obvious for k = 1. Let us assume that x2 = 1 + p + · · · + pk−1 + (apk + · · · ) and gcd(x, p) = 1. Then we have (x + cpk)2 = x2 + 2cxpk + c2p2k = 1 + p + · · · + pk−1 + (a + 2cx)pk + · · · . If p is odd, then gcd(2x, p) = 1. Moreover, the congruence 2xc + a ≡ 1 (mod p) has at least one solution c ∈ {0, 1, . . . , p − 1}. If c0 is a solution, then (x + c0pk)2 has its last k + 1 digits equal to 1. Comments 4 The result can be extended to higher exponents r ≥ 2 and prime numbers p that do not divide r , or more generally to p such that gcd(p, r) = 1. Moreover, the last digits can be arbitrarily prescribed. Problem 1.9. Any natural number greater than 6 can be written as a sum of two numbers that are relatively prime. Solution 1.9. For odd n ≥ 7, we can write n = 2k + 1 = k + (k + 1), where gcd(k, k+1) = 1. Consider now n = 2m, where m ≥ 3. Recall Chebyshev’s theorem that for any m ≥ 2 there exists a prime number p such that m < p < 2m. Therefore, we have 0 < 2m − p < m. Further, gcd(p, 2m − p) = 1, since 2m > p > m and gcd(m, p) = 1. Problem 1.10. Prove that it is impossible to extract an infinite arithmetic progression from the sequence S = {1, 2k, 3k, . . . , nk, . . .}, where k ≥ 2. Solution 1.10. Let r be the ratio of an arithmetic progression A ⊂ S. If n > r , then (n + 1)k − nk ≥ 2n + 1 > r . Thus the difference between consecutive terms of S grows higher than r , and hence A is finite. Comments 5 The result is also true when we replace S with the sequence an+2 = pan+1 + qan, for 1 ≤ p ≤ q + 1. Problem 1.11. Prove that ba−j+1 divides Cjba if a, b ≥ 2, j ≤ a + 1. 5 Number Theory Solutions 41 Solution 1.11. An easy induction on n shows that bn ≥ n+ 1, and hence ba ≥ j . We write the combinations (also called binomial coefficients) as C j ba = 1 j ! j−1∏ i=0 (ba − i). Let i = brm, where gcd(m, b) = 1, so that ba − i = br(ba−r − m). This shows that i and ba − i are divisible by the same power of b. Therefore, the exponent of b in Cjba equals the result of subtracting from a the exponent of b in j !. The latter is at most j − 1 because bj−1 ≥ j , and hence the claim. Problem 1.12. Solve in integers the following equations: (1) x2 = y2 + y3; (2) x2 + y2 = z2. Solution 1.12. 1. The plane curve determined by the first equation has the following form: x y 0 D T P We will look for a parameterization of this curve by projecting from the origin 0 onto the line D given by y = 1. Thus, to each point P of the curve we associate the point T , which is the intersection of the half-line 0P with the line D. Denote by t the natural parameter of D. This amounts to setting y = 1 t x within the equation, which immediately yields x = t3 − t and y = t2 − 1. If x, y ∈ Z, then t ∈ Q. Moreover, t2 = y + 1 ∈ Z, but a rational number whose square is an integer must be an integer itself. Thus the integer solutions are given by the family x = t3 − t and y = t2 − 1, where t ∈ Z. 2. Dividing by z2, we reduced the problem of finding the rational solutions of the equation x2 + y2 = 1. This is a circle of unit radius in the plane. We now use the stereographic projection from the point P(−1, 0) onto the vertical line x = 1 in order to find a more-convenient parameterization of the circle. P Q X T 42 5 Number Theory Solutions If t is the new parameter on the vertical line, then x2 + y2 = 1 and x+12 = yt . This yields x = 1−s21+s2 and y = 2s1+s2 , where s = t2 . Therefore the integer solutions of our equation are given by x = (r2 − s2)v, y = 2vrs, and z = (r2 + s2)v, where s, r, v ∈ Z. The primitive solutions, for which gcd(x, y, z) = 1, are those for which v = 1 and gcd(r, s) = 1. The triples (x, y, z) are called Pythagorean triples. Comments 6 The two Diophantine equations above correspond to rational algebraic curves, i.e., to curves in the plane that admit a natural parameterization by means of rational functions. This situation is quite exceptional. Comments 7 The Pythagorean equation x2 +y2 = z2 can be generalized by adding more variables and still keeping its rational behavior to x21 + x22 + · · · + x2n+1 = z2, n ≥ 1. The general (primitive) solution of this equation in integers can be obtained by the same method as above and reads x1 = s21 + s22 + · · · + s2n − u2, x2 = 2s1u, x3 = 2s2u, ... ... xn+1 = 2snu, z = s21 + s22 + · · · + s2n + u2, for arbitrary integer parameters s1, s2, . . . , sn, u ∈ Z. Comments 8 Another famous generalization of the Pythagorean equation is xk + yk = zk, k ≥ 3, which was conjectured by Fermat to have no nontrivial solutions for k ≥ 3. Several particular cases were solved over the years (k = 4 by Fermat, k = 3 by Euler, k = 5, 14 by Dirichlet, k = 7 by Lamé, etc). However, it took more than three hundred years for the conjecture be eventually settled by A.Wiles in 1995. Comments 9 Euler considered the more general equation xk1 + xk2 + · · · + xkn = zk. He conjectured that there are no nontrivial solutions unless n ≥ k and proved it for n = 3. However, Lander and Parkin found counterexamples for n = 5, for example, 275 + 845 + 1105 + 1335 = 1445, 5 Number Theory Solutions 43 and later, N. Elkies settled the case n = 4 by finding another counterexample: 2 682 4404 + 15 365 6394 + 18 796 7604 = 20 615 6734. A subsequent computer search by R. Frye found the following minimal solution for n = 4 (which is unique in the range z ≤ 1 000 000): 95 8004 + 217 5194 + 414 5604 = 422 4814. See also • N.D. Elkies: On A4 + B4 + C4 = D4, Math. Comp. 51 (1988), 184, 825–835. • J.L. Lander and T.R. Parkin: A counterexample to Euler’s sum of powers conjec- ture. Math. Comp. 21 (1967), 101–103. Problem 1.13. Let a, b, c, d ∈ Z+ be such that at least one of a and c is not a perfect square and gcd(a, c) = 1. Show that there exist infinitely many natural numbers n such that an + b, cn + d are simultaneously perfect squares if one of the following conditions is satisfied: 1. b and d are perfect squares; 2. a + b, c + d are perfect squares; 3. a(d − 1) = c(b − 1). Moreover, there do not exist such numbers if a = 1, b = 0, c = 4k2 − 1, d = 1. Solution 1.13. There exists n such that an + b, cn + d are perfect squares iff the equation ax2 − cy2 = ad − bc has integer solutions. In fact, √ cn + d,√an + b are solutions of the equation. Con- versely, if (x, y) is a solution, then a(x2 − d) = c(y2 − b). Since gcd(a, c) = 1, we find that c divides x2−d, while a divides y2−b. In particular, we obtain that x2 − d = kc and y2 − b = ka. Further, ac is not a perfect square and so the Pell equation u2 − acv2 = 1 has infinitely many solutions (un, vn). If (x, y) is a solution of our equation, then xn = unx + dvny, yn = uny + vnx form an infinite sequence of solutions. Thus it suffices to determine one solution for the equation in order to find infinitely many. The first two cases are immediate, and the third corresponds to the solution (1, 1). Finally, note that the Pell-type equation x2 − (4k2 − 1)y2 = −1 has no integer solutions, by modulo 4 considerations. See also: 44 5 Number Theory Solutions • T. Andreescu and D. Andrica, Quadratic Diophantine Equations, Springer Mono- graphs in Math., 2006. Problem 1.14. If N = 2 + 2√28n2 + 1 ∈ Z for a natural number n, then N is a perfect square. Solution 1.14. If √ 28n2 + 1 = k, then N = 2 + 2k. If N ∈ Z, then k ∈ 12Z. But k is the square root of an integer, and whenever such a square root is a rational number, then it is actually an integer. This proves that k ∈ Z. Now, 28n2 = (k − 1)(k + 1), and therefore k is odd, k = 2t + 1. Thus, 7n2 = t (t + 1). We have two cases: 1. If 7 divides t + 1, then, t = 7s − 1 and so (7s − 1)s = n2, for s ∈ Z. But gcd(s, 7s − 1) = 1 and thus 7s − 1 = n21 and s = n22, where n1n2 = n. Moreover, any square, in particular n21, is congruent to 1, 2, or 4 (mod 7). In particular, it cannot be equal to 7s − 1. 2. If 7 divides t , i.e., t = 7s, then n2 = s(7s + 1). Since gcd(s, 7s + 1) = 1, we have 7s + 1 = n21 and s = n22. This implies that N = 2 + 2k = 2 + 2(14n22 + 1) = 4(7n22 + 1) = 4(7s + 1) = 4n21. Problem 1.15. Let n ≥ 5, 2 ≤ b ≤ n. Prove that [ (n − 1)! b ] ≡ 0 (mod b − 1). Solution 1.15. We have four cases to consider: 1. If b < n, then (n−1)! is divisible by b(b−1) and therefore (n−1)! b is an integer divisible by b − 1. 2. Suppose b = n, where n is not prime and not the square of a prime number, hence n = rs, where 1 < r < s < n. Since gcd(n, n − 1) = 1 and s < n − 1, we derive that (n − 1)! contains the factors r, s, and n − 1, and therefore it is divisible by rs(n − 1) = b(b − 1). 3. If b = n is the square of a prime, i.e., n = p2, then p ≥ 3 and hence 1 < p < 2p < 2p2 −1 = n−1. This implies that (n−1)! is divisible by the product of p, 2p, and n − 1, i.e., by 2p2(n − 1) = 2b(b − 1). 4. If b = n is a prime number p, then by Wilson’s theorem we have (p−1)!+1 ≡ 0 (mod p), [ (p − 1)! p ] = [ (p − 1)! + 1 p − 1 p ] = (p − 1)! + 1 p −1 = (p−1) ( (p − 2)! − 1 p ) . Now gcd(p, p − 1) = 1, so p − 1 divides [ (p−1)! p ] . Problem 1.16. Prove that for every natural number n, there exists a natural number k such that k appears in exactly n nontrivial Pythagorean triples. Solution 1.16. Let us show that 2n+1 appears in exactly n triples. If n = 0, it amounts to saying that 2 does not appear in any such triple, which is immediate. 5 Number Theory Solutions 45 Let us now use induction on n. All primitive Pythagorean triples are given by x = u2 − v2, y = 2uv, z = u2 + v2, where gcd(u, v) = 1 and u, v are not both odd and u > v. If u, v are both odd, then y = 2uv = 2n+1 and therefore u = 2n and v = 1. The nonprimitive triples in which 2n+1 appears are those divisible by 2, and thus they are in bijection with the Pythagorean triples in which 2n appears. By the recurrence hypothesis we have n such triples. Therefore 2n+1 appears in n nonprimitive triples and one primitive triple, and thus in n + 1 such triples. Problem 1.17. Let n, q ∈ Z+ be such that all prime divisors of q are greater than n. Show that (q − 1)(q2 − 1) · · · (qn−1 − 1) ≡ 0 (mod n!). Solution 1.17. Let p ≤ n, prime. Since p is not a divisor of qk , we have qk(p−1) ≡ 1 (mod p), for all k. Therefore at least [ (n−1) (p−1) ] factors from the left-hand side are divisible by p. Let us now compute the exponent of p in n!, which is expn!(p) = ∞∑ k=1 [ n pk ] < ∞∑ k n pk = n p − 1 . Now expn!(p) ∈ Z, and so the strict inequality above implies that expn!(p) ≤ [ n−1 p−1 ] . Thus the left-hand side is divisible by pexpn!(p). Since this holds for all p ≤ n, the claim follows. Comments 10 Let q = pm, where p > n is a prime number. Let G denote the group of nonsingular matrices over the Galois field with q elements. Then the order of the groupG is |G| = qn(n−1)/2(q−1) · · · (qn−1). Let nowG∗ be the subgroup generated by the diagonal matrices and their permutations. One finds that |G∗| = (q − 1)nn!. Since the order of a subgroup divides the order of the larger group, we derive (q − 1) · · · (qn − 1) ≡ 0 (mod (qn − 1)nn!), improving the relation above. Problem 1.18. Every natural number n ≥ 6 can be written as a sum of distinct primes. Solution 1.18. Let n ≥ 15. One knows by Chebyshev’s theorem that there exists a prime p between n2 and n. Then 0 < n− p < n2 . Further, there exists a prime q such that n−p2 < q < n−p, and the rest n−p− q satisfies therefore 0 < n−p− q < n4 . We continue this process until there remains either a prime number or a number from the set {7, 8, . . . , 14}. Since 7 = 5 + 2, 8 = 5 + 3, 9 = 7 + 2, 10 = 7 + 3, 11 = 11, 12 = 7 + 5, 13 = 13, and 14 = 2 + 5 + 7, the claim follows. 46 5 Number Theory Solutions Problem 1.19. Every number n ≥ 6 can be written as a sum of three numbers which are pairwise relatively prime. Solution 1.19. By Chebyshev’s theorem , there exists a prime p such that n2 < p < n. Note next that a natural number m such that m is divisible by all prime numbers strictly smaller than m must be 2. In fact, otherwise, again by Chebyshev’s theorem, there exists a prime bigger than m2 , which cannot divide m. Thus there exists a prime q such that n − p is not divisible by q, and so gcd(n − p, q) = 1. If gcd(p, n−p−q) > 1, then n−p−q = ap, which cannot happen, since p > n2 and a �= 0, since n − p − q > 0. If gcd(q, n − p − q) > 1, then n − p − q = aq, which contradicts our choice of q. Thus p, q, n − p − q satisfy the claim. Problem 1.20. Find all pairs of integers (m, n) such that Cnm = 1984, where Cnm = m!n!(m−n)! denotes the usual binomial coefficient. Solution 1.20. We consider 0 ≤ n ≤ m2 . We have 1984 = 26 · 31. Since 1984 = (n+1)(n+2)···(m−1)m (m−n)! , then either 31 divides (n+ 1)(n+ 2) · · · (m− 1)m (and does not divide (m − n)!), so that m ≥ 31, or else 31 also divides (m − n)! and hence again m ≥ 31. If n ≥ 3, then Cnm ≥ C3m ≥ C331 = 16 (29 · 30 · 31) > 64 · 31 = 1984. Thus n ∈ {0, 1, 2}. Finally, if n = 1, then m = 1984, while C2m = 1984 does not have any solutions in natural numbers. Thus, the solutions are (1984, 1), (1984, 1983). Comments 11 More generally, if N has a prime divisor p > k ( k √ N + 1 ) , k < p 2 , then the equation Cnm = N has at most 2k solutions. One derives, furthermore, that the number F(N) of such solutions can be estimated from above: F(N) = O(logN/ log logN). Recall that F(N) = O(g(N)) if limN→∞ F(N)g(N) < ∞. It is conjectured that F(N) is uniformly bounded, independently of N . Problem 1.21. Find the set A consisting of natural numbers n that are divisible by all odd natural numbers a with a2 < n. Solution 1.21. Let p1 = 3, p2, . . . , pm be the first m odd primes. Choose k such that (2k − 1)2 < n ≤ (2k + 1)2 and m maximal such that pm ≤ 2k − 1. By hypothesis, pm+1 > 2k + 1 and hence, n < p2m+1. Assume that n is in A. Then the pi divide n for all i ≤ m, and thus p1p2 · · ·pm divides n. Moreover, the Bonse–Pósa inequality states that p1 · · ·pm > p2m+1, m ≥ 4. 5 Number Theory Solutions 47 Thus m ≤ 3. If m = 3, then n ≤ 169 and 105 divides n. However, n = 105 does not belong to A, since it is not divisible by 9. If m = 2, then n ≤ 49 and n is divisible by 15. If m = 1, then n ≤ 25 is divisible by 3. If m = 0, then n ≤ 9. The answer is thereforeA = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 24, 30, 45}. Comments 12 Panaitopol recently gave an elementary proof of a more-general Bonse–Pósa inequality p1p2 · · ·pn > pn−π(n)n+1 , for n ≥ 2, which implies that p1p2 · · ·pn > pkn+1 for n > 2k. This has been further improved by Alzer and Berg. • L. Panaitopol: An inequality involving prime numbers, Univ. Beograd. Publ. Elek- trotehn. Fak. Ser. Mat. 11 (2000), 33–35. • H. Alzer, C. Berg: Some classes of completely monotonic functions II, The Ra- manujan Journal 11(2006), 225–248. Problem 1.22. Prove that 21092 − 1 is divisible by 10932. Solution 1.22. Set p = 1093 and thus p2 = 1194649. Observe that 37 = 2187 = 2p + 1, 314 ≡ 4p + 1 (mod p2), 214 = 16348 = 15p − 11, 228 ≡ −330p + 121 (mod p2), 32 · 228 ≡ −2970p + 1089 ≡ −1876p − 4 (mod p2), and therefore 32 · 226 ≡ −469p − 1 (mod p2). Therefore we infer that 314 · 2182 ≡ −(489p + 1)7 ≡ −3283p − 1 ≡ −4p − 1 ≡ −314 (mod p2), and using the fact gcd(314, p2) = 1, we obtain 2182 ≡ −1 (mod p2), yielding 21092 − 1 ≡ 0 (mod 10932). Problem 1.23. Let n ≥ 0, r > 1, and 0 < a ≤ r be three integers. Prove that the number n, when written in base r , has precisely ∞∑ k=1 [ nr−k + ar−1 ] − [ nr−k ] digits that are greater than or equal to r − a. Solution 1.23. We write n = n1 + n2r + n3r2 + · · · in base r . Then the digits nj are computed by the formulas n1 = n − [n r ] r, n2 = [n r ] − [ n r2 ] r, . . . , nk = [ n rk−1 ] − [ n rk ] r. For a fixed k, we have nk ≥ r − a if and only if [ n rk−1 ] − [ n rk ] r ≥ r − a, which is equivalent to asking that [ 1 r ([ n rk−1 ] + a )] ≥ [ n rk ] + 1. Now let n = mrk−1 + s, where 0 ≤ s < rk−1. Then 48 5 Number Theory Solutions [ 1 r ([ n rk−1 ] + a )] = [m r + a r ] = [m r + s rk + a r ] = [ n rk + a r ] and therefore nk ≥ r − a if and only if [ n1 rk + a r ] > [ n rk ] . However, we have − [ n rk ] + [ n rk + a r ] = { 0, if nk < r − a, 1, if nk ≥ r − a. Thus, the total number of digits greater than r − a is ∞∑ k=1 ([ n rk + a r ] − [ n rk ]) as claimed. Comments 13 The same argument shows that given a sequence ak ∈ {1, 2, . . . , r}, the number of digits nk of n written in base r that satisfy the inequalities nk ≥ r − ak is ∞∑ k=1 ([ n rk + ak r ] − [ n rk ]) . Problem 1.24. Find a pair (a, b) of natural numbers satisfying the following proper- ties: 1. ab(a + b) is not divisible by 7; 2. (a + b)7 − a7 − b7 is divisible by 77. Solution 1.24. Since a, b, a + b �≡ 0 (mod 7), there exists the inverse a−1 of a in Z/77Z; moreover, k = ba−1 �≡ 0,−1 (mod 77). Then 77 divides (a−1)7((a + b)7 − a7 − b7) = (1 + k)7 − 1 − k7. Developing the latter, we find that 7k(k5 + 3k4 + 5k3 + 5k2 + 3k + 1) ≡ 0 (mod 77); thus (k + 1)(k4 + 2k3 + 3k2 + 2k + 1) ≡ 0 (mod 76). Since k �≡ −1 (mod 77), we obtain (k2 + k + 1)2 ≡ 0 (mod 76), and so we have to solve in integers the equation k2 +k+1 = 73s. The discriminant is D = 4·73s−3, which has to be a squarep2. We need then to findp such that 73 divides p2 + 3. We note that 72 divides p2 + 3 only if either p = 12 + 49q or p = 37 + 49q. By replacing above, we find in the first case p2 + 3 = 49(24q + 3) ≡ 0 (mod 343), which yields q ≡ −1 (mod 7) and thus k = −19 + 343r . The second case yields the solutions k = 18 + 343r . In particular, (1, 18) and (1, 324) are solutions for the problem. 5 Number Theory Solutions 49 Problem 1.25. Let 0 < a < b < c < d be odd integers such that 1. ad = bc 2. a + d = 2k, b + c = 2m, for some integers k and m. Prove that a = 1. Solution 1.25. Let a b = c d = x y , where gcd(x, y) = 1. One obtains then, from the other equations, the value a = x2 m(y − 2k−mx) y2 − x2 . Since a is odd, we have y2 − x2 = 2ms for some integer s. Recall that x, y are odd, since they divide a and b, respectively. We obtain, then, the following types of solutions: 1. { y = 2m−2s1 + s2, x = 2m−2s1 − s2, 2. { y = s1 + 2m−2s2, x = s1 − 2m−2s2, where the si are both integers. Moreover, when replacing x and y by these values in the expression of a, we obtain that s1 must divide 2k−m + 1 and s2 must divide 2k−m − 1. Further, we have the equivalences k > m iff a + d > b + c iff a + bc a > b + c iff (a − b)(a − c) > 0. The inequality a + bc a ≤ 1 + bc implies that k < 2m − 2. Now, in the first type of solutions, we have s1 > 2m−2s2, and s1 divides 2k−m+1, hence s2 = 1, s1 = 2k−m + 1, k = 2m − 2. In the second case, we have y ≤ b < 2m−1 and thus s1 < 2, which yields s1 = 1. Since 3y > 2m−1, we find that x = a, y = b, and replacing in the expression of a, we obtain s2 = 2m−2 − 1, so that a = 1. Problem 1.26. Find those subsets S ⊂ Z+ such that all but finitely many sums of elements from S (possibly with repetitions) are composite numbers. Solution 1.26. Let S∗ = {x =∑ai∈S ai} be the semigroup generated by S. We prove that S∗ contains only composite numbers iff there exists a prime number p, with p /∈ S, such that p divides all elements of S. This condition is obviously sufficient. In order to prove the converse, let us assume the contrary, i.e., that gcd(S) = 1. Then there exists a finite subset T = {c1, . . . , cn} ⊂ S such that gcd(T ) = 1. This implies that there exist integers xi ∈ Z such that n∑ i=1 cixi = 1. 50 5 Number Theory Solutions Separating the terms with xi > 0 and those with xi < 0, we obtain two elements a, b ∈ S∗ with the property that a − b = 1. In particular, gcd(a, b) = 1. From Dirichlet’s theorem, the arithmetic progression {a+kb}k∈Z+ contains infinitely many primes; but a + kb ∈ S∗, contradicting our assumptions. The claim follows. Comments 14 Another proof can be given by using the Chinese remainder theorem and showing that if S is such that gcd(S) = 1, then S∗ is Z+ −A, where A is a finite subset. The problem of determining the maximal element of A is called the Frobenius coin problem. If S = {a1, a2}, then the maximal number is F(S) = a1a2 − a1 − a2, as was proved by Sylvester. If card(S) = 3, the problem was solved by Selmer and Beyer, but for card(S) ≥ 4, there is no closed formula for F(S). Erdo˝s and Graham proved that F({a1 < a2 < · · · < an}) ≤ 2an−1 [an n ] − an, and Brauer that F({a1 < a2 < · · · < an}) ≤ n∑ i=1 ai ( gcd(a1, a2, . . . , ai) gcd(a1, a2, . . . , ai+1) − 1 ) . Curtis proved that there is no closed formula expressing F(S) for n ≥ 4 that involves only polynomials. • A. Brauer: On a problem of partitions, Amer. J. Math. 64 (1942), 299–312. • F. Curtis: On formulas for the Frobenius number of a numerical semi-group, Math. Scand. 67 (1990), 2, 190–192. • P. Erdo˝s and R.L. Graham: On a linear Diophantine problem of Frobenius, Acta Arith. 21 (1972), 399–408. • E.S. Selmer and Ö. Beyer: On the linear Diophantine problem of Frobenius in three variables, J. Reine Angew. Math. 301 (1978), 161–170. Problem 1.27. Prove that for any natural number n > 1, the number 2n − 1 does not divide 3n − 1. Solution 1.27. Set An = 2n−1 and Bn = 3n−1, where n > 1. For even n, 3 divides An, while 3 does not divide Bn, and so An does not divide Bn. Now let n > 1 be odd, of the form n = 2m−1. Using 24 ≡ 22 (mod 12), we infer An ≡ −5 (mod 12). Since any prime greater than three is congruent to either ±5, or ±1 modulo 12, there is at least one prime divisor p of An such that p ≡ ±5 (mod 12). IfAn dividesBn, thenp divides 3Bn and thus 3 ≡ 3n+1 ≡ 32m (modp), which implies that 3 is a quadratic residue modulo p. But this contradicts the quadratic reciprocity theorem (an easy consequence of which is the fact that 3 cannot be a quadratic residue modulo p), since p ≡ ±5 (mod 12). Problem 1.28. Let un be the least common multiple of the first n terms of a strictly increasing sequence of positive integers. Prove that ∞∑ n=1 1 un ≤ 2. 5 Number Theory Solutions 51 Find a sequence for which equality holds above. Solution 1.28. Let ak be the increasing sequence, and uk = lcm(a1, a2, . . . , ak). We have ∞∑ k=1 1 uk = ∞∑ k=1 1 lcm(a1, a2, . . . , ak) ≤ 1 a1 + ∞∑ k=2 1 lcm(ak, ak+1) = 1 a1 + ∞∑ k=2 gcd(ak, ak+1) akak+1 ≤ 1 a1 + ∞∑ k=2 ak+1 − ak akak+1 = 1 a1 + ∞∑ k=2 ( 1 ak − 1 ak+1 ) = 2 a1 ≤ 2. Examples of sequences for which equality holds are ak = 2k−1, and a1 = 1, a2 = 2, a3 = 6, ak = 3 · 2k−2, for k ≥ 4. Problem 1.29. Let ϕn(m) = ϕ(ϕn−1(m)), where ϕ1(m) = ϕ(m) is the Euler totient function, and set ω(m) the smallest number n such that ϕn(m) = 1. If m < 2α , then prove that ω(m) ≤ α. Solution 1.29. If we consider the minimal j with ϕj (m) = 1, then ϕj−1(m) = 2. Also, we have ϕ1(m) < m < 2α , but ϕ1(m) is an even number and thus ϕ2(m) ≤ 1 2ϕ1(m) < 2 α−1 . By induction, we obtain ϕj−1(m) < 2α−(j−2), whence the claim. For even m, we have ω(m) ≤ α − 1. Problem 1.30. Let f be a polynomial with integer coefficients and N(f ) = card{k ∈ Z; f (k) = ±1}. Prove that N(f ) ≤ 2 + deg f , where deg f denotes the degree of f . Solution 1.30. Set N+(f ) = card{k; k ∈ Z, f (k) = 1} and, respectively, N−(f ) = card{k; k ∈ Z, f (k) = −1}. We claim first that either N+(f ) ≤ 2 or N−(f ) ≤ 2. Assume the contrary. Then there exist a, b, c, d, e, g distinct integers such that f (a) = f (b) = f (c) = −1 and f (d) = f (e) = f (g) = +1. Thus we can write f (x) = Q(x)(x − a)(x − b)(x − c) − 1 = R(x)(x − d)(x − e)(x − g) + 1, where Q and R are polynomials with integer coefficients. Consequently, we have the polynomial identity (x − a)(x − b)(x − c)Q(x) = (x − d)(x − e)(x − g)R(x) + 2. Let x take the values a, b, and c, respectively. Then we find that a − d, a − e, a − g divide 2 and therefore a − d, a − e, a − g ∈ {1,−1, 2,−2}. In a similar way, b − d, b − e, b − g ∈ {1,−1, 2,−2} and c − d, c − e, c − g ∈ {1,−1, 2,−2}. We can assume, from symmetry, that a < b < c and g < e < d. We have then the strict inequalities a − d < b− d < e− d < c− e < c− g, and hence at least five distinct difference numbers that belong to {1,−1, 2,−2}, which is absurd. Thus our starting assumption is false. In particular, if N+(f ) ≤ 2, then N(f ) ≤ N(f ) + 2 ≤ deg f + 2. 52 5 Number Theory Solutions Comments 15 The result can be improved as follows: N(f ) − deg f = ⎧ ⎪⎪⎨ ⎪⎪⎩ 1, if deg f = 1, ≤ 2, if deg f ∈ {2, 3}, ≤ 1, if deg f = 4, ≤ 0, if deg f ≥ 5. The solution is analogous. The equality for degree 2 can be attained, for instance, by taking f (x) = x2 − (2k + 3)x + k2 + 3k + 1. Problem 1.31. Prove that every integer can be written as a sum of 5 perfect cubes. Solution 1.31. We have 6n = 03 + (n + 1)3 + (n − 1)3 + (−n)3 + (−n)3, and this is written as a sum of four perfect cubes. Thus 1. 6n + 1 = 6n + 13, 2. 6n + 2 = 6(n − 1) + 23, 3. 6n + 3 = 6(n − 4) + 33, 4. 6n + 4 = 6(n + 2) + (−2)3, 5. 6n + 5 = 6(n + 1) + (−1)3. Comments 16 Observe that we did not require the cubes to be positive. It is still unknown whether every integer is the sum of four cubes, although three cubes do not suffice. For instance, numbers that are congruent to ±4 modulo 9 cannot be written as sums of three cubes, and moreover, it is not yet known whether they can be written as sums of four cubes. If we are looking for positive cubes, then the minimal number needed is 9, this being a particular case of the Waring problem. Notice that any sufficiently large number is a sum of seven positive cubes. The proof is more involved; for instance, see: • G.L. Watson: A proof of the seven cube theorem. J. London Math. Soc. 26 (1951), 153–156. Problem 1.32. If n ∈ Z, then the binomial coefficient Cn2n = (2n)!(n!)2 has an even number of divisors. Solution 1.32. The number of divisors of x, usually denoted by τ(x), is odd if and only if x is a perfect square. In fact, once we have a divisor d , there is a complementary divisor associated with it, namely x d . These divisors are distinct unless x is a square and the divisor is its square root. Further, it is known that for every natural number n, there exists a prime p satis- fying n < p ≤ 2n, and therefore 2n < 2p ≤ 4n. Writing Cn2n = (2n)!n!n! , we find that the prime p appears with exponent 1 in the development of Cn2n and thus it cannot be a perfect square. 5 Number Theory Solutions 53 Problem 1.33. Prove that every n ∈ Z+ can be written in precisely k(n) different ways as a sum of consecutive integers, where k(n) is the number of odd divisors of n greater than 1. Solution 1.33. We have x + (x + 1)+ (x + 2)+ · · · + (x + y) = 12 (y + 1)(2x + y). Therefore, we must find the number of natural solutions (x, y) of the Diophantine equation 12 (y + 1)(2x + y) = n. We have two cases to consider: 1. If y = 2z is even, then (2z + 1)(x + z) = n. If we let u = 2z + 1, then u is an odd divisor of n. There is also associated the complementary divisor v, so that uv = n. Now v = x + z, and thus we find that z = 12 (u− 1) and x = 12 (2v − u+ 1). Therefore (x, z) is an acceptable solution if and only if 2v − u ≥ 1. 2. If y = 2z − 1 is odd, then z(2x + 2z − 1) = n. Let u = 2x + 2z − 1, which is an odd divisor of S, and let v be the complementary divisor: uv = n. Then z = v; hence x = 12 (u − 2v + 1). This implies that (x, z) is an acceptable solution if and only if 2v − u ≤ −1. Now, given u > 1, an odd divisor of n, then either the first inequality or the second (but not both) could hold. Therefore, the number of solutions is that claimed. Problem 1.34. Let π2(x) denote the number of twin primes p with p ≤ x. Recall that p is a twin prime if both p and p + 2 are prime. Show that π2(x) = 2 + ∑ 7≤n≤x sin ( π 2 (n + 2) [ n! n + 2 ]) sin ( π 2 n [ (n − 2)! n ]) for x > 7. Solution 1.34. If n > 5 is composite, then (n−2)! n is an even integer, and therefore sin ( π 2 n [ (n−2)! n ]) = 0. If p is prime, then by Wilson’s theorem, (p − 2)! ≡ −(p − 1)! ≡ 1 (mod p), which implies that [ (p − 2)! p ] = (p − 2)! − 1 p . If p > 5, then 4 divides (p − 2)!, and therefore sin ( π 2 p [ (p − 2)! p ]) = sin (π 2 [(p − 2)! − 1] ) = −1. Consequently, the nth term of the sum is zero if at least one among n, n + 2 is composite, and 1 if both n and n + 2 are prime. Finally, one adds 2 units in order to count the pairs (3, 5), (5, 7). Problem 1.35. 1. Find all solutions of the equation 3x+1 + 100 = 7x−1. 2. Find two solutions of the equation 3x + 3x2 = 2x + 4x2 , and prove that there are no others. 54 5 Number Theory Solutions Solution 1.35. 1. x = 4 is a solution. If x < 4, then 3x+1 + 100 > 7x−1, and if x > 4, then 3x+1 + 100 < 7x−1. Therefore, this solution is unique. 2. Let ρ(x) = 4x2−3x23x−2x . We have ρ′(x) = (2 · 4 x2 ln 4 − 2 · 3x2 ln 3)x(3x − 2x) − (4x2 − 3x2)(3x ln 3 − 2x log 2) (3x − 2x)2 . Therefore ρ′(x) ≥ (4x2 −3x2)(3x −2x)(x−1) ln 3 = μ(x). Next, we have μ(x) = 0 if and only if x ∈ {0, 1}. These values are isolated roots of ρ′, and thus ρ is strictly increasing forx ∈ R+. Therefore, the only two solutions of the equation arex ∈ {0, 1}, because, for other values of x, we have ρ(x) > 1. Problem 1.36. Let σ(n) denote the sum of the divisors of n. Prove that there exist infinitely many integersn such thatσ(n) > 2n, or, even stronger, such thatσ(n) > 3n. Prove also that σ(n) < n(1 + log n). Solution 1.36. 1. Recall that if n = pa11 . . . pakk is the factorization of n into prime factors, then we have the formula σ(n) =∏ki=1 p ai+1 i −1 pi−1 . Observe next that n = 2p3q satisfies the inequality σ(n) > 2n for p, q > 2. Moreover, if n = 2p3q5s , then σ(n) > 3n. 2. We have σ(n) =∑d/n d; hence σ(n) n = ∑ d/n d n = ∑ d/n 1 d = ∑ d/n 1 d < 1 + 1 2 + 1 3 + · · · + 1 n < 1 + log n. Therefore σ(n) < n(1 + log n). Problem 1.37. Let ai be natural numbers such that gcd(ai, aj ) = 1, and ai are not prime numbers. Show that: 1 a1 + · · · + 1 an < 2 Solution 1.37. Each ai has at least one proper prime divisor pi . Thus one can write ai = piqisi , where pi and qi are, not necessarily distinct, primes. Now, gcd(ai, aj ) = 1 implies that pi, qi �= pj , qj . Therefore, the n numbers min(pi, qi) are distinct. We derive that ∑ 1 ai ≤ ∑ 1 piqi ≤ ∑ 1 (min(pi, qj ))2 ≤ ∑ k≥1 1 k2 = ζ(2) = π 2 6 < 2. Comments 17 We denoted above by ζ the Riemann zeta function, defined by ζ(s) = ∑ k≥1 k−s . One knows the values of the zeta function at specific points. Let Bn denote the nth Bernoulli number, determined by the Taylor expansion 5 Number Theory Solutions 55 x ex − 1 = ∞∑ n=0 Bn n! xn. The Bernoulli numbers are rational for all n and have the property that B2n+1 = 0 if n ≥ 1. Their first values are B0 = 1, B1 = − 12 , B2 = 16 , B4 = − 130 , B6 = 142 . Using contour integration, one finds the following expression for the Riemann zeta function in terms of Bernoulli numbers: ζ(2n) = (−1)n−1 (2π) 2n 2(2n)! B2n for all natural n. In particular, ζ(n) is transcendental for even n. It is considerably more difficult to show the irrationality at odd values. The arithmetic nature of the zeta values at odd positive integers has remained a mystery since Euler’s time even though one now conjectures that the numbersπ, ζ(3), ζ(5), . . . are algebraically independent over the rationals. Apéry finally proved in 1979 that ζ(3) is irrational. Recent progress was made by Rivoal in 2000, who showed that there are infinitely many integers n such that ζ(2n + 1) is irrational. This result was subsequently tightened by Zudilin, who showed that one of ζ(5), ζ(7), ζ(9), and ζ(11) is irrational. The Riemann zeta function has an analytic continuation to the entire complex plane except for a simple pole at s = 1. We have another marvelous identity concern- ing its values at negative integers and Bernoulli numbers: Bn = (−1)nnζ(1 − n) for natural n. • T. Rivoal: La fonction zêta de Riemann prend une infinité de valeurs irrationnelles aux entiers impairs, C. R. Acad. Sci. Paris Sér. I Math. 331 (2000), 4, 267–270. • V.V. Zudilin: One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational, Uspekhi Mat. Nauk 56 (2001), 4 (340), 149–150; translation in Russian Math. Surveys 56 (2001), 4, 774–776. Problem 1.38. Let σ(n) denote the sum of divisors of n. Show that σ(n) = 2k if and only if n is a product of Mersenne primes, i.e., primes of the form 2k −1, for k ∈ Z+. Solution 1.38. If n = ∏i paii , then σ(n) = ∏ i σ (p ai i ) = ∏ i (1 + pi + · · · + paii ). Moreover, if each prime factor is a Mersenne number, then pi = 2bi − 1 and ai = 1, which implies that σ(n) =∏i 2bi = 2k . Conversely, each prime factor p of n must satisfy 1 + · · · + pa = 2s for some s ∈ Z. This implies that a = 2q + 1 for q ∈ Z and furthermore, 2s = (1 + p)(1 + p2 + p4 + · · · + p2q). In particular, 1 + p = 2e (thus p is a Mersenne prime) and 1 + p2 + · · · + (p2)q = 2n. If q > 0, then the last equation implies again that q is odd, i.e., q = 2v + 1. We obtain next (1 + p2)(1 + p4 + · · · + p4v) = 2n, and so 1 + p2 = 2w. In particular, 1+p = 2e divides 2w = 1+p2. But 1+p divides 1−p2 and so 1+p2 +1−p2 = 2. 56 5 Number Theory Solutions Since 1+p ≥ 1+2 = 3, we obtain a contradiction. Therefore, q = 0 and the exponent of the prime is a = 1, as claimed. Problem 1.39. Find all integer solutions of the equation |pr − qs | = 1, where p, q are primes and r, s ∈ Z \ {0, 1}. Solution 1.39. The only solutions are p = 3, q = 2, r = 2, s = 3 and p = 2, q = 3, r = 3, s = 2. Obviously, one prime number should be even, hence 2. Let us assume that q = 2 and p is odd. Then the equation reads pr ± 1 = 2s . If r > 1 is odd, then: 1. (pr + 1)/(p + 1) is the odd integer pr−1 − pr−2 − pr−3 − · · · − 1 > 1, contradiction. 2. (pr − 1)/(p − 1) is the odd integer pr−1 + pr−2 + pr−3 + · · · + 1 > 1, contradiction again. Therefore r = 2t . Further, we have two cases for the equation above: 1. 2s = (pt)2 + 1 = (2n + 1)2 + 1 = 4n2 + 4n + 2. This is impossible since 4 divides 2s since s > 1, while 4 does not divide the right-hand side 4n2 + 4n + 2. 2. 2s = (pt )2 − 1 = 4n2 + 4n = 4n(n+ 1). Now, either n or n+ 1 is odd, which would lead us to a contradiction as soon as n ≥ 2. Thus n = 1 and the solutions are those claimed above. Comments 18 Catalan conjectured in 1844 that 32 − 23 = 1 is the only solution of the Diophantine equation xm − yn = 1, in the integer unknowns x, y,m, n ≥ 2. This was confirmed in 2002 by P. Mihaˇilescu, who gave a brilliant solution using cyclotomic fields. Note that it suffices to consider the case in which the exponents m, n are prime numbers. About one hundred years before Catalan, Euler had proved that there are no other solutions to the equation |x3 − y2| = 1, using the descent method.V.A. Lebesgue showed in 1850 that the equation xm−y2 = 1 has no solutions. More than one century later, in 1961, Chao Ko showed that the equation x2 −yn = 1 has no solutions if n ≥ 3. His proof was improved by E. Z.Chein in 1976. More details concerning the history of the conjecture and various partial solutions can be found in the book of Ribenboim. • T. Metsänkylä: Catalan’s conjecture: another old Diophantine problem solved, Bull. Amer. Math. Soc. (N.S.) 41(2004), 1, 43–57. • P. Mihaˇilescu: Primary cyclotomic units and a proof of Catalan’s conjecture, J. Reine Angew. Math. 572 (2004), 167–195. • P. Ribenboim: Catalan’s Conjecture. Are 8 and 9 the Only Consecutive Powers? Academic Press, Inc., Boston, MA, 1994. Problem 1.40. Consider an arithmetic progression with ratio between 1 and 2000. Show that the progression does not contain more than 10 consecutive primes. 5 Number Theory Solutions 57 Solution 1.40. Suppose that there are 11 consecutive prime terms in the progression. Let r be the ratio. The respective terms are: h + nr, h + nr + r, h + nr + 2r, . . . , h + nr + 10r. Let p be a prime number less than 12. If r �= 0 (mod p), then there exists some i ≤ 10 such that h+ ir ≡ 0 (mod p), and hence h+ ir will not be prime. This shows that r is divisible by 2, 3, 5, 7, 11 and thus greater than 2000, contradicting our assumptions. Comments 19 More generally, if the ratio r is less than p1 · · ·pn, then the maximum number of consecutive primes is at most pn − 1. Problem 1.41. Let a1 = 1, an+1 = an + [√ an ] . Show that an is a perfect square iff n is of the form 2k + k − 2. Solution 1.41. Letnk = 2k+k−2.We will prove by induction first thatank = (2k−1)2 and second that nk−1 < m < nk implies that am is not a perfect square. The assention is true for k = 1. One proves by induction on i that the formulas ank+2i = (2k−1 + 1 − i)2 + 2k, ank+2i+1 = (2k−1 + i)2 + 2k−1 − i, hold for 0 ≤ i ≤ 2k−1. In particular, if i = 2k−1, we obtain ank+1 = (2k)2. The other values are not perfect squares, since they are located between two consecutive squares. Problem 1.42. Recall that ϕ(n) denotes the Euler totient function (i.e., the number of natural numbers less than n and prime to n), and that σ(n) is the sum of divisors of n. Show that n is prime iff ϕ(n) divides n − 1 and n + 1 divides σ(n). Solution 1.42. If n is prime, then ϕ(n) = n − 1, σ (n) = n + 1. Conversely, let us assume n ≥ 3. Then ϕ(n) is even and thus n must be odd. If p is an odd prime such that pr divides n and r ≥ 2, then pr−1 divides ϕ(n) and thus pr−1 divides n − 1, which is a contradiction, since n and n − 1 are coprime, i.e., gcd(n, n − 1) = 1. Therefore, we have n = p1 · · ·pk , where pi are odd primes. We have further, ϕ(n) = (p1 − 1) · · · (pk − 1) and σ(n) = (p1 + 1) · · · (pk + 1). This implies that 2k divides both ϕ(n) and σ(n). If k ≥ 2, then 4 divides ϕ(n) and hence n−1. This implies that gcd(4, n+1) = 2. Since 2k divides σ(n), we find that 2k−1 must divide σ(n) n+1 . In particular, 2 k−1 < σ(n) n , but σ(n) n = ( 1 + 1 p1 ) · · · ( 1 + 1 pk ) < ( 4 3 )k , which leads to a contradiction when k ≥ 2. Therefore, k = 1 and so n is prime. Comments 20 An old conjecture due to Lehmer states that if ϕ(n) divides n−1, then n is prime. P. Hagis Jr. proved that the number of prime divisors τ(n) is either 1 or τ(n) ≥ 298 848 and n > 101937042 if 3 divides n, and τ(n) ≥ 1991 and n > 108171 in general. The reader may consult: 58 5 Number Theory Solutions • P. Hagis Jr.: On the equation M · ϕ(n) = n− 1, Nieuw Arch. Wisk. (4) 6 (1988), 3, 255–261. Problem 1.43. A number is called ϕ-subadditive if ϕ(n) ≤ ϕ(k)+ ϕ(n− k) for all k such that 1 ≤ k ≤ n−1, andϕ-superadditive if the reverse inequality holds. Prove that there are infinitely many ϕ-subadditive numbers and infinitely many ϕ-superadditive numbers. Solution 1.43. Let p ≥ 3 prime and 1 < k < p. We have then ϕ(k) + ϕ(p − k) ≤ k − 1 + (p − k − 1) = p − 2 < ϕ(p). Now p − 1 is composite; thus ϕ(p − 1) ≤ p − 1, and hence ϕ(1) + ϕ(p − 1) ≤ 1 + p − 3 < ϕ(p). Therefore, every prime number p is ϕ-superadditive. Let r ≥ 2 and set nr = p1 · · ·pr for the product of the first r prime numbers. Consider some natural number m, such that 1 < m < nr . Let q1, . . . , qs be the prime divisors of m. We have qj ≥ pj and thus s ≤ r , since otherwise we would have m ≥ q1 · · · qs > p1 · · ·pr = nr . Further, one computes ϕ(m) m ≥ s∏ i=1 ( 1 − 1 qi ) ≥ r∏ i=1 ( 1 − 1 pi ) = φ(nr) nr . This implies that ϕ(m) + ϕ(nr − m) ≥ mϕ(nr) nr + (nr − m)ϕ(nr) nr = ϕ(nr). Problem 1.44. Find the positive integers N such that for all n ≥ N , we have ϕ(n) ≤ ϕ(N). Solution 1.44. 1. Let N ≥ 5. According to Chebyshev’s theorem, there exists a prime number p such that p [ N + 3 3 ] < p < 2 [ N + 3 2 ] − 2 ≤ N + 2. We haveϕ(p) = p−1 > N+12 ≥ ϕ(N+1) for oddN andϕ(p) = p−1 > N2 ≥ ϕ(N) for even N . This means that N ≤ 5, and since ϕ(1) = ϕ(2) = 1, ϕ(3) = ϕ(4) = 2, we have N ∈ {1, 2, 3, 4}. Problem 1.45. A number n is perfect if σ(n) = 2n, where σ(n) denotes the sum of all divisors of n. Prove that the even number n is perfect if and only if n = 2p−1(2p −1), where p is a prime number with the property that 2p − 1 is prime. Solution 1.45. Since n is even, one can write n = 2qm, where q ≥ 1 and m is odd. One knows that σ(pa) = pa+1−1 p−1 if p is prime, and σ(p a1 1 p a2 2 · · ·pass ) = σ(p a1 1 )σ (p a2 2 ) · · · σ(pass ) if pj are distinct primes. 5 Number Theory Solutions 59 We have then σ(m) > m, and we write σ(m) = m + r for some natural number r ≥ 1. The hypothesis σ(n) = 2n is equivalent then to 2q+1m = (2q+1−1)(m+r) = 2q+1m − m + (2q+1 − 1)r . Thus m = (2q+1 − 1)r . We have found therefore that r divides m and r < m. But σ(m) = m + r is the sum of all divisors of m, and thus r is the sum of all divisors of m that are strictly smaller than m. Since r itself is such a divisor, it follows that it cannot be any other divisor of m smaller than r and so r = 1. In particular, this means that m is prime. This means that σ(m) = m + 1 = 2q+1 − 1 and 2q+1m = (2q+1 − 1)(m + 1). Since m is odd and relatively prime to m + 1, it follows that m = 2q+1 − 1. Further, m has to be prime and so p = q + 1 is a prime number. In fact, if p is composite, say p = ab, then 2p − 1 = (2a − 1)(1 + 2a + · · · + 2a(b−1)). This proves the claim. Comments 21 Perfect numbers were known in antiquity. Euclid stated in Book IX of his Elements that a number of the form 2p−1(2p − 1) for which the odd factor is prime is a perfect number, and noticed that 6, 28, 496, and 8128 are perfect. Descartes claimed in 1638, in a letter to Mersenne, that every even perfect number should be of this form, and this was proved later by Euler to be true. Euler also considered the problem of the existence of odd perfect numbers. Although numbers up to 10300 have been checked, no odd perfect numbers have been found to this day, supporting the conjecture that there are no odd perfect numbers. This is one of the oldest open problems in mathematics, along with the congruent number problem. Dickson proved in 1913 that for each k, there are at most finitely many odd perfect numbers with k distinct prime factors, and the best estimate known (due to Pace P. Nielsen) is that such a number is smaller than 24k . Here are some properties that an odd perfect number must satisfy: • P. Hagis Jr. and, independently, J.E.Z. Chein proved that it has at least 8 distinct prime factors (at least 11 if it is not divisible by 3); • H.G. Hare showed that it has at least 47 prime factors in total, including repeti- tions; • P. Jenkins proved that it has at least one prime factor greater than 107, and D.E. Iannucci showed that it has two prime factors greater than 104, and three prime factors greater than 100; • it is divisible by the power of a prime number that is greater than 1020; • it is a perfect square multiplied by an odd power of a single prime. Primes of the form 2p − 1 (so p is also prime) are called Mersenne primes. It is still unknown whether there exist infinitely many Mersenne primes. The first few such primes correspond to p = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89. As of October 2006, only 44 Mersenne primes were known; the largest known prime number, 232 582 657 − 1, is a Mersenne prime having 9 808 358 digits. It was discovered in September 2006 by Curtis Cooper and Steve Boone. • R.K. Guy: Unsolved Problems in Number Theory, section B1, 2nd ed. New York, Springer-Verlag, 44–45, 1994. 60 5 Number Theory Solutions • P. Hagis, Jr.: An outline of a proof that every odd perfect number has at least eight prime factors, Math. Comput. 34 (1980), 1027–1032. • D.E. Iannucci: The second largest prime divisor of an odd perfect number exceeds ten thousand, Math. Comput. 68 (1999), 1749–1760. • D.E. Iannucci: The third largest prime divisor of an odd perfect number exceeds one hundred, Math. Comput. 69 (2000), 867–879. • P. Jenkins: Odd perfect numbers have a factor that exceeds 107, Math.Comput. 72 (2003), 1549–1554. • P.P. Nielsen: An upper bound for odd perfect numbers, Integers 3(2003), A14, (electronic). Problem 1.46. A number n is superperfect if σ(σ(n)) = 2n, where σ(k) is the sum of all divisors of k. Prove that the even number n is superperfect if and only if n = 2r , where r is an integer such that 2r+1 − 1 is prime. Solution 1.46. If n = 2r and 2r+1 − 1 is prime, then σ(σ(n)) = σ(2r+1 − 1) = 2r+1 = 2n. Conversely, let us consider n = 2rq, where q is odd and r ≥ 1. Then we have the identity 2r+1q = 2n = σ(σ(n)) = σ((2r+1 − 1)σ (q)). Moreover, if q > 1, then the number (2r+1−1)σ (q) has at least three distinct divisors, namely (2r+1 − 1)σ (q), σ (q), and 2r+1 − 1. Thus σ((2r+1 − 1)σ (q)) ≥ 2r+1σ(q) + 2r+1 − 1 > 2r+1q, which is a contradiction. Therefore q = 1 and 2r+1 = σ(2r+1 − 1); thus 2r+1 − 1 is prime, as claimed. Comments 22 It is still unknown whether there exist odd superperfect numbers. If one exists, it must be a square and greater than 7 · 1024, as was proved by J.L. Hunsucker and C. Pomerance. Earlier, in 1967, D. Suryanarayana proved that there do not exist odd superperfect numbers that are powers of a prime. • J.L. Hunsucker and C. Pomerance: There are no odd super perfect numbers less than 7 · 1024, Indian J. Math. 17 (1975), 3, 107–120. Problem 1.47. Ifa, b are rational numbers satisfying tan aπ = b, thenb ∈ {−1, 0, 1}. Solution 1.47. Write a = m n , where gcd(m, n) = 1. De Moivre’s identities read (cos aπ+i sin aπ)n = cosmπ+i sin mπ, (cos aπ−i sin aπ)n = cosmπ−i sin mπ. Therefore ( 1 + ib 1 − ib )n = ( 1 + i tan aπ 1 − i tan aπ )n = ( cos aπ + i sin aπ cos aπ − i sin aπ )n = 1. 5 Number Theory Solutions 61 Consider the polynomial P = xn − 1 ∈ Z[i][x]. Since P is a monic polynomial and Z[i] is a factorial ring, any root of P that lies in Q[i] actually belongs to the ring of integers Z[i]. This means that 1 + ib 1 − ib = 1 − b2 1 + b2 + i 2b 1 + b2 ∈ Z[i] and hence 1−b21+b2 ∈ Z, 2b1+b2 ∈ Z. Now it is immediate that b ∈ {−1, 0, 1}. Problem 1.48. Let An = run + svn, n ∈ Z+, where r, s, u, v are integers, u �= ±v, and Pn be the set of prime divisors of An. Then P =⋃∞n=0 Pn is infinite. Solution 1.48. We suppose that rsuv �= 0. We can assume, without loss of generality, that gcd(ru, sv) = 1, and so the prime numbers in P do not divide rsuv. Let us suppose that P is finite, say {p1, p2, . . . , pm}. We have ui �= vi , for any i �= 0, because u �= ±v. Then there exists some a such that ui �≡ vi (mod pak ) for all 1 ≤ i ≤ m + 1, 1 ≤ k ≤ m. Moreover, if 1 ≤ i < j ≤ m + 1, then Aj − vj−iAi = ruj ( uj−i − vj−i ) , and therefore we cannot have Ai ≡ Aj (mod pak ), for all k ≤ m. Thus there exists t within the range 1 ≤ t ≤ m + 1 such that At �≡ 0 (mod pak ) for 1 ≤ k ≤ m. We choose now some b such that ub ≡ vb (mod pak ) for all 1 ≤ k ≤ m. We can take, for instance, b = ϕ(ca), where ϕ is Euler’s totient function and c = p1 · · ·pm. We now choose n large enough that At+nb > ca . Since we have At+nb ≡ At �≡ 0 (mod pak ) for all 1 ≤ k ≤ m, we obtain that At+nb must also have a prime divisor different from p1, . . . , pm. Comments 23 This problem is a particular case of a result of G. Pólya, see also: • G. Pólya: Arithmetische Eigenschaften der Reihenentwiclungen rationalen Func- tionen, J. Reine Angew. Math. 151 (1921), 1–31. Problem 1.49. Solve in natural numbers the equation x2 + y2 + z2 = 2xyz. Solution 1.49. If x, y, z are odd, then the left-hand side is odd and therefore nonzero. Thus 2xyz ≡ 0 (mod 4). Each square is congruent to either 0 or 1 modulo 4, and thus x, y, z must be even. Set x = 2x1, y = 2y1, z = 2z1. The equation becomes x21 + y21 + z21 = 4x1y1z1. The same argument shows that x1, y1, z1 are even. By induction, we find a sequence of triples with xn = 2xn+1, yn = 2yn+1, zn = 2zn+1, satisfying x2n+1 + y2n+1 + z2n+1 = 2n+2xn+1yn+1zn+1. Moreover, if n ≥ max(|x|, |y|, |z|), then |xn| = | 12n+1 x| < 1 and similarly for the other unknowns. Since xn is an integer, we obtain xn = 0 and thus x = y = z = 0. 62 5 Number Theory Solutions Problem 1.50. Find the greatest common divisor of the following numbers:C12n, C32n, C52n, . . . , C 2n−1 2n . Solution 1.50. We have the identity C12n + · · · + C2n−12n = 22n−1, which can be obtained by writing the binomial expansion of (x + 1)n and evaluating it at x = 1 and x = −1. Therefore, the greatest common divisor g should be of the form 2p, for some natural number p. If n = 2kq, where q is odd, then C12n = 2k+1q and hence g ≤ 2k+1. But now we can compute Cp2n = Cp2k+1q = 2 k+1q p C p−1 2k+1q−1. Since p is odd and the binomial coefficient is an integer, it follows that 2k+1 divides Cp2n for all odd p. This proves that the number we are seeking is g = 2k+1. Problem 1.51. Prove that if Sk =∑ni=1 kgcd(i,n), then Sk ≡ 0 (mod n). Solution 1.51. 1. ∑n i=1 kgcd(i,n) = ∑ d/n ϕ ( n d ) kd . If gcd(m, n) = 1, then we have the following identity: mn∑ i=1 kgcd(i,mn) = ∑ d/m ϕ (m d )∑ τ/n ϕ (n τ ) (kd)τ . Thus it suffices to prove the claim for n = pα , where p is prime. The case α = 1 or p divides k is trivial. Let α > 1 and gcd(p, k) = 1. We will prove the claim by induction on α, as follows: ∑ ϕ ( pα pi ) kp i = kpα + ϕ(p)kpα−1 + ϕ(p2)kpα−2 + · · · + ϕ(pα)k ≡ kpα−1 + (p − 1)kpα−1 + · · · + ϕ(pα)k ≡ p(kpα−1 + · · · + ϕ(pα−1))k ≡ p α−1∑ i=0 ϕ ( pα−1 pi ) kp i (mod pα). This proves the induction step. 2. One observes that 1 n Sk counts the number of circular permutations of length k on n elements. Hence, Sk ≡ 0 (mod n). Problem 1.52. Prove that for any integer n such that 4 ≤ n ≤ 1000, the equation 1 x + 1 y + 1 z = 4 n has solutions in natural numbers. Solution 1.52. It suffices to consider the case n prime, since a solution for n yields a solution for a multiple of n. Let a, b, c, d ∈ Z+. If 5 Number Theory Solutions 63 an + b + c = 4abcd, then (x, y, z) = (bcd, abdn, acdn) is a solution. In particular, take a = 2, b = 1, c = 1. Then, if n = 4d − 1, we obtain a solution as above. This works actually for any n ≡ −1 (mod 4). Taking a = 1, b = 1, c = 1, we solve the case n ≡ 2 (mod 4). If n ≡ 0 (mod 4), then we have the solutions x = y = z = 3n/4. It remains to check the case n ≡ 1 (mod 4). Further, take a = 1, b = 1, c = 2, and thus we solve the case n ≡ 3 (mod 8). The only case remaining is then n ≡ 1 (mod 8). If n = 3, we have the solutions (x, y, z) = (3, 2, 2). Write an + b = c(4abd − 1) = cq, where q ≡ −1 (mod 4ab). Take q = 3, a = b = 1; then n ≡ −1 (mod 3) is solved. There remains the case n ≡ 1 (mod 3), which we will suppose from now on. If n = 7, then we have the solution (x, y, z) = (2, 28, 28). Next take q = 7 and ab ∈ {1, 2}. This leads to solutions for n ≡ −1,−2,−4 (mod 7). Thus it suffices to consider n ≡ 1, 2, 4 (mod 7). Take q = 15 and a = 2, b = 1 or a = 1, b = 2. This solves the equation for n ≡ −2,−8 (mod 15). Since n ≡ 1 (mod 3), we have n ≡ −2,−3 (mod 5). Hence, we may suppose that n ≡ 1, 4 (mod 5). By asking that n satisfy all these congruences, it follows that the problem is solved for all n such that n �≡ 1, 121, 169, 289, 361, 529 (mod 840). Since the first prime number not satisfying this condition is 1009, the claim follows. Comments 24 Erdo˝s and Straus conjectured that the equation 1 x + 1 y + 1 z = 4 n has solutions for every n ≥ 4. Mordell proved that this conjecture is true, except possibly for the primes in six particular residue classes modulo 840. Vaughan showed that a sufficient condition for the more general conjecture in which 4 n is replaced by m n is the existence of positive integers a, b for which an + b has a divisor congruent to −1 (mod mab). The conjecture was verified by I. Kotsireas, via computer search, for n ≤ 1010. Schinzel conjectured further that the equation m n = 1 x1 + · · · + 1 xk has positive integer solutions xi if m > k ≥ 3; actually, the truth of the conjecture for m > k = 3 implies its validity for every m > k ≥ 3. See also the book of R.K. Guy which contains more partial results on this topic. • P. Erdo˝s: On a Diophantine equation (Hungarian, English summary) Matematikai Lapok 1 (1950), 192–210. • R.K. Guy: Unsolved Problems in Number Theory, Section D11, Springer, New York, 1981. 64 5 Number Theory Solutions • L.J. Mordell: Diophantine Equations, Pure and Applied Mathematics, Vol. 30 Academic Press, London-New York 1969. • A. Schinzel: Sur quelques propriétés des nombres 3/n et 4/n, où n est un nombre impair, Mathesis 65 (1956), 219–222. • W. Sierpin´ski: Sur les décompositions de nombres rationnels en fractions pri- maires, Mathesis 65 (1956), 16–32. Problem 1.53. For a natural number n, we set S(n) for the set of integers that can be written in the form 1 + g + · · · + gn−1 for some g ∈ Z+, g ≥ 2. 1. Prove that S(3) ∩ S(4) = ∅. 2. Find S(3) ∩ S(5). Solution 1.53. 1. Let g, h ≥ 2 be integers satisfying the equation 1 + g + g2 = 1 + h + h2 + h3 and therefore h3 = (g − h)(g + h + 1). Any prime divisor p of h divides g − h or g + h + 1. In the first case, p divides g, and in the second case, p divides g+1. Furthermore, if pm divides h then either p3m divides g − h or p3m divides g + h + 1, because gcd(g, g + 1) = 1. Thus we can write h = h1h2, where gcd(h1, h2) = 1, so that h31 = g − h, h32 = g + h + 1. In particular, we have 2h1h2 = 2h = h32 − h31 − 1. We will show that this Diophantine equation does not have any solutions h1, h2 ≥ 0. Let fh1(h2) = h32 −h31 − 2h1h2 − 1. First, we have fh1(h2) ≤ −1 if 0 ≤ h2 ≤ h1. Moreover, if h2 > h1, then fh1 is increasing and so fh1 does not have any natural zeros, since fh1(h1 + 1) = h21 + h1 > 0. This proves that S(3) ∩ S(4) = ∅. 2. Let f (x) = 1 + x + x2 and g(x) = 1 + x + x2 + x3 + x4. Then f and g are strictly increasing functions on Z+. Moreover, we have the following inequalities: [f (4n2 + 5n + 1) < g(2n + 1) < f (4n2 + 9n + 4) < g(2n + 2) < f (4n2 + 9n + 5) < f (4(n + 1)2 + 5(n + 1) + 1), for all n ≥ 1. This implies that g(n) �= f (k) if n ≥ 3, k ≥ 1 are natural numbers. Finally, if n ∈ {1, 2}, then f (1) = 3 < g(1) = 5 < 7 = f (2), and g(2) = 31 = f (5). Therefore S(3) ∩ S(5) = {31}. Problem 1.54. If k ≥ 202 and n ≥ 2k, then prove that Ckn > nπ(k), where π(k) denotes the number of prime numbers smaller than k. 5 Number Theory Solutions 65 Solution 1.54. 1. Let us show first that Ck2k > 4k k if k ≥ 4. If k = 4, then Ck2k = 70 ≥ 64 = 4k k . We will use next a recurrence on k. We have Ck+12k+2 = (2k + 2)(2k + 1) (k + 1)2 C k 2k > 2(2k + 1) k + 1 · 4k k > 4k · 4k k · (k + 1) = 4k+1 k + 1 . 2. We will use the well-known estimate π(k) < k log k − 32 , for k > 5. The inequality 4k k > (2k)k/(log k−3/2) is equivalent to k log 4 − log k > k log k+log 2log k−3/2 , which is true as soon as k ≥ 1414. Moreover, Ck2k > 4k k > 2k k/(log k−3/2) > (2k)π(k), for k ≥ 1414. By direct computer verification, one has Ck2k > (2k)π(h) for 202 ≤ h ≤ 1413. 3. Now if s ≥ 1, we have k > π(k) ≥ 1 + π(k) − 1 s = π(k) + s − 1 s . The product of these inequalities for 1 ≤ s ≤ r gives us kr > Crπ(k)+r−1. Then Ckn+1 = ( 1 − k n + 1 )−1 Ckn > ( 1 − k n + 1 )−1 nπ(k) = nπ(k) · ∞∑ r≥0 kr (n + 1)r > nπ(k) · ∞∑ r≥0 Crπ(k)+r−1 (n + 1)r = n π(k) ( 1 − 1 n + 1 )−π(k) = (n + 1)π(k). Problem 1.55. Let Sm(n) be the sum of the inverses of the integers smaller than m and relatively prime to n. If m > n ≥ 2, then show that Sm(n) is not an integer. Solution 1.55. Let us write Sm(n) = 1 + 1 a1 + · · · + 1 as , where 1 < a1 < a2 < · · · < as < m and gcd(ai, n) = 1. Observe first that a = a1 is prime, since otherwise, any proper divisor of a would be smaller and still relatively prime to n. Let k be the largest integer such that ak ≤ m. Then there exists some i ≤ s such that ak = ai , because ak is also a number relatively prime to n. Moreover, if ak divides aj for some j , then j = i. In fact, let us assume that aj = cak for some integer c > 1. If c < a, then by hypothesis, gcd(c, n) > 1, and 66 5 Number Theory Solutions hence aj would not be relatively prime to n. If c ≥ a, then m > aj = cak ≥ ak+1, which would contradict the choice of k, supposed to be maximal with the property that ak ≤ m. Let us suppose now that Sm(n) is an integer, and set L = lcm(a1, . . . , as). One knows from above that ak divides L, while ak+1 does not divide L. We have also LSm(n) − s∑ j=1,j �=i L aj = L ai . If Sm(n) were an integer, then the left-hand side would be a multiple of a. On the other hand, the term on the right-hand side is L ai = L ak , which cannot be a multiple of a, by the choice of k. This contradiction proves that Sm(n) is not an integer. Problem 1.56. Solve in integers the following equations: 1. x4 + y4 = z2; 2. y3 + 4y = z2; 3. x4 = y4 + z2. Solution 1.56. 1. Suppose that there exists a nontrivial solution for which x or y is nonzero and hence a solution with x, y, z > 0. Consider the solution for which z is minimal among all possible solutions. We may suppose that gcd(x, y, z) = 1. Then the triple (x2, y2, z) is a primitive Pythagorean triple and so gcd(x, y) = gcd(y, z) = gcd(x, z) = 1. We can write x2 = m2 − n2, y2 = 2mn, z = m2 + n2, where m, n are integers. Since the solutions are primitive, it follows that x and z must be odd. In particular, m �≡ n (mod 2), since otherwise, x would be even. We have two cases: (i). If m is even and n is odd, then x2 = m2 − n2 ≡ −1 (mod 4), which is impossible, since any square is congruent to either 0 or 1 modulo 4. (ii). If m is odd and n is even, then we have y2 = 2mn and gcd(m, n) = 1. Thus m = m21 and n = 2n21. Moreover, x2 = m41 − 4n41, and so (m21 − x)(m21 + x) = 4n41. If d divides both m21 −x and m21 +x, then d divides 2m and n and so d = 2 or d = 1. But m21 − x and m21 + x are congruent modulo 2, and they cannot be both odd, since their product is even. Thus m21 − x = 2a4 and m21 + x = 2b4. In particular, we find that a4 + b4 = m21, where a, b are nonzero and m1 < m21 = m < z. Thus we have obtained a solution with z smaller than the initial one, which is a contradiction. Therefore there are no nontrivial solutions. 2. We have (y3 + 4y)2 = y6 + 8y4 + 16y2 = z4. Also, (y3 − 4y)2 = y6 + 16y2 − 8y4. Adding these equations, we obtain (2y)4 = z4 − (y3 − 4y)2. It is known that the equation u4 − v4 = w2 does not have nontrivial integer solutions (see the previous equation). Therefore either 2y = 0 and z4 = (y3 − 4y)2 = 0, which leads to y = z = 0, or else z = 2y, y3 − 4y = 0, which implies y = 2, z = 4. 5 Number Theory Solutions 67 3. Consider a positive solution (x, y, z) with gcd(x, y) = 1 and minimal x. Then (y2, z, x2) is a primitive Pythagorean triple, and hence we have two possible situations: (i). Either y2 = m2 −n2, x2 = m2 +n2, z = 2mn, where m > n and gcd(m, n) = 1. Then a4 = b4 + (xy)2 and a < x, contradicting the fact that x was chosen minimal. (ii). Or z = m2 −n2, x2 = m2 +n2, y2 = 2mn, where m > n and gcd(m, n) = 1. Then (m, n, x) is also a primitive Pythagorean triple and thus m is odd and n is even. Since y2 = 2mn and n is odd, we can write m = 2a2, n = b2, so that y = 2ab. Using the fact that (m, n, x) is a primitive Pythagorean triple, where m is even and n is odd, we derive that m = 2rs, n = r2 − s2 = b2, with gcd(r, s) = 1. Thus a2 = rs, which yields r = u2, s = v2, and thus, expressing n as above, we obtain: u4 = v4 + b2 with u < r < m < x, contradicting the minimality of x. Thus there are no nontrivial solutions. Comments 25 In particular, the Fermat equation x4 + y4 = z4 has no nontrivial solutions. The same method, called infinite descent, can be used successfully to prove that the equations x4 + dy4 = z2 have no solutions, when d is prime and d ≡ 7, 11 (mod 16), or d2 is prime and d 2 ≡ ±3 (mod 8), or −d is prime and −d ≡ ±3, 11 (mod 16), or d4 is prime and d4 ≡ ±3, 11 (mod 8). For more information about Diophantine equations see • L.J. Mordell: Diophantine Equations, Academic Press, Pure and Applied Math. vol. 30, London, New York, 1969. Problem 1.57. Show that the equation x3 + y3 = z3 + w3 has infinitely many integer solutions. Prove that 1 can be written in infinitely many ways as a sum of three cubes. Solution 1.57. 1. We make the change of variables x+y = X, x−y = Y , z+w = Z, z − w = W . The equation now reads X ( X2 + 3Y 2 ) = Z ( Z2 + 3W 2 ) . We use the identity ( a2 + 3b2 ) ( A2 + 3B2 ) = (aA + 3bB)2 + 3(bA − aB)2 and derive 68 5 Number Theory Solutions X ( X2 + 3Y 2 )2 = Z ( (XZ + 3YW)2 + 3(YZ − XW)2 ) . Set p = XZ + 3YW X2 + 3Y 2 , q = YZ − XW X2 + 3Y 2 , and thus X Z = p2 + 3q2. Choose Z = 1, for simplicity. We have then pX + 3qY = 1, pY − qX = W, from which one can obtain the values of Y and W . Finally, we obtain x = 1 − (p − 3q) ( p2 + 3q2 ) , y = −1 + (p + 3q) ( p2 + 3q2 ) , z = p + 3q − ( p2 + 3q2 )2 , w = −p + 3q + ( p2 + 3q2 )2 . These are all rational solutions of our equation. If p, q ∈ Z, then we obtain infinitely many integer solutions. However, it is not clear that these formulas will provide all integer solutions of our equations. 2. Set p = 3q, t = 2q. We obtain 1 = ( 1 − 9t3 )3 + ( 3t − 9t4 )3 + ( 9t4 )3 . Comments 26 It is unknown whether 3 can be written in infinitely many ways as a sum of three cubes. Problem 1.58. Prove that the equation y2 = Dx4 + 1 has no integer solution, except x = 0, if D �≡ 0,−1, 3, 8 (mod 16) and there is no factorization D = pq, where p > 1 is odd, gcd(p, q) = 1, and either p ≡ ±1 (mod 16), p ≡ q ± 1 (mod 16), or p ≡ 4q ± 1 (mod 16). Solution 1.58. Let x, y > 0 be a nontrivial solution with minimal x. If x is odd, then Dx4 +1 �≡ 0, 1, 4, 9 (mod 16), and thus it is not a quadratic residue modulo 16. Thus x should be even and y odd. Further, we have y + 1 = 2pa4, y − 1 = 8qb4, x = 2ab or y − 1 = 2pa4, y + 1 = 8qb4, x = 2ab, where D = pq, gcd(p, q) = 1, and a is odd. If p > 1, then pa4 − 4qb4 = ±1. If b is even, then p ≡ ±1 (mod 16), contradiction. If b is odd, then p ≡ 4q ± 1 (mod 16), again false. Thus p = 1 and a4 − 4Db4 = ±1. 5 Number Theory Solutions 69 The equation a4 −4Db4 = −1 has no solution modulo 4; thus a4 −1 = 4Db4. Then D = rs, b = cd , gcd(r, s) = 1, and a2 + 1 = 2rc4, a2 − 1 = 2sd4. Thus c, r are odd, and we have rc4 − sd4 = 1. If r = 1, then c4 −Dd4 = 1 and x = 2acd, so there exists a solution of our equation with d ≤ x/2, contradicting the minimality of x. If r > 1, then d cannot be even because r �≡ ±1 (mod 16). But if d is odd, then r ≡ s + 1 (mod 16), contradicting our assumptions. Problem 1.59. Find all numbers that are simultaneously triangular, perfect squares, and pentagonal numbers. Solution 1.59. We have the Diophantine equations n = p2 = 1 2 q(q + 1) = 1 2 r(3r − 1). We derive that (2q + 1)2 = 1 + 8p2and (6r − 1)2 = 1 + 24p2. In particular, we obtain (6r − 1)2 = (4p)2 + (2q + 1)2. Since gcd(4p, 2q+1) = 1, the triple (4p, 2q+1, 6r −1) is a primitive Pythagorean triple. This means that the integer solutions are determined as functions of two integer parameters m, n ∈ Z by means of the formulas 4p = 2mn, 2q + 1 = m2 + n2, 6r − 1 = m2 + n2. Substituting in the previous equation, we obtain the m4 − 4m2n2 +n4 = 1 constraint on the parameters, which can also be written as (m2 − 2n2)2 = 3n4 + 1. We now use a theorem that says that the Ljunggren equation x2 = Dy4 + 1 has at most two positive integer solutions if D > 0 is not a perfect square. In our case, the equation x2 = 3y4 + 1 has the obvious solutions (7, 2) and (2, 1) and thus these are all the solutions in natural numbers. Therefore there is only one solution (m, n) = (2, 1). 70 5 Number Theory Solutions Comments 27 The proof above is based on a deep result concerning the equation x2 = Dy4 + 1, proved by Ljunggren in 1942. Ljunggren also gave an algorithm that computes the nontrivial solution when it exists. The previous problem is a particular case in which a short elementary proof (due to L. Mordell) is available. Ljunggren extended his methods to deal with the more-delicate equation x2 = Dy4 − 1. He proved that this Diophantine equation has at most two positive solutions for D > 0, not a perfect square. In particular, x2 = 2y4 − 1 has only the solutions (1, 1) and (13, 239), which have been known for two centuries. Recently, Chen and Voutier proved that there is at most one solution in positive integers when D ≥ 3. • W. Ljunggren: Einige Eigenschaften der Einheiten reeller quadratischer und rein- biquadratischer Zahlkörper mit Anwendung auf die Lösung einer Klasse unbe- stimmter Gleichungen vierten Grades, Skr. Norske Vid. Akad. Oslo 1936, 12, 1–73. • W. Ljunggren: Über die Gleichung x4 − Dy2 = 1, Arch. Math. Naturvid. 45 (1942) 5, 61–70. • W. Ljunggren: Zur Theorie der Gleichung x2 +1 = Dy4, Avh. Norske Vid. Akad. Oslo. I. 1942. • J.H. Chen and P.Voutier: Complete solution of the Diophantine equationX2+1 = dY 4 and a related family of quartic Thue equations, J. Number Theory 62 (1997), 1, 71–99. Problem 1.60. Find all inscribable integer-sided quadrilaterals whose areas equal their perimeters. Solution 1.60. We have 16S2 = ∏cyclic(a + b + c − d), and so the problem is equivalent to the Diophantine equation 16(a + b + c + d)2 = ∏ cyclic (a + b + c − d). Assume that 0 < d ≤ c ≤ b < a ≤ b + c + d, for the sake of simplicity. If we write x = −a + b + c + d , y = a − b + c + d, z = a + b − c + d, t = a + b + c − d , then the above equation becomes (x + y + z + t)2 = xyzt. If this equation has a real solution, then the quadratic equation (X + y + z + x)2 − Xxyz = 0 has two real roots. The product of these real roots is (x + y + z)2. Since the solution t satisfies 0 < t < x + y + z, it follows that t is the smallest of the two roots, and thus it is determined by the formula 2t = xyz − 2(x + y + z) −√xyz(xyz − 4x − 4y − 4z) with the obvious constraint xyz > 4(x + y + z). Therefore, we need that λ = x+y+z xyz < 14 . Now, 3t ≥ x + y + z, since t is the longest side, and thus 83λ + √ 1 − 4λ ≤ 1, implying that λ ≥ 316 . 5 Number Theory Solutions 71 We have obtained the inequalities 14 > λ ≥ 316 , which immediately imply that 5 ≤ xy ≤ 16. Since λ < 14 , we get z > 4(x + y)/(xy − 4). On the other hand, we assumed that z ≤ t , which is equivalent to (xy − 4)z2 − 4(x + y)z − (x + y)2 ≤ 0, and therefore z ≤ (x + y)/(√xy − 2). Consequently, 4(x + y) xy − 4 ≤ z ≤ x + y√ xy − 2 . One can list all integer values of x, y, z, t satisfying the inequalities above and that are solutions to the Diophantine equation. This leads to (x, y, z, t) ∈ {(1, 9, 10, 10), (2, 5, 5, 8), (3, 3, 6, 6), (4, 4, 4, 4)} and their permutations. Problem 1.61. Every rational number a can be written as a sum of the cubes of three rational numbers. Moreover, if a > 0, then the three cubes can be chosen to be positive. Solution 1.61. Assume that we know how to write a = x3 + y3 + z3 and let us introduce u = x+y+z and v = x+y. Then a = (x+y+z)3−3(x+y)(y+z)(z+x), and we derive easily that a = u3 − 3v(u − x)(u − v + x). We will require furthermore that u and v satisfy the additional constraint u3 = 3v(u2 − x2). Then a = 3v2(u − x). We introduce the new variable w = x u ∈ Q, so that u = 3v(1 − w2). Therefore a = 3v2(u− x) = 3v2u(1 −w) = 9v3(1 −w)2(1 +w) = 27v3(1 −w)3 · 13 · 1+w1−w . Set m = 3v(1 − w) and introduce it above in order to obtain a = m33 1+w1−w . Now we reverse the line of thought by letting m ∈ Q be an arbitrary parameter. The previous formulas yield w = 3a−m33a+m3 and also the following solutions: v = 3a + m 3 6m2 , u = 6am 3a + m3 . This yields the rational solutions x = wu, y = v − wu, z = u − v. Let us prove the positivity part. We have that x = uw ≥ 0 is equivalent to 0 < m < 3 √ 3a. But let us assume for a moment that m = 3√3a. Then y = 3 √ 3a 3 , z = 2 3 √ 3a 3 , x = 0. Regarding the solutions y, z as functions y = y(m), z = z(m) of the parameter m, we immediately see that these are continuous functions, which are strictly positive at m = 3√3a. In particular, there exists a small neighborhood W of 3√3a such that y(m), z(m) > 0. This means that for m ∈ Q+ ∩ (0, 3 √ 3a) ∩ W, we have x, y, z ∈ Q+, as claimed. 72 5 Number Theory Solutions Problem 1.62. 1. Every prime number of the form 4m + 1 can be written as the sum of two perfect squares. 2. Every natural number is the sum of four perfect squares. Solution 1.62. 1. Let p be such a prime number. The congruence z2 +1 ≡ 0 (mod p) has at least one solution, because −1 is a quadratic residue when p is of the form 4m+ 1. This follows from the more general Euler criterion computing the Legendre symbol ( a p ) as follows: ( a p ) = a p−12 (mod p) for arbitrary prime p. Recall that the Legendre symbol ( a b ) equals 1 if a is a quadratic residue modulo b (i.e., there exists some integer n such that n2 ≡ a(mod b)), and −1 otherwise. Let then m ∈ Z+, m �= 0, and z ∈ Z be such that mp = z2 + 1. We can suppose that −p2 < z < p2 , by using a translation multiple of p. Then m = 1 p (z2 + 1) < 1 p ( p2 4 + 1 ) < p. Set x = z, y = 1, so that mp = x2 + y2. We shall prove that if m > 1, then there exists m′ < m with the same properties. This process stops when m = 1, where we find the wanted sum of two squares. Let us consider u, v such that −m2 ≤ u, v ≤ m2 , satisfying u ≡ x (mod m), v ≡ y (mod m). Thus u2 + v2 = x2 + y2 ≡ 0 (mod m), and so there exists some r ∈ Z+ for which u2 + v2 = mr. Observe that r �= 0, since otherwise x, y would be multiples of m, and so m would divide p. Further, r = 1 m ( u2 + v2) < 1 m ( m2 4 + m 2 4 ) < m. We will use the identity ( a2 + b2 ) ( A2 + B2 ) = (aA + bB)2 + (aB − bA)2 and multiply the two equalities above, obtaining m2rp = ( x2 + y2 ) ( u2 + v2 ) = (xu + yv)2 + (xv − yu)2. Moreover, xu + yv ≡ x2 + y2 ≡ 0 (mod m) and xv − yu ≡ xy − yx ≡ 0 (mod m), 5 Number Theory Solutions 73 and thus, dividing the equation by m2, we obtain rp = a2 + b2, where r < m. Thus, m′ = r satisfies all required properties. This proves the first claim. 2. The identity (a2 + b2 + c2 + d2)(A2 + B2 + C2 + D2) = (aA + bB + cC + dD)2 +(aB − bA − cD + dC)2 + (aC + bD − cA − dB)2 + (aD − bC + cB − dA)2 reduces the problem to the case of prime numbers. Moreover, we know that primes of the form 4m+ 1 can be written as a sum of two squares, and 2 = 12 + 12. Thus it suffices to consider the primes of the form 4m + 3. The method of proof is the same as above. We show first that the congruence x2 + y2 + 1 ≡ 0 (mod p) has solutions. By the Euler criterion, −1 is a quadratic nonresidue modulo p and thus −y2 is a quadratic nonresidue modulo p. Moreover, each quadratic nonresidue is of the form −y2 (mod p). Then we have to find a quadratic nonresidue n and a quadratic residue r such that r + 1 = n (mod p). Consider in the sequence {1, 2, 3, . . . , p − 1, p} the smallest quadratic nonresidue n. Then n > 1 and n − 1 must be a quadratic residue by the minimality of n, and this yields our pair (r, n). We write now mp = x2 + y2 + 12 + 02, where m ∈ Z+ is nonzero. We choose x, y such that −p2 < x, y < p2 , using a translation multiple of p. Then m = 1 p ( x2 + y2 + 1) < 1 p ( p2 4 + p 2 4 + 1 ) < p. Consider a = x, b = y, c = 1, d = 0. If m > 1, we will find another m′ < m such that m′p is still a sum of four squares. Choose A,B,C,D such that −m2 < A,B,C,D ≤ m2 and a ≡ A (mod m), b ≡ B (mod m), c ≡ C (mod m), d ≡ D (mod m). Then A2 + B2 + C2 + D2 ≡ 0 (mod m) and thus there exists r ∈ Z+ such that mr = A2 + B2 + C2 + D2. Furthermore, r �= 0, since otherwise,A = B = C = D = 0, and thus a, b, c, d would be multiples of m, and so m would divide p. Moreover, r = 1 m (A2+B2+C2+D2) ≤ m, with equality if and only if A = B = C = D = m2 , m even. The equality would imply that a2 = m24 (mod p) and thus pm ≡ 0 (mod m2). Thus m divides p, which is false. Thus r < m. Further, multiplying the two identities, we obtain m2rp = ( a2 + b2 + c2 + d2 ) ( A2 + B2 + C2 + D2 ) = X2 + Y 2 + Z2 + W 2. The terms X, Y,Z,W are multiples of m, since, for example, 74 5 Number Theory Solutions X = aA + bB + cC + dD ≡ a2 + b2 + c2 + d2 ≡ 0 (mod m), and similarly for Y,Z,W . Therefore, one can divide by m2 and find that rp = X′2 + Y ′2 + Z′2 + W ′2 with r < m. Then take m′ = r . This proves the claim. Comments 28 The proof given above was obtained by Lagrange, and all elementary proofs known until now are variations of the present one. The numbers of the form 4m(8k + 7) cannot be represented as sums of three squares. Thus 4 is the minimal number with this property. However, all numbers that are not of the form 4m(8k + 7) are sums of three perfect squares. Problem 1.63. Every natural number can be written as a sum of at most 53 integers to the fourth power. Solution 1.63. Consider the identity 6 ( x21 + x22 + x23 + x24 )2 = ∑ 1≤i≤j≤4 (xi − xj )4 + ∑ 1≤i i. 78 5 Number Theory Solutions Solution 1.66. Fix an arbitrary i. The set Sn = {n − ai, n − ai + 1, . . . , n − 1} is called admissible for aj (where j > i) if no element of Sn is divisible by aj . It is obvious that Sn is admissible for aj if aj divides n. Moreover, Sn is not admissible for aj if there exists some e, with 1 ≤ e ≤ ai , such that n ≡ e (mod aj ). Consider ε > 0 sufficiently small and k sufficiently large such that ∑ j>k a−1j < ε. Let β = ai+1 · · · ak and Sn with the property that β divides n. Observe first that the sets Sn are admissible for those aj with i ≤ j ≤ k. Now assume that j > k. Let Cj (x) denote the set of those c ≤ x for which Scβ is not admissible for aj . Recall that c ∈ Cj (x) iff there exists e ∈ {1, 2, . . . , ai} such that cβ ≡ e (mod aj ). If gcd(β, aj ) = d, then d should divide e, and we can write β = dβ ′, e = de′, aj = da′j . Moreover, the condition above is equivalent to cβ ′ ≡ e′ (mod a′j ). If aj > βx, then obviously Cj (x) = ∅. Further, for a fixed value of e′, there are at most x a′j + 1 solutions c for the previous congruence in the given range. Since e′ can take at most ai d values, we derive that the cardinality of Cj (x) is bounded by ai ( x aj + 1 ) . Therefore, among the multiples of β satisfying n < βx, there are at most ∑ j>k aix aj + ai card{j |aj < βx} = εaix + o(x) integers n such that Sn is not admissible. In fact, the estimate card{j |aj < βx} = o(x) is a consequence of the convergence of the series. This asymptotic estimate implies that for large x there exist infinitely many values of n that do not belong to any Cj , so that Sn is admissible with respect to all aj . Problem 1.67. Denote by Cn the claim that there exists a set of n consecutive integers such that no two of them are relatively prime. Prove that Cn is true for every n, such that 17 ≤ n ≤ 10000. Solution 1.67. We suppose that both p and 2p + 1 are primes and consider those n for which 3p+2 ≤ n ≤ p2. Set q for the product∏pi≤n,pi �=p,2p+1 pi of those prime numbers smaller than n and distinct from p and 2p+ 1. From the Chinese remainder theorem, there exists a natural number x satisfying the congruences x ≡ 0 (mod q), x ≡ −3p − 1 (mod p(2p + 1)). Now let Sn = {x, x+1, . . . , x+n−1}. Then Sn contains x+1, x+2p+1, . . . , x+ kp + 1, . . ., which have p as a common factor. Also, x + p has the factor 2p + 1 in common with x + 3p + 1 ∈ Sn. All other numbers from Sn have a common divisor with q, therefore with x. Therefore the set Sn satisfies the claim Cn, and we have to look for specific values of p. 5 Number Theory Solutions 79 Ifp = 5, thenSn holds for 17 ≤ n ≤ 25.We choose furtherp = 11, 29, 251, 1013, 49919, 5008193, and this yields the claim for all n such that 35 ≤ n ≤ (5008193)2. The case 26 ≤ n ≤ 34 must be treated separately. There one considers x satisfying x ≡ 0 (mod 2 · 5 · 11 · 17), x ≡ −1 (mod 3), x ≡ −2 (mod 7 · 19), x ≡ −3 (mod 13), x ≡ −4 (mod 23). Comments 32 The claim Cn is false for all n ≤ 16, and true for all n ≥ 17, as proved in: 1. S.S. Pillai: On m consecutive integers, Proc. IndianAcad. Sci., Sect.A. 11 (1940), 6–12, 13 (1941), 530–533. 2. A. Brauer: On a property of k consecutive integers, Bull. Amer. Math. Soc. 47 (1941), 328–331. The proof, which is elementary, is based on the estimate π(2n) − π(n) ≥ 2 [ log n log 2 ] + 2, for n ≥ 75. Problem 1.68. Prove that a natural number n has more divisors that can be written in the form 3k + 1, for k ∈ Z, than divisors of the form 3m − 1, for m ∈ Z. Solution 1.68. Let f (n) (and respectively g(n)) be the number of divisors of the form 3k + 1 (and respectively 3m − 1), for k,m ∈ Z. Let n = uv, where gcd(u, v) = 1. By making use of the obvious congruences (−1)(−1) ≡ 1 (mod 3), (−1)1 ≡ −1 (mod 3), we obtain the formulas f (n) = f (u)f (v) + g(u)g(v), g(n) = g(u)f (v) + f (u)g(v). Set further h(n) = f (n) − g(n). The identities above prove that h(n) = h(u)h(v). In particular, if n = pα11 · · ·pαrr is the factorization into prime factors of n, then h(n) = r∏ j=1 h ( p αj j ) . Now we have h(3α) = 1. If the prime number p ≡ −1 (mod 3), then the divisors of pα of the form 3k + 1 are 1, p2, p4, . . . and the ones of the form 3k − 1 are p, p3, . . . . Therefore 80 5 Number Theory Solutions h(pα) = { 1, if α is even 0, otherwise. If the prime number p satisfies p ≡ 1 (mod 3), then all divisors of pα are of the form 3k + 1, and therefore h(pα) = 1 + α. Therefore if n = pα11 · · ·pαrr , h(n) = {∏ ν;pν≡1 (mod 3)(1 + αν), if all αν for which pν ≡ 1 (mod 3) are even, 0, otherwise. Comments 33 It can be proved that 6h(n) is the number of integer solutions of the equation x2 − xy + y2 = n. In the same way, we can prove that the difference between the number τ4k+1(n) of divisors of the form 4k + 1 and the number τ4k−1(n) of divisors of the form 4k − 1 is also positive. Moreover, 4(τ4k+1(n) − τ4k−1(n)) is the number of solutions of the Diophantine equation x2 + y2 = n. Comments 34 The positivity result holds more generally when one compares the numbers of divisors of the form 4k ± 1, 6k ± 1, 8k ± 1, 12k ± 1, 24k ± 1. However, the analogous result do not hold for any other congruences. Comments 35 The number of solutions in integers of quadratic equations appears in many arithmetic questions. The oldest and perhaps one of the most famous unsolved problems is the congruent numbers problem. Specifically, one asks to determine all integers that are the areas of right triangles with rational sides, which are called congruent numbers. For instance, n = 6 is the area of the right triangle of sides 3, 4, and 5. It is less obvious that 157 is a congruent number, where the simplest solution for the sides of the associated right triangle reads 6803298487826435051217540 411340519227716149383203 , 411340519227716149383203 21666555693714761309610 , 224403517704336969924557513090674863160948472041 8912332268928859588025535178967163570016480830 . It is known that congruent numbers are those d for which the equation dy2 = x3 − x has infinitely many rational solutions. Deep results of Tunnell imply the following conjecture, which would be a complete characterization of congruent numbers. Let d be square-free. Denote by N(d) (and respectively M(d)) the number of integer solutions of 2x2 + y2 + 8z2 = d, and respectively 2x2 + y2 + 32z2 = d, for odd d, 5 Number Theory Solutions 81 4x2 + y2 + 8z2 = d 2 , and respectively 4x2 + y2 + 32z2 = d 2 , for even d. Then d is a congruent number if and only if N(d) = 2M(d). This is a particular case of the Birch–Swinnerton-Dyer conjecture, considered one of the most important unsolved problems in arithmetic, and included as one of the seven Millennium Problems (for which the Clay Mathematics Institute will award one million dollars for a solution). The interested reader might consult • N. Koblitz: Introduction to Elliptic Curves and Modular Forms, second edition, Graduate Texts in Mathematics, 97, Springer-Verlag, New York, 1993. Problem 1.69. A number N is called deficient if σ(N) < 2N and abundant if σ(N) > 2N . 1. Let k be fixed. Are there any sequences of k consecutive abundant numbers? 2. Show that there are infinitely many 5-tuples of consecutive deficient numbers. Solution 1.69. 1. Let pn denote the nth prime number. It is known that the infinite product ∞∏ n=1 ( 1 + 1 pn ) is infinite. Thus, for any m, there exists an integer N(m) such that N(m)∏ n=m ( 1 + 1 pn ) > 2. In particular, this shows that the number t =∏N(m)n=m pn is abundant, because σ(t) = t ∏N(m) n=m ( 1 + 1 pn ) . Consider now the sequence defined by m1 = 1, mj+1 = N(mj ) + 1, for j ≥ 1, and tj = N(mj )∏ n=mj pn. Then the numbers tj are pairwise relatively prime, and therefore for every k there exists some integer L (depending on k) with the property that L + j ≡ 0 (mod tj ), for all j ∈ {1, . . . , k}. Observe that any multiple of an abundant number is abundant itself. This shows that each of the numbers L + 1, . . . , L + k is abundant. 2. It is well known that lim n→∞ ( 1 + 2 n log n )n = 1 and limn→∞ pnn log n = 1. Take n large enough so that 82 5 Number Theory Solutions ( 1 + 2 n log n )n < 1.01, and pn > 1 2 n log n > 60. Let us next consider Qn = n−1∏ i=1 pj , and Mn = 60pn + 1. We set xi = Mn+i−1i for i ∈ {1, 2, 3, 4, 5}. Because n > 3, it is easy to see that every xi is relatively prime to Qn. Therefore each xi can be factored into prime factors xi = ti∏ j=1 pnji , where each nji ≥ n (because otherwise, it would have common factors with Qn) and the number of factors ti is less than or equal to n (otherwise, xi would be greater than Mn). Since 1 + 1 pnji ≤ 1 + 1 pn and for prime p we have σ(ps)/ps ≤ (σ (p)/p)s , we conclude that σ(xi) xi ≤ ( 1 + 1 pn )n < ( 1 + 2 n log n )n < 1.01 and therefore σ(Mn + i − 1) Mn+i = σ(i) i · σ(xi) xi < 1.75 · 1.01 < 2 . Thus Mn,Mn + 1,Mn + 2,Mn + 3,Mn + 4 are deficient numbers. Problem 1.70. Does there exist a nonconstant polynomial an2 + bn+ c with integer coefficients such that for any natural numberm, all its prime factorspi are congruent to to 3 modulo 4? Prove that, for any nonconstant polynomial f with integer coefficients and any m ∈ Z there exist a prime number p and a natural number n such that p divides f (n) and p ≡ 1 (mod m). Solution 1.70. 1. We write 4af (n) = (2an+b)2+4ac−b2. If a = 0 or 4ac−b2 = 0, it is obvious that there are infinitely many prime numbers 4m+1 that divide f (n). Let us then assume that a(4ac − b2) �= 0. Set p1, . . . , pm for the odd prime divisors of a(4ac − b2). The Chinese remainder theorem and Dirichlet’s theorem on arithmetic progressions imply the existence of a prime p satisfying p ≡ 1 (mod pi), for all i and p ≡ 1 (mod 8). The Legendre symbol satisfies the identities ( pi p ) = ( p pi ) = ( 2 p ) = (−1 p ) = 1, which imply that there exists an integer m such that m2 ≡ b2−4ac (mod p). Since gcd(2a, p) = 1, the congruence 2an+b ≡ m(mod p) has some solution n and then p divides f (n). The answer is therefore negative. 5 Number Theory Solutions 83 2. Assume that f is irreducible. Let α1, . . . , αk be the roots of f and let ω be a primitive mth root of unity. The field K = Q(ω, α1, . . . , αk) is a normal extension of Q. The ˇCebotarev density theorem says that the density of rational primes that completely split in K is 1/[K : Q], and thus nonzero. Let p be such a prime that moreover, does not divide the discriminant of K . Then p splits completely in every subfield of K , in particular in Q(α1). From Dirichlet’s unit theorem, it follows that the congruence f (x) ≡ 0 (mod p) has k distinct solutions, since p does not divide the discriminant of Q(α1). Let �m be the cyclotomic polynomial of degree m. A similar argument concerning Q(ω) shows that the congruence �m(x) ≡ 0 (mod p) has exactly φ(m) roots. Let r be one of them. Then the order of r in Z/pZ is exactly m and thus m divides p − 1. Therefore p has the required properties. Comments 36 If we required the prime factors to be congruent to 1 modulo 4, then the answer would be positive. In fact, for any integer n, the prime divisors of 4n2 + 1 are of the form 4m + 1. 6 Algebra and Combinatorics Solutions 6.1 Algebra Problem 2.1. Set Sk,p = ∑p−1i=1 ik , for natural numbers p and k. If p ≥ 3 is prime and 1 < k ≤ p − 2, show that Sk,p ≡ 0 (mod p). Solution 2.1. Consider the symmetric polynomial P(x) = xk1 + · · · + xkp−1. Ac- cording to the fundamental theorem of symmetric polynomials, there exists G ∈ Z[x1, . . . , xp−1] such that P(x) = G(σ1, . . . , σp−1), where σi(x1, . . . , xp−1) are the fundamental symmetric polynomials σi(x1, . . . , xp−1) = ∑ 1≤j1<··· 2005, and thus it cannot belong to the given set. Therefore the number of elements to be deleted is at most 43. Consider now the triples (2, 87, 2 · 87), (3, 86, 3 · 86), . . . , (44, 45, 44 · 45) con- sisting of elements of the given set. We must delete at least one number from each triple in order that the remaining set be free of products. Thus the minimum number of elements to be deleted is 43. 2. Yes, it does. Take for instance 1 k! , 2 k! , . . . , k k! . Problem 2.6. Consider a set S of n elements and n + 1 subsets M1, . . . ,Mn+1 ⊂ S. Show that there exist r, s ≥ 1 and disjoint sets of indices {i1, . . . , ir}∩{j1, . . . , js} = ∅ such that r⋃ k=1 Mik = s⋃ k=1 Mjk . 88 6 Algebra and Combinatorics Solutions Solution 2.6. Assume that S = {a1, a2, . . . , an}. We associate to the subset Mj the vector vj ∈ Rn having the components vj = (v1j , v2j , . . . , vnj ), given by vij = { 1, if ai ∈ Mj, 0, if ai �∈ Mj. Since we have n + 1 vectors v1, . . . , vn+1 in Rn, there exists a nontrivial linear combination of them that vanishes. We separate the positive coefficients and the negative coefficients and write down this combination in the form ∑ i∈I λivi = ∑ j∈J λj vj , where I, J are disjoint sets of indices and λk > 0, for k ∈ I ∪ J . We claim now that ∪i∈IMi = ∪j∈JMj . In fact, assume that ak ∈ ∪i∈IMi and thus ak ∈ Mi for some i ∈ I . Then the kth coordinate of the vector vi is nonzero. All coordinates of the vectors vs are non- negative and the coefficients λi are positive, which implies that the kth coordinate of the vector ∑ i∈I λivi is nonzero. Since the kth coordinate of ∑ j∈J λj vj is nonzero, there should exist some j ∈ J for which the kth coordinate of vj is nonzero and hence ai ∈ Mj ⊂ ∪j∈JMj . This proves that ∪i∈IMi ⊂ ∪j∈JMj , and by symmetry, we have equality. Problem 2.7. Let p be a prime number, and A = {a1, . . . , ap−1} ⊂ Z∗+ a set of integers that are not divisible by p. Define the map f : P(A) → {0, 1, . . . , p − 1} by f ({ai1 , . . . , aik }) = k∑ p−1 aip (mod p), and f (∅) = 0. Prove that f is surjective. Solution 2.7. Let C0 = ∅, Cn = {a1, . . . , an}, for n ≤ p − 2. Set Pn = {f (β)|β ⊂ Cn}, and bn = f ({an}) �≡ 0. One obtains easily the following inductive description of the sets Pn: Pn+1 = {r + bn+1|r ∈ Pn ∪ {0}} ∪ Pn. Furthermore, if Pn = Pn+1, then bn+1 + r ∈ Pn for all r ∈ Pn. In particular, bn+1 ∈ Pn, and by induction, kbn+1 ∈ Pn, for any k. Since p is prime and bn+1 �= 0, we obtain Pn = {0, 1, . . . , p − 1}. Assume now that Pn �= Pn+1 for all n. Since card P0 = 1 and Pn ⊂ Pn+1, we derive that card Pn ≥ n + 1, and hence Pp−1 = P . Problem 2.8. Consider the function Fr = xr sin rA + yr sin rB + zr sin rC, where x, y, z ∈ R, A + B + C = kπ , and r ∈ Z+. Prove that, if F1(x0, y0, z0) = F2(x0, y0, z0) = 0, then Fr(x0, y0, z0) = 0, for all r ∈ Z+. 6.1 Algebra 89 Solution 2.8. Consider the complex numbers u = x0(cosA + i sin A), v = y0(cosB + i sin B), w = z0(cosC + i sin C). We denote the argument of the complex number z by arg z and its imaginary part by �z. We have arg u + arg v + arg w = kπ, � u + � v + � w = 0, and � u2 + � v2 + � w2 = 0. Now � z = 0 means that z is real, so that � z2 = 0. Thus 2�(uv + vw + wu) = �((u + v + w)2) − �(u2 + v2 + w2) = 0. Next, arg uvw = arg u + arg v + arg w = kπ , and so � uvw = 0. Let us consider the polynomial Pn(x1, x2, x3) = xn1 + xn2 + xn3 . According to the fundamental theorem of symmetric polynomials, we can write Pn as a polynomial in the fundamental symmetric polynomials: Pn(x1, x2, x3) = G(σ1, σ2, σ3), whereG ∈ R[σ1, σ2, σ3] is a real polynomial andσi , i ∈ {1, 2, 3}, are the fundamental symmetric polynomials in three variables, namely σ1(x1, x2, x3) = x1 + x2 + x3, σ2(x1, x2, x3) = x1x2 + x2x3 + x3x1, σ3(x1, x2, x3) = x1x2x3. Notice now that σ1(u, v,w) = u + v + w ∈ R, σ2(u, v,w) = uv + vw + wu ∈ R, σ3(u, v,w) = uvw ∈ R, which implies that Pn(u, v,w) ∈ R, and therefore Fr(x0, y0, z0) = �(Pn(u, v,w)) = 0. Problem 2.9. Let T (z) ∈ Z[z] be a nonzero polynomial with the property that |T (ui)| ≤ 1 for all values ui that are roots of P(z) = zn − 1. Prove that either T (z) is divisible by P(z), or else there exists some k ∈ Z+, k ≤ n − 1, such that T (z) ± zk is divisible by P(z). The same result holds when instead of P(z), we consider zn + 1. Solution 2.9. Let us write T (z) = A(z)P (z) + B(z), where deg B ≤ n − 1 and A,B ∈ Z[x]. Assume that T (z) is not divisible by P(z), and so B �= 0. Set k for the multiplicity of the root 0 in B, meaning that B(z) = zkC(z), where C(0) �= 0. We have then |C(ui)| = |B(ui)| = |T (ui)| ≤ 1. If we set Qd(z) = zd , then we can compute easily the sum of roots of unity as d∑ i=1 Qd(ui) = 0, if 1 ≤ d ≤ n − 1. We want to compute this kind of sum for the polynomial C. Since degC ≤ n − 1, the previous identities show that only the degree-zero part of C has a nontrivial contribution, and thus n∑ i=1 C(ui) = nC(0). This implies that 90 6 Algebra and Combinatorics Solutions 0 < |C(0)| ≤ 1 n ∣∣ ∣∣∣ n∑ i=1 C(ui) ∣∣∣ ∣∣ ≤ 1 n n∑ i=1 |C(ui)| ≤ 1. Now C(0) ∈ Z and thus C(0) = ±1, which implies that either C(ui) = 1, for all i, or else C(ui) = −1, for all i. Since deg C ≤ n − 1, we derive that either C(z) = 1 or else C(z) = −1. In particular, T (z) ± zk is divisible by P(z). The case in which P(z) is replaced by zn + 1 is similar. Problem 2.10. 1. If the map x �→ x3 from a group G to itself is an injective group homomorphism, then G is an abelian. 2. If the map x �→ x3 from a group G to itself is a surjective group homomorphism, then G is an abelian. 3. Find an abelian group with the property that x �→ x4 is an automorphism. 4. What can be said for exponents greater than 4? Solution 2.10. 1. We have a3b3 = (ab)3, and thus a2b2 = (ba)2, whence a4b4 = (a2)2(b2)2 = (b2a2)2 = (ab)4. Using again that a3b3 = (ba)3, we get (ab)3 = (ba)3. Since the map is injective, we have ab = ba. 2. We continue from above, a3b3 = (ba)3 = (ba)2ba = a2b2(ba); thus ab3 = b3a. Since the homomorphism is surjective, b3 can take any value in G; hence the relation above shows that G is abelian. 3. Take any nonabelian group of order 8, for instance the dihedral group D4. 4. For m > 3, there exists a nonabelian group Gm such that x �→ xm is the identity. For instance, if p is not of the form 2n + 1, then let p be an odd prime factor of m − 1 and Gm the group of matrices of the form M = ⎛ ⎝ 1 a b 0 1 c 0 0 1 ⎞ ⎠ with arbitrary a, b, c ∈ Z/pZ. Then Gm is a nonabelian group of order p3 such that the order of every element �= e is p. Problem 2.11. Let V be a vector space of dimension n > 0 over a field of character- istic p �= 0 and let A be an affine map A : V → V . Prove that there exist u ∈ V and 1 ≤ k ≤ np such that Aku = u. Solution 2.11. We write Ax = Bx + a, and thus Akx = Bkx + Cka, where Ck = Bk−1 + · · · +B + I . We use (B − I )Ck = Bk − I to rewrite the equation to solve as (B − I )Cku = Cka. There exists a solution if and only if a ∈ ker Ck + Im(B − I ). We suppose B − I is singular. The restriction of B to the eigenspace S corresponding to the eigenvalue 1 can be written BS = I + N , where N is nilpotent. Therefore Ck|S = k−1∑ i=0 (I + N)i = k−1∑ i=0 i∑ j=0 C j i N j = k−1∑ j=0 (k−1∑ i=j C j i ) C j i N j = k−1∑ j=0 C j+1 k N j . 6.1 Algebra 91 If p divides k, then Ck|S = Nk−1, and thus if k − 1 ≥ dim S, then Ck|S = 0. Therefore, S ⊂ ker Ck if k = lp ≥ 1 + dim S. Since Im(B − I ) ⊃ Im(B − I )n, which is complementary to S, we can take k = (n − 1)p if (n − 1)p ≥ dim S + 1, in which case Im(B − I ) + ker Ck = V . Otherwise, we have either n = 1, or n = 2, p = 2, and dim S = 2. If n = 1, then take k = p. If n = 2 = p, then take k = 2 if dim S = 1 or B = I , and k = 4 otherwise. In this situation, k = np is convenient, while in the former case k = (n−1)p suffices. This proves the claim. Problem 2.12. Find the cubic equation the zeros of which are the cubes of the roots of the equation x3 + ax2 + bx + c = 0. Solution 2.12. It is known that if the matrix M has eigenvalues λ1, . . . , λn and if F is a polynomial, then F(M) has eigenvalues F(λ1), . . . , F (λn). We consider now the matrix M = ⎛ ⎝ 0 0 −c 1 0 −b 0 1 −a ⎞ ⎠ , which has the characteristic polynomial x3 + ax2 + bx + c. The characteristic poly- nomial of M3 will then have as roots the cubes of the given polynomial. We now compute easily the characteristic polynomial of M3 as x3 + (a3 −3ab+c)x2 + (b3 − 3ab + 3c2)x + c3 = 0. Problem 2.13. Assume that the polynomials P,Q ∈ C[x] have the same roots, pos- sibly with different multiplicities. Suppose, moreover, that the same holds true for the pair P + 1 and Q + 1. Prove that P = Q. Solution 2.13. Let A and B be the sets of roots of P and P +1, respectively. Suppose that deg P = m ≥ n = deg Q. Obviously, A and B should be disjoint. It follows that P ′ has at least m− r roots from the set A and at least m− s roots from the set B, because multiple roots of P are roots of P ′. Since deg P ′ = m − 1, we derive that (m − r) + (m − s) ≤ m − 1 and thus r + s > m. Now, the union A ∪ B has r + s elements, because these sets are disjoint. On the other hand, each element of A ∪ B is a root of the polynomial P −Q, of degree m. Thus P −Q has at least m+ 1 roots, and thus it vanishes identically. Problem 2.14. Determine r ∈ Q, for which 1, cos 2πr, sin 2πr are linearly depen- dent over Q. Solution 2.14. We consider 0 ≤ r < 1. We suppose that there exists a linear relation a + b cos 2πr + c sin 2πr = 0, with a, b, c, r ∈ Q. This implies that (b2 + c2) cos2 2πr + 2ab cosπr + a2 − c2 = 0, i.e., cos 2πr satisfies a quadratic equation with rational coefficients. The same holds for sin 2πr . 92 6 Algebra and Combinatorics Solutions Consider now ξ = cos 2πr + i sin 2πr . If n is the smaller denominator of r , then ξ is a primitive root of nth order. On the other hand, since both cos 2πr and sin 2πr satisfy quadratic equations, then ξ satisfies an equation of order 4, with rational coefficients. We know that ξn = 1. Further, the irreducible polynomial of smallest degree that divides xn − 1 is 1 + x + · · · + xϕ(n), of degree ϕ(n), where ϕ(n) is Euler’s totient function. Our previous discussion implies that ϕ(n) must be a divisor of 4, and hence ϕ(n) ∈ {1, 2, 4}. Therefore, we have the following cases: 1. ϕ(n) = 1 and hence n ∈ {1, 2}, and so r ∈ {0, 12 } . 2. ϕ(n) = 2, and thus n ∈ {3, 4, 6}, whence r ∈ { 1 4 , 3 4 , 1 3 , 2 3 , 1 6 , 5 6 } . 3.ϕ(n) = 4, and thus n ∈ {5, 8, 10, 12}, whence r ∈ { 1 8 , 3 8 , 5 8 , 7 8 , 1 12 , 5 12 , 7 12 , 11 12 } . We omitted in the list above those r having denominator 5 or 10 because sin 2π5 = √ 1 2 (5 + √ 5) does not satisfy any quadratic equation with rational coef- ficients. Therefore, r ∈ {x|x = m24 , gcd(m, 24) �= 1,m ∈ Z } . Problem 2.15. 1. Prove that there exista, b, c ∈ Z, not all zero, such that |a|, |b|, |c| < 106 ∣∣∣a + b√2 + c√3 ∣∣∣ < 10−11. 2. Prove that if 0 ≤ |a|, |b|, |c| < 106, a, b, c ∈ Z, and at least one of them is nonzero, then ∣∣∣a + b√2 + c√3 ∣∣∣ > 10−21. Solution 2.15. 1. Consider the set S = { r + s√2 + t√3 } r,s,t∈{0,1,...,106−1}. Then card S = 1018. Set d = (1 + √2 + √3)106. Then all elements x ∈ S are bounded by d , i.e., 0 ≤ x < d. One divides the interval [0, d] into 1018 − 1 equal intervals [(k − 1)e, ke], where e = 10−18d and k = 1, . . . , 1018 − 1. By the pigeonhole principle, there exist two elements x, y ∈ S lying in the same interval, and hence satisfying |x − y| ≤ e < 10−11 2. Let F1 = a + b √ 2 + c√3, F2 = a + b √ 2 − c√3, F3 = a − b √ 2 + c√3, F4 = −a + b √ 2 + c√3. Now, 1,√2,√3 are linearly independent over Z, and thus Fi �= 0, for all 1 ≤ i ≤ 4. Further, F = F1F2F3F4 ∈ Z, since the product of conjugates contains no more square roots. This implies that |F1| ≥ 1/|F2F3F4| > 10−21, because 1/|Fi | > 10−7. Problem 2.16. Prove that if n > 2, then we do not have any nontrivial solutions for the equation xn + yn = zn, where x, y, z are rational functions. Solutions of the form x = af, y = bf, z = cf , wheref is a rational function anda, b, c are complex numbers satisfyingan+bn = cn, are called trivial. 6.1 Algebra 93 Solution 2.16. Any rational solution yields a polynomial solution, by clearing the denominators. Let further (f, g, h) be a polynomial solution for which the degree r = max{deg(f ), deg(g), deg(h)} is minimal among all polynomial solutions, and where r > 0. Consider the root of unity ξ = exp ( 2πi n ) and assume that f, g, and h are relatively prime. The equation can be written as n−1∏ j=0 (f − ξjh) = gn. Now, gcd(f, h) and gcd(f, g) are relatively prime, so that f − ξjh and f − ξkh are relatively prime when k �= j . Since the factorization of polynomials is unique, we must have g = g1 · · · gn−1, where gj are polynomials satisfying gnj = f − ξjh. Consider now the set {f − h, f − ξh, f − ξ2h}. Since n > 2, these elements belong to the two-dimensional space generated by f and h over C. Thus there exists a vanishing linear combination with complex coefficients in these three elements. Thus, there exist ai ∈ C so that a0gn0 + a1gn1 = a2gn2 . We then set hj = n√ajgj , and observe that hn0 + hn1 = hn2 . Moreover, the polynomials hi and hj are relatively prime if i �= j and max deg(hi) < r , which contradicts our choice of r . This proves the claim. Problem 2.17. A table is an n × k rectangular grid drawn on the torus, every box being assigned an element from Z/2Z. We define a transformation acting on tables as follows. We replace all elements of the grid simultaneously, each element being changed into the sum of the numbers previously assigned to its neighboring boxes. Prove that iterating this transformation sufficiently many times, we always obtain the trivial table filled with zeros, no matter what the initial table was, if and only if n = 2p and k = 2q , for some integers p, q. In this case, we say that the respective n× k grid is nilpotent. Solution 2.17. Consider the m × m square matrices (m ≥ 2) Dm = ⎛ ⎜⎜⎜⎜⎜⎜⎜ ⎝ 0 1 0 0 · · · 0 1 1 0 1 0 · · · 0 0 0 1 0 1 · · · 0 0 ... ... ... ... ... ... ... 0 0 0 0 · · · 0 1 1 0 0 0 · · · 1 0 ⎞ ⎟⎟⎟⎟⎟⎟⎟ ⎠ . We will write ≡ below if the equality holds (only) for matrices of entries modulo 2. Then the transformation T from the statement acts on the vector space Mn,k(Z/2Z) of n × k matrices with entries in Z/2Z as follows: T (X) = DnX + XDk, for X ∈ Mn,k(Z/2Z). 94 6 Algebra and Combinatorics Solutions By induction on s, we have that T 2 s (X) ≡ D2sn X + XD2 s k . Thus there exists N such that T N(X) ≡ 0 if and only if there exists some s such that D2 s n X ≡ XD2sk . Moreover, we can prove easily that for two matrices A and B of appropriate sizes, we have AX ≡ XB for all X ∈ Mn,k(Z/2Z) if and only if A and B are multiples of the identity by the same scalar. This follows, for instance, if we take X having all but one column (or line) trivial. This assertion is a special case of Schur’s lemma. Thus T is nilpotent if and only if there exist s and β ∈ Z/2Z such thatD2sn = β ·1n and D2sk = β · 1k . Thus the n × k grid is nilpotent if and only if the n × 1 and 1 × k grids are nilpotent. Let us analyze the case of the n× 1 grid. Set a = (a1, . . . , an) and a[k] = Dkna =( a [k] 1 , . . . , a [k] n ) . By induction, we have a[2 k] s ≡ as+2k−1 + as−2k−1 , where the indices are modulo n. In particular, if n = 2q , then D2qn = 0 and the grid is nilpotent. This proves the “if” part of the statement. Assume next that n is odd and there exists some r such that a[r] ≡ 0, for any a. Take such an r that is minimal. If n ≥ 3, then r ≥ 3. We claim now that a[r−1]s ≡ 1 for all s. This is true at least for one s, since r is minimal. Since a[r]s ≡ a[r−1]s+1 + a[r−1]s−1 , we find that the terms corresponding to even indices (modulo n) are all equal. But n is odd and thus all terms are equal. Further, we have a[r−1]s = a[r−2]s−1 + a[r−2]s+1 . We obtain then that n ≡ n∑ s=1 a[r−1]s = n∑ s=1 ( a [r−2] s−1 + a[r−2]s+1 ) = 2 n∑ s=1 a[r−2]s ≡ 0 (mod 2), which is a contradiction. This proves actually that a[r] ≡ 0 when n is odd if and only if a ∈ {(0, 0, . . . , 0), (1, 1, . . . , 1)}. Finally, consider n = 2mh, with h odd. For a vector a of length n, we set �s(a) = (as, as+2m, . . . , as+(h−1)2m). We observe now that �s(T 2m(a)) = T (�s(a)) because a[2 m+1] s = as + as+2m . Since the vector �s(a) is of length h (which is odd), the previous claim shows us that T N(�s(a)) ≡ 0 for some N only if �s(a) ∈ {(0, 0, . . . , 0), (1, 1, . . . , 1)}. Moreover, this happens for all values of s. This condition is trivial only when h = 1. This proves the “only if” part. Comments 37 This result was stated in the 1980s as an open question in the journal Kvant. The same method can be used to show that the n × k grids in the plane are 6.2 Algebraic Combinatorics 95 nilpotent if and only if n = 2p − 1 and k = 2q − 1 for some natural numbers p, q. A nice corollary is the following. Define the sequence (αk) by the recurrence relations α1 = 0, αk+1 = min(2αk + 2, 2N − 2αk − 1). Then there exists some k such that αk = N − 1 if and only if the parameter N has the form 2q − 1 for some natural number q. The solution given here is from: • V. Boju, L. Funar: Iterative processes for Zn2 , Analele Univ. Craiova 15 (1987), 33–38. 6.2 Algebraic Combinatorics Problem 2.18. Let us consider a four-digit number N whose digits are not all equal. We first arrange its digits in increasing order, then in decreasing order, and finally, we subtract the two obtained numbers. Let T (N) denote the positive difference thus obtained. Show that after finitely many iterations of the transformation T , we obtain 6174. Solution 2.18. Let N have the digits a, b, c, d. 1. Assume that a ≥ b = c ≥ d. Then in T (N), the sum of the first and the fourth digits is 9, as well as the sum of the second and the third. These give five combinations of four digits that give 6174, by applying T once more. 2. Otherwise, a ≥ b > c ≥ d. Then in T (N), the sum of the digits placed at extremities is 10, while the sum of the middle ones is 8. This gives 25 combinations all leading to the desired result, namely 6174. Comments 38 Let a be a positive integer having r digits, not all equal, in base g. Let a′ be the g-adic integer formed by arranging the digits of a in descending order and let a′′ be that formed by ascending order. Define T (a) = a′ − a′′. Then a is said to be self-producing if T (a) = a. There exists an algorithm for obtaining self-producing numbers in any base. Such a fixed point k of T is called a (g-adic) Kaprekar constant if it has the further property that every r-digit integer a eventually yields k on repeated iteration of T and moreover, k is fixed by T . D.R. Kaprekar found that 6174 has this nice property in a short note published in 1949. Ludington proved that for large r ≥ ng there does not exist a Kaprekar constant. Here ng is given by ng = ⎧ ⎨ ⎩ g − 1 + (k − 1)δ, if g = 2k, 2k + 1 + (g − k − 2)δ, if g = 2t k + 2t−1 + 1, 3k − 4 + (g − 1)δ, if g = 2t + 1, and δ ∈ {0, 1} is equal to g (mod 2). Further improvements by Lapenta, Ludington, and Prichett showed that there is no Kaprekar constant in base 10 when the number of digits is r > 4. 96 6 Algebra and Combinatorics Solutions Hasse and Prichett found that there exists a 4-digit Kaprekar constant in base g if and only if g = 2n · 5, where n is either 0 or an odd number. Moreover, the situation is completely understood for 3-digit numbers. If the base g is odd, then there is a Kaprekar constant given by ( r−2 2 , r − 1, r2 ) , which is reached within r+22 iterations of T (respectively 1, if r = 2). For instance, 495 is a Kaprekar constant if g = 10. If the base g is even, then there is no Kaprekar constant. More precisely, iterations of T will eventually reach the loop of length 2 formed by the pair of numbers ( r−32 , r − 1, r+12 ) , ( r−1 2 , r − 1, r−12 ) , after at most r+12 steps (respectively 1 if r = 3). • K.E. Eldridge and S. Sagong: The determination of Kaprekar convergence and loop convergence of all three-digit numbers, Amer. Math. Monthly 95 (1988), 105–112. • J.F. Lapenta, A.L. Ludington, and G.D. Prichett: An algorithm to determine self- producing r-digit g-adic integers, J. Reine Angew. Math. 310 (1979), 100–110. • G.D. Prichett, A.L. Ludington, and J.F. Lapenta: The determination of all decadic Kaprekar constants, Fibonacci Quart. 19 (1981), 45–52. • A.L. Ludington: A bound on Kaprekar constants, J. Reine Angew. Math. 310 (1979), 196–203. • H. Hasse and G.D. Prichett: The determination of all four-digit Kaprekar con- stants, J. Reine Angew. Math. 299/300 (1978), 113–124. Problem 2.19. Find an example of a sequence of natural numbers 1 ≤ a1 < a2 < · · · < an < an+1 < · · · with the property that every m ∈ Z+ can be uniquely written as m = ai − aj , for i, j ∈ Z+. Solution 2.19. We consider the sequence a1 = 1, a2 = 2, a2n+1 = 2a2n, a2n+2 = a2n+1 + rn, where rn is the smallest natural number that cannot be written in the form ai − aj , with i, j ≤ 2n + 1. Comments 39 One does not know the minimal growth of such a sequence ak . Problem 2.20. Consider the set of 2n integers {±a1,±a2, . . . ,±an} and m < 2n. Show that we can choose a subset S such that 1. The two numbers ±ai are not both in S; 2. The sum of all elements of S is divisible by m. Solution 2.20. Let S1, . . . , S2n−1 be the 2n − 1 nonempty distinct subsets of the set {a1, . . . , an}, where ai ≥ 0. Let F(Si) denote the sum of the elements of Si . Then, by the pigeonhole principle, there exist i, j such that F(Si) ≡ F(Sj ) (mod m). Consider next the set S = {Si \ Sj } ∪ −{Sj \ Si}. We derive that F(S) ≡ 0 (mod m). 6.2 Algebraic Combinatorics 97 Problem 2.21. Show that for every natural number n there exist prime numbers p and q such that n divides their difference. Solution 2.21. Consider the following arithmetic progression: 1, 1+n, 1+2n, . . . , 1+ rn, . . . .According to Dirichlet’s theorem, there exist infinitely many prime numbers among the terms of this progression. Let p, q be two of these prime numbers; then p = 1 + nr and q = 1 + ns, where r �= s. This yields n(r − s) = p− q, as claimed. An alternative solution is to consider the set of n arithmetic progressions of ratio n starting respectively at 0, 1, 2, . . . , n− 1. Since the set of prime numbers is infinite (this is elementary), there exists at least one progression having infinitely many prime numbers among its terms. The argument above settles the claim. Problem 2.22. An even number, 2n, of knights arrive at King Arthur’s court, each one of them having at most n − 1 enemies. Prove that Merlin the wizard can assign places for them at a round table in such a way that every knight is sitting only next to friends. Solution 2.22. 1. We consider the friendship graph G defined below: its vertices are in bijection with the knights and the edges are joining pairs of vertices whose respective knights are not enemies. The degree of each vertex is at least n. According to Dirac’s theorem, such a graph admits a Hamiltonian cycle, and this yields the wanted assignment of places. 2. Choose an arbitrary assignment of places in which we have two neighbors, A and B, who are enemies. Let us assume that A lies on the right-hand side of B. According to the pigeonhole principle, there exists another pair of enemies, say A˜, B˜, who are neighbors, and moreover, A˜ is on the right of B˜. Let us switch all the places starting at A (and lying on the right side of A) and ending at B˜, using a symmetry. After such a transformation, the number of enemy pairs (A,B) is diminished by 2. Applying such transforms iteratively, one obtains a position in which all neighbors are friends. Problem 2.23. Let r, s ∈ Z+. Find the number of 4-tuples of positive integers (a, b, c, d) that satisfy 3r7s = lcm(a, b, c) = lcm(a, b, d) = lcm(a, c, d) = lcm(b, c, d). Solution 2.23. The numbers a, b, c, d are of the form 3mi 7ni , 1 ≤ i ≤ 4, where 0 ≤ mi ≤ r and 0 ≤ ni ≤ s. Also, mi = r for at least two values of i, and ni = s for at least two values of i. We have then: (1) one possibility that mi = r for all four i; (2) 4r possibilities that precisely one mi ∈ {0, . . . , r − 1}; (3) C24r2 possibilities that exactly two mi ∈ {0, . . . , r − 1}. Therefore, there are ( 1 + 4r + 6r2) possibilities for the mi’s and a similar number for the ni’s, yielding a total of ( 1 + 4r + 6r2)(1 + 4s + 6s2) possibilities. Problem 2.24. 1. Let n ∈ Z+ and p be a prime number. Denote by N(n, p) the number of binomial coefficients Csn that are not divisible by p. Assume that n is written in base p as n = n0 + n1p + · · · + nmpm, where 0 ≤ nj < p, for all j ∈ {0, 1, . . . , m}. Prove that N(n, p) = (n0 + 1)(n1 + 1) · · · (nm + 1). 98 6 Algebra and Combinatorics Solutions 2. Write k in base p as k = k0 + k1p + · · · + ksps , with 0 ≤ kj ≤ p − 1, for all j ∈ {0, 1, . . . , s}. Prove that Ckn ≡ Ck0n0Ck1n1 · · ·Cksns (mod p). Solution 2.24. 1. It is clear that Ckp ≡ 0 (mod p) for all k ∈ {1, 2, . . . , p}. Thus (1 + x)p ≡ 1 + xp in (Z/pZ)[x]. By induction, we find that (1 + x)pn ≡ 1 + xpn in (Z/pZ)[x] for any natural number n. We have then the following congruences in (Z/pZ)[x]: (1+x)n = (1+x)n0(1+x)n1p · · · (1+x)nmpm ≡ (1+x)n0(1+xp)n1 · · · (1+xpm)nm. When developing factors on the right-hand side, we obtain a sum of factors xa0+a1p+a2p2+···+ampm , with a coefficient that is nonzero modulo p. Since every number can be uniquely written in base p, all these factors are distinct and their coefficients modulo p are nonzero. There are exactly (n0 + 1)(n1 + 1) · · · (nm + 1) such factors, and therefore as many binomials not divisible by p. 2. As observed above, the factor xk appears in the form xa0+a1p+a2p2+···+ampm ; since k can be uniquely written in base p, we have ai = ki . This implies that the coefficient of xk is Ck0n0C k1 n1 · · ·Ckjnj , and hence the claim. Problem 2.25. Define the sequence Tn by T1 = 2, Tn+1 = T 2n − Tn + 1, for n ≥ 1. Prove that if m �= n, then Tm and Tn are relatively prime, and further, that ∞∑ i=1 1 Ti = 1. Solution 2.25. We prove by induction that Tn+1 = 1 + T1T2 · · · Tn. In fact, we have Tn+1 = T 2n − Tn + 1 = Tn(Tn − 1)+ 1 = Tn(Tn−1Tn−2 · · · T1)+ 1 = 1 + T1 · · · Tn. Then takem < n. Therefore, Tm divides T1 · · · Tn−1 = Tn−1 and thus gcd(Tm, Tn) = 1. Further, we prove by induction that n∑ i=1 1 Ti = 1 − 1 Tn+1 − 1 . In fact, k+1∑ i=1 1 Ti = 1 − 1 Tk+1 − 1 + 1 Tk+1 = 1 − 1 Tk+1(Tk+1 − 1) = 1 − 1 Tk+2 − 1 . Since Tn tends to infinity, we therefore obtain ∑∞ i=1 1Ti = 1. 6.2 Algebraic Combinatorics 99 Problem 2.26. Let α, β > 0 and consider the sequences [α], [2α], . . . , [kα], . . . ; [β], [2β], . . . , [kβ], . . . , where the brackets denote the integer part. Prove that these two sequences taken together enumerate Z+ in an injective manner if and only if α, β ∈ R \ Q and 1 α + 1 β = 1. Solution 2.26. Set AN = {1, 2, . . . , N}; take k maximal such that kα < N + 1 and l maximal such that lβ < N + 1. Therefore, the following inequalities hold: [ N α ] ≤ k ≤ [N+1 α ] and [ N β ] ≤ l ≤ [ N+1 β ] . Since AN is injectively enumerated by the two sequences, we have k + l = N , and hence [ N α ] + [ N β ] ≤ N ≤ [ N + 1 α ] + [ N + 1 β ] . Letting N go to infinity, we obtain 1 α + 1 β = 1. If α ∈ Q, then also β ∈ Q. In this case, write α = m n and β = p q . It follows that [αnqp] = [βmqn], which contradicts the injectivity assumption. Therefore α �∈ Q and β �∈ Q. Conversely, we have N+1 = N + 1 α +N + 1 β > [ N + 1 α ] + [ N + 1 β ] > N + 1 α +N + 1 β −2 = N−1, whence [ N + 1 α ] + [ N + 1 β ] = N. In particular, using the notation from above, we have k + l = N . If the two sequences enumerate AN in a surjective manner, then they will also enumerate AN injectively, because k+l = N . It suffices then to prove the surjectivity. Let us assume the contrary. Then there exist u, x, y ∈ Z+ such that xα < u < u + 1 < (x + 1)α and yβ < u < u + 1 < (y + 1)β. Dividing by α, β respectively and adding up the inequalities, we obtain x + y < u α + u β < u + 1 α + u + 1 β < x + 1 + y + 1, which amounts to x + y < u < u + 1 < x + 2 + y. This is false, because x, y, u are integers, and our claim follows. Problem 2.27. We say that the sets S1, S2, . . . , Sm form a complementary system if they make a partition of Z+, i.e., every positive integer belongs to a unique set Si . Let m > 1 and α1, . . . , αm ∈ R+. Then the sets 100 6 Algebra and Combinatorics Solutions Si = {[nαi], where n ∈ Z+} form a complementary system only if m = 2, α−11 + α−12 = 1, and α1 ∈ R \ Q. Solution 2.27. If all αj < 1, then the collection Sj contains twice some number. Let 1 ∈ S1, and thus [α1] = 1, α1 �= 1. We prove first that S1 = {[nα1]; n ∈ Z+} and T = {[nβ − �]; n ∈ Z+} form a complementary system, where β = α1 α1 − 1 , and � = { (2(a − c))−1, if α1 = ac ∈ Q, 0, otherwise. Let � = {nα1, nβ − �; n ∈ Z+}. It is sufficient to show that for any integer M > 1, we have the formula N = card({x ∈ �; x < M}) = M − 1. Now, nα1 < M is equivalent to nα1 + δ < M , where δ = (2c)−1 if α1 = a/c ∈ Q, while δ = 0, if α ∈ R \Q. The maximum value for n is therefore [ M−δ α1 ] . In a similar way, the number of elements of the form nβ − � that are less than M is [ M+� β ] . This implies that N = [(M − δ)/α1] + [(M + �)/β]. Now we have M − δ α1 − 1 < [ M − δ α1 ] < M − δ α1 , M + � β − 1 < [ M + � β ] < M + � β , and by summing up the two inequalities, we obtain M − 2 < N < M , and therefore N = M − 1, as claimed. Now let k be the smallest integer k �∈ S1. Then there exists some α2 such that k = [α2] = [β − �] ≥ 2. We have nβ − 1 ≤ [nβ − �] < nβ − �, and therefore [(n + 1)β − �] − [nβ − �] < (n + 1)β − � − nβ + 1 = β + 1 − � ≤ k + 2 − �, [(n + 1)β − �] − [nβ − �] > (n + 1)β − 1 − nβ + � = β − 1 + � > k − 1 + 2�. Therefore k ≤ [(n + 1)β − �] − [nβ − �] = [(n + 1)α2] − [nα2] ≤ k + 1. The difference between two consecutive terms of the sequence [nβ − �] is equal to the difference between consecutive missing terms from the sequence [nα1]; that is, k or k + 1 is the same as the difference between two consecutive terms of the sequence [nα2]. This implies the fact that the j th term that does not belong to S1 is precisely [jα2] = [jβ − �]. This implies that m = 2. If α1 ∈ Q, then α2 = b/d, and for j = d, we obtain [jα2] > [jβ − �], which is false. 6.2 Algebraic Combinatorics 101 Comments 40 The fact that the two sequences from the previous problem are com- plementary is known as the Rayleigh–Beatty theorem, since S. Beatty proposed it as a problem in Amer. Math. Monthly in 1926. Presumably, this was known to Rayleigh, who mentioned it without proof in 1894. In 1927, J.V. Uspensky proved that if the m sequences are complementary, then m ≤ 2; his proof was simplified in several papers by Skolem, Graham, and Fraenkel, who also provided a far-reaching generalization in 1969. • S. Beatty: Problem 3173, Amer. Math. Monthly 33 (1926), 3, 156. • J. Lambek, L. Moser: Inverse and complementary sequences of natural numbers, Amer. Math. Monthly 61 (1954), 454–458. • A.S. Fraenkel: Complementary systems of integers, Amer. Math. Monthly 84 (1977), 114–115. • A.S. Fraenkel: The bracket function and complementary sets of integers, Canad. J. Math. 21 (1969), 6–27. • R.L. Graham: On a theorem of Uspensky, Amer. Math. Monthly 70 (1963), 407– 409. • Th. Skolem: On certain distributions of integers in pairs with given differences, Math. Scand. 5 (1957), 57–68. • J.V. Uspensky, M.A. Heaslet: Elementary Number Theory, McGraw-Hill, New York, 1939. Problem 2.28. Let f : Z+ → Z+ be an increasing function and set F(n) = f (n) + n, G(n) = f ∗(n) + n, where f ∗(n) = card({x ∈ Z+; 0 ≤ f (x) < n}). Then {F(n); n ∈ Z+} and {G(n); n ∈ Z+} are complementary sequences. Conversely, any two complemen- tary sequences can be obtained this way using some nondecreasing function f . Solution 2.28. We compute the number N of integers from {F(n),G(n), n ∈ Z+} that are smaller than M . Suppose that k terms of the form G(n) are smaller than M . Then f ∗(k) + k < M ≤ f ∗(k + 1) + k + 1, and so f ∗(k) < M − k < f ∗(k + 1) + 1. Since f ∗(k) = card{x ∈ Z+; f (x) < k} < M − k, we derive f (M − k) ≥ k. Further, f ∗(k + 1) = {x ∈ Z+; f (x) < k + 1} ≥ M − k − 1 implies that f (M − k − 1) < k + 1. Therefore F(M − k) = f (M − k) + M − k ≥ M and F(M − k − 1) < M . This means that exactly M − k terms of the form F(n) are smaller than M; therefore N = M . The converse is immediate by defining f (n) = F(n) − n. Comments 41 Since f ∗ is nondecreasing, it follows that it makes sense to define f ∗∗. The sequences G(n) and H(n), where H(n) = f ∗∗(n)+n, are complementary, and thus f ∗∗ = f . The result is due to A. Frankel. 102 6 Algebra and Combinatorics Solutions Problem 2.29. Let M denote the set of bijective functions f : Z+ → Z+. Prove that there is no bijective function between M and Z. Solution 2.29. We will find an injection ρ : R \ (Q ∪ [0, 2]) → M , as follows. To any α ∈ R \ (Q∪ [0, 2]) there is associated an irrational β < 2 such that 1 α + 1 β = 1. Let ρ(α) : Z+ → Z+ be the function defined by ρ(α)(n) = {[ α n2 ] , for even n,[ β n+12 ] , for odd n. According to the previous problem, ρ(α) : Z+ → Z+ is a bijection. Moreover, the map ρ is easily seen to be injective. Thus card(M) ≥ card(R \ Q) = card(R) > card(Z), and the result follows. Problem 2.30. Let F ⊂ Z be a finite set of integers satisfying the following proper- ties: 1. For any x ∈ F , there exist y, z ∈ F such that x = y + z. 2. There exists n such that, for any natural number 1 ≤ k ≤ n, and any choice of x1, . . . , xk ∈ F , their sum x1 + · · · + xk is nonzero. Prove that card(F ) ≥ 2n + 2. Solution 2.30. By hypothesis, 0 �∈ F . Let F+ = F ∩ Z+, F− = F ∩ Z−, so that F = F+ ∪ F−. Consider the unoriented graph � whose vertices are the elements of F+ and whose edges are defined below: x and y are adjacent if there exists z ∈ F such that x = y + z. By the first hypothesis, each vertex is adjacent to at least one edge. This implies that the graph � contains a cycle [x1, . . . , xk]. Assume that k is minimal with this property. This means that there exist zi ∈ F such that: x1 = x2 + z1 = x3 + z2 + z1 = · · · = x1 + zk + · · · + z1, which implies that z1 + · · · + zk = 0. The second hypothesis implies that k ≥ n + 1, and thus card(F+) ≥ k ≥ n + 1. A similar argument shows that card(F−) ≥ n + 1, and the claim follows. Problem 2.31. For a finite graph G we denote by Z(G) the minimal number of colors needed to color all its vertices such that adjacent vertices have different colors. This is also called the chromatic number of G. Prove that the inequality Z(G) ≥ p 2 p2 − 2q holds if G has p vertices and q edges. 6.2 Algebraic Combinatorics 103 Solution 2.31. If we fix N = χ(G) and the number p of vertices of the graph G, then we have to show that the number q of edges allowed q ≤ p 2 2 ( 1 − 1 N ) . By hypothesis, there exists a partition (according to the colors) of the set of vertices into N classes such that two vertices in the same class are not adjacent. Let n1, n2, . . . , nN be the number of vertices in the respective classes. The only possibility to get an edge is to join two vertices lying in different classes, and thus the total number q of edges is at most ∑ 1≤i σ(2) > · · · > σ(n). Let us prove first the following intermediate result. If a < c, b < d, then the inequality |a − b| + |c − d| ≤ |a − d| + |b − c| holds with equality when either a < c < b < d or b < d < a < c. From the symmetry of the previous inequality, we can assume that a < b, d, and using a translation followed by a homothety on the real axis, we can, moreover, assume that a = 0, c = 1. If b < 1, d < 1, the inequality reduces to 2b < 2d. If b < 1, d > 1, then the inequality is 2b < 2. If b > 1, d > 1, we obtain equality. This proves the claim. 6.2 Algebraic Combinatorics 107 Now let σ be a permutation such that S(σ) is maximal. Suppose that there exists i < j such that σ(i) < σ(j). Let us define another permutation σ ∗ = σ ◦ (i, j), where (i, j) denotes the transposition interchanging i and j . According to our claim, we have S(σ ∗) ≥ S(σ). By modifying iteratively σ by taking the product with the transpositions deter- mined by all pairs (i, j) as above, we will finally obtain γ : γ : ( 1 2 · · · n n n − 1 · · · 1 ) . Moreover, the inequality above shows that S(σ) ≤ S(γ ) = n∑ i=1 |n − 2i + 1|. Problem 2.37. On the set Sn of permutations of {1, . . . , n} we define an invariant distance function by means of the formula d(σ, τ ) = n∑ i=1 |σ(i) − τ(i)|. What are the values that d could possibly take? Solution 2.37. We have d(ρσ, ρτ) = d(σ, τ ), for any ρ, σ, τ ∈ Sn. Therefore, it suffices to consider the values of d(1, σ ), where 1 is the identity permutation. If m > 0 is a value of d, we will show below that m − 2 is also a value of d, and hence d takes all even values from 0 to some 2t , where t has to be determined. Notice first that d takes only even values, because d(σ, τ ) ≡ n∑ i=1 (σ (i) − τ(i)) ≡ n∑ i=1 σ(i) − n∑ i=1 τ(i) ≡ 0 (mod 2). Let m = d(1, σ ) > 0. There then exist 1 ≤ r < s ≤ n such that σ(r) > r, σ (s) < s, and σ(i) = i, if r < i < s. Let ρr,s be the cycle (r, r + 1, . . . , s). We claim that d(1, ρr,sσ ) = m − 2. First, by hypothesis, σ(r) > s and σ(s) < r . Looking at the contribution of the elements from r to s to the respective distances, we obtain d(1, σ )−d(1, ρr,sσ ) = (σ (r)−r+s−σ(s))−(r−σ(s)+s−r−1+σ(r)−r−1) = 2. Let us show now that t = [ n2 4 ] (see also the previous problem) and the maximum distance d(1, σ ) is realized for the permutation θ(i) = n + 1 − i. We have d(1, θ) = 2 [ n2 4 ] = [ n2 2 ] . Let k < n−1, such that σ(i) = n+1− i, for all i < k, while σ(k) �= n+1−k. We claim that there exists some permutation σ˜ ∈ Sn with the property that d(1, σ ) ≤ d(1, σ˜ ) and that satisfies σ˜ (i) = n + 1 − i, i ≤ k. Then, using induction k, we find a sequence of permutations reaching θ such that 108 6 Algebra and Combinatorics Solutions d(1, σ ) ≤ d(1, σ˜ ) ≤ · · · ≤ d(1, θ). If σ(k) �= n + 1 − k, we have σ(k) < n + 1 − k because every number greater than n + 1 − k is the image of some i < k. But this implies that σ(r) = n + 1 − k, for some r > k. Let τ be the transposition (k, r). Set σ˜ = τσ ; we have d(1, σ˜ ) − d(1, σ ) = |r − kσ | + |m + 1 − 2k| − |k − kσ | + |n + 1 − k − r| = w. Let us suppose that k ≤ σ(k) ≤ r ≤ n + 1 − k (the other cases are similar). Then w = r−σ(k)+n+1−2k−σ(k)+k−n−1+k+r = 2(r−kσ) ≥ 0. Analogously, for all k, r we have d(1, σ˜ ) − d(1, σ ) ≥ 0, which proves the claim. Problem 2.38. The set M = {1, 2, . . . , 2n} is partitioned into k sets M1, . . . ,Mk , where n ≥ k3 + k. Show that there exist i, j ∈ {1, . . . , k} for which we can find k + 1 distinct even numbers 2r1, . . . , 2rk+1 ∈ Mi with the property that 2r1 − 1, . . . , 2rk+1 − 1 ∈ Mj . Solution 2.38. There exists a set Ms that contains at least 2nk ≥ 2(k2 + 1) elements. We have to consider two cases: 1. Either Ms contains at least 2(k 2+1) 2 = k2 +1 even numbers. Then the set of odd numbers O = {r − 1,where r is even, and r ∈ Ms} has k2 +1 elements. Then there exists some set Ma containing at least k2+1k elements from O. We choose therefore i = s and j = a. Notice that i might be equal to j . 2. Or else Ms contains at least k2 + 1 odd numbers. The solution is similar to that from above, considering the set of even numbers. Problem 2.39. Let S be the set of odd integers not divisible by 5 and smaller than 30m, where m ∈ Z∗+. Find the smallest k such that every subset A ⊂ S of k elements contains two distinct integers, one of which divides the other. Solution 2.39. Consider the subset N = {1, 7, 11, 13, 17, 19, 23, 29, . . . , 30m − 1} of elements of S that are not divisible by 3. There are 8m elements in N , which are written in increasing order a1 < a2 < · · · < a8m. Every element of S can be uniquely written as x = ai · 3t , where t ∈ Z+ and ai ∈ N . If k ≥ 8m + 1, then according to the pigeonhole principle, any subset A ⊂ S of cardinality card A = k contains two distinct elements x, y ∈ A for which x = ai3t and y = ai3q . In this case, either x divides y, or y divides x. Next, for all i, choose the maximal t (i) with the property that ai3t (i) < 30m < ai3t (i)+1, and set bi = ai3t (i). We have therefore 10m < bi < 30m. The set {b1, b2, . . . , b8m} contains 8m elements from S, and also we have the inequalities 0 < bi/bj < 3, for any i, j . Since all numbers bi are odd, we derive that bi/bj �∈ Z. Therefore {b1, . . . , b8m} does not contain a number and one divisor of it. This shows that the required value of k is 8m + 1. 6.2 Algebraic Combinatorics 109 Problem 2.40. Prove that ∏ 1≤jj ai − aj i − j = ∏ i>j (ai − aj ) (n − 1)!(n − 2)! · · · 1! . We now consider the determinant D of the matrix ⎛ ⎜⎜ ⎜ ⎝ 1 1 · · · 1 C1a1 C 1 a2 · · · C1an ... ... ... Cn−1a1 C n−1 a2 · · · Cn−1an ⎞ ⎟ ⎟⎟ ⎠ . It is obvious that D ∈ Z+. We now write the binomial coefficient Ckm using factorials, and we arrange the factors in the lines. We obtain D = 1 1!2! · · · (n − 1)! det ⎛ ⎜⎜⎜ ⎝ 1 1 · · · 1 a1 a2 · · · an ... ... ... an−11 a n−1 2 · · · an−1n ⎞ ⎟⎟⎟ ⎠ = ∏ i>j ai − aj∏ i=1(i − 1)! = S, and the claim follows. Problem 2.41. Is there an infinite set A ⊂ Z+ such that for all x, y ∈ A neither x nor x + y is a perfect power, i.e., ak , for k ≥ 2? More generally, is there an infinite set A ⊂ Z+ such that for any nonempty finite collection xi ∈ A, i ∈ J , the sum∑i∈J xi is not a perfect power? Solution 2.41. 1. Yes. Let p1 = 2, p2 = 3, . . . , pn, . . . be the set of primes in in- creasing order. We then set A = {223, 22325, . . . , 223252 · · ·p2npn+1, . . .}. Obviously, no element of A is a perfect power. Moreover, if x ≤ y ∈ A, then we can write x = 22 · · ·p2kpk+1 and x = 22 · · ·p2npn+1. If k < n, then pk+1 divides x+y = 22 · · ·p2kpk+1 ( 1+pk+1p2k+2 · · ·p2npn+1 ) , while p2k+1 does not divide x+y, and thus it cannot be a perfect power. If k = n, the same argument works, unless k = 0, which was excluded from A. 2. Let us prove that if C ⊂ Z+ is a set of density d(C) = 0, then there exists an infinite set A ⊂ Z+ such that for any nonempty finite collection xi ∈ A, i ∈ J , the sum ∑ i∈J xi is not in C. Recall that the density of the subset C ⊂ Z+ is defined as d(C) = lim n→∞ card(C ∩ {1, 2, . . . , n}) n . Since d(C) = 0, there exists a natural number a1 �∈ C. Then consider C1 = C ∪ (C − a1) ∩ Z+, where we set X − λ = {x − λ|x ∈ X}. The density d has the following fundamental property that results from the definition: 110 6 Algebra and Combinatorics Solutions d(A ∪ B) ≤ d(A) + d(B). This implies that d(C1) = 0, and therefore there exists a natural number a2 �∈ C1. Assuming that both the subsetCk of density zero and the sequencea1, a2, . . . , ak+1, with ak+1 �∈ Ck , are defined, we set Ck+1 = Ck k⋃ p=1 ⋃ i1≤i2≤···≤ip≤k+1 (Ck − (ai1 + ai2 + · · · + aip )) ∩ Z+. It follows that d(Ck+1) = 0 and thus there exists at least one integer ak+2 > ak+1 such that ak+2 �∈ Ck+1. The sequence (ak) forms an infinite subset A with the required property. It suffices now to compute the density of the set C = {ak|a ≥ 2, k ≥ 2}. We have card(C ∩ {1, 2, . . . , n}) n = [ √ n] + [ 3√n] + · · · n ≤ √ n log2 n n = log2 n√ n . Therefore d(C) = 0. Comments 44 There is an analogous result when products are used instead of sums of elements in A. For instance, the set A = { a1, a2 = Ca12a1 , Ca2a1+a2 , . . . , C an an−1+an } has the property that no product of its elements is a perfect power. Is it true that d(A) = 0? Problem 2.42. Let f (n) = max AA A ... Ak 3 2 1 , where n = A1 +· · ·+Ak . Thus, f (1) = 1, f (2) = 2, f (3) = 3, f (4) = 4, f (5) = 9, f (6) = 27, f (7) = 512, etc. Determine f (n). Solution 2.42. It is clear that if f (n) = A··· Ak 1 , then Ai > 1. Also, f is an increasing function, and thus f (n+ 1) ≥ f (n)+ 1. If A1 = k, then A·· ·Ak 2 ≤ f (n− k), and this implies that whenever n ≥ 4, we have f (n) = max 2≤k≤n−2 k f (n−k). It is easy now to see that f (n + 1) ≥ 2f (n). In fact, let f (n) = kf (n−k), for some k ≥ 2; then f (n + 1) ≥ kf (n−k+1) ≥ kf (n−k)+1 ≥ 2f (n). The next step is to compare af (b+1) with (a + 1)f (b), that is, f (b+1) f (b) with log(a+1)log(a) . If a ≥ 2, then a + 1 < a2 and hence log(a + 1)/ log a < 2. Thus, if b ≥ 4, then 6.2 Algebraic Combinatorics 111 f (b + 1)/f (b) ≥ 2, which yields af (b+1) > (a + 1)f (b), as soon as a ≥ 2, b ≥ 3. For small values of the arguments, we have log 4 log 3 < f (4) f (3) < f (3) f (2) < log 3 log 2 . These show that kf (n−k) has a maximum when k = 2 if n > 6. Therefore, the final answer is f (n) = ⎧ ⎪⎨ ⎪⎩ 2·· ·23 2 , for odd n > 3, 2·· ·23 3 , for even n > 4. Problem 2.43. Consider a set M with m elements and A1, . . . , An distinct subsets of M such that card(Ai ∩ Aj) = r ≥ 1 for all 1 ≤ i �= j ≤ n. Prove that n ≤ m. Solution 2.43. Observe first that ifA is anm×nmatrix andm > n, then det(A·A�) = 0, where A� denotes the transpose matrix. In fact, let us border the matrix A by a null matrix of size (m− n)× n, in order to get an m×m matrix A′. It is obvious that A′ · A′� = A · A�, and thus det(A · A�) = det(A′ · A′�) = 0. Let D be the n × n matrix whose entries are Dij = { r ≥ 1, if i �= j, dii ≥ r, otherwise. Assume that there exists at most one i such that dii = r . We claim then that det(D) �= 0. This follows by subtracting the last line from the first n − 1 and an inductive argument. Consider now the m × n matrix A whose entries are Aij = { 1, if i ∈ Aj , 0, otherwise. It is now immediate that D = A · A� is given by Dij = n∑ j=1 aikakj = card(Ai ∩ Aj). By hypothesis, Dij = r when i �= j , and since the Ai are distinct, there exists at most one i such that Dii = r . The claim from above tells us that det(D) = det(A · A�) �= 0, while the first observation implies that m ≤ n. Problem 2.44. Setπ(n) for the number of prime numbers less than or equal ton. Prove that there are at most π(n) numbers 1 < a1 < · · · < ak ≤ n with gcd(ai, aj ) = 1. 112 6 Algebra and Combinatorics Solutions Solution 2.44. If X is a set, we denote by P(X) the set of nonempty subsets of X. Define the map h : {1, . . . , n} → P({p1, . . . , pπ(n)}) that associates to the integer x the set of prime divisors of x. One then has card(h(ai)) ≥ 1 and furtherh(ai)∩h(aj ) = ∅, from our assumption. Since the maximal number of disjoint subsets of {p1, . . . , pπ(n)} is bounded by π(n), the claim follows. Problem 2.45. Prove that for every k, there exists n such that the nth term of the Fibonacci sequenceFn is divisible by k. Recall thatFn is determined by the recurrence Fn+2 = Fn + Fn+1, for n ≥ 0, where the first terms are F0 = 0, F1 = 1. Solution 2.45. Using the pigeonhole principle, there exists an infinite sequence ni such that Fn1 ≡ Fn2 ≡ · · · ≡ Fnm ≡ · · · (mod k). Moreover, there exists an infinite subsequence nis such that Fni1+1 ≡ Fni2+1 ≡ · · · ≡ Fnim+1 ≡ · · · (mod k) by the same argument. Thus one knows that Fnis ≡ Fnit (mod k) and Fnis+1 ≡ Fnit+1 (mod k). The recurrence relation of the Fibonacci numbers shows that Fnis+m ≡ Fnit+m (mod k), for any m ∈ Z. Since F0 = 0, we find that there exists n (actually infinitely many such integers) such that k divides Fn. Problem 2.46. Consider a set of consecutive integers C + 1, C + 2, . . . , C + n, where C > nn−1. Show that there exist distinct prime numbers p1, p2, . . . , pn such that C + j is divisible by pj . Solution 2.46. Let k ≤ h and consider the factorization of C + k into prime factors. If C + k has at least n distinct prime divisors, then we can find one prime divisor that is not yet associated with the other n − 1 numbers C + i. Assume then that C + k has at most j ≤ n − 1 prime factors, which are p1k, . . . , pjk for 1 ≤ j ≤ n − 1. We write down the factorization as p a1k 1k · · ·p ajk jk = C + k > C > nn−1. The pigeonhole principle implies that there exists some prime power qk = pajkjk such that qk > n. We then associate the prime number pjk to C+k. Let us show that this procedure yields distinct prime numbers. Suppose that there exist j and k such that the associated numbers are the same. We know that qk divides C + k and qj divides C + j , and they are both prime powers of the same prime. Since qk > n and qj > n, we obtain gcd(qk, qj ) > n. Moreover, gcd(qk, qj ) divides both C + k and C + j and hence their difference |j − k| < n, which is a contradiction. This proves the claim. 6.2 Algebraic Combinatorics 113 Problem 2.47. Let p be a prime number and f (p) the smallest integer for which there exists a partition of the set {2, 3, . . . , p} into f (p) classes such that whenever a1, . . . , ak belong to the same class of the partition, the equation k∑ i=1 xiai = p does not have solutions in nonnegative integers. Estimate f (p). Solution 2.47. 1. Suppose that a1, a2 are in the same class and gcd(a1, a2) = 1. One knows then that any natural number greater than or equal to a1a2 can be written as x1a1 + x2a2, where xi ∈ Z+. Therefore, the prime numbers smaller than √p should to different classes of our partition, and hence f (p) > π ( p 1 2 ) ∼ 2 √ p logp , where π(n) denotes the number of primes smaller than n. 2. If t ∈ Z+, let us consider the sets of consecutive elements At ={[ p t+1 ] + 1, . . . , [p t ]} . We claim that the equation ∑k i=1 aixi = p has no integer solutions if ai ∈ At for 1 ≤ i ≤ k. In fact, if we had a solution xi , then we would have p t + 1 k∑ i=1 xi < k∑ i=1 aixi < p t k∑ i=1 xi, and so t < ∑k i=1 xi < t + 1, which is false, since xi are integers. Let us consider the classes A1, A2, . . . , AL−1, where L is an integer to be fixed later. Consider now, for every prime q < p/L, the classesBq = {q, 2q, . . . , αq, . . .}. It is immediate that the associated linear equation has no solutions if the coefficients be- long to some Bq , since q does not divide p. Further, A1, . . . , AL−1, B2, B3, . . . , B[ p L ] form a partition of {2, 3, . . . , p}. The total number of classes of this partition is then L − 1 + π(p/L). Setting L = √ 2p logp , we obtain f (p) < √ 8p logp (1 + O(1)). Problem 2.48. Consider m distinct natural numbers ai smaller than N such that lcm(ai, aj ) ≤ N for all i, j . Prove that m ≤ 2 [√ N ] . Solution 2.48. Assume that the numbers ai are ordered as 1 ≤ am < · · · < a1 ≤ N . We will prove by induction on k that ak ≤ Nk . If k = 1, then it is obvious. Moreover, if k ≥ 1, then ak − ak+1 ≥ gcd(ak, ak+1) = akak+1lcm(ak, ak+1) ≥ akak+1 N . This is equivalent to ak+1 + akak+1N ≤ ak , which yields 114 6 Algebra and Combinatorics Solutions ak+1 ≤ Nak ak + N = N − N2 ak + N ≤ N − N2 N k + N = N k + 1 , and our claim follows. Furthermore, one has a[√ N ] +1 ≤ N [√ N ]+ 1 ≤ [√ N ] . The sequence (aj ) contains then at most [ √ N ] terms between 1 and [ √ N ] . On the other hand, between √ N and N , there are no more than [√ N ] terms of our sequence, and thus we have a maximum number of m ≤ 2[√N] terms. Comments 45 If m(N) denotes the maximum number of terms of a sequence (ai), as in the statement, then Erdo˝s conjectured the following asymptotic behavior of the function m(n): m(N) = 3 23/2 √ N + O(1). See also: 1. P. Erdo˝s: Remarks on number theory. IV. Extremal problems in number theory, Matematikai Lapok 13 (1962), 228–255. Problem 2.49. The set M ⊂ Z+ is called A-sum-free, where A = (a1, a2, . . . , ak) ∈ Z k+, if for any choice of x1, x2, . . . , xk ∈ M we have a1x1 + a2x2 + · · · + akxk �∈ M . If A,B are two vectors, we define f (n;A,B) as the greatest number h such that there exists a partition of the set of consecutive integers {n, n + 1, . . . , h} into S1 and S2 such that S1 is A-sum-free and S2 is B-sum-free. Assume that B = (b1, b2, . . . , bm) and that the conditions below are satisfied: a1 + a2 + · · · + ak = b1 + b2 + · · · + bm = s, and min 1≤j≤k aj = min1≤j≤m bj = 1, k,m ≥ 2. Prove that f (n;A,B) = ns2 + n(s − 1) − 1. Solution 2.49. First consider the sets S1 = {n, n + 1, . . . , ns − 1} ∪ {ns2, ns2 + 1, . . . , ns2 + n(s − 1) − 1}, S2 = {ns, ns + 1, . . . , ns2 − 1}. If xj ∈ S2, then b1x1 + b2x2 + · · · + bmxm ≥ ns2 and thus S2 is B-sum-free. If x1, x2, . . . , xk ∈ S1, then we have two cases: 1. If xi ≤ ns − 1 for all i, then ns ≤ a1x1 + a2x2 + · · · + akxk ≤ s(ns − 1) < ns2. 6.2 Algebraic Combinatorics 115 2. If some xi belongs to {ns2, ns2 + 1, . . . , h}, then a1x1 + a2x2 + · · · + akxk ≥ ns2 + n(s − 1). Thus S1 is A-sum-free. This proves that f (n;A,B) ≥ ns2 + n(s − 1) − 1. Assume now that the set {n, n + 1, . . . , ns2 + n(s − 1)} can be partitioned into two sets Si that are A (respectively B) sum-free. Let n ∈ S1. If we take x1 = x2 = · · · = xk = n, then ns = a1x1 + a2x2 + · · · + akxk �∈ S1, so ns ∈ S2. Taking now x1 = x2 = · · ·+ = xm = ns ∈ S2, we find that ns2 ∈ S1. Suppose that a1 = b1 = 1. If we consider the elements x1 = ns2, x2 = · · · = xk = n of S1, we derive that n(s2 + s − 1) ∈ S2. 1. If n(s+1) ∈ S1, then take x1 = n, x2 = · · · = xk = n(s+1), and since ns2 ∈ S1, we find that the set S1 is not A-sum-free. 2. If n(s + 1) ∈ S2, then take x1 = ns, x2 = · · · = xm = n(s + 1), and since n(s2 + s − 1) ∈ S2, we obtain that the set S2 is not B-sum-free. This contradiction shows that f (n;A,B) = ns2 + n(s − 1) − 1. Comments 46 Generalized sum-free sets of integers were considered first by Rado, who in 1933 gave the upper bound f (1;A,A) ≤ max ( (bmc2 − 1)(c − 1) + bmc, bmc 2(c − 1) a ) , where A = (a, b), m = agcd(a,b) and c = max(x0, y0, z0), where (x0, y0, z0) is the minimal solution of the Diophantine equation ax + by = z. There are several partial results known for particular forms of A and B. If we denote the vector (d, d, . . . , d) with k components by k〈d〉, then Kasá proved that f (1; 2〈1〉, k〈1〉) = { 3k − 3, for odd k, 3k − 2, for even k, f (n; 2〈1〉, k〈1〉) = (2k + 1)n − 1, for even n. Further, Seress improved this by showing that f (n;m〈1〉, k〈1〉) = (mk + m − 1)n − 1, for n > 1,m ≥ 3. The result from the problem is due to L. Funar, who proved also that f (n; k〈d〉, k〈d〉) = k2d3 + kd − d − 1 for d even and k ≥ d and in several other cases. Abbott showed that this formula holds for all k, d ≥ 2. Other estimates for f (n;A,B) in the case in which A,B have not necessarily the same sum of components have been obtained by P. Moree. These suggest that there is no such simple closed formula for arbitrary A,B. However, recent progress has lead to the determination of f (1;A,A) for arbitrary A (also 116 6 Algebra and Combinatorics Solutions known as the 2-color Rado number), by Guo and Sun (improving previous results by B. Hopkins and D. Schaal). Specifically, we have f (1;A,A) = a(s − a)2 + (2a2 + 1)(s − a) + a3, where a = min1≤j≤k aj and s = a1 + · · · + ak . The general case, when we consider partitions into n ≥ 3 subsets Sj such that each Sj is Aj -sum-free, seems to be much more difficult, and no general estimates from above are known. • H.L. Abbott: On a conjecture of Funar concerning generalized sum-free sets, Nieuw Arch. Wisk. (4) 9 (1991), 249–252. • L. Funar: Generalized sum-free sets of integers, Nieuw Arch. Wisk. (4) 8 (1990), 49–54. • Song Guo and Zhi-Wei Sun: Determination of the two-color Rado number for a1x1 + · · · + amxm = x0, math.CO/0601409. • B. Hopkins and D. Schaal: On Rado numbers for ∑m−1i=1 aixi = xm, Adv. Appl. Math. 35 (2005), 433–441. • P. Moree: On a conjecture of Funar, Nieuw Arch. Wisk. (4) 8 (1990), 55–60. • R. Rado: Studien zur Kombinatorik, Math. Zeitschrift 36(1933), 424–480. • A. Seress: k-sum-free decompositions, Matematikai Lapok 31 (1978/83), 191– 194. Problem 2.50. Let 1 ≤ a1 < a2 < · · · < an < 2n be a sequence of natural numbers for n ≥ 6. Prove that min i,j lcm(ai, aj ) ≤ 6 ([n 2 ] + 1 ) . Moreover, the constant 6 is sharp. Solution 2.50. We assume that among the ai’s we can find both a and 2a. Then mini,j lcm(ai, aj ) ≤ lcm(a, 2a) ≤ 2n < 6 ([ n 2 ]+ 1). Let a ≤ n be an element of the sequence, if it exists. If 2a also belongs to the sequence, the claim follows from above. If 2a is not in the sequence, then we will replace the element a with 2a. This way the value of mini,j lcm(ai, aj ) is not diminished. We continue this process as far as possible. At some point, all the elements of the sequence will be greater than n, and thus the last sequence is forced to be n + 1, n + 2, . . . , 2n. If n = 2k + 1, then take a = 2k + 2, b = 3k + 3, so that lcm(a, b) = 6k + 6 = 6 ([ n 2 ]+ 1). Now if c, d are such that n + 1 < c < d ≤ 2n, then we have lcm(c, d) ≥ 6 ([ n 2 ]+ 1). In fact, lcm(c, d) = pd = qc for some q > p > 1, and thus either p = 2 and q ≥ 3, or else q ≥ 4. In the first case, c is an even number; hence c ≥ 2 ([n2 ]+ 1) and lcm(c, d) = qc ≥ 3c ≥ 6 ([n2 ]+ 1). In the second case, lcm(c, d) = qc ≥ 4(n + 1) > 6 ([n2 ]+ 1). Therefore, if we choose the set n + 1, . . . , 2n, we have mini,j lcm(ai, aj ) = 6 ([ n 2 ]+ 1). 6.2 Algebraic Combinatorics 117 Comments 47 This result is due to P. Erdo˝s, who claimed also that under the same hypothesis, we have max i �=j gcd(ai, aj ) > 38n 147 − C, where C is a constant independent of n. Problem 2.51. Let 1 ≤ a1 < a2 < · · · < ak < n be such that gcd(ai, aj ) �= 1 for all 1 ≤ i < j ≤ k. Determine the maximum value of k. Solution 2.51. Let f (n) denote the maximum of k as a function of n. We have then f (2) = 1. Assume from now on that n ≥ 3. Define li (for 1 ≤ i ≤ k) to be the smallest integer satisfying the inequality n/2 < ai2li . If there exists a pair of distinct indices i and j such that ai2li = aj2lj , then either ai divides aj , or conversely, aj divides ai . Therefore, we can replace ai by ai2li without diminishing the value of k. Thus we can assume that ai ≥ n2 , for all i. Further, the condition gcd(ai, aj ) �= 1 tells us that we cannot find two consecutive numbers that both belong to the sequence ai . Thus, if n is of the form 4m − 1, 4m, or 4m + 2, then f (n) ≤ m. Moreover, if n = 4m + 1, then f (n) ≤ m + 1. If we have equality above, then the sequence ai is forced to be a1 = 2m+ 1, a2 = 2m+ 3, . . . , am+1 = 4m+ 1, and therefore gcd(a1, a2) = 1, which is a contradiction. Thus f (n) ≤ m if n = 4m + 1. By considering the set of all even numbers in the interval [n/2, n] we obtain that for any n ≥ 3, we have f (n) = [n+14 ] . Problem 2.52. Consider the increasing sequence f (n) ∈ Z+, 0 < f (1) < f (2) < · · · < f (n) < · · · . It is known that the nth element in increasing order among the positive integers that are not terms of this sequence is f (f (n))+ 1. Find the value of f (240). Solution 2.52. 1. One knows that the nth absent number is N = f (f (n))+ 1. How- ever, the numbers smaller than N that belong to the sequence are f (1), f (2), . . . , f (f (n)). Moreover, there are n numbers less than or equal to N that do not belong to the sequence. This means that N = f (n) + n; thus f (f (n)) = f (n) + n − 1. Therefore f (1) = 1, f (2) = 3. It follows by induction that f (n) = n + card {m|f (m) < n}. Set a0 = 2, an+1 = f (an). We compute an+1 = an + card {m|f (m) < an} = an + card {m|f (m) < an−1} = an + an−1 − 1, and therefore the sequence bn = an − 1 satisfies the Fibonacci recurrence bn+1 = bn + bn−1. We will prove by induction that f (bn + x) = bn+1 + f (x), for 1 ≤ x ≤ bn−1. 118 6 Algebra and Combinatorics Solutions If n = 0, then b1 = 1 and x = 1 and the claim is verified. Further f (bn + x) = bn + x + card {y|f (y) < f (bn−1)}+ card {y|bn+1 < f (y) < bn + x}, which implies that f (bn − 1) < f (an−1) = an and thus f (bn−1) ≤ bn ≤ bn + x. In particular, f (y) ≤ bn + x ≤ bn + bn−1 ≤ bn+1 < f (an), from which we obtain y < an, y ≤ bn. We now write y = bn+1 + z, where z ≤ bn − bn−1 = bn−2. Hence f (bn + x) = bn + x + bn−1 + card {z|f (z) < x} = bn+1 + f (x). This proves our claim. Recall now that every natural number can be uniquely written as a sum of Fibonacci numbers, i.e., as x = 1 + bk1 + bk2 + · · · + bkp , where 1 ≤ k1 < k2 < · · · < kp. If x is written as above, the previous inductive formula yields the value f (x) = 1 + bk1+1 + bk2+1 + · · · + bkp+1. In particular, f (240) = 1 + 2 + 8 + 377 = 388, since 240 = 1 + (1 + 5 + 233). 2. We can use in a clever manner the recursion law f (f (n)) = f (n) + n − 1 in order to find, step by step, f (3) = 3 + 1 = 4, f (4) = 4 + 2 = 6, f (6) = 6 + 3 = 9, f (9) = 9 + 5 = 14, f (14) = 22 f (22) = 35, f (35) = 56, f (56) = 90, f (90) = 145, f (145) = 234, f (234) = 378. One knows that f (f (35))+ 1 = 91 is absent, so that f (57) = 92 and thus f (92) = 148, f (148) = 239, f (239) = 386. Next f (f (148)) + 1 = 387 is absent, and hence f (240) = 388. Problem 2.53. We define inductively three sequences of integers (an), (bn), (cn) as follows: 1. a1 = 1, b1 = 2, c1 = 4; 2. an is the smallest integer that does not belong to the set {a1, . . . , an−1, b1, . . . , bn−1, c1, . . . , cn−1}; 6.2 Algebraic Combinatorics 119 3. bn is the smallest integer that does not belong to the set {a1, . . . , an−1, an, b1, . . . , bn−1, c1, . . . , cn−1}; 4. cn = 2bn + n − an. Prove that 0 < n ( 1 + √3 ) − bn < 2 for all n ∈ Z+. Solution 2.53. We will actually prove that α < n ( 1 + √3)− bn < β, where α = ( 9 − 5√3)/3, β = (12 − 4√3)/3. Observe that the sequence (cn) does not contain two consecutive numbers. Thus an+1 might jump at most two units ahead from bn, and bn two units ahead from an. By induction, we obtain 1 ≤ bn − an ≤ 2 and 1 ≤ an+1 − bn ≤ 2. Using the equality cn+1 − cn = 2(bn+1 − an+1) + (an+1 − bn) − (bn − an) + 1 we deduce that 2 ≤ cn+1 − cn ≤ 6. In particular, there are not six consecutive integers among the sequences aj and bj . Set γn = n ( 1 + √3)−bn. We will prove by induction that α < γn < β, which is trivially verified for n = 1. Suppose now that this holds true for all n < k. We consider the truncated sequence ak−2 < bk−2 < ak−1 < bk−1 < ak < bk . As remarked above, there should be at least one element of the form cj , which can be inserted between ak−2 and bk , since bk − ak−2 ≥ 6. Let us choose the greatest such cn, which is cn ≤ bk − 1. It follows that the set {a1, a2, . . . , ak, b1, b2, . . . , bk, c1, c2, . . . , cn} is precisely the set of the first consecutive numbers contained between 1 and bk . In particular, we have bk = 2k + n = cn + r, where 1 ≤ r ≤ 5. On the other hand, we know that cn = 2bn + n − an = bn + n + (bn − an) and bn − an ≤ 2, and so 2k + n = bk = bn + n + s, where 2 ≤ s ≤ 7. However, one can improve the upper bound for s as follows. We have equality only if r = 5, but in this case we have cn+1 − cn = 6. Using the formula for cn+1 − cn, we derive that bn − an = 1, and thus one has s ≤ 6. 120 6 Algebra and Combinatorics Solutions Let us now estimate γk , which reads γk = (1+ √ 3)k−bk = (1+ √ 3) ( bn + s 2 ) − (bn +n+ s) = (s−γn) (√ 3 − 1 2 ) . If s ≤ 6 we use γn ≥ α to obtain γk ≤ (6 − α) (√ 3 − 1 2 ) < β. If s = 6, then we must have bn+1−an+1 = an+1−bn = 2, and therefore bn+1−bn = 4. Thus γn+1 −γn = 1+ √ 3− (bn+1 −bn) = √ 3−3. Since n+1 < k, one uses the induction hypothesis γn+1 > α in order to get γn = γn+1 + 3 − √ 3 > α + 3 − √3. Using this inequality above, we derive that γk ≤ (6 − α − 3 + √ 3) (√ 3 − 1 2 ) = β. If s = 5, then bn+1 −bn ∈ {3, 4}. Thus γn+1 −γn ≤ √ 3−2, and so γn > α+2− √ 3. Introducing this above, we obtain γk < (5 − α − 2 + √ 3) (√ 3 − 1 2 ) = β. The other inequality, γk > α, follows along the same lines. 6.3 Geometric Combinatorics Problem 2.54. We consider n points in the plane that determine C2n segments, and to each segment one associates either +1 or −1. A triangle whose vertices are among these points will be called negative if the product of numbers associated to its sides is negative. Show that if n is even, then the number of negative triangles is even. Moreover, for odd n, the number of negative triangles has the same parity as the number p of segments labeled −1. Solution 2.54. Let Pi denote the points in the plane, and let aij be the label of the segmentPiPj . Then the signature of the trianglePiPjPk is given byPijk = aij ajkaki . In particular, ∏ i �=j �=k �=i Pijk = (−1)m, where m is the number of negative triangles. Now every segment belongs to n − 2 triangles, and thus ∏ i �=j �=k �=i Pijk = ∏ aij ajkaki = ⎛ ⎝ ∏ i �=j aij ⎞ ⎠ n−2 = (−1)p(n−2). Therefore m ≡ pn (mod 2), and the claim follows. 6.3 Geometric Combinatorics 121 Problem 2.55. Given n, find a finite set S consisting of natural numbers larger than n, with the property that, for any k ≥ n, the k × k square can be tiled by a family of si × si squares, where si ∈ S. Solution 2.55. Each of the sets S = {s ∈ Z| n ≤ s ≤ n2}, and S = {s ∈ Z|s is prime, n2 < s < 2n2 + n} is convenient, although neither one is minimal. We will prove this by induction on the size of the square, for the second set S. Consider a k × k square, where k ≥ n2. By hypothesis, the m × m square is tiled by squares from S, for any m with n ≤ m < k. If k is composite, then write k = pq, where k > p ≥ n. We cover the k × k square by means of q2 p × p squares and we proceed inductively. If k is prime, then k > 2n2 + n. We divide the square into two squares, one m×m and the other (k−m)× (k−m), and two m× (k−m) rectangles, where m = n(n+ 1). We have k −m > n2 and thus each m× (k −m) rectangle can be covered with m× n or m× (n+ 1) pieces and each of these pieces can be further divided into squares. Comments 48 The problem whether some region can be tiled by a given set of tiles is difficult and unsolved in general. The interested reader might consult the survey of Ardila and Stanley. The simpler problem of whether an n×m rectangle was tiled by a×b rectangles was solved by de Bruijn and, independently, by Klarner in 1969. Specifically, this happens to be possible if and only if mn is divisible by ab, both m and n can be written as sums of a’s and b’s, and either m or n is divisible by a or else m (or n) is divisible by b. More generally, the m1 ×m2 ×· · ·×mn box is tiled by a1 × a2 ×· · ·× an bricks if and only if each of the ai has a multiple among the mj ’s. Conway and Lagarias gave necessary conditions for the existence of tilings of rectilinear polygons (i.e., those having sides parallel to the axes) by a set of finite rectilinear tiles using boundary invariants, which are combinatorial group-theoretic invariants associated with the boundaries of the tile shapes and the regions to be tiled. Some new invariants were discovered recently by Pak. Notice that for n ≥ 2,m > 2, the problem whether there exists a tiling of a given rectilinear polygon having only horizontal n× 1 tiles and only vertical 1 ×m tiles is an NP -complete question, as shown by Beauquier et al. • F. Ardila, R. Stanley: Tilings, math.CO/0501170. • D. Beauquier, M. Nivat, E. Rémila, M. Robson: Tiling figures of the plane with two bars, Comput. Geom. 5 (1995), 1–25. • N. de Bruijn: Filling boxes with bricks, Amer. Math. Monthly 79 (1969), 37–40. • J.H. Conway, J.C. Lagarias: Tiling with polyominoes and combinatorial group theory, J. Combin. Theory Ser. A 53 (1990), 183–208. • D. Klarner: Packing a rectangle with congruent n-ominoes, J. Combin. Theory 7 (1969), 107–115. • I. Pak: Ribbon tile invariants, Trans.Amer. Math. Soc. 352 (2000), 12, 5525–5561. Problem 2.56. We consider 3n points A1, . . . , A3n in the plane whose positions are defined recursively by means of the following rule: first, the triangle A1A2A3 is 122 6 Algebra and Combinatorics Solutions equilateral; further, the points A3k+1, A3k+2, and A3k+3 are the midpoints of the sides of the triangle A3kA3k−1A3k−2. Let us assume that the 3n points are colored with two colors. Show that for n ≥ 7 there exists at least one isosceles trapezoid having vertices of the same color. Solution 2.56. Any subset of three points has two points of the same color. Therefore, there are two points having the same color (which we denote by Aik and Ajk ) among {A3k+1, A3k+2, A3k+3}. Consider now the lines determined by these monocolor pairs of points: {Ai1Aj1 , Ai2Aj2 , . . . , AinAjn}. We know that all these directions are parallel to the three sides of the initial equilateral triangle. Thus there exist at least [n3 ] + 1 ≥ 3 lines in the set above that are parallel. Moreover, each line comes with one of the two colors, and thus there exist at least[ [ n3 ]+1 2 ] +1 ≥ 2 parallel lines of the same color. The four points that determine these lines form a monocolor isosceles trapezoid. Problem 2.57. Is there a coloring of all lattice points in the plane using only two colors such that there are no rectangles with all vertices of the same color, whose side ratio belongs to { 1, 12 , 1 3 , 2 3 } ? Solution 2.57. No. According to Van der Waerden’s theorem, there exist four lattice (equidistant) points Ai = (xi, 0) of the same color (say red) on the axis y = 0 such that x2 − x1 = x3 − x2 = x4 − x3 = k. Consider the points Bi with coordinates (xi, k), lying on a parallel line, at distance k. Then the points Bj contain either two red points—in which case we are done, by considering the respective Aj—or else at least three black points, Bj1 , Bj2 , and Bj3 . Consider next the points Ci = (xi, 2k). Then there exist two points among Cj1 , Cj2 , and Cj3 that have the same color. If this color is black, then using the respective Bji and Cji , we obtain a black rectangle as needed. If the color is red, then the respective Aji and Cji form a red rectangle, as claimed. Comments 49 Van der Waerden’s theorem from 1927 states that if the set of nat- ural numbers Z+ is partitioned into finitely many color classes, then there exist monochromatic arithmetic progressions of arbitrary length. Erdo˝s and Turán conjec- tured in 1936 the existence of arbitrarily long arithmetic progressions in sequences of integers with positive density, and their conjecture was successfully confirmed by Szemerédi in 1975. Many extensions have been made since that time. Actually, Sze- merédi proved that for a given length l and density δ > 0, there is an L(l, δ) such that if L ≥ L(l, δ), any subset A ⊂ {1, 2, 3, . . . , L} with more than δL elements, will contain an arithmetic progression of length l. This makes precise Van der Waerden’s theorem in that there exists W(l, r) such that if W ≥ W(l, r) and {1, 2, 3, . . . ,W } is the union of r subsets, then one of these necessarily contains an l-term arithmetic progression, and moreover, W(l, r) can be taken as L(l, 1/r). 6.3 Geometric Combinatorics 123 The next breakthrough was made recently by Gowers, who provided the first effective upper bound to the function L(l, δ). One consequence is a significant improvement of the upper bound for W(l, r); for example, for some constant c, W(4, r) can be taken as exp(exp(rc)). Generally, we can take L(l, δ) = exp(δ−c(l)), where c(l) = 22l+9 . These estimates fall short of providing a proof to another conjecture of Erdo˝s claiming that any set A ⊂ N with∑a∈A 1a = ∞ contains arbitrarily long arithmetic progressions. Another far-reaching generalization was provided by Bergelson and Leibman, as follows. Suppose that p1, p2, . . . , pm are polynomials with integer coefficients and no constant term. Then, whenever Z+ is finitely colored, there exist natural numbers a and d such that the point a and all the points a + pi(d), for 1 ≤ i ≤ m, have the same color. Finally, Szemerédi’s theorem was proved to hold for the primes in a spectacular recent paper by Ben Green and Terence Tao. Specifically, if A is a subset of positive density within the set of primes (for instance the set of all primes), then A contains infinitely many arithmetic progressions of length k (for any k ≥ 3). An exposition of this result can be found in the survey of Kra. • V. Bergelson, A. Leibman: Polynomial extensions of Van der Waerden’s and Sze- merédi’s theorems, J. Amer. Math. Soc. 9 (1996), 725–753. • W.T. Gowers: A new proof of Szemerédi’s theorem, Geometric FunctionalAnalysis 11 (2001), 465–588. • B. Kra: The Green–Tao theorem on arithmetic progressions in the primes: an ergodic point of view, Bull. Amer. Math. Soc. (N.S.) 43 (2006), 1, 3–23. • E. Szemerédi: On sets of integers containing no k elements in arithmetic pro- gression, Collection of articles in memory of J.V. Linnik, Acta Arith. 27 (1975), 199–245. • B.L. Van der Waerden: Beweis einer Baudetschen Vermutung, Nieuw Arch. v. Wiskunde 15 (1927), 212–216. Problem 2.58. Let G be a planar graph and let P be a path in G. We say that P has a (transversal) self-intersection in the vertex v if the path has a (transversal) self- intersection from the curve-theoretic viewpoint. Let us give an example. Take the point 0 in the plane and the segments 01, 02, 03, 04 going counterclockwise around 0. Then a path traversing first 103 and then 204 has a (transversal) self-intersection at 0, while a path going first along 102 and further on 304 does not have a (transversal) self-intersection. Prove that any connected planar graph G, with only even-degree vertices, admits an Eulerian circuit without self-intersections. Recall that an Eulerian circuit is a path along the edges of the graph, that passes precisely once along each edge of the graph. Solution 2.58. We will proceed by double induction, first on the maximum degree of the vertices and, for the class of graphs with several vertices of maximum degree, on the number of such vertices. If the degree is d = 2, then the obvious Eulerian circuit does not have self-intersections. Now let v be a vertex of maximal degree d > 2. 124 6 Algebra and Combinatorics Solutions We split v into two vertices, one of degree 2 and the other of degree d − 2; the first among the new vertices is incident to two vertices that were previously incident to v, by means of two consecutive edges (this makes sense since the graph is planar), and the latter is incident to the remaining ones. If the obtained graph is connected, then there exists an Eulerian path without self-intersections, and then we simply restore the graph and see that this way the circuit remains without self-intersections. If the graph is not connected, then we get two Eulerian circuits. When restoring the graph G, we glue together the two circuits into one Eulerian circuit, still without self-intersections. Problem 2.59. Let us consider finitely many points in the plane that are not all collinear. Assume that one associates to each point a number from the set {−1, 0, 1} such that the following property holds: for any line determined by two points from the set, the sum of numbers associated to all points lying on that line equals zero. Show that, if the number of points is at least three, then to each point one associates the number 0. Solution 2.59. Let us assume the contrary, i.e., that there exist points with nontrivial numbers associated to them. Then there should exist points labeled both 1 and −1. Choose first a point x0 labeled 1. For any other point x �= x0, the line d = x0x has nε(d) points labeled by ε ∈ {−1, 1}. Moreover, the assumptions imply that n−1(d) = n+1(d) and n+1(d) ≥ 1, because x0 is labeled 1. Furthermore, the total number of negative points is n−1 = ∑d n−1(d), the sum being taken on all lines d passing through x0. Next, the total number of positive points is n+1 = 1 +∑d(n+1(d) − 1) = n−1 − l + 1, where l is the number of distinct lines passing through x0. We know that l ≥ 2 by hypothesis; hence we obtain that n+1 ≤ n−1 − 1. Choose now a point y0 labeled −1. Then the same argument implies that the reversed inequality n−1 ≤ n+1 − 1 holds, contradicting the former inequality. Problem 2.60. If one has a set of squares with total area smaller than 1, then one can arrange them inside a square of side length √ 2, without any overlaps. Solution 2.60. We put the squares in decreasing order with respect to their side lengths. The first square is put in the left corner of the square S (the side of which is√ 2), the second square will be set to the right of S, and we go on until it becomes impossible to put another square without surpassing the borders of S. Then, we start another line of squares above the first square and so on. Assume that the side lengths are S1 ≥ S2 ≥ · · · and we haven1, n2, . . . , nk squares on the lines number 1, 2, . . . , k respectively. It follows that √ 2 − Sn1+1 ≤ S1 + S2 + · · · + Sn1 ≤ √ 2,√ 2 − Sn1+n2+1 ≤ Sn1+1 + · · · + Sn1+n2 ≤ √ 2, and so on. We want to show that S1 + Sn1+1 + Sn1+n2+1 + · · · ≤ √ 2, 6.3 Geometric Combinatorics 125 which will prove our claim. We know from above that S2 + S3 + · · · + Sn1+1 ≥ √ 2 − S1, from which we infer S22 + S23 + · · · + S2n1+1 ≥ (√ 2 − S1 ) Sn1+1, and by the same trick, we obtain S2n1+2 + · · · + S2n1+n2+n3 ≥ (√ 2 − S1 ) Sn1+n2+1, and so on. By summing up these inequalities, we obtain 1 − S21 ≥ S22 + S23 + · · · ≥ (√ 2 − S1 ) (Sn1+1 + Sn1+n2+1 + · · · ), and hence S1 + Sn1+1 + Sn1+n2+1 + · · · ≤ 1 − S21√ 2 − S1 + S1 = √ 2 − ( 1 − √2S1 )2 √ 2 − S1 ≤ √2. Comments 50 The problem of packing economically unequal rectangles into a given rectangle has recently received a lot of consideration. L. Moser asked in 1968 to find the smallest packing default ε in each of the following situations: 1. the set of rectangles of dimensions 1 × 1 n , for n ∈ Z+, n ≥ 2, whose total area equals 1 can be packed into the square of area 1 + ε without overlap; 2. the set of squares of side lengths 1 n , for n ∈ Z+, can be packed without overlap into a rectangle of area π2/6 − 1 + ε; 3. the (infinite) set of squares with sides of lengths 1/(2n + 1), n ∈ Z+, can be packed in a rectangle of area π2/8 − 1 + ε. The best results known to this day yield ε very small, at least smaller than 10−9, using a packing algorithm due to M. Paulhus. Further computational investigations support the apparently weaker claim that the packings in the problems above are possible for every positive number �. However, G. Martin showed that if this weaker claim holds, then one can also find a packing for ε = 0, in which case the packing is called perfect. The general case of the problem above is still open, but a variation of it has been solved in the meantime. J. Wästlund proved that if 1/2 < t < 2/3, then the squares of side n−t , for n ∈ Z+, can be packed into some finite collection of square boxes of the same area ζ(2t) as the total area of the tiles. On the other hand, Chalcraft proved that there is a perfect packing of the squares of side n−3/5 into a square, and presumably, his technique works for packing the squares of side n−t into a square, where 1/2 < t ≤ 3/5; this is true for packing into a rectangle for all t in the range 0.5964 ≤ t ≤ 0.6. • A. Chalcraft: Perfect square packings, J. Combin. Theory Ser. A 92 (2000), 158– 172. 126 6 Algebra and Combinatorics Solutions • H.T. Croft, K.J. Falconer, R.K. Guy: Unsolved Problems in Geometry, Springer- Verlag, New York, 1994. • G. Martin: Compactness theorems for geometric packings, J. Combin. Theory Ser. A 97 (2002), 225–238. • A. Meir, L. Moser: On packing of squares and cubes, J. Combin. Theory 5 (1968), 126–134. • M. Paulhus: An algorithm for packing squares, J. Combin. Theory Ser. A 82 (1998), 147–157. • J. Wästlund: Perfect packings of squares using the stack-pack strategy, Discrete Comput. Geom. 29 (2003), 625–631. Problem 2.61. Prove that for each k there exist k points in the plane, no three collinear and having integral distances from each other. If we have an infinite set of points with integral distances from each other, then all points are collinear. Solution 2.61. 1. Consider the point P1 on the unit circle having complex coordinates z = 35 + 45 √−1. If θ denotes the argument of P1, then θ is noncommensurable with π , i.e., θ/π ∈ R − Q. Consider the points Pn of coordinates zn. Then the Pn form a dense set of points in the unit circle. Moreover, we can compute the distance |PnPk| = 2| sin(n − k)θ |. Since sin nθ is a polynomial with rational coefficients in sin θ and cos θ , we derive that the distances between all these points are rational. Given k, choose P1, . . . , Pk and a homothety of large ratio in order to clear the denominators in these distances. We obtain then k points with integral pairwise distances. 2. Given three noncollinear pointsP,Q, andR, with |PQ|, |PR| ≤ k, the number of points X such that the distances |XP |, |XQ|, |XR| are integral is bounded by 4(k + 1)2. In fact, we have, by the triangle inequality, ∣∣|XP | − |XQ|∣∣ < |PQ| = k, and thus ∣∣|XP | − |XQ|∣∣ ∈ {1, 2, . . . , k}. Moreover, the geometric locus of thoseX for which ∣∣|XP |−|XQ|∣∣ = j is a hyperbola with foci P and Q. On the other hand, ∣∣|XP | − |XR|∣∣ ∈ {1, 2, . . . , k}, and thus X belongs to one of the k + 1 hyperbolas having P and R as foci. Thus X belongs to the intersections of (k + 1)2 pairs of hyperbolas, which have at most 4(k + 1)2 intersection points. Comments 51 This construction of a dense set of points in the unit circle whose pairwise distances are rational is due to A. Muller. Then W. Sierpin´ski found such a dense set in the circle of radius r , provided that r2 is rational, this condition being necessary as soon as we have three points at rational distances. The second result is due to Erdo˝s. The question of Ulam whether there exists a dense set in the plane such that the distances between any two of its points is rational is still unsolved. 6.3 Geometric Combinatorics 127 • P. Erdo˝s: Integral distances, Bull. Amer. Math. Soc. 51 (1945), 996. • W. Sierpin´ski: Sur les ensembles de points aux distances rationnelles situés sur un cercle, Elem. Math. 14 (1959), 25–27. Problem 2.62. Let O,A be distinct points in the plane. For each point x in the plane, we write α(x) = x̂OA (counterclockwise). Let C(x) be the circle of center O and radius |Ox| + α(x)|Ox| . If the points in the plane are colored with finitely many colors, then there exists a point y with α(y) > 0 such that the color of y also belongs to the circle C(y). Solution 2.62. Let G be the graph of vertices {x ∈ E2, α(x) > 0} in which points x and y are adjacent if y ∈ C(x). If C1(ρ1) and C2(ρ2) are two circles of radii ρ1 < ρ2 < 1 centered at the origin, then there exists a pointM1 ∈ C1(ρ1) that is adjacent to all points ofC2(ρ2). The point M1 has α(M1) = θ uniquely determined, as follows. The condition C(M1) = C2(ρ2) is equivalent to ρ2 = ρ1 + θρ1 , yielding θ = (ρ2 − ρ1)ρ1 ∈ (0, 1) ⊂ (0, 2π), and therefore M1 is well determined. Assume now that the claim from the statement is not true. It will follow that the color of M1 is distinct from the colors appearing on the circle C2(ρ2). Let now C1, C2, . . . , Ck, . . . be an infinite sequence of circles centered at the origin whose respective radii are 0 < ρ1 < ρ2 < · · · < ρk < · · · < 1. Then let the set of colors that we encounter on Ck be denoted by τk. The previous argument, when applied to the pair of circles Ci and Cj , shows that there exists a point on Ci that is colored by a color that does not exist on Cj , if i < j , and thus τi \τj �= ∅. Since the number of colors is finite, this is impossible as soon as k is large enough (e.g., k > 2p, where p is the number of colors). This contradiction proves the claim. Problem 2.63. Let k, n ∈ Z+. 1. Assume thatn−1 ≤ k ≤ n(n−1)2 . Show that there existn distinct points x1, . . . , xn on a line, that determine exactly k distinct distances |xi − xj |. 2. Suppose that [ n 2 ] ≤ k ≤ n(n−1)2 . Then there exist n points in the plane that determine exactly k distinct distances. 3. Prove that for any ε > 0, there exists some constant n0 = n0(ε) such that for any n > n0 and εn < k < n(n−1)2 , there exist n points in the plane that determine exactly k distinct distances. Solution 2.63. 1. Let m be an integer between 1 and n − 1 such that n − 1 = C2n − C2n−1 ≤ k ≤ C2n − C2m. 128 6 Algebra and Combinatorics Solutions Set p = k− (m− 1)−C2n +C2m+1; in particular, we have 1 ≤ p ≤ m. Consider then the following set of points of the real line: X = {1, 2, . . . , m,m + p} ∪ {πm+2, πm+3, . . . , πn}. The distances between the firstm+1 points correspond to the set {1, 2, . . . , p+m−1}. The remaining points are independent transcendental points, and thus the distances be- tween them are all distinct transcendental numbers. Moreover, the distances between points of the second set and points of the first set are also all distinct (translates of the former set of transcendental distances by integers). Thus the number of distances obtained so far is p + m − 1 + ( n∑ i=m+2 (i − 1) ) = k. 2. If [ n 2 ] ≤ k ≤ C2n , we put the n points in the consecutive vertices of a regular (2k+ 1)-gon. There are exactly k distinct distances in a regular (2k+ 1)-gon, and all of them are realized at least once, since k ≥ [n2 ] . 3. It suffices to check the case εn ≤ k ≤ [n2 ] . Let m = 3 + [√n]. For m ≤ a ≤ n − 2m, 1 ≤ b ≤ m, we consider the following subset of Z × Z: X = C ∪ {(i, 1)|1 ≤ i ≤ a − 1} ∪ {(1, j)|2 ≤ j < m} ∪ {(0, h)|1 ≤ h ≤ b}, where C is a subset of {(1, j), 2 ≤ i ≤ a − 1, 2 ≤ j ≤ m} with the property that the distances between the pairs of points in C are all distinct. The set of squares of the distances between points in X is D(a, b) = {(i2 + j2)|0 ≤ i ≤ a − 1, 0 ≤ j ≤ m − 1} ∪ {a2 + j2|0 ≤ j ≤ b − 1}. We have D(m, 1) ⊂ D(m, 2) ⊂ · · · ⊂ D(m,m) ⊂ D(m + 1, 1) ⊂ D(m + 1, 2) ⊂ · · · ⊂ D(n − 2m,m), and each set from above is obtained from the previous one in the sequence by adjoining at most one more distance; thus the difference between consecutive sets is either empty or a singleton. Therefore, card(D(a, b)) is an increasing sequence of consecutive elements that lie between card(D(m, 1)) and card(D(n− 2m,m)). Observe now that the upper bound card(D(n − 2m,m)) is at least n − 2m − 1 ≥ [n2 ] . Further, the set D(m, 1) consists of integers between 1 and 2m2 (note that we have the asymptotic behavior m2 ∼ n) that are sums of two perfect squares. It is known that whenever μ = λ2 + δ2 and p is a prime number p ≡ 3 (mod 4) if p divides μ, then p2 also divides μ. Thus, the density of the set of numbers that are sums of two perfect squares is bounded by the product over the primes p ∏ p≡3 (mod 4) ( 1 − 1 p + 1 p2 ) , which tends to 0. Therefore, for n large enough, we have card(D(m, 1)) < εn, which proves the claim. 6.3 Geometric Combinatorics 129 Comments 52 The result is due to P. Erdo˝s, who proposed it in 1980 as a problem in Amer. Math. Monthly, related to another question that he considered long ago. In 1946, Erdo˝s posed the problem of determining the minimum number d(n) of different distances determined by a set of n points in R2, proved that d(n) ≥ cn1/2, and conjectured that d(n) ≥ cn/√log n. If true, this inequality is best possible, as shown by the lattice points of the plane. The best result known is due to J. Solymosi and Cs.D. Tóth, and yields d(n) > cn6/7. • P. Erdo˝s: On the set of distances of n points, Amer. Math. Monthly 53 (1946), 248–250. • P. Erdo˝s: Problem 6323, Amer. Math.Monthly 87 (1980), 826. • J. Solymosi, Cs.D. Tóth: Distinct distances in the plane, The Micha Sharir birthday issue, Discrete Comput. Geom. 25 (2001), 629–634. Problem 2.64. Show that it is possible to pack 2n(2n + 1) nonoverlapping pieces having the form of a parallelepiped of dimensions 1 × 2 × (n + 1) in a cubic box of side 2n + 1 if and only if n is even or n = 1. Solution 2.64. Let us consider the cube composed of (2n + 1)3 cells of dimensions 1×1×1, which are labeled using the coordinate system (a1, a2, a3), ai ∈ {1, . . . , 2n+ 1}. 1. Case n = 1. The 6 pieces are arranged as follows. The first three pieces are given by x1 = {(1, 1, 2), (1, 1, 3), (1, 2, 2), (1, 2, 3)}, x2 = {(1, 2, 1), (1, 3, 1), (2, 2, 1), (2, 3, 1)}, x3 = {(2, 1, 1), (3, 1, 1), (2, 1, 2), (3, 1, 2)}. The next three pieces x4, x5, x6 are located as above, by permuting the indices 3 and 1. The cells (1, 1, 1), (2, 2, 2), (3, 3, 3) remain empty. 2. Suppose now that n is even. We have then (2n+1)3−2n(2n+1)·1·2·(n+1) = 2n+1 cases that remain empty. Each slice P1,i = {(i, a, b)|a, b ∈ {1, 2, . . . , 2n+1}} can be divided in four smaller boxes P1,i;1 = {(i, a, b)|a ≤ n + 1, b ≤ n}, P1,i;2 = {(i, a, b)|n + 2 ≤ a, b ≤ n + 1}, P1,i;3 = {(i, a, b)|n + 1 ≤ a, n + 2 ≤ b}, P1,i;4 = {(i, a, b)|a ≤ n, n + 1 ≤ b} and one extra cell, namely {(i, n + 1, n + 1)}. Each box P1,i;s has dimensions 1 × n × (n + 1) and can be covered by n2 pieces 1 × 2 × (n + 1). The remaining cells of coordinates (i, n + 1, n + 1) remain empty. 3. Assume further that n is odd, n ≥ 3, and that there exists a packing as in the statement. One associates to a cell (a1, a2, a3) the number p, which counts the number of i ∈ {0, 1, 2, 3} such that ai = n + 1. One colors the cell blue if p = 3, red if p = 2, yellow when p = 1, and leaves it colorless when p = 0. Then, we have one blue cell and 6n red cells. Now, for each i ≤ 2n + 1, k ∈ {1, 2, 3}, we set Pk,i = {a1, a2, a3), ak = i}. Then each slice Pk,i contains an odd number of cells (2n+ 1)2. In the meantime, every piece contains an even number of cells from a given slice Pk,i , which might be 130 6 Algebra and Combinatorics Solutions 2, n+ 1, or 2(n+ 1) cells, respectively. This means that in each slice Pk,i there exists at least one empty cell. The total number of empty cells is 2n+1, and for fixed k the slicesPk,1, . . . , Pk,2n+1 are disjoint, so that each slice Pk,i contains exactly one empty cell. In particular, the set consisting of the blue cell and the red ones contains at most one empty cell. Furthermore, each piece S contains colored cells because n+1 > 12 (2n+1), and it fits into one of the cases below: Type 1. S contains the blue cell, n + 1 red cells, and n yellow cells. Type 2. S does not contain the blue cell, and it contains 2 red and 2n yellow cells. Type 3. S does not contain the blue cell, and it contains 1 red, n + 1 yellow, and n colorless cells. Type 4. S does not contain either the blue cell or red cells, but 2 yellow and 2n colorless cells. Let us analyze now the total number of colored cells which can be covered by all the pieces. • The blue cell is empty. Then there are nonempty red cells and we compute that the 2n(2n+ 1) pieces contain 2n(2n+ 1) · 2 + 6n · n = 14n2 + 4n colored cells. • One red cell is empty. Then there exists a piece S of type 1 that contains n+1 red cells and 4n2 + 2n− 1 pieces of type 2. Altogether they contain 5n− 2 red cells. Then the pieces contain 2n + 2 + (4n2 + 2n − 1) · 2 + (5n − 2)n = 13n2 + 4n colored cells. • The blue and red cells are not empty. Then the pieces of soap will contain n more colored cells than in the previous case. However, the total number of colored cells within the cube is 1 + 6n+ 12n2, and for n ≥ 3, we have the inequality 13n2 + 4n > 12n2 + 6n+ 1, which is a contradiction, because the pieces are supposed to be nonoverlapping. 4. Second proof when n > 3. For k ∈ {1, 2, 3}, let us denote by mk the number of pieces whose longest side is orthogonal to the plane of the slice Pk,i . Each such piece has exactly 2 cells in common with Pk,n+1. Every piece from the remaining ones has either n+ 1 or 2(n+ 1) cells in common with Pk,n+1. Therefore, n+ 1 should divide the number of remaining cells, which is 4n(n + 1) − 2mk = (2n + 1)2 − 1 − 2mk . This implies that n + 1 divides mk for all k. Moreover, the total number of cells is m1 + m2 + m3 = 2n(2n + 1). Since gcd(n, n + 1) = gcd(n, 2n + 1) = 1, we derive that n ∈ {1, 3}. Problem 2.65. Let F be a finite subset of R with the property that any value of the distance between two points from F (except for the largest one) is attained at least twice, i.e., for two distinct pairs of points. Prove that the ratio of any two distances between points of F is a rational number. Solution 2.65. Let F = {s1, . . . , sn} be a finite set from a vector space V over Q. Consider the n(n − 1)/2 difference vectors si − sj , where i < j . Some difference vectors appear several times in the difference vectors sequence, and we call them D vectors. We will prove a more general result: The vector space generated by those 6.3 Geometric Combinatorics 131 differences that are not D vectors coincides with the vector space generated by all difference vectors. Proof of the claim. Let us assume the contrary. By the Hahn–Banach lemma, there exists a linear functional f : V → Q that vanishes on those vectors that are not D vectors, although f is nonzero on all difference vectors. Observe that we can replace V with a finite-dimensional linear subspace that contains F . Let M = f (si),m = f (sj ) be the greatest and respectively the smallest values from f (F ), where of course, m �= M . The set of linear functionals f : V → Q that map F one-to-one into Q is dense in the set of linear functionals. Consequently, there exists g : V → Q that injects F into Q such that |f (s) − g(s)| ≤ 1 5 (M − m), for all s ∈ F. Consider g(sp) the maximal value and g(sq) the minimal value within the set g(F ) ⊂ Q. Therefore sp − sq is not a D vector and thus f vanishes: f (sp − sq) = 0. This implies that g(si) − g(sj ) ≤ g(sp) − g(sq) ≤ f (sp) − f (sq) + 25 (M − m) = 2 5 (M − m). On the other hand, g is closely approximated by f ; hence g(si) − g(sj ) ≤ g(sp) − g(sq) ≤ f (sp) − f (sq) + 25M − n = 2 5 (M − n), and also g(si) − g(sj ) ≥ f (si) − 15 (M − m) − ( f (sj ) + 15 (M − m) ) = 3 5 (M − m), contradicting the previous equality. This proves our general claim. Assume now that F ⊂ R, and the extreme points of F are 0 and 1. Consider then V = R as a vector space over Q. From the hypothesis, 1 is the only distance that is not attained twice. The claim above tells us that the Q-linear space generated by 1 contains all difference vectors. Thus all distances are rational. Comments 53 The Hahn–Banach lemma, alluded to above, says that for any vector subspace W of the vector space V and any vector v ∈ V that does not belong to W , there exists a linear map f : V → R such that f (W) = 0 but f (v) �= 0. The result from the statement is due to Mikusin´ski and Schinzel, and the proof given above is due to Straus. • J. Mikusin´ski, A. Schinzel: Sur la réductibilité de certains trinômes, Acta Arith. 9 (1964), 91–95. • E.G. Straus: Rational dependence in finite sets of numbers, Acta Arith. 11 (1965), 203–204. 7 Geometry Solutions 7.1 Synthetic Geometry Problem 3.1. Let I be the center of the circle inscribed in the triangle ABC and consider the points α, β, γ situated on the perpendiculars from I on the sides of the triangle ABC such that |Iα| = |Iβ| = |Iγ |. Prove that the lines Aα,Bβ,Cγ are concurrent. Solution 3.1. The lines Aα,Bβ,Cγ are concurrent if and only if they satisfy Ceva’s theorem. Let us draw A1A2 ‖ BC, where A1 ∈ |AB|, A2 ∈ |AC|, and α ∈ |A1A2|. We also draw their analogues. Using Thales’ theorem, the Ceva condition is reduced to ∏ cyclic |αA2| |αA1| = 1. Now I is situated on all three bisectors and |Iγ | = |Iα|. By symmetry we have |C1γ | = |A2α| and their analogues. This shows that the Ceva condition is satisfied. Problem 3.2. We consider the angle xOy and a pointA ∈ Ox. Let (C) be an arbitrary circle that is tangent to Ox and Oy at the points H and D, respectively. Set AE for the tangent line drawn from A to the circle (C) that is different from AH . Show that the line DE passes through a fixed point that is independent of the circle (C) chosen above. Solution 3.2. Let A′ be the point on Oy such that |OA′| = |OA|. Let us further consider the intersection points AE ∩ Oy = {F } and DE ∩ |AA′| = {P }. 134 7 Geometry Solutions O A A' D F E P H We use Menelaus’s theorem in the triangle AFA′ with the line DEP as transversal. We have |PA′| |PA| · |EA| |EF | · |DF | |DA′| = 1. But one knows that |EA| = |AH |, |EF | = |DF |, and |A′D| = |AH |. Then |PA′||PA| = 1, and therefore P is the midpoint of |AA′|. Consequently, P is a fixed point. Problem 3.3. Let C be a circle of center O and A a fixed point in the plane. For any point P ∈ C, let M denote the intersection of the bisector of the angle ÂOP with the circle circumscribed about the triangle AOP . Find the geometric locus of M as P runs over the circle C. Solution 3.3. One knows that OM is the bisector of ÂOP , and therefore OM is the mediator of the segment PB. Also, M is the midpoint of the arc of the circle ÂP , and therefore M belongs to the mediator of the segment |AP |. In the triangle PAB, M is the intersection of two mediators, and hence it also belongs to the mediator of segment |AB|. It is immediate now that the geometric locus of M is the entire mediator. Problem 3.4. Let ABC be an isosceles triangle having |AB| = |AC|. If AS is an interior Cevian that intersects the circle circumscribed about ABC at S, then describe the geometric locus of the center of the circle circumscribed about the triangle BST , where {T } = AS ∩ BC. Solution 3.4. We have B̂ST = B̂CA = ĈBA. Thus the line AB is tangent to the circle circumscribed about BST . Consequently, the center of this circle is situated on the line that passes through B that is perpendicular to AB. 7.1 Synthetic Geometry 135 Problem 3.5. Let AB,CD,EF be three chords of length one on the unit circle. Then the midpoints of the segments |BC|, |DE|, and |AF | form an equilateral triangle. Solution 3.5. Leta, b, c, d, e, f be the complex affixes (coordinates) of the respective points on the unit circle and ε = exp ( 2πi3 ) . Then the triangles OAB,OCD,OEF are equilateral, and therefore we have the following complex identities: a + �b = 0, �c + �2d = 0, �2e + f = 0. These imply that a+f2 + � b+c2 + �2 d+c2 = 0. Now the midpoint of |AF | has the affix a+f 2 , and the other midpoints are similarly associated to b+c 2 and d+e 2 . The relation above is equivalent to the claim of the statement. Problem 3.6. Denote by P the set of points of the plane. Let � : P × P → P be the following binary operation: A � B = C, where C is the unique point in the plane such that ABC is an oriented equilateral triangle whose orientation is counterclock- wise. Show that � is a nonassociative and noncommutative operation satisfying the following “medial property”: (A � B) � (C � D) = (A � C) � (B � D). Solution 3.6. Let A �= B,A � B = C. Then (A � B) � C = C � C = C. But A � (B � C) = A �A = A �= C; therefore the operation � is nonassociative. Because A � B,B � A are symmetric with respect to AB, the operation � is noncommutative. Now identify the points of the plane with complex numbers. Set ε = exp ( 2πi3 ) . We have then A � B = A + ε(B − A). It is now easy to verify that (A � B) � (C � D) = (1 − ε)2A + ε(1 − ε)(B + C) + ε2D. Moreover, this expression is symmetric in B and C. Comments 54 It is possible to take an arbitrary ε ∈ C and to define A �ε B = C, where C is the point such that the triangle ABC is similar to the triangle determined by the points 0, 1, and ε. Again, the medial property holds for this binary law. Problem 3.7. Consider two distinct circles C1 and C2 with nonempty intersection and let A be a point of intersection. Let P,R ∈ C1 and Q,S ∈ C2 be such that PQ and RS are the two common tangents. Let U and V denote the midpoints of the chords PR and QS. Prove that the triangle AUV is isosceles. Solution 3.7. Let B be the other intersection point of the two circles. It is known that the line AB is the radical axis of the two circles, i.e., the geometric locus of the points having the same power with respect to the two circles. Let us then consider the intersection point AB ∩ |PQ| = X. The power of X with respect to either circle is PC1(X) = |XA| · |XB|. On the other hand, PC1(X) = |XP |2 and PC2(X) = |XQ|2, 136 7 Geometry Solutions and so we obtain that X is the midpoint of |PQ|. Thus the line AB passes through the midpoints of |PQ| and |RS|, and thus it is also passes through the midpoint of |UV |, being also orthogonal to the latter. This shows that AUV is isosceles. Problem 3.8. If the planar triangles AUV, VBU , and UVC are directly similar to a given triangle, then so is ABC. Recall that two triangles are directly similar if one can obtain one from the other using a homothety with positive ratio, rotations and translations. Solution 3.8. Using complex numbers, each similarity (i.e., homothety) has the form z �→ αiz + βi , for some αi, βi ∈ C. The fixed triangle has its vertices located at z1, z2, and z3, and the pointsA,B,C,U, V are located at a, b, c, u, v respectively. By hypothesis, u = αizi+1 + βi and v = αizi+2 + βi , where the indices are considered mod 3. This implies that αizi(zi+1 − zi+2) = zi(u − v) and βi(zi+1 − zi+2) = zi+1v − zi+2u, so that 3∑ i=1 (αizi + βi)(zi+1 − zi+2) = 0. But from 3∑ i=1 (zi+1 − zi+2) = 0, 3∑ i=1 zi(zi+1 − zi+2) = 0 we derive det ⎛ ⎝ a z1 1 b z2 1 c z3 1 ⎞ ⎠ = det ⎛ ⎝ α1z1 + β1 z1 1 α2z2 + β2 z2 1 α3z3 + β3 z3 1 ⎞ ⎠ = 0, which is equivalent to saying that ABC is similar to the fixed triangle. Problem 3.9. Find, using a straightedge and a compass, the directrix and the focus of a parabola. Recall that the parabola is the geometric locus of those points P in the plane that are at equal distance from a point O (called the focus) and a line d called the directrix. Solution 3.9. Let � be an arbitrary direction in the plane. We draw two distinct lines �1 and �2 cutting the parabola in two nontrivial chords X1X2 and Y1Y2, both being parallel to �. Let X, Y be the midpoints of |X1X2| and respectively |Y1Y2|. It is simple to show that the line XY (called the diameter conjugate to the direction �) is parallel to the symmetry axis of the parabola. Moreover, let T� be the intersection point between the line XY and the parabola. Then the line that passes through T� and is parallel to � is tangent (at T�) to the parabola. Now we will use a well-known theorem in the geometry of conics, which claims that if P is a point in the plane (outside the convex hull of the parabola), PA, PB are the two tangents to the parabola issued from P (where A and B belong to the parabola) and if the angle ÂPB = π2 , then AB is a focal chord, i.e., the focus F lies on the chord |AB|. 7.1 Synthetic Geometry 137 Choose now the direction �⊥ that is orthogonal to � and construct as above the point T�⊥ on the parabola. Then, according to the previous result, the chord T�T�⊥ is focal. Choose next another couple of orthogonal directions, δ and δ⊥, and derive the new focal chord TδTδ⊥ . Then the two focal chords intersect at the focus T�T�⊥ ∩TδTδ⊥ = {O}. Moreover, we can draw the axis of symmetry of the parabola as the line passing through the focus that is parallel to (any) diameter conjugate to XY . Finally, let S be the intersection of the axis with the parabola and let O ′ be the symmetric of the focus with respect to S. Then the directrix d is the line passing through O ′ that is orthogonal to the axis OSO ′. Then �˜, the conjugate of �, meets � at a point on the axis. We represent then �, �˜ and we find two points of the axis. Problem 3.10. Prove that if M is a point in the interior of a circle and AB ⊥ CD are two chords perpendicular at M , then it is possible to construct an inscribable quadrilateral with the following lengths: ∣∣|AM| − |MB|∣∣, |AM| + |MB|, ∣∣|DM| − |MC|∣∣, |DM| + |MC|. Solution 3.10. Let us construct AF ‖ CD and CE ‖ AB, where the points E,F lie on the given circle. 138 7 Geometry Solutions Then the intersection point BF ∩ ED = O is the center of the circle. Moreover, we can compute |CE| = ∣∣|MB| − |MA|∣∣ and |AF | = ∣∣|MD| − |MC|∣∣. We rotate the triangle CED around the center O in order to superpose ED onto FB. Call C′ the new position of the point C. Then, the length of the fourth side of the quadrilateral ABC′F is |BC′| = |DM| + |MC|. Problem 3.11. If the Euler line of a triangle passes through the Fermat point, then the triangle is isosceles. Solution 3.11. The Euler line OG is determined by the circumcenter O and the cen- ter of gravity G. The trilinear coordinates of the points involved are G(sin B sin C, sin A sin C, sin A sin B), O(cosA, cosB, cosC), and the Fermat point F(sin(B + 60), sin(C + 60), sin(A + 60)). Therefore, G,O, and F are collinear only if the determinant of their trilinear coordinates vanishes. By direct calculation, this deter- minant can be computed as sin(A − B) sin(B − C) sin(C − A) = 0, and hence the triangle must be isosceles. Problem 3.12. Consider a point M in the interior of the triangle ABC, and choose A′ ∈ AM,B ′ ∈ BM , and C′ ∈ CM . Let P,Q,R, S, T , and U be the intersections of the sides of ABC and A′B ′C′. Show that PS, TQ, and RU meet at M . Solution 3.12. The triangles ABC and A′B ′C′ are in perspective with center M , and hence A′C′ ∩ AC,B ′C′ ∩ BC, and B ′A′ ∩ BA are collinear. Let xy be the line on which these points lie. We make a projective transformation that sends the line xy into the line at infinity. This way, the sides of the two triangles become pairwise parallel. Consider A′ fixed and let the points B ′′, C′′ be mobile points sitting on AB ′ and AC′, respectively. If we move the line B"C" by keeping it parallel to BC, then the intersection points Q′ = AB ∩B ′′C′′ and P ′ = AC ∩B ′′C′′ determine homographic divisions on AB and AC. When we reach the position where A ∈ B ′′C′′, we see that Q′T ∩ SP ′ = A; also, when we reach the position for which A′ ∈ B ′′C′′, we find 7.1 Synthetic Geometry 139 that Q′T ∩ SP ′ = A′. Therefore, since Q′, P ′ determine homographic divisions, we obtain that for any position of B ′′C′′, we have Q′T ∩ SP ′ ∈ AA′. In particular, this happens when B ′′C′′ = B ′C′, and thus QT ∩ SP ∈ AA′. Analogously, we have QT ∩ RU ∈ CC′, SP ∩ RU ∈ BB ′. According to Pascal’s theorem, the diagonals of the hexagon PQRSTU are concurrent, since the edges are pairwise parallel. This shows that QT ∩ SP ∩ RU ∈ AA′ ∩ BB ′ ∩ CC′ = M . Problem 3.13. Show that if an altitude in a tetrahedron crosses two other altitudes, then all four altitudes are concurrent. Solution 3.13. If the altitude hA fromA intersects hB , thenAB⊥CD. Similarly, if hA intersects hC , then AC⊥BD. These can be written vectorially as 〈 → AB, → AD − →AC〉 = 0, and 〈 →AC, →AD − →AB〉 = 0, where 〈, 〉 denotes the Euclidean scalar product. These imply that 〈 →AB − →AC, →AD〉 = 0, which is equivalent to AD⊥BC. This implies that all four altitudes are concurrent. Problem 3.14. Three concurrent Cevians in the interior of the triangle ABC meet the corresponding opposite sides at A1, B1, C1. Show that their common intersection point is uniquely determined if |BA1|, |CB1|, and |AC1| are equal. Solution 3.14. Let |BA1| = |CB1| = |AC1| = δ. From Ceva’s theorem, we obtain (a − δ)(b − δ)(c − δ) = δ3. We may assume that a ≤ b ≤ c, and hence δ ∈ [0, a]. The left-hand side is a decreasing function of δ ranging from the value abc, obtained for for δ = 0, to 0, when δ = a. Further, the right-hand-side function δ3 is strictly increasing. Therefore, the two functions can have only one common value on the interval [0, a]. This determines uniquely the intersection point. Comments 55 Also, if we ask that |CA1|, |BC1|, and |AB1| be equal, their common value δ˜ coincides with the δ we found above. In particular, the two intersection points are isotomic to each other. Problem 3.15. Let ABCD be a convex quadrilateral with the property that the circle of diameter AB is tangent to the line CD. Prove that the circle of diameter CD is tangent to the line AB if and only if AD is parallel to BC. Solution 3.15. Let M be the midpoint of |AB| and draw the perpendicular MM ′ ⊥ CD, where M ′ ∈ CD. Then M ′ is the contact point of the circle C1 of diameter AB and the line CD. Thus ÂM ′B = π2 , and so |MM ′| = |AB|2 . Conversely, if M,M ′ are as above and satisfy the condition |MM ′| = |AB|2 , then the circle of diameter AB is tangent to CD. 140 7 Geometry Solutions Let now O = AB∩CD, assuming that the two lines are not parallel. If |OM| = λ and ÂOD = θ , then |MM ′| = λ sin θ and so |MB| = |MA| = λ sin θ . This turns to |OB| |OA| = |OM| + |MA| |OM| − |MB| = λ(1 + sin θ) λ(1 − sin θ) = 1 + sin θ 1 − sin θ . Conversely, |OB||OA| = 1+sin θ1−sin θ implies that |MM ′| = |MA| = |MB|. Let now P be the midpoint of CD and draw the perpendicular PP ′ ⊥ AB, with P ′ ∈ AB. The previous argument shows that |PP ′| = |CD|2 is equivalent to|OC| |OD| = 1−sin θ1+sin θ , which, according to Thales’ theorem, is equivalent to saying that AD is parallel to BC. But |PP ′| = |CD|2 iff the circle of diameter CD is tangent to AB, as claimed. The last case occurs when AB is parallel to CD. Then |MM ′| is the distance between the lines AB and CD, and hence |MM ′| = |AB|2 = |CD|2 . Thus ABCD is a parallelogram, and hence AD is parallel to BC. Problem 3.16. Let A′, B ′, C′ be points on the sides BC,CA,AB of the triangle ABC. Let M1,M2 be the intersections of the circle A′B ′C′ with the circle ABA′ and let N1, N2 be the analogous intersections of the circle A′B ′C′ with the circle ABB ′. 1. Prove that M1M2, N1N2, AB are either parallel or concurrent, in a point that we denote by A′1; 2. Prove that the analogously defined points A′1, B ′1, C′1 are collinear. Solution 3.16. 1. More precisely, M1M2, N1N2, AB are the radical axes of the three pairs of circles formed by A′B ′C′, ABB ′, and ABC′. The first part follows. 2. Further, A′1 is the radical center of A′B ′C′, ABB ′, ABA′, and thus |A′1A| · |A′1B| = p A′1 ABB ′ = p A′1 A′B ′B ′ , where p denotes the power of the point with respect to the given circle. Therefore pA ′ 1 A′B ′C′ = p A′1 ABC , and their analogues, which imply that the three points A′1, B ′1, and C′1 belong to the radical axis of the pair of circles A′B ′C′ and ABC. Problem 3.17. A circumscribable quadrilateral of area S = √abcd is inscribable. Solution 3.17. By the circumscribability hypothesis, we have a + c = b + d . Let k be the length of the diagonal that leaves a, b on one side and c, d on the other side. 7.1 Synthetic Geometry 141 Let α, β be the angles of the quadrilateral between a and b, and c and d, respectively. We have then the law of cosines computing k: k2 = a2 + b2 − 2ab cosα, k2 = c2 + d2 − 2cd cosβ. Subtracting the terms (a − b)2 = (c − d)2, we obtain 2ab(1 − cosα) = 2cd(1 − cosβ). Now compute the area S of the quadrilateral S = 12 (ab sin α + cd sin β), which is by hypothesis √ abcd. Therefore 4S2 = 4abcd = a2b2(1 − cos2 α) + c2d2(1 − cos2 β) + 2abcd sin α sin β. From the former identity, we obtain 4abcd = ab(1+cosα)cd(1−cosβ)+cd(1+cosβ)ab(1−cosα)+2abcd sin α sin β. After simplification of the terms, we obtain 4 = 2 − 2 cos(α + β), and therefore α + β = π . This implies that the quadrilateral is inscribable. Problem 3.18. Let O be the center of the circumcircle, Ge the Gergonne point, Na the Nagel point, and G1, N1, the isogonal conjugates of G and N , respectively. Prove that G1, N1, and O are collinear (see also the Glossary for definitions of the important points in a triangle). Solution 3.18. We will use the trilinear coordinates with respect to the given trian- gle. These are triples of numbers (see the triangle compendium from the Glossary) proportional to the distances from the given point to the edges of the triangle. The trilinear coordinates behave very nicely when we consider the isogonal conjugate. If P has the coordinates (x, y, z), then its isogonal conjugate has coordinates ( 1 x , 1 y , 1 z ) . The trilinear coordinates of the points considered above are then O(cosA, cosB, cosC), Ge ( 1 a(p − a) , 1 b(p − b) , 1 c(p − c) ) , G1 (a(p − a), b(p − b), c(p − c)) , Na ( p − a a , p − b b , p − c c ) , N1 ( a p − a , b p − b , c p − c ) . Further, three points are collinear if the determinant of the 3 × 3 matrix of their coordinates vanishes. The claim from the statement is then equivalent to the following identity: 142 7 Geometry Solutions det ⎛ ⎝ cosA cosB cosC a(p − a) b(p − b) c(p − c) a p−a b p−b c p−c ⎞ ⎠ = 0. We develop the determinant and obtain ∑ cyclic cosA bc ( p − b p − c − p − c p − b ) = ∑ cyclic cosA bc (p − b)(p − c) a(c − b). This vanishes iff ∑ cyclic cosA (p − a)(c − b) = 0, which is equivalent to ∑ cyclic (b2 + c2 − a2)(c − b)(−a + b + c)a = 0, which is immediate. Note that we could use instead the barycentric coordinates as well. These coordi- nates are triples of numbers proportional to the areas of the three triangles determined by the point and two of the vertices of the initial triangle. The barycentric coordinates of the points considered above are given by O(a, b, c), Ge ( 1 p − a , 1 p − b , 1 p − c ) , G1 ( a2(p − a), b2(p − b), c2(p − c) ) , Na (p − a, p − b, p − c) , N1 ( a2 p − a , b2 p − b , c2 p − c ) . Finally, three points are collinear if and only if their coordinate matrix has vanishing determinant. 7.2 Combinatorial Geometry Problem 3.19. Consider a rectangular sheet of paper. Prove that given any ε > 0, one can use finitely many foldings of the paper along its sides in either 2 equal parts or 3 equal parts to obtain a rectangle whose sides are in ratio r for some r satisfying 1 − � ≤ r ≤ 1 + �. Solution 3.19. Observe that log3 2 �∈ Q and thus the residues modulo 1 of the ele- ments n log3 2, with n ∈ Z+, form a dense subset of (0, 1). This implies that the set of numbers of the form 2a3b, with a, b ∈ Z, are dense in R. Problem 3.20. Show that there exist at most three points on the unit disk with the distance between any two being greater than √ 2. 7.2 Combinatorial Geometry 143 Solution 3.20. Consider the points Ai on the unit disk of radius 1 and center O. Then |OAi | ≤ 1, and if we assume that |AiAj | > √ 2, then 2 < |OAi |2 + |OAj |2 − 2|OAi | · |OAj | cos ÂiOAj . Thus, cos ÂiOAj < 0 and hence ÂiOAj > π2 , which implies that n ≤ 3. Comments 56 H.S.M. Coxeter proposed this problem in 1933. The same argument shows that if we have k points Ai within the unit disk, then there exists a pair of points such that |AiAj | ≤ max ( 1, 2 sin π k ) , and this is sharp for 2 ≤ k ≤ 7. The result was generalized to higher dimensions by Davenport, Hajós, and, indepen- dently, Rankin, as follows: There are at most n + 1 points in the unit ball of Rn such that the distance between any two points is greater than √ 2. Here is a proof sketch. Let Ai denote the m points satisfying the hypothesis, so that they are different from the origin O. Then we have m ≥ n + 2 unit vectors OAi|OAi | in Rn with the property that 〈vi, vj 〉 < 0. We claim next that there exist pairwise orthogonal subspaces W1,W2, . . . ,Wm−n ⊂ Rn that span Rn such that each space Wj contains 1+dim Wj vectors among the unit vectors above, v1, . . . , vm, that span Wj . Since m − n ≥ 2, there exist two vectors among the vi that are orthogonal to each other, which contradicts our assumptions. The claim follows by induction on the dimension n. If n = 2, it is immediate. Now, if vi = −vn, for some i < n, then all other vj should be orthogonal to vn (and to vi). Otherwise, we can write for all i < n, vi = ui + xivn, for some scalars xi < 0 and some nonzero vector ui orthogonal to vn. This implies that 〈ui, uj 〉 < 0, when i �= j . We can apply therefore the induction hypothesis for the subspace orthogonal to vn and the set of vectors vi , i < n, and obtain the subspaces W ′j . We define then Wi = W ′i , for i ≥ 2, and let W1 be the span of W ′1 and vn. This proves the claim. The claim shows that if we have m ≥ n + 2 points on the unit sphere in Rn whose pairwise distances are at least √ 2, then m ≤ 2n, and moreover, there exist two points at distance precisely √ 2. • H.S.M. Coxeter: Amer. Math. Monthly 40 (1933), 192–193. • R.A. Rankin: The closest packing of spherical caps in n dimensions, Proc. Glas- gow Math.Assoc. 2 (1955), 139–144. Problem 3.21. A convex polygon with 2n sides has at least n diagonals not parallel to any of its sides. The equality is attained for the regular polygon. Solution 3.21. Let l be a side of the polygon. Then we have at most n − 2 diagonals parallel to l. In fact, let d1, d2, . . . , dp be these diagonals. Then the pair of seg- ments (d1, l) determines a quadrilateral (actually a trapezoid) having three common sides with the polygon. Moreover, all the pairs (d2, d1), . . . , (dp−1, dp) determine quadrilaterals, each one containing two sides of the initial polygon. This implies that p ≤ n − 2. The total number of diagonals that are parallel to at least one side is at most 2n(n − 2). But there are 2n(2n−3)2 diagonals, and hence n(2n − 3) − 2n(n − 2) = n diagonals with the desired property. 144 7 Geometry Solutions Problem 3.22. Let d be the sum of the lengths of the diagonals of a convex polygon P1 · · ·Pn and p its perimeter. Prove that for n ≥ 4, we have n − 3 < 2 d p < [n 2 ] [n + 1 2 ] − 2. Solution 3.22. Set αk =∑ni=1 |PiPi+k|. We will prove first that p = α1 < α2 < α3 < · · · < α[ n2 ] and α[ n2 ]+1 > · · · > αn−2 > αn−1 = p. Let us show first that p < α2. Let AiAi+2 ∩ Ai+1Ai+3 = {Oi+1}. Observe that |Ai+1Oi+1| ≤ |Ai+1Oi+2|, since ̂Ai+1Ai+2Oi+1 ≤ ̂Ai+1Ai+2Oi+2. The triangle inequality in AiOiAi+1 further yields |Ai+1Oi+1| + |Oi+1Ai+2| > |Ai+1Ai+2|. Summing up these inequalities, and using the remark, it follows that p < α2, as claimed. Let now k < [ n 2 ] − 1, and set AiAi+k ∩ Ai+1Ai+k+1 = {O}. The triangle inequality again shows that |AiAi+k| + |Ai+1Ai+k+1| = |AiO| + |OAi+k| + |Ai+1O| + |OAi+k+1| > |Ai+1Ai+k| + |AiAi+k+1|. By summing up over i, we obtain 2αk > αk+1 + αk−1; thus αk+1 − αk < αk − αk−1 < · · · < α2 − α1, and hence the claim. Moreover, if n = 2m, then 2αm > αm+1 + αm−1 = 2αm+1, and so αm > αm−1. Thus, we have proved that αk+1 > αk for all k. By summing up all these inequalities over k, we find that 2d > (n − 3)p. 7.2 Combinatorial Geometry 145 Further, consider the triangle inequality |AiAj | < |AiAi+1| + · · · + |Aj−1Aj | and sum up over all i and j ; this yields 2d < (([n 2 ] [n + 1 2 ]) − 2 ) p. Problem 3.23. Find the convex polygons with the property that the function D(p), which is the sum of the distances from an interior point p to the sides of the polygon, does not depend on p. Solution 3.23. Let K be our polygon, p an interior point, a1, . . . , an the distances from p to the sides of K , and let u1, . . . , un be unit vectors perpendicular to the sides. Choose another point, say q, in the interior of K . Denote by v the vector pq. The distance from q to the side with label i is therefore di = ai +〈ui, v〉, where 〈, 〉 is the scalar product. It follows that D(q) = D(p) + 〈 n∑ i=1 ui, v 〉 . Therefore, a necessary and sufficient condition for D(p) to be constant is that n∑ i=1 ui = 0. Observe that the solution can be extended to convex polyhedra in Rm. Problem 3.24. Prove that a sphere of diameter 1 cannot be covered by n strips of width li if ∑n i=1 li < 1. Prove that a circle of diameter 1 cannot be covered by n strips of width li if ∑n i=1 li < 1. Solution 3.24. 1. The lateral area of the body obtained by cutting off the sphere with two planes at distance h is 2πRh. Let us assume that we have a covering by means of the strips Bi . Then the sum of areas cut open by the strips Bi is at least the area of the sphere, hence π ∑n i=1 li ≥ π , contradicting our assumptions. 2. If we had such a covering, let B∗i be the spatial strip that cuts the planar strip Bi , orthogonal to the plane of the circle. Then the union of the strips B∗i would cover the cylinder over the disk, in particular the sphere of diameter 1. This would contradict the first part. Problem 3.25. Consider n points lying on the unit sphere. Prove that the sum of the squares of the lengths of all segments determined by the n points is less than n2. Solution 3.25. Let O be the center of the sphere and let Xi denote the given points. We denote by vj the vector −−→ OXj , and by 〈, 〉 the Euclidean inner product. We have then |XiXj |2 = 〈vj − vi, vj − vi〉. 146 7 Geometry Solutions Set S =∑i,j |XiXj |2. We have then 2S = n∑ i=1 〈vi − v1, vi − v1〉 + n∑ i=1 〈vi − v2, vi − v2〉 + · · · + n∑ i=1 〈vi − vn, vi − vn〉 = 2n(|v1|2 + · · · + |vn|2) − 2 ∑ i,j 〈vj , vi〉 = 2n n∑ i=1 |vi |2 − 2〈v, v〉 = 2n2 − 2|v|2 ≤ 2n2, where v =∑ni=1 vi . Problem 3.26. The sum of the vectors −−→OA1, . . . ,−−→OAn is zero, and the sum of their lengths is d . Prove that the perimeter of the polygon A1 . . . An is greater than 4d/n. Solution 3.26. If A is an arbitrary point in the plane, then n−→OA = ∑ni=1( −−→ OAi +−−→ AiA) = ∑ni=1 −−→ AiA, which implies that n|OA| ≤ ∑ |AiA|. Let us put A = Ai for i = 1, 2, . . . , n and sum up the inequalities so obtained: d = |OA1| + · · · + |OAn| ≤ 2 n ∑ i,j |AiAj |. Using the triangle inequality, |AiAj | < |AiAi+1| + · · · + |Aj−1Aj |, we infer that d ≤ 2 n ( p + 2p + · · · + n − 1 2 p ) = n 2 − 1 4n p. Therefore p ≥ 4d n n2−1 > 4d n . Problem 3.27. Find the largest numbers ak , for 1 ≤ k ≤ 7, with the property that for any point P lying in the unit cube with vertices A1 . . . A8, at least k among the distances |PAj | to the vertices are greater or equal than ak . Solution 3.27. At least eight distances are greater than or equal to 0, and this is sharp by taking P = Aj , and hence a8 = 0. By taking P to be the midpoint of an edge, we get a7 ≤ 12 . Assume that a7 < 12 . Then there exists some P for which at least two distances |PAj | are strictly smaller than 12 , which is obviously false, since vertices are distance 1 apart. Thus a7 = 12 . Consider P as the center of some face of the cube. We obtain then a6 ≤ a5 ≤ √ 2 2 . If equality does not hold above, then there exists P for which three distances are smaller than √ 2 2 . However, given three vertices of the cube, there are at least two of them at distance greater than or equal to the diagonal of a face, namely √ 2. Thus our assumption fails and so a6 = a5 = √ 2 2 . 7.2 Combinatorial Geometry 147 Now let P be the center of the cube. We derive that a4 ≤ a3 ≤ a2 ≤ a1 ≤ √ 3 2 . Assume that a4 < √ 3 2 . Then there exist P and at least five distances strictly smaller than √ 3 2 . However, for any choice of five vertices of a cube, one finds a pair of opposite vertices that are distance √ 3 apart. This contradicts our assumptions and thus a4 = a3 = a2 = a1 = √ 3 2 . Problem 3.28. The line determined by two points is said to be admissible if its slope is equal to 0, 1,−1, or ∞. What is the maximum number of admissible lines determined by n points in the plane? Solution 3.28. We join two points by an edge if the line determined by them is ad- missible. We obtain a graph with n vertices. By hypothesis, each vertex has degree at most 4. Since the sum of degrees over all vertices is twice the number of edges, we find that there are at most 2n edges, thus at most 2n admissible lines. Moreover, there are only four directions in the plane, and the same computation shows that there are at most 12n edges associated to every direction. This implies that for odd n, the maximum number of admissible lines is 2n − 2. It is easy to construct examples in which we have precisely 2n admissible lines if n is even and n ≥ 12. For odd n, the associated graph is obtained by adding an isolated vertex to a maximal graph of order n − 1, and this works for any n ≥ 3. Finally, if f (n) denotes the maximum number of admissible lines, then explicit inspection shows that f (2) = 1, f (3) = 3, and f (4) = 6, and so the upper bound 2n is not attained. It can be shown that f (6) = 11 and f (10) = 19. Therefore, for all n but those from {2, 3, 4, 6, 10}, we have f (n) = 4[n2 ] . Comments 57 More generally, assume that the admissible directions are k in num- ber. The same proof shows that there therefore exist at most [ kn2 ] admissible lines determined by n points. Moreover, the minimum number of directions (not necessar- ily admissible) defined by n points is at least n − 1. See also the article: • P.R. Scott: On sets of direction determined by n points, Amer. Math. Monthly 77 (1970), 502–505. Problem 3.29. If A = {z1, . . . , zn} ⊂ C, then there exists a subset B ⊂ A such that ∣∣∣∣ ∑ z∈B z ∣∣∣∣ ≥ π−1 n∑ i=1 |zi |. Solution 3.29. We reorder the numbers z1, . . . , zn,−(z1 + · · · + zn) in a sequence w1, . . . , wn+1 such that the sequence arg w1, . . . , arg wn+1 is monotonic. In the complex plane, the points w1, w1 + w2, . . . , w1 + · · · + wn+1, are the vertices of a convex polygon P . Set d(P ) for the diameter of P . Since the diameter of a polygon is the distance between two vertices, there exists B ⊂ A such that d(P ) = ∣∣∣ ∣ ∑ zi∈B zi ∣ ∣∣∣. 148 7 Geometry Solutions Let p(P ) be the perimeter of P . We have then p(P ) = ∑ |wi | = n∑ i |zi | + ∣∣∣ ∣ n∑ i zi ∣∣ ∣∣. The main isoperimetric inequality states that for any convex planar polygon Q we have the inequality πd(Q) ≥ p(Q). Applying this to the polygon P settles the claim. Comments 58 Using a refined isoperimetric inequality for polygons with a given number of vertices, one finds that there exists B ⊂ {1, . . . , n} such that ∣∣ ∣∣ ∑ i∈B zi ∣∣ ∣∣ ≥ sin(kπ/(2k + 1)) (2k + 1) sin(π/2k + 1) n∑ i=1 |zi |. Problem 3.30. Let A1, . . . , An be the vertices of a regular n-gon inscribed in a circle of center O. Let B be a point on the arc of circle A1An and set θ = ÂnOB. If we set ak = |BAk|, then find the sum n∑ k=1 (−1)kak in terms of θ . Solution 3.30. We have ak = 2r sin ( kπ n − θ2 ) . Then n∑ k=1 (−1)kak cos π2n = ∑ (−1)kr ( sin ( (2k − 1)π 2n − θ 2 ) + sin ( (2k + 1)π 2n − θ 2 )) = ( −1 + (−1)n+1 ) r sin ( π 2n − θ 2 ) , which leads us to n∑ k=1 (−1)kak = { 2r sin ( θ 2 − π2n ) ( cos π2n )−1 , for even n, 0, for odd n. Problem 3.31. Consider n distinct complex numbers zi ∈ C such that min i �=j |zi − zj | ≥ maxi≤n |zi |. What is the greatest possible value of n? Solution 3.31. Let |zn| = maxi≤n |zi |; then the zi belong to the disk of radius R = |zn| centered at the origin. Moreover, the distance between two points of complex coordinates zi and zj is given by |zi − zj |, which by hypothesis is greater than R. In particular, we have n points whose pairwise distances are at least R within the circle of radius R. Let O be the origin and Zi the points. We order the points Zi in 7.2 Combinatorial Geometry 149 increasing order with respect to their arguments. If the origin is among the Zi , then we choose it to be the first point. Further, there remain at least n − 1 points distinct from the origin. One knows that n∑ i=2 ̂ZiOZi+1 = 2π, and thus there exists at least one angle ̂ZiOZi+1 ≤ 2πn−1 . In particular, if n ≥ 8, then |ZiZi+1| ≤ R √ 2 − 2 cos ̂ZiOZi+1 < R, contradicting our assumptions. Thus n ≤ 7 and the maximal value n = 7 is reached by the configuration of six vertices of a regular hexagon together with the origin. Problem 3.32. The interior of a triangle can be tiled by n ≥ 9 pentagonal convex surfaces.What is the minimal value ofn such that a triangle can be tiled byn hexagonal strictly convex surfaces? Solution 3.32. There is an obvious recurrence, showing how to go from k to k + 3. We still have to find the appropriate tiling for n ∈ {9, 10, 11}. If n = 9, then we have the following tiling of the triangle: A slight deformation of the left corner pentagon shows that a concave quadrilateral can also be tiled with nine pentagons. Moreover, a tiling by n pentagons of a concave quadrilateral can be adjusted as below to a tiling of a triangle with n + 1 pentagons. This method furnishes a tiling with any number n ≥ 9 of pentagons. 150 7 Geometry Solutions 2. We will show that there are no hexagonal tilings of triangles with convex tiles. Assume the contrary. We call a tiling good if no hexagon tile has sides in common with (at least) two sides of the triangle. Given an arbitrary tiling, we can construct a good tiling of a triangle as follows. First, the triangle can be assumed to be a right triangle by means of the following transformation. Consider the cone in R3 over the initial tiled triangle with a cone vertex far away, and cut it by a convenient plane so that one of the angles of the section become right angled. The trace of the coned tiling in the planar section is still a hexagonal tiling with convex polygons. Further, we double the right triangle along one of its sides and obtain a tiling that has no hexagons with common edges with both of the sides opposite to the doubled side. Applying again the same procedure yields a good tiling. The main feature of a good tiling by convex polygons is that any hexagon having a common side with the triangle (called a boundary hexagon) has precisely one side in common, excepting the case in which the hexagon might have a common angle with the triangle (called an exceptional boundary hexagon). In the last case, we have a vertex of the hexagon and its incident sides that are common with the vertex of a triangle and its incident sides. Thus there are λ ∈ {0, 1, 2, 3} exceptional boundary hexagons. Let a∂ be the total number of boundary hexagons and ain the hexagons that are interior, i.e., they have no sides in common with the triangle. Construct the dual graph associated to the tiling. This is a planar graph whose vertices have degree 6. The edges going outside of the triangle will be called exterior edges of the graph, and the remaining ones interior edges. The number of vertices of this graph is a = ain + a∂ . Counting the sum of degrees of all vertices, we obtain 6a on one side, and 2ein + eo, where ein (respectively eo) is the number of interior (respectively exterior) edges. Moreover, each exterior edge comes from a unique boundary hexagon, and two different edges come from different hexagons, excepting the exceptional hexagons, which have two outgoing edges. Thus eo = a∂ + λ. This implies that ein = 3a − 12 (a∂ + λ). If we erase the exterior edges of our graph, then we obtain a planar polygon partitioned into polygons (duals of the hexagons). The total number of vertices is a, the number of edges is ein and the number of faces f can be estimated from above as follows. We have ain vertices in the interior of a polygon with a∂ sides. Each polygon has at least three sides (because our initial hexagons were supposed to be 7.2 Combinatorial Geometry 151 strictly convex). Consider the sum of angles of all polygons involved. On the one hand this yields 2πain + π(aδ − 2), and on the other hand this is at least πf . Thus f ≤ 2ain + a∂ − 2. The Euler–Poincaré theorem requires that 1 = a − ein + f ≤ a − ( 3a − 1 2 (a∂ + λ) ) + 2ain + a∂ − 2 = λ2 − 2 − a∂ 2 . But λ ≤ 3 and a∂ ≥ 0 lead us to a contradiction. Comments 59 The same proof shows that there does not exist a tiling of the quadri- lateral by strictly convex hexagons. Problem 3.33. We say that a transformation of the plane is a congruence if it preserves the length of segments. Two subsets are congruent if there exists a congruence sending one subset onto the other. Show that the unit disk cannot be partitioned into two congruent subsets. Solution 3.33. Let us assume that the disk D can be partitioned as D = A∪B, where A and B are congruent. Let p denote the center of D, and assume that p ∈ A. Let F : A → B be a congruence. Let rs be the diameter that is perpendicular to the linepF(p). For any point x ∈ D, we have |px| ≤ 1. Thus |F(p)F (x)| ≤ 1, for any x ∈ A, and thus |F(p)y| ≤ 1, for any y ∈ B. Now |F(p)r| > 1 and |F(p)s| > 1, and thus r, s ∈ A. Fur- ther, |F(r)F (s)| = |rs| = 2 and thus F(r)F (s) is a diameter of the disk. Since |F(r)F (p)| = |F(s)F (p)| = 1, we derive that F(p) is the midpoint of the diameter F(r)F (s), and thus it coincides with the centerp. This contradiction proves the claim. Comments 60 Actually, the same argument shows that the disk cannot be partitioned into finitely many congruent sets. The problem is due to B.L. Van der Waerden. More generally, D. Puppe showed that the same holds for any convex compact set in the plane. See also: • H. Hadwiger and H. Debrunner: Combinatorial Geometry in the Plane, translated by V. Klee. With a new chapter and other additional material supplied by the translator, New York, 1964. Problem 3.34. Prove that the unit disk cannot be partitioned into two subsets of diameter strictly smaller than 1, where the diameter of a set is the supremum distance between two of its points. 152 7 Geometry Solutions Solution 3.34. Observe first that the diameter of a planar set A coincides with the diameter of its closure A in R2. Assume that the disk is partitioned into two sets A and B of smaller diameter. The points of the boundary circle cannot belong to only one of the partition sets, since opposite points are at distance 1. One the other hand, the unit circle cannot be written as the union of two disjoint nontrivial closed subsets (i.e., closed intervals) because it is connected. This implies that there exists some point M of the unit circle that belongs to both A and B. Consider now the opposite point M ′ on the circle. If M ′ ∈ A, then A has diameter 1; otherwise, M ′ ∈ B and so B has diameter 1, which gives us a contradiction. Problem 3.35. A continuous planar curve L has extremities A and B at distance |AB| = 1. Show that, for any natural number n there exists a chord determined by two points C,D ∈ L that is parallel to AB and whose length |CD| equals 1 n . Solution 3.35. Assume that L does not admit chords CD parallel to AB either of length a or of length b. We will show that L does not admit chords CD parallel to AB of length a + b. Let Tx be the planar translation along the line AB of the number of x units. We have, by hypothesis, TaL ∩ L = ∅ and TbL ∩ L = ∅. This implies thatTa+bL∩TaL = ∅. On the other hand, let d(U, V ) denote the distance between the sets U and V in the plane, given by infP∈U,Q∈V |PQ|. Then obviously d(Ta+bL,L) ≤ d(TaL,L) + d(Tb+aL, TaL). This implies that d(Ta+bL,L) = 0 if d(TaL,L) = d(Tb+aL, TaL) = 0. Moreover, L and Ta+b(L) are closed subsets, and thus the distance is attained, i.e., there exist P ∈ L and Q ∈ Ta+b(L) such that |PQ| = d(L, Ta+b(L)). In particular, P = Q and thus Ta+bL∩L �= ∅, proving our claim. Suppose now that there is no parallel chord of length a = b = 1 n . Then we derive that there is no such chord of length 2 n . Take a = 1 n and b = 2 n . The previous claim shows that there is no chord of length 3 n . An easy induction shows that there is no chord of length k n for k ≤ n. This contradicts the fact that AB has length 1. Comments 61 The result is due to H. Hopf, who characterized all the possible lengths of chords arising as above. • H. Hopf: Über die Sehnen ebener Kontinuen und die Schleifen geschlossener Wege, Comment. Math. Helv. 9 (1937), 303–319. Problem 3.36. The diameter of a set is the supremum of the distance between two of its points. Prove that any planar set of unit diameter can be partitioned into three parts of diameter no more than √ 3 2 . Solution 3.36. The main ingredient is to show that a planar set F of unit diameter is contained within a regular hexagon whose opposite sides are at distance 1. Given a vector v = v1, let us consider the thinest strip Sv parallel to v that contains F . Then the strip Sv is bounded by two parallel lines, each line containing 7.2 Combinatorial Geometry 153 at least one point of F , by the minimality of the strip. In particular, the width of the strip is bounded by the distance between two points of F , and so by the diameter of F . Now let v2 be the image of v1 by a rotation of angle π3 , and let v3 be the image of the rotation of angle 2π3 . The intersection of the corresponding strips Hv = Sv1 ∩Sv2 ∩Sv3 is a semiregular (i.e., all opposite edges are parallel) hexagon having all angles 2π3 . However, in general, this hexagon is not regular. We choose two opposite vertices, say A and A′, of the parallelogram Sv1 ∩ Sv2 . For instance, we choose A such that the vectors spanning the parallelogram issued from it are positive multiples of both v1 and v2. We denote by x the distance from A to the strip Sv3 and by x′ the distance from A′ to the strip Sv3 . We observe that the hexagon Hv is regular if x = x′. Moreover, let us set f (v) = x − x′. Obviously, the function that associates to the vector v the quantity x − x′ is continuous. Let us compute f (−v), by rotating v through π . This amounts to interchanging A with A′, and thus f (−v) = −f (v). A continuous function taking both positive and negative values must have a zero. Thus there exists some position of v for which f (v) = 0 and so Hv is regular. Further, the regular hexagon of width 1 can be partitioned into three pieces having diameter √ 3 2 , as in the figure below: This proves the claim. Comments 62 K. Borsuk proved in 1933 that a planar set of diameter 1 can be partitioned into subsets of diameter d < 1. The main ingredient above is a lemma due to J. Pál, from 1920. B. Grünbaum proved (improving on ideas of D. Gale) that a set of unit diameter in R3 is contained in a regular octahedron of width 1, with three of its vertices (one vertex on each diagonal) cut off by planes orthogonal to the diagonals, 154 7 Geometry Solutions at distance 12 from the center. Using a suitable dissection of this polyhedron, one finds that the set F can be partitioned into four pieces having diameter not bigger than √ 6129030 − 937419√3 1518 √ 2 ≈ 0.9887. Gale’s conjecture from 1953, that one can always get a partition into four pieces of diameter no bigger than √ 3+√3 6 , is still open. • D. Gale: On inscribing n-dimensional sets in a regular n-simplex, Proc. Amer. Math. Soc. 4 (1953), 222–225. • B. Grünbaum: A simple proof of Borsuk’s conjecture in three dimensions, Math. Proc. Cambridge Philos. Soc. 53 (1957), 776–778. Problem 3.37. 1. Prove that a finite set of n points in R3 of unit diameter, can be covered by a cube of side length 1 − 23n(n−1) . 2. Prove that any planar set of n points having unit diameter can be partitioned into three parts of diameter less than √ 3 2 cos 2π 3n(n−1) . Solution 3.37. 1.A direction will mean below a line without specified orientation. The angle between two directions is the smallest among the four angles they define, and thus it is always within the range [ 0, π2 ] . Consider the set of n points {x1, x2, . . . , xn}, which determine at most N = n(n−1)2 distinct directions xixj , which we denote by l1, l2, . . . , lN . We want to prove first that there exist three mutually orthogonal directions y1, y2, y3, so that min 1≤i≤3,1≤s≤N ∠(yi, ls) ≥ arccos ( 1 − 1 3N ) . Consider an orthogonal frame in R3 given by three mutually orthogonal vectors e1, e2, e3, and choose the rotation R1 and R2 of the space with the property that R1e1 = e2, R1e3 = e3, R2e1 = e3, R2e2 = e2. Using these rotations, we have min 1≤i≤3,1≤s≤N ∠(yi, ls) = min1≤s≤N min(∠(y1, ls),∠(y1, R1ls),∠(y1, R2ls)). Consider the set of 3N directions D = {lj , R1lj , R2lj , 1 ≤ j ≤ N}. We will prove the following estimate from below: max y min l∈D ∠(y, l) ≥ arccos ( 1 − 1 3N ) = θ, for an arbitrary set D of 3N directions. Assume the contrary. The set of directions y making an angle smaller than θ with a given direction l form a symmetric cone Cl,θ of angle 2θ with axis l. If there is no direction y satisfying the inequality above, then the union of all cones Cl,θ , over l ∈ D will cover the set of all directions in space. Each symmetric cone Cl,θ intersects the unit sphere along a symmetric pair of 7.2 Combinatorial Geometry 155 spherical caps of radius θ . In particular, these spherical caps cover the surface of the whole sphere, since a point not covered yields a direction not belonging to the union. Furthermore, this implies that the total area of the spherical caps is (strictly) greater than the area of the sphere. Now, the area of each spherical cap is 2π(1 − cos θ), and thus the total area covered by the cones is 12πN(1− cos θ) = 4π . But 4π is the total area of the sphere, which contradicts the claim. Observe that the inequality must be strict, since N ≥ 1 and any covering of the sphere by nontrivial spherical caps will have some overlaps. Consider now three mutually orthogonal directions yj satisfying the claim above. Let further ξj ⊂ yj be the image of the given set of points under orthogonal projection ontoyj . The distance between the images ofxi andxk onyj is |xixk| cos∠(yj , xixk) ≤ sin θ . Thus each subset ξj is contained in some interval of length less than cos θ , and thus the set of points is contained in the strip of width sin θ , orthogonal to yj . The intersection of the three orthogonal strips associated to the direction yj is a cube of side length cos θ , and the claim follows. 2. The second part follows using the same idea. This time, we consider three directions y1, y2, y3 in the plane such that y2 is obtained by a counterclockwise rotation R1 of angle π3 from y1, and y3 by a counterclockwise rotation R2 of angle π 3 from y2. We claim that for any N directions in the plane, we have min 1≤i≤3,1≤s≤N ∠(yi, ls) ≥ π 6N = α. The same trick as above shows that the 3N symmetric angle sectors of angle 2α cover the set of all directions and thus the unit circle. Now, the total length of arcs covered by these symmetric cones is 12Nα = 2π , and the claim follows. Therefore, the set of points is contained within the intersection of three strips of width w = cos ( 2π 3n(n−1) ) , which are orthogonal, respectively, to the directions yj . The intersection is a semiregular hexagon H . Since all three strips have the same width, we find that there are only two different side lengths, as in the figure below: This means that H is the intersection of two equilateral triangles of the same incenter O. Let A1, . . . , A6 be the vertices of H . We have then |A1A2| = |A3A4| = |A5A6| = a and |A2A3| = |A4A5| = |A6A1| = b. Moreover, by computing the 156 7 Geometry Solutions width of H , we find that the lengths a, b are constrained to satisfy a + b = 2√ 3 w. Assume from now on that a ≤ b.We divide the hexagonH by means of three segments OP,OQ,OR orthogonal to the three longest sides of the hexagon. The diameter of the pentagonOPA1A2Q is one of its diagonals, and one computes easily |PQ| = a+ b 2 , |OA1| = √ 3b2+(2a+b)2 2 √ 3 , and |PA2| ≤ |PQ| since ∠(PA2Q) ≥ ∠(PQA2) = π3 . Using the relation between a and b, we obtain that |PQ| ≤ 34 (a + b) ≤ √ 3 2 w, with equality only when a = b and the hexagon is regular. On the other hand, |OA1| ≤ 2w3 , with equality when a = 0. This implies that the diameter of each piece is at most√ 3 2 w, where w has the value found above. Comments 63 This version of Borsuk’s problem for finite sets and its higher- dimensional generalization for covering by cubes was given in: • L. Funar: A minimax problem in geometry (in Romanian), Proc. 1986 Nat. Conf. Geometry and Topology - Tîrgoviste, Univ. Bucharest, 91–96, 1988. Problem 3.38. Prove that any convex body in Rn of unit diameter having a smooth boundary can be partitioned into n + 1 parts of diameter d < 1. Solution 3.38. The first step is to observe that the claim holds for the unit ball Bn. We consider the regular spherical simplex obtained by projecting the regular simplex onto the sphere from its incenter. The n + 1 faces of the spherical simplex form a dissection into pieces of diameter smaller than 1, since there are no opposite points lying in the same face. Otherwise, one can compute explicitly the diameter of each piece. Consider the convex body F with smooth boundary W ⊂ Rn. The only point where the smoothness is used is when we assume that there exists a unique tangent hyperplane TpW at each point p of W . Moreover, there exists a unique unit vector vp in Rn, based at p, that is orthogonal to the tangent hyperplane TpW and points toward the half-space that does not contain F . This permits us to define a smooth map G : W → Sn−1 to the unit sphere centered at the origin O, usually called the Gauss map in differential geometry. We associate to the point p ∈ W the unique point G(p) ∈ Sn−1 for which the vectors OG(p) and vp are parallel, and so the tangent space TpW is parallel to the tangent hyperplane TG(p)Sn−1 at the point G(p) on the sphere. Since F is convex and W is smooth, the map G is a smooth diffeomorphism. Consider now the preimage by G of a partition of the unit sphere into pieces of diameter smaller than 1. This yields a partition of W into n + 1 pieces, which we denote by A1, A2, . . . , An+1. We claim that each piece Aj has diameter smaller than 1. Assume the contrary. We can replace the Aj by their closures, without modifying their diameters. Now there should exist two points x, y ∈ Aj at distance 1, which is the diameter of F . Let TxW and TyW be the tangent planes at W at x, y respectively. ThenTxW andTyW are orthogonal to the segment xy. Otherwise, a small perturbation x′ ∈ W of x in some tangent direction from TxW , making an obtuse angle with xy, will increase the length of |x′y|, but the diameter of W is no larger than 1. 7.2 Combinatorial Geometry 157 Therefore G(x) and G(y) are opposite points on the unit sphere, and moreover, they belong to the same set of the partition, but this is impossible, since the pieces of the partition are of diameter strictly smaller than 1. Comments 64 The question whether any set F in Rn can be partitioned into n + 1 pieces of strictly smaller diameter is known as Borsuk’s problem. Since the diameter of a set equals the diameter of its convex hull, it suffices to consider the problem the convex sets. The result in the problem above was obtained by H. Hadwiger in 1945. The case left open is that in which the convex body has corners. There was no significant progress in the subject until 1993, when J. Kahn and G. Kallai gave an unexpected counterexample by showing that the minimal number f (n) of pieces needed for a dissection into parts of strictly smaller diameter satisfies f (n) ≥ (1.1)√n for large n and that f (n) > n + 1 for n ≥ 1825. The subsets that they considered are finite sets of points. This bound was subsequently lowered by N. Alon and then, by A. Hinrichs, and C. Richter to n = 298. A thorough treatment of the subject is in the recent book of Boltyanski, Martini, and Soltan. However, the 4-dimensional case is still open. • V. Boltyanski, H. Martini, and P.S. Soltan: Excursions into Combinatorial Geom- etry, Springer, Universitext, 1997. • H. Hadwiger: Überdeckung einer Menge durch Mengen kleineren Durchmesser, Commentarii Math. Helvet. 18 (1945/1946), 73–75, 19 (1946/1947), 71–73. • A. Nilli: On Borsuk’s problem, Jerusalem combinatorics ’93, 209–210, Contemp. Math., 178, Amer. Math. Soc., Providence, RI, 1994. • J. Kahn and G. Kalai: A counterexample to Borsuk’s conjecture, Bull.Amer. Math. Soc. 29 (1993), 1, 60–62. Problem 3.39. Let D be a convex body in R3 and let σ(D) = supπ area(π ∩ D), where the supremum is taken over all positions of the variable plane π . Prove that D can be divided into two parts D1 and D2 such that σ(Di) < σ(D). Solution 3.39. Choose a point in the interior ofD and a sequence of chords l1, l2, . . . , ln, . . . of D passing through this point, the chords being dense in D. Choose also a sequence ε of positive numbers decreasing rapidly to zero. Let Cn be the set of points at distance less than εn from ln. By choosing ε small enough, we can ensure that σ(Cn) < 2−nσ (D). Set D1 = ∪∞n=1Cn and D2 = D − D1. Then σ(D1) <∑∞ n=1 σ(Cn) < σ(D). It is also clear that for any plane π , area(π ∩ D2) < σ(D). But D2 is a compact and area(π ∩ D2) is a continuous function on the plane π . The set of planes that intersect D can be parameterized by a point in D (which will be considered the origin) times the projective space RP 2, and thus by a compact set. Thus area(π ∩ D2) achieves its minimum, which will therefore be strictly smaller than σ(D). Comments 65 The problem was proposed by L. Funar and the solution above was given by C.A. Rogers. A more interesting question is to consider dissections into convex subsets. The additional convexity assumption implies that the boundary structure of 158 7 Geometry Solutions each piece (at the interior of the body) should be polyhedral. In particular, the minimal number of convex pieces with smaller σ , which the unit ball requires, is 3. If the body D is strictly convex (meaning that the open segment determined by two of its boundary points is contained in the interior of D) or D is a polyhedron without parallel edges, then we can show that four pieces suffice. By a recent result of M. Meyer, two plane sections of maximal area cannot be disjoint. Fix a maximal-area cross-section F and let F(ε) be the set of points of D, at distance less than ε from F . The complement of F(ε) has two components, A+ and A−, which are of strictly smaller σ , because there is no maximal-area plane section contained completely in D − F(ε). Further, F(ε) can be subdivided, by a hyperplane orthogonal to F , into two pieces of smaller σ if ε is small enough. However, we don’t know whether three pieces suffice for an arbitrary convex body. A related problem is whether we have σ(D) ≤ 14 area(∂D), where ∂D denotes the boundary of D. This would imply that the number of maximal area planar sections with disjoint interiors is at most 4, with equality for the regular simplex. • L. Funar: Problem E 3094, Amer. Math. Monthly 92 (1985), 427. • M. Meyer: Two maximal volume hyperplane sections of a convex body generally intersect, Period. Math. Hungar. 36 (1998), 191–197. Problem 3.40. If we have k vectors v1, v2, . . . , vk in Rn and k ≤ n + 1, then there exist two vectors making an angle θ with cos θ ≥ − 1 k−1 . Equality holds only when the endpoints of the vectors form a regular (k − 1)-simplex. Solution 3.40. Assume that the vectors vj are unit vectors. Then ∑ 1≤i 0 by a simple computation. Let ε2 = 1 − 6 √ 3 (1+√5)2 . 1. If 1 ≤ b ≤ 14 , set f (x) = F(x, b). Then we observed above that f ′(x) is increasing, and since f ′(0) = (1 + √5)y − 1 y2−2y+2 > 0, then f ′(x) > 0 for any x, and thus f (x) is increasing. In particular, F(a, b) ≥ F(0, b) ≥ 0, provided that 1 4 ≤ b ≤ 1. 2. If 0 ≤ a ≤ 14 and 0 ≤ y ≤ ε, we have f ′(x) = (1 + √5)b + x√ 1 + x2 − 1 − x √ (1 − x)2 + (1 − b)2 ≤ (1 + √5)ε + x√ 1 + x2 − 1 − x√ x2 − 2x + 2 . Thus f ′( 14 ) < 0. As f ′ is increasing, we have f ′(x) < 0, for x ∈ [0, 14 ] , and so f is decreasing on [ 0, 14 ] . But f ( 1 4 ) = F ( 14 , b ) = F (b, 14 ) ≥ 0, by the previous case. Thus F(a, b) ≥ 0, for 0 ≤ a ≤ 14 and 0 ≤ y ≤ ε. 3. Since F is symmetric, it suffices to check now the case ε ≤ a ≤ 14 and ε ≤ b ≤ 14 . Actually, we will use only the fact that ab ≥ ε2. This amounts to saying 162 7 Geometry Solutions that the area S of the triangle OAB is S = 1−ab2 ≤ 1−ε 2 2 . Any triangle satisfies the inequality S ≤ p2 3 √ 3 , obtained from the Heron formula for the area by the means inequality. In our case, we have r = S p ≤ S√ 3 √ 3S ≤ √5 + 1. This settles our claim. Comments 67 The solution presented here is due to Abi-Khuzam and Barbara: • L. Funar: Problem 6477, Amer. Math.Monthly 81 (1984), 588. • F.Abi-Khuzam, R. Barbara: A sharp inequality and the inradius conjecture, Math. Inequalities and Appl. 4 (2001), 323–326. Problem 3.44. Let P be a point in the interior of the tetrahedron ABCD, with the property that |PA| + |PB| + |PC| + |PD| is minimal. Prove that ÂPB = ĈPD and that these angles have a common bisector. Solution 3.44. Let fA(x) = |AX|. Then fA : E3 → E3 is a smooth function on E3 − {A}. Moreover, the gradient function (∇fA)(X) is the unit vector of direction AX, namely AX|AX| . Consider now the function g = fA + fB + fC + fD for P ∈ int(ABCD). Then g having a minimum at P implies that (∇g)(P ) = ∑ (∇fA)(P ) + ∑ (∇fB)(P ) + ∑ (∇fC)(P ) + ∑ (∇fD)(P ) = 0. If a, b, c, d are the unit vectors of directions AP,BP,CP,DP , then the previous conditions reads a+ b+ c+ d = 0, or alternatively, −(a+ b) = c+ d. Observe now that (a+b) is the direction of the bisector of the angle ÂPB, while c+d is the direction of the bisector of ĈPD. Thus these two angles have a common bisector. On the other hand, we have 〈a + b, a + b〉 = 〈c + d, c + d〉, and because |a| = |b| = |c| = |d|, by simplifying the terms, we obtain that 〈a, b〉 = 〈c, d〉. This is equivalent to saying that ÂPB = ĈPD. Problem 3.45. Let OA1, . . . , OAn be n linearly independent vectors of lengths a1, . . . , an. We construct the parallelepiped H having these vectors as sides. Then consider the n altitudes in H as a new set of vectors and further, construct the paral- lelepiped E associated with the altitudes. If h is the volume of H and e the volume of E, then prove that he = (a1 . . . an)2. Solution 3.45. Let vi = OAi and wi be the altitude vectors. We have the relations 〈wj , vi〉 = δij‖vi‖2, where δij is Kronecker’s symbol and 〈, 〉 is the scalar product. The linear functionals fi(vj ) = δij‖vi‖2 are linearly independent, and thus the dual vectors wj are linearly independent. Thus E is well defined. 7.2 Combinatorial Geometry 163 Moreover, let A be the matrix consisting of the components of the vi’s and let B be the matrix consisting of the components of the wj ’s. It follows that AB� = diag (‖v1‖2, . . . , ‖vn‖2 ) , where diag(c1, . . . , cn) denotes the diagonal matrix with given entries. Therefore vol(E)vol(H) = | det A| · | det B�| = (a1 . . . an)2. Problem 3.46. Let F be a symmetric convex body in R3 and let AF,λ denote the family of all sets homothetic to F in the ratio λ that have only boundary points in common with F . Set hF (λ) for the greatest integer k such that AF,λ contains k sets with pairwise disjoint interiors. Prove that hF (λ) ≤ (1 + 2λ) 3 − 1 λ3 . Solution 3.46. Set Bλ = ∪H∈AF,λH . We prove first that Bλ ⊂ (1 + 2λ)F, where αF denotes the homothety of F in ratio α. If v is a point of the boundary ∂F , let v′ be given by |ov′| = (1 + λ)|ov|, where o is the symmetry center of F . Set Fv for the set in AF,λ having center at v′. Let x be a point on the boundary of Fv , ox ∩ ∂F = {a}, ov ∩ ∂Fv = {q}, and let vx′′ be parallel to qx, with x′′ ∈ ∂F . Then q̂v′x = v̂′xo + v̂′ox ≥ v̂′ox, so that v̂ox ≤ v̂ox′′ = q̂v′x. Since F is convex, the intersection of the two segments |oa| ∩ |vx′′| is nonempty and contains at least one point, say b. Then |vx′′| ⊂ F , b ∈ F , b ∈ |ox|. Since ‖oa‖ ‖ox‖ ≥ ‖ob‖ ‖ox‖ = ‖ov‖ ‖oq‖ = 1 (1 + 2λ) , the point x belongs to (1 + 2λ)F . Now assume that we have k subsetsFi ∈ AF,λ that have pairwise disjoint interiors. Then their union is contained in Bλ, and all of them are disjoint from F . This means that their union is contained in (1 + 2λ)F − F . Their total volume cannot therefore exceed vol((1 + 2λ)F − vol(F ), and thus their number is bounded by 1 vol(F ) ( vol(1 + 2λ)F − vol(F )) = (1 + 2λ) 3 − 1 λ3 . Comments 68 Hadwiger considered the problem of finding estimates for hF (1) (which is called the Hadwiger number of F ) and proved in 1957 that 164 7 Geometry Solutions n2 + n ≤ hF (1) ≤ 3n − 1 for a convex body F in Rn. Grünbaum conjectured in 1961 (and proved for n = 2) that for any r satisfying these inequalities there exists a convex body F such that hF (1) = r . More generally, the proof used above shows that hF (λ) ≤ (1 + 2λ) n − 1 λn . We have equality above only if 1 λ ∈ Z+ and F is a parallelohedral body. The result is due to V. Boju and L. Funar. There are no good estimates from below except in dimension 2. In fact, if n = 2, the function λhF (λ) approaches an interesting invariant of the oval F , called the intrinsic perimeter. Define the norm ‖ ∗ ‖F by means of ‖xy‖F = ‖xy‖‖oz‖ , where z is a point of the boundary ∂F such that xy and oz are parallel. This is called the Minkowski norm (or metric) defined by the oval F . The intrinsic perimeter p(F) of ∂F is the limit of the perimeters of polygons inscribed in F that approach ∂F . For instance, the intrinsic perimeter of a circle is 2π and that of a square is 8. Golab in 1932 and Reshetnyak in 1953 showed that the intrinisc perimeter satisfies the inequalities 6 ≤ p(F) ≤ 8 with equality in the left side only when F is a hexagon, and on the right side only when F is a rectangle. Then it was proved by V. Boju and L. Funar that the intrinisc perimeter is related to Hadwiger numbers by means of the formula p(F) = 2 lim λ→0 λhF (λ), and moreover, we have 3 + 3 λ ≤ hF (λ) ≤ 4 + 4 λ , with equality on the left side only if 1 λ ∈ Z+ and F is an affine regular hexagon. More about packing and covering invariants of convex bodies can be found in the recent book of Böröckzy, and a complete survey on the Minkowski geometry in the book of Thompson. • V. Boju and L. Funar: Generalized Hadwiger numbers for symmetric ovals, Proc. Amer. Math. Soc. 119 (1993), 931–934. • K. Böröczky Jr.: Finite packing and covering, Cambridge Tracts in Mathematics, 154, Cambridge University Press, Cambridge, 2004. • H. Hadwiger: Über Treffanzahlen be. translationsgleichen Eikörpern, Archiv Math. 8 (1957), 212–213. 7.2 Combinatorial Geometry 165 • A.C. Thompson: Minkowski geometry, Encyclopedia of Mathematics and Its Ap- plications, 63. Cambridge University Press, Cambridge, 1996. Problem 3.47. Let � denote the square of equations |xi | ≤ 1, i = 1, 2, in the plane, and let A = (a1|a2) be an arbitrary nonsingular 2 × 2 matrix partitioned into two columns. We identify each column with a vector in R2. Prove that the following inequality holds: min A max x∈� ∣∣ ∣∣ 〈a1, x〉〈a2, x〉 det A ∣ ∣∣∣ = 1 2 , where 〈x, y〉 = x1x2 + y1y2 is the usual scalar product. Solution 3.47. Let f (A, x) be the function under minimax. Taking A0 = ( 1 1 −1 1 ) , we obtain max x∈� |f (A0, x)| = 1 2 . Thus it suffices to show that for general A, we have max x∈� |f (A, x)| ≥ 1 2 . This is equivalent to showing that | det A| ≤ 1, provided that max x∈� |〈a1, x〉〈a2, x〉| = 1 2 . Assume that the contrary holds, and so | det A| > 1. The set of points in the plane with coordinates (〈a1, x〉, 〈a2, x〉) ∈ R2, for x ranging in the square �, is a parallelogram π centered at the origin O and having area 4| det A| > 4. According to our claim, the parallelogram π is contained in the planar region Z = {(u, v) ∈ R2; |uv| ≤ 12 } . Let us consider the point P = (0, s), where s > 0, is a boundary point of π , and let L ⊂ R2 be the lattice generated by the vectors (1/s, s/2) and (−1/s, s/2). The lattice has area 12 . Then, according to Minkowski’s theorem, the interior of π contains at least one point of the lattice L, different from the origin. Let this point be given by m(1/s, s/2) + n(−1/s, s/2) = ((m − n)/s, (m + n)s/2),where m, n ∈ Z. We can also suppose gcd(n,m) = 1. Since the interior of π is contained in Z, it follows that |m2 − n2| < 1 and thus m2 = n2. The only possibilities are (0,±s) and (±2/s, 0). The points (0,±s) are on the boundary of π and they are not convenient. If Q = (2/s, 0) ∈ int(π), then the midpoint S of the segment |PQ| should also belong to the interior of π . However, S = (1/s, s/2) belongs to the boundary curve of equation |uv| = 1/2 of the region Z containing π , so it cannot lie within the interior of π . This contradiction proves our claim. Problem 3.48. We denote by δ(r) the minimal distance between a lattice point and the circle C(O, r) of radius r centered at the origin O of the coordinate system in the plane. Prove that lim r→∞ δ(r) = 0. 166 7 Geometry Solutions Solution 3.48. Let G be a lattice point on the axis Ox such that |OG| = [r]. Draw a perpendicular line at G to Ox that intersects the circle C(O, r) at the point C situated between the lattice points A and B, in the upper half-plane. We choose B to be the point that lies in the interior of the disk D(O, r) of radius r . Notice that |AB| = 1, because these are neighboring lattice points on the same axis line. Let us now draw BD⊥AB, where D is again a lattice point, |BD| = 1, and D is situated outside the disk D(O, r). Since the problem asks for the value at the limit of δ(r) when r → ∞, we can assume that r > 4. Let now consider the line CF that is tangent to the circle C(O, r) and such that F ∈ BD. We show first that |BF | ≤ |BD|. Assume the contrary, namely that |BF | > |BD|. Then ĈFB < ĈDB ≤ ÂDB = π 4 . Furthermore, sin ĈFB = sin ÔCG = sin ( arcsin [r] r ) > r − 1 r > √ 2 2 = sin π 4 , if r > 4. Thus ĈFB > π4 , which is false. Consequently, |BF | ≤ |BD| = 1. The triangles CBF and OCG are similar, and therefore |BC| |OG| = |CG| |OG| = √ r2 − [r]2 2 < √ r2 − (r − 1)2 r < √ 2r r = √ 2√ r . Now it is clear that δ(r) ≤ |BC| ≤ |BC||BF | = √ 2√ r , for r > 4. Thus limr→∞ δ(r) = 0. Problem 3.49. Consider a curve C of length l that divides the surface of the unit sphere into two parts of equal areas. Show that l ≥ 2π . Solution 3.49. Let C′ be the symmetric (opposite, or antipodal) of C with respect to the center O of the sphere. If C∩C′ = ∅, then int C∩ int C′ = ∅ and thus the area of the sphere is strictly greater than the sum of areas of int C and int C′. By hypothesis, the latter is just the area of the sphere, which is a contradiction. Therefore, C ∩C′ contains at least one point, say P . The point P ′, the symmetric ofP with respect to the symmetry center ofC∩C′, should also belong toC∩C′, since the later is symmetric. Finally, note that the shortest curve on the sphere that joins two opposite points is an arc of a great circle, and since C is made of two different arcs joining P and P ′, it follows that l ≥ 2π . Comments 69 More generally, let K be a body with a symmetry center O and set r = minP∈∂K |OP |, where ∂K denotes the boundary surface of K . If C ⊂ ∂K is a curve that separates two regions on ∂K of the same area, then the length l of C satisfies l ≥ 2πr . 7.2 Combinatorial Geometry 167 Problem 3.50. Let K be a planar closed curve of length 2π . Prove that K can be inscribed in a rectangle of area 4. Solution 3.50. Let bK(θ) be the width of the figure K in the direction θ , identified with a vector θ ∈ S1. If R(K, θ) denotes the area of the rectangle circumscribing K and having one side parallel to θ , then R(K, θ) = bK(θ)bK ( θ + π 2 ) . Thus min θ R(K, θ) 1 2 ≤ 1 2π ∫ 2π 0 bK(θ) 1 2 bK ( θ + π 2 )1/2 dθ, with equality holding iff bK(θ)bK ( θ + π2 ) is constant. From the Cauchy–Schwartz inequality, one derives min θ R(K, θ) ≤ 1 4π2 (∫ 2π 0 bK(θ) dθ )2 , the equality holding iff K is a curve of constant width. We are able now to apply the following theorem of Cauchy, which expresses the length L(K) of the curve K in terms of the integral of its width with respect to all directions: L(K) = 1 2 ∫ 2π 0 bK(θ) dθ. This implies that min θ R(K, θ) ≤ 1 π2 L(K)2 ≤ 4, where the second inequality uses the hypothesis L(K) ≤ 2π . Comments 70 We can consider the parallelograms with a given acute angleφ instead of rectangles. The same argument shows that min θ Rφ(K, θ) ≤ cosφ π2 L(K)2. See also: • E. Lutwak: On isoperimetric inequalities related to a problem of Moser, Amer. Math. Monthly 86 (1979), 476–477. Problem 3.51. 1. Consider a family of plane convex sets with area a, perimeter p, and diameter d. If the family covers the area A, then there exists a subfamily with pairwise disjoint interiors that covers at least area λA, where λ = a a+pd+πd2 . 2. Assume that any two members of the family have nonempty intersection. Prove that there exists then a subfamily with pairwise disjoint interiors that covers area at least μA, where μ = a πd2 . 168 7 Geometry Solutions Solution 3.51. 1. If K is a member of the family and K(d) the set of points at dis- tance less than d from K , then it is known that area(K(d)) = a + pd + πd2. Let {K1,K2, . . . , Kn} be a maximal subfamily with disjoint interiors. Then every mem- ber of the family intersects some Ki , and thus it is contained in some Ki(d). Thus the family is contained within ⋃n i=1 Ki(d). Thus the area covered by all members of the family is at most n · area(Ki(d)) ≤ A, while the subfamily above covers na ≥ λA. 2. The diameter of the union of all sets is at most 2d, by the assumption. Its area is therefore at most πd2, by the well-known isoperimetric inequality. Thus λA ≤ a, and one single set will cover at least λA. Comments 71 It is still unknown whether the second claim holds without any addi- tional assumption on the sets. Problem 3.52. Let C be a regular polygon with k sides. Prove that for every n there exists a planar set S(n) ⊂ R2 such that any subset consisting of n points of S(n) can be covered by C, but S(n) itself cannot be included in C. Solution 3.52. Let r be the radius of the circle inscribed in C. For n ∈ Z+, let S(n) be the circle of radius r sec(π/2kn). 1. C cannot cover S(n). In fact, there is no disk of radius greater than r that can be included in C. Here is a proof for this somewhat obvious assertion. For a fixed radius R, let X(R) be the set of centers of circles of radius R included in C. Then X(R) is convex and invariant of rotations centered at the center O of C, of angle 2π/k. Thus X(R) is either empty or contains O. Thus R ≤ r . 2. Let P1, . . . , Pn ⊂ S(n) be arbitrary points of S(n). We show that C can cover P1, . . . , Pn. We fix a point Q in C at a distance of r sec π2kn from O. We place C such that their centers coincide and we rotate C around its center O such that Q runs through S(n). While Q travels along the circle, some of the points Pi will be in the interior of C and the others outside. However, we see that Pi /∈ C if and only if Q ∈ ⋃nj=1 Aj , where each Aj is an arc of a circle from outside C, which is of length π kn . More precisely, the arc Aj is determined by the inequalities 2 πj k ≤ P̂iOQ ≤ 2πj k + π kn . Therefore ⋃n j=1 Aj is a set of arcs of total length smaller than nkπ/kn · π and thus S −⋃nj=1 Aj has length π , and thus it is nonempty. Now rotate C until Q lies inside S −⋃nj=1 Aj . Then all the Pj belong to C, as claimed. 7.2 Combinatorial Geometry 169 Problem 3.53. Let M be a convex polygon and let S1, . . . , Sn be pairwise disjoint disks situated in the interior of M . Does there exist a partition M = D1 ∪ · · · ∪ Dn such that Di are convex disjoint polygons, each of which contains precisely one disk? Solution 3.53. Let Oi, ri be the center and the radius of the disk Si . For a point X in the plane, let us define hi(X) = XO2i − r2i . We consider Dk = {x ∈ M;hk(X) ≤ hi(X), for all i = 2, 3, . . . , n}. We will prove that the Di are convex and Si ⊂ Di . Fix k, for instance k = 1. Define Hi = {X ∈ R2;hk(X) ≤ hi(X)}. Then we claim that Hi are half-planes. Let i = 2 for simplicity. Choose the point P ∈ |O1O2| such that |PO1|2 − r21 = |PO2|2 − r22 . Take for instance |PO1| = |O1O2|2−(r21 −r22 ) 2|O1O2| . If l is the line perpendicular to O1O2 at P and X ∈ l, then we have |XO1|2 − r21 = |PO1|2 + |PX|2 − r21 = |PO2|2 + |PX|2 − r22 = |XO2|2 − r22 . If H˜2 is the half-plane that contains O1, then |XO1|2 − r21 ≤ |XO2|2 − r22 , for all X ∈ H˜2, |XO1|2 − r21 ≥ |XO2|2 − r22 for all X /∈ H˜2. Therefore, H2 is the half-plane H˜2. Consequently, Dk = M n⋂ i=1,i �=k Hn is a convex set as an intersection of convex sets, and Sk ⊂ Dk . It is clear that⋃ Di = M. Problem 3.54. Consider an inscribable n-gon partitioned by means of n − 2 nonin- tersecting diagonals into n− 2 triangles. Prove that the sum of the radii of the circles inscribed in these triangles does not depend on the particular partition. Solution 3.54. If we have a triangle inscribed in a circle of center O and radius R, of inradius r , then the distances from O to the sides satisfy the well-known formula ∑ cyclic d∗(O, side) = R + r. Here d∗ is the signed distance, which is positive on one half-plane determined by the side and negative on the other half-plane. We write the corresponding formulas for each triangle of the partition and sum up all the terms. Let d1, . . . , dn be the distances from O to the sides of the n-gon. When looking at the signed distances from O to a diagonal, we observe that each signed distance is considered twice: once with the plus sign and once with the opposite sign. When these terms are summed, they will cancel. We then obtain d1 + · · · + dn = (n − 2)R + r1 + · · · + rn, and thus the sum of inradii r1 + r2 + · · · + rn does not depend on the partition. 170 7 Geometry Solutions Problem 3.55. Prove that in an ellipse having semiaxes of lengths a and b and total length L, we have L > π(a + b). Solution 3.55. We cut the ellipse along the axes into four congruent pieces. We can further rearrange the four pieces by adding a square of side length b − a as in the figure below: We want to use the well-known Bonnesen’s isoperimetric inequality, which states that L2 ≥ 4πA for any planar convex domain of perimeter L and area A. The area of the new domain is A+ (b− a)2, where A was the area of the ellipse. Moreover, one knows that the area of the ellipse is A = πab. Applying this inequality to the domain pictured in the figure above, we obtain L2 > 4π2ab + 4π(b − a)2 > π2(a + b)2, whence the claim. Problem 3.56. Let F be a convex planar domain and F ′ denote its image by a homo- thety of ratio − 12 . Is it true that one can translate F ′ in order for it to be contained in F ? Can the constant 12 be improved? Generalize to n dimensions. Solution 3.56. Let ABC ⊂ F be a triangle of maximal area inscribed in F . Let O denote the geometric centroid of ABC. We use the homothety of center O and ratio − 12 that transforms F into F ′ and ABC into A′B ′C′. Since O is the centroid of ABC, we find that A′ is the midpoint of the segment |BC|, B ′ is the midpoint of |CA|, and C′ is the midpoint of |AB|. 7.2 Combinatorial Geometry 171 We will show that F ′ ⊂ ABC. Assume the contrary, namely that there exists a point X of F ′ that does not belong to the triangle ABC. By symmetry, it suffices to consider that X belongs to the half-plane determined by the line BC, which does not contain A. Since BC and B ′C′ are parallel, it follows that the area of XB ′C′ is strictly greater than the area of A′B ′C′. Let us consider the inverse image Y of X by the homothety of center O and ratio − 12 . Then Y belongs to F and the area of YBC is strictly greater than the area of ABC, which contradicts our assumptions. This proves that F ′ ⊂ ABC ⊂ F . The constant 12 is sharp in the case that F is a triangle. The same proof shows that the image F ′ of a convex domain F in Rn by the homothety of ratio − 1 n can be translated in order to be contained in F . The constant 1 n is sharp, with equality for the simplex. Problem 3.57. A classical theorem, due to Cauchy, states that a strictly convex poly- hedron in R3 whose faces are rigid, must be globally rigid. Here, rigidity means continuous rigidity, in the sense that any continuous deformation of the polyhedron in R3 that keeps the lengths of edges fixed is the restriction of a deformation of rigid Euclidean motions of three-space. Prove that a 3-dimensional cube immersed in Rn remains rigid for all n > 3. Solution 3.57. Three sides are common to a vertex and determine a three-dimensional space R3 ⊂ Rn. The 3-dimensional spaces determined by two adjacent vertices have two sides in common, and therefore they coincide. Thus, the entire cube is immersed in a 3-dimensional subspace of Rn, and therefore it remains rigid. The same argument shows that a rigid polyhedron with the property that exactly three sides meet at any vertex remains rigid after immersion in Rn, for all n > 3. Problem 3.58. Consider finitely many great circles on a sphere such that not all of them pass through the same point. Show that there exists a point situated on exactly two circles. Deduce that if we have a set of n points in the plane, not all of them lying on the same line, then there must exist one line passing through precisely two points of the given set. Solution 3.58. Assume that the contrary holds. Consider the partition of the sphere into (spherical) polygons, induced by the circles, and identify it with a polyhedron. Each vertex will have degree at least 6, since at least three great circles pass through each point. If V is the number of vertices, E the number of edges, and F the number of faces of the spherical polyhedron, then Euler’s formula states that V − E + F = 2. Moreover, by counting the edges in each face and summing over the faces, we obtain 2E = 3F3 + 4F4 + · · · ≥ 3F, where Fk is the number of faces with k sides. Finally, counting the edges adjacent to each vertex and summing over the vertices, we get 172 7 Geometry Solutions 2E = 6V6 + 7V7 + · · · ≥ 6V, where Vk is the number of vertices of degree k, because Vk = 0 when k ≤ 5. Thus 6(E + 2) = 6(V + F) ≤ 2E + 4E = 6E, which is absurd. Therefore, there exists a vertex of valence smaller than 6 in which exactly two great circles intersect. The second claim can be reformulated using projective duality: if we have n lines in the projective plane, not all passing through the same point, then there exists a point lying on exactly two of the lines. This is equivalent to the problem considered above. Comments 72 The problem was posed by Sylvester in 1839, and it was not settled until 1930, when T. Gallai gave a solution. The solution above is due to N. Steenrod. Moreover, it can be proved that the number of points where exactly two circles intersect is at least 3n/7; see for instance: • H.S.M. Coxeter: The classification of zonohedra by means of projective diagrams, J. Math. Pures Appl. (9) 41 (1962), 137–156. • G.D. Chakerian, Sylvester problem on collinear points and a relative,Amer. Math. Monthly 77 (1970), 164–167. Problem 3.59. Given a finite set of points in the plane labeled with +1 or −1, and not all of them collinear, show that there exists a line determined by two points in the set such that all points of the set lying on that line are of the same sign. Solution 3.59. We use projective duality as above and reformulate the question as follows: given n great circles on a sphere, to each one being assigned either +1 or −1, and not all passing through the same point, show that there exists a point lying on at least two circles such that all circles containing that point have the same sign. The circles determine a spherical polyhedron whose edges are labeled by +1 and −1. Assume that the result is not true. Then the edges around each vertex must have both the labels +1 and −1. By symmetry, the number of sign changes in the labels of the edges that one meets when traveling-cyclically around each vertex is at least 4. Denote by N the sum of all such sign changes over all vertices. Then N ≥ 4V . On the other hand, by counting the sign changes as we travel around the faces and summing over all faces, we should again obtain N . Further, the number of sign changes in a k-sided face is even and at most k, so that N ≤ 2F3 + 4F4 + 4F5 + 6F6 + 6F7 + · · · . Moreover, using Euler’s formula and 2E = 3F3 + 4F4 + · · · we obtain 4V − 8 = 4E − 4F = 2(3F3 + 4F4 + 5F5 + · · · ) − 4(F3 + F4 + · · · ) = 2F3 + 4F4 + 6F5 + · · · ≥ N ≥ 4V, which is a contradiction. Thus our claim follows. 7.2 Combinatorial Geometry 173 Comments 73 The result concerning the sign changes is valid more generally when some edges are allowed to be unlabeled, and as such it is known as Cauchy’s lemma. This is an essential ingredient in proving the famous rigidity theorem of Cauchy, which states that a convex polyhedron in R3 whose faces are rigid is globally rigid. Problem 3.60. If Q is a given rectangle and ε > 0, then Q can be covered by the union of a finite collection S of rectangles with sides parallel to those of Q in such a way that the union of every nonoverlapping subcollection of S has area less than ε. Solution 3.60. Observe first that every rectangle Q has a subset H that is the union of a finite collection � of rectangles with parallel sides such that: 1. any two members of � overlap; 2. area(H) > δ area(Q), where δ = exp(−1/ε); 3. area(A) < ε2 area(H), for any A ∈ �. If Q = [0, a] × [0, b], we consider rectangles Ai that are parallel and have one vertex at (0, 0) and the other vertex on the curve of equation xy = δ area(Q). Thus each rectangle has area δ. The curved triangle determined by the inequalities xy ≤ δ area(Q), x, y ≥ 0, will be denoted by T . Moreover, we consider a collection of such rectangles such that their union H satisfies area(H) > 1+ε ε area(T ) = δ area(Q). Further, area(A) = δ area(Q) < ε area(T ). Set Q0 = Q, H0 = H , �0 = � and assume that area(Q) = 1, for simplicity. Then Q0 − H0 is a finite union of parallel rectangles Qi,0 and area(Q0 − H0) < (1 − δ). Choose in each rectangle Qi a set of type Hi,0 as above. The union of all such Hi,0 and H0 will be denoted by H1, and the collection of rectangles will be denoted by �1. Then Q − H1 is also a finite union of parallel rectangles Qi,1 and area(Q − H1) < (1 − δ)2 area(Q). We continue this process by defining the sets Hj and the collections �j of rectangles such that Q − Hj is a finite union of rectangles Qi,j whose complement has total area area(Q − Hj) < (1 − δ)j area(Q). We stop this process at the stage n, where n is large enough that (1 − δ)n area(Q) < ε/2. Consider then the union � of the collections �n obtained so far with the set of rectangles in which Q − Hn decomposes. If we consider a subfamily of disjoint rectangles from �, then there is at most one element Ai from any collection �i,j covering Hi,j , because all elements in �i,j overlap each other. Thus the total area covered by these Ai is at most ∑ i,j ε 2 area(Hi,j ) < ε 2 . Since the rectangles from Q − Hn contribute at most ε2 , we are done. 174 7 Geometry Solutions Problem 3.61. Prove that the 3-dimensional ball cannot be partitioned into three sets of strictly smaller diameter. Solution 3.61. Assume that we can divide the unit ball into three pieces of smaller diameter and let X1, X2, X3 be the partition that it induces on the boundary sphere. Thus the diameter of each piece Xi is smaller than 1 − 10h, for some h > 0. We choose a very thin triangulation of the sphere with triangles that have edges smaller than h. Denote by Nj the union of those spherical triangles of the triangulation above that intersect the set Xj nontrivially. It follows that N1, N2, N3 are closed subsets of the sphere that are domains, i.e., they are closures of open subsets. In particular, the boundary of each domain Nj is the union of several piecewise linear circles. Moreover, the union N1 ∪ N2 ∪ N3 covers the sphere. Since each triangle has a small size, the diameter of Nj is at most 1 − 8h < 1. In particular, opposite points cannot belong simultaneously to the same set Nj . Let us denote by x∗ the point of the sphere opposite the point x. We know that the set N1 and its symmetric N∗1 (which is the set of those points x∗ for which x ∈ N1) should be disjoint; otherwise, there would exist in N1 a pair of points at distance 1. Consider the circles determined by the boundary circles of both N1 and N∗1 . The circles associated to Nj are disjoint and distinct from the circles associated to N∗1 , because N1 ∩ N∗1 = ∅. Thus, there is an even number of such circles. Observe that k disjoint circles on the sphere divide the surface of the sphere into k + 1 connected complementary regions. Thus, the collection of boundary circles associated to N1 and N∗1 will split the surface of the sphere into an odd number of connected regions, which we denote by R1, R2, . . . , R2m+1. Consider the symmetric R∗j of the region Rj . There are two possibilities: either R∗j is another connected region Rk that is disjoint from Rj , or else R∗j = Rj , meaning that Rj is symmetric. The second case happens, for instance, when the region Rj is bounded by two circles, one coming from N1 and the other from N∗1 . If there were no symmetric regions Rj , then the total number of regions would be even, which would contradict our previous claim. Thus there must be at least one connected symmetric region Rj . We consider a point x ∈ Rj and its symmetric x∗, which must also belong to Rj , since the region is symmetric. If x ∈ N1, then all points of the sphere that are connected to x by a path not hitting a boundary circle of N1 must also belong to N1. Thus, the connected component Rj will be contained in N1, and so N1 will contain pairs of symmetric points, contradiction. This implies that x �∈ N1 and hence x ∈ N2 ∪ N3. Let us assume that x ∈ N2. Then its symmetric x∗ is not in N2, since N2 has diameter smaller than 1 and thus x∗ ∈ N3. Further, Rj is connected and thus there exists a path γ within Rj that connects x to x∗. One knows that N2 and N3 are domains covering the path γ . Moreover, there exists a point z ∈ γ that belongs to both N2 and N3. In fact, consider the point z ∈ γ closest to x∗ and that belongs to N2. There exists such a point since N2 is closed. If z = x∗, then both x and x∗ belong to N2, contradiction. If z �= x∗, then the points on the right of z should belong to N3 and thus z belongs to the closure of N3 and thus to N3. 7.3 Geometric Inequalities 175 Now z ∈ Rj and thus its symmetric z∗ also belongs to Rj . This means that z, z∗ �∈ N1. But z∗ can belong neither to N2, since N2 does not contain opposite points, nor to N3, for the same reason. This contradiction proves the claim. Comments 74 More generally, it is known that the unit n-ball (or equivalently, the (n−1)-sphere in Rn) cannot be partitioned into n pieces of strictly smaller diameter. This is related to Borsuk’s problem discussed in comments to Problem 3.38. The result was proved by Lusternik and Shnirelman in 1930 and, independently, by Borsuk in 1932, but the higher-dimensional case n ≥ 4 requires deeper methods from topology, and this opened a new chapter in algebraic topology. The interested reader might consult the survey of Steinlein, which presents both historical remarks and a large number of references. • K. Borsuk: Über die Zerlegung einer n-dimensionalen Vollkugel in n-Mengen, Verh. International Math. Kongress Zürich 1932, 192. • L.A. Lusternik and L.G. Shnirelman: Topological Methods for Variational Prob- lems, Moscow, 1930. • H. Steinlein: Borsuk’s antipodal theorem and its generalizations and applications: a survey, Topological Methods in Nonlinear Analysis, 166–235, Sém. Math. Sup., 95, Presses Univ. Montreal, Montreal, 1985. • H. Steinlein: Spheres and symmetry: Borsuk’s antipodal theorem. Topol. Methods Nonlinear Anal. 1 (1993), 1, 15–33. 7.3 Geometric Inequalities Problem 3.62. If a, b, c, r, R are the usual notations in the triangle, show that 1 2rR ≤ 1 3 (∑ 1 a )2 ≤ ∑ 1 a2 ≤ 1 4r2 . Solution 3.62. We derive from ∑ ( 1 a − 1 b )2 ≥ 0 that ∑ 1 bc ≤ 1 3 (∑ 1 a )2 ≤ ∑ 1 a2 . Moreover, the usual Cauchy–Schwarz inequality shows that (p − a)(p − b) ≤ c24 , and its analogues. Then we have 1 2rR = 2p abc = a + b + c abc = 1 ab + 1 bc + 1 ca ≤ 1 3 (∑ 1 a )2 ≤ ∑ 1 a2 ≤ 1 4 (∑ 1 (p − b)(p − c) ) = p 4(p − a)(p − b)(p − c) = 1 4r2 . Problem 3.63. If a, b, c are the sides of a triangle, then prove that (b+c) 2 4bc ≤ mawa and b2+c2 2bc ≤ maka , where ma,wa, ka denotes the respective lengths of the median, bisector, and altitude issued from A. 176 7 Geometry Solutions Solution 3.63. We have the following formulas computing ma,wa , and ka : ma = 12 √ 2 ( b2 + c2)− a2, wa = √ bc b + c √ (b + c)2 − a2, ka = bc b2 + c2 √ 2 ( b2 + c2)− a2. From the triangle inequality, we have a2 > (b − c)2, which implies furthermore that (b + c)2 − a2 < 4bc (b − c) 2 4bc ≤ (b − c) 2 (b + c)2 − a2 , with equality only if b = c. Adding 1 in both members yields (b + c)2 4bc ≤ 2b 2 + 2c2 − a2 (b + c)2 − a2 ; hence b + c 2 √ bc ≤ √ 2b2 + 2c2 − a2 (b + c)2 − a2 . This leads to (b + c)2 4bc ≤ b + c 2 √ bc √ 2(b2 + c2) − a2 (b + c)2 − a 2 = ma wa . For the second inequality, it suffices to observe that ma ha ≥ ma ka = b 2 + c2 2bc . Notice that equality in the first case implies b = c, while for the second case either b = c or a2 = b2 + c2. Problem 3.64. If S(x, y, z) is the area of a triangle with sides x, y, z, prove that √ S(a, b, c) +√S(a′, b′, c′) ≤ √S(a + a′, b + b′, c + c′). Solution 3.64. We make the following change of variables: s = (a + b + c)/2, t = s − a, u = s − b, and v = s − c. Using Heron’s formula, the inequality becomes 4√ stuv + 4√s′t ′u′v′ ≤ 4√(s + s′)(t + t ′)(u + u′)(v + v′). We prove first that for positive x, x′, y, y′, we have √ xy +√x′y′ ≤ √(x + x′)(y + y′). In fact, this follows from xy + x′y′ + 2√x′y · y′x ≤ xy + x′y′ + xy′ + yx′, and equality holds when the geometric means of x′y and y′x are equal. By applying the last inequality to x = √st and y = √uv twice, we obtain the claim. Moreover, the equality is attained only when t/t ′ = u/u′ = v/v′ = s/s′, which amounts to a/a′ = b/b′ = c/c′. 7.3 Geometric Inequalities 177 Problem 3.65. It is known that in any triangle we have the inequalities 3 √ 3r ≤ p ≤ 2R + (3√3 − 4)r, where p denotes the semiperimeter. Prove that in an obtuse triangle, we have (3 + 2√2)r < p < 2R + r. Solution 3.65. We have a + b c = cos A−B 2 sin C2 . Since C > π2 , we have sin C 2 > √ 2 2 and thus c √ 2 > a + b. Therefore 2p √ 2 > (a + b)(√2 + 1) ⇒ 8p2 > (3 + 2√2)(a + b)2 > 4(3 + 2√2)ab > 8(3 + 2√2)S, where S is the area. This implies that p > ( 3 + 2√2)S p = (3 + 2√2)r. Finally, recall that r = (p − c) tan C2 > (p − c) and so p < c+ r = 2R sin C + r < 2R + r . Problem 3.66. Prove the Euler inequality R ≥ 2r. Solution 3.66. |OI |2 = R2 − 2Rr. Problem 3.67. Prove that in a triangle we have the inequalities 36r2 ≤ a2 + b2 + c2 ≤ 9R2. Solution 3.67. We have p 3 = 1 3 (p − a + p − b + p − c) ≥ 3√(p − a)(p − b)(p − c) = 3 √ S2 p = 3 √ r2p, and hence p2 ≥ 27r2. Thus a2 + b2 + c2 ≥ 4 3 p2 ≥ 36r2. The second inequality is a consequence of the following identity: |OH |2 = 9R2 − (a2 + b2 + c2), where O is the circumcenter and H the orthocenter. 178 7 Geometry Solutions Problem 3.68. 1. Let ABC and A′B ′C′ be two triangles. Prove that a2 a′ + b 2 b′ + c 2 c′ ≤ R2 (a ′ + b′ + c′)2 a′b′c′ . 2. Derive that a2 + b2 + c2 ≤ 9R2, cosA cosB cosC ≤ 1 8 . Solution 3.68. 1. The inequality from the statement is equivalent to a′2 + b′2 + c′2 − 2(b′c′ cosA′′ + a′c′ cosB ′′ + b′a′ cosC′′) ≥ 0, where A′′ = π − 2A,B ′′ = π − 2B,C′′ = π − 2C. Moreover, this can be written as (a′ − b′ cosC′′ − c′ cosB ′′)2 + (b′ sin C′′ − c′ sin B ′′)2 ≥ 0. A necessary condition for equality is the vanishing of the second square, i.e., a′ sin A′′ = b ′ sin B ′′ = c ′ sin C′′ . But A′′ ∈ (−π, π); therefore A′′, B, C′′ are the angles of a triangle that is similar to A′B ′C′. Therefore, we obtain equality if and only if ABC is an acute triangle and A′B ′C′ is similar to the orthic triangle of the triangle ABC. 2. If A′B ′C′ is equilateral, then a2 + b2 + c2 ≤ 9R2, which is equivalent to sin2 A+ sin2 B + sin2 C ≤ 94 . Observe that ∑ sin2 A = 2+2 cosA cosB cosC, and therefore we obtain cosA cosB cosC ≤ 18 . Problem 3.69. Prove that the following inequalities hold in a triangle: 4 ∑ cyclic hAhB ≤ 12S √ 3 ≤ 54Rr ≤ 3 ∑ cyclic ab ≤ 4 ∑ cyclic rArB. Solution 3.69. 1. We have 4 ∑ cyclic rArB = ⎛ ⎝ ∑ cyclic a ⎞ ⎠ 2 ≥ 3 ∑ cyclic ab. 2. In order to prove that 18Rr ≤ ∑ cyclic ab, we start from the identity 1 ha + 1 hb + 1 hc = 1 r . 7.3 Geometric Inequalities 179 Thus 2 abc 4R ( 1 a + 1 b + 1 c ) = 2S ( 1 a + 1 b + 1 c ) = ha + hb + hc ≥ 9∑ 1 ha ≥ 9r, leading to the inequality we wanted. 3. Recall that Jensen’s inequality states that if f is a concave (i.e., f ′′(x) ≤ 0) smooth real function on some interval J , then λ1f (x1) + λ2f (x2) + · · · + λnf (xn) ≤ f (λ1x1 + λ2x2 + · · · + λn + xn) for all xj ∈ J and λi ≥ 0 such that λ1 + λ2 + · · · + λn = 1. Notice now that the function cos is concave on [ 0, π2 ] , and by Jensen’s inequality we have cos A 2 + cos B 2 + cos C 2 ≤ 3 √ 3 2 . Therefore, by the means inequality, cos A 2 cos B 2 cos C 2 ≤ ( cos A2 + cos B2 + cos C2 3 )3 ≤ 3 √ 3 8 . We have then S = abc 4R = 4rR cos A 2 cos B 2 cos C 2 ≤ 3 √ 3 2 rR. 4. We have ∑ cyclic hAhB = hAhBhC ( 1 ha + 1 hb + 1 hc ) = 2S 2 R 1 r ≤ 3√3S. Problem 3.70. Prove that in an any triangle ABC, we have √ 1 + 8 cos2 B sin A + √ 1 + 8 cos2 C sin B + √ 1 + 8 cos2 A sin C ≥ 6. Solution 3.70. By applying the means inequality, the left-hand side is greater than 3 √√√√√3 ⎛ ⎝ ∏ cyclic ( √ 1 + 8 cos2 A ⎞ ⎠ ∏ cyclic 3 √ 1 sin A , and it is sufficient to prove that F(A,B,C) = ∏ cyclic 1 + 8 cos2 A sin2 A ≥ 64. 180 7 Geometry Solutions Let us find the extremal points of F , that satisfy ∂F ∂A = ∂F ∂B = ∂F ∂C . We have 1 F · ∂F ∂A = − 18 cot A 1 + 8 cos2 A, and thus we have sin B cosC(1 + 8 cos2 B) = sin C cosB(1 + 8 cos2 C). This implies that sin(B − C) + 4 cosB cosC(sin 2B − sin 2C) = 0 and hence sin(B − C)(1 − 8 cosA cosB cosC) = 0. We claim that this happens only if A = B = C. In fact, if the triangle is obtuse, then cosA cosB cosC ≤ 0 < 1 8 and hence B = C = A. Otherwise, cos is positive and cosA + cosB + cosC = 1 + 4 sin A 2 sin B 2 sin C 2 = 1 + 4 ∏ cyclic √ (p − b)(p − c)bc = 1 + r R ≤ 3 2 , and so by the means inequality, cosA cosB cosC ≤ ( cosA + cosB + cosC 3 )3 ≤ 1 8 . Equality holds above only if the triangle is equilateral. This shows that the only extremal points are A = B = C = π3 , and one verifies that it corresponds to a minimum, so that F(A,B,C) ≥ 64, as claimed. Problem 3.71. Let P be a point in the interior of the triangle ABC. We denote by Ra,Rb, Rc the distances from P to A,B,C and by ra, rb, rc the distances to the sides BC,CA,AB. Prove that ∑ cyclic R2a sin2 A ≤ 3 ∑ cyclic r2a , with equality if and only if P is the Lemoine point (i.e., the symmedian point). 7.3 Geometric Inequalities 181 Solution 3.71. Let G be the centroid of the poder triangle DEF of P , obtained by projecting P to the sides of ABC. We have then 3 ∑ cyclic r2a = 3(|DG|2 + |EG|2 + |FG|2 + 3|PG|2) ≥ 3(|DG|2 + |EG|2 + |FG|2) = |DE|2 + |EF |2 + |FD|2 = ∑ cyclic R2a sin2 A. We obtain equality when P is the centroid of its poder triangle. Therefore P is the symmedian point. Comments 75 We can generalize this type of inequality. We have, for instance, xR21 + yR21 + zR21 ≥ a2yz + b2xz + c2xy x + y + z for every x, y, z ∈ R such that x + y + z > 0. The equality holds above if and only if x F1 = y F2 = z F3 , where F1, F2, F3 denote the areas of BPC, APC, and APB, respectively. Comments 76 With every inequality �(Ra,Rb, Rc, a, b, c) ≥ 0, valid for any tri- angle of sides a, b, c, we can associate a dual inequality in a new set of variables, �(ra, rb, rc, Ra sin A,Rb sin B,Rc sin C) ≥ 0, obtained by replacing the original triangle by the poder triangle associated to P . Then the distances from P to the vertices of the poder triangle are ra, rb, rc and the sides of the poder triangle are Ra sin A,Rb sin B,Rc sin C. For instance, the dual of the inequality from the previous remark is xr2a + yr2b + zr2c ≥ yzR2a sin2 A + xzR2b sin2 B + xyR2c sin2 C x + y + z . If x = y = z, then this is the inequality of our problem. Applying the inequality from the previous remark to the right-hand side, we also obtain ∑ yzR21 sin 2 A x + y + z ≥ 4S2 a2 x + b2 y + c2 z . Problem 3.72. Prove the inequalities 16Rr − 5r2 ≤ p2 ≤ 4R2 + 4Rr + 3r2. Solution 3.72. Let I be the incenter of the triangle and H be the orthocenter. Direct calculations show that |IH |2 = 2r2 − 4R2 cosA cosB cosC, 182 7 Geometry Solutions and replacing the term 4R2 cosA cosB cosC = p2 − (2R + r)2, we find that |IH |2 = 4R2 + 4Rr + 3r2 − p2. The next step is to use Euler’s theorem, which states that O,G, and H are collinear and |OG| |GH | = 1 2 , where G denotes the centroid of the triangle. We compute the length of |GI | using the triangle HIO as follows: |GI |2 = 2 3 |IO|2 + 1 3 |IH |2 + 2 9 |OH |2. We substitute |OI |2 = R2 − 2Rr, |OH |2 = 9R2 − (a2 + b2 + c2) and derive that |GI |2 = r2 − 1 3 p2 + 2 9 ( a2 + b2 + c2). Moreover, 4p2 = 2(a2 + b2 + c2)+ 16Rr + 4r2, and so p2 − 16Rr + 5r2 = 8|GI |2. Thus the inequality follows. Comments 77 Using 2p2 = 2r(4R+ r)+ a2 + b2 + c2 and S = rp, we derive also the inequalities 12r(2R − r) ≤ a2 + b2 + c2 ≤ 4r2 + 8R2, r3(16R − 5r) ≤ S2 ≤ r2(3r2 + 4rR + 4R2). • O. Bottema, R.Ž. Djordjevic´, R.R. Janic´, D.S. Mitrinovic´, and P.M. Vasic´: Geo- metric Inequalities, Wolters–Noordhoff Publishing, Gröningen 1969. Problem 3.73. Prove the following inequalities, due to Roché: 2R2 + 10Rr − r2 − 2(R − 2r) √ R2 − 2Rr ≤ p2 ≤ 2R2 + 10Rr − r2 + 2(R − 2r) √ R2 − 2Rr. Solution 3.73. Suppose that a ≥ b ≥ c. By direct calculation, we compute the area of this triangle as follows: area(HIO) = 2R2 sin ( B − C 2 ) sin ( A − C 2 ) sin ( A − B 2 ) = (b − c)(a − c)(a − b) 8r . 7.3 Geometric Inequalities 183 Expressing R and r in terms of a, b, c, we can conclude with the identity (area(HIO))2 = −p4 + 2(2R2 + 10Rr − r2)p2 − r(4R + r)3. Since this binomial in p2 is positive, the value of p2 lies between the two real roots of the polynomial, which yields the claim. Comments 78 W.J. Blundon gave another proof, using the following identity: −p4 + 2(2R2 + 10Rr − r2)p2 − r(4R + r)3 = 1 4r2 (a − b)2(b − c)2(c − a)2. • D.S. Mitrinovic´, J.E. Pecˇaric´, V. Volenec: Recent Advances in Geometric Inequal- ities, Math. Appl. (East European Series), 28, Kluwer Academic Publ., Dordrecht, 1989. 8 Analysis Solutions Problem 4.1. Prove that z ∈ C satisfies |z| − �z ≤ 12 if and only if z = ac, where|c − a| ≤ 1. We denote by �z the real part of the complex number z. Solution 4.1. We have the identity |ac| − � ac = 1 2 |c − a|2 − 1 2 (|c| − |a|)2, and the “if” part follows. For the converse, we consider a, c such that |a| = |c| = |z|1/2. Problem 4.2. Let a, b, c ∈ R be such that a+2b+3c ≥ 14. Prove that a2+b2+c2 ≥ 14. Solution 4.2. From (a − 1)2 + (b − 2)2 + (c − 3)2 ≥ 0, we obtain a2 + b2 + c2 ≥ 2(a + 2b + 3c) − 14 ≥ 14. Comments 79 More generally, if wi ∈ R+ and ai ∈ R are such that ∑ni=1 aiwi ≥∑n i=1 w2i , then ∑n i=1 a2i ≥ ∑n i=1 w2i . Problem 4.3. Let fn(x) denote the Fibonacci polynomial, which is defined by f1 = 1, f2 = x, fn = xfn−1 + fn−2. Prove that the inequality f 2n ≤ ( x2 + 1)2(x2 + 2)n−3 holds for every real x and n ≥ 3. Solution 4.3. Since f3(x) = x2 + 1, the inequality is trivially satisfied for n = 3. We proceed by induction on n: 186 8 Analysis Solutions f 2n+1(x) = [xfn(x) + fn−1(x)]2 ≤ [x(x2 + 1)(x2 + 2)(n−3)/2 + (x2 + 1)(x2 + 2)(n−4)/2]2 ≤ (x2 + 1)2(x2 + 2)n−2[x(x2 + 2)−1/2 + (x2 + 2)−1]2 < (x2 + 1)2(x2 + 2)k−2. We have used above that x ( x2 + 2)−1/2 + (x2 + 2)−1 < 1, which is a consequence of x4 + 2x2 < (x2 + 1)2. Second proof. One proves by induction that fn(x) = det ⎛ ⎜⎜⎜ ⎜⎜ ⎝ x −1 0 0 · · · 0 0 1 x −1 0 · · · 0 0 0 1 x −1 · · · 0 0 ... ... ... ... ... ... 0 0 0 0 · · · 1 x ⎞ ⎟⎟⎟ ⎟⎟ ⎠ ⎫ ⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎭ n − 1. Let us recall now the Hadamard inequality, which gives an upper bound for the determinant of an arbitrary k × k matrix, as follows: (det(aij ))2 ≤ k∏ j=1 ( k∑ i=1 a2ij ) . This yields the claimed inequality. Problem 4.4. Prove the inequality min ( (b − c)2, (c − a)2, (a − b)2) ≤ 1 2 ( a2 + b2 + c2). Generalize to min1≤k μ > 0 for all i ≤ n − 1, then (aj − ai)2 > (i − j)2μ2 and hence ∑ (ai − aj )2 > μ2 ∑ 1≤i 0, we can apply the intermediate value theorem to find that P has a real zero in (−1, 1). Problem 4.9. Find the minimum of β and the maximum of α for which ( 1 + 1 n )n+α ≤ e ≤ ( 1 + 1 n )n+β holds for all n ∈ Z+. 8 Analysis Solutions 189 Solution 4.9. Considering the logarithms, we have αmax = inf n { 1 log(1 + 1/n) − n } , βmin = sup n { 1 log(1 + 1/n) − n } . The function F(x) = 1log(1+1/x) − x is monotonically increasing for positive x, since the derivative F ′(x) is positive. Therefore αmax = 1log 2 − 1 and βmin = lim n→∞F(n). Using the Maclaurin series for the function log(1 + 1/x), we obtain F(n) = 11 n − 12n2 + O( 1n3 ) − n, whence limn→∞ F(n) = 12 = βmin. Problem 4.10. Prove that for nonnegative x, y, and z such that x + y + z = 1, the following inequality holds: 0 ≤ xy + yz + zx − 2xyz ≤ 7 27 . Solution 4.10. We have x, y, z ∈ [0, 1] and thus 0 ≤ xyz ≤ 1, which turns into xy + yz+ zx − 2xyz ≥ 3 3√x2y2z2 − 2xyz ≥ 3xyz− 2xyz = xyz ≥ 0. This yields the first inequality. Also, if one of the three variables x, y, z is at least 12 , then (1 − 2x)(1 − 2y)(1 − 2z) ≤ 0 < 127 . Observe that it is not possible for two among the three variables to be greater than 12 , because this would imply that the third is negative. Further, if x, y, z ≤ 12 , then, by the means inequality, (1 − 2x)(1 − 2y)(1 − 2z) ≤ 127 . This implies that 1 − 2(x + y + z) + 4(xy + yz + zx) − 8xyz ≤ 1 27 and thus xy + yz + zx − 2xyz ≤ 727 . Second Solution: Let f (x, y, z) = xy + yz + zx − 2xyz and the associated Lagrange multiplier F(x, y, z, λ) = f − λ(x + y + z − 1). 190 8 Analysis Solutions The critical points of F are given by the formula ∂F ∂x = ∂F ∂y = ∂F ∂z = ∂F ∂λ = 0. Therefore, we obtain the system of equations x + y − 2xy − λ = 0, x + z − 2xz − λ = 0, y + z − 2yz − λ = 0, x + y + z − 1 = 0. If one variable, say z, is not equal to 12 , then x = z−λ1−2z = y. If x = y �= 12 , then x = z = y−λ1−2y . Thus x = y = z = 13 . If x = y = 12 , then z = 0. Therefore, we have the solutions ( 1 3 , 1 3 , 1 3 ) , ( 1 2 , 1 2 , 0 ) and their circular permutations. The Jacobian matrix reads J (x, y, z) = det ⎛ ⎜⎜ ⎝ ∂2F ∂x2 ∂2F ∂x∂y ∂2F ∂x∂z ∂2F ∂y∂x ∂2F ∂y2 ∂2F ∂y∂z ∂2F ∂z∂x ∂2F ∂z∂y ∂2F ∂z2 ⎞ ⎟⎟ ⎠ = det ⎛ ⎝ 0 1 − 2z 1 − 2x 1 − 2z 0 1 − 2y 1 − 2y 1 − 2x 0 ⎞ ⎠ , and one computes J ( 1 3 , 1 3 , 1 3 ) = 2 27 > 0, J ( 1 2 , 1 2 , 0 ) = 0. Now f ( 1 2 , 1 2 , 0 ) = 14 < f ( 1 3 , 1 3 , 1 3 ) = 727 , and so the only extremal point is a maximum at ( 1 3 , 1 3 , 1 3 ) . Problem 4.11. Consider the sequence of nonzero complex numbers a1, . . . , an, . . . with the property that |ar − as | > 1 for r �= s. Prove that ∞∑ n=1 1 a3n converges. Solution 4.11. Let Sk = {n ∈ Z+; k < |an| ≤ k+1}. The disks Dn = { z; |z−an| ≤ 1/2 } are pairwise disjoint, according to the hypothesis. If one considers those disks Dn for which n ∈ Sk , then these disks are contained in the annulus Ck = { z ∈ C; k − 1 2 ≤ |z| ≤ k + 3 2 } . Thus the total area covered by these disks is bounded from above by the area of the annulus, i.e., 8 Analysis Solutions 191 card(Sk) π 4 ≤ π (( k + 3 2 )2 − ( k − 1 2 )2) = 2π(k + 1), and thus card(Sk) ≤ 8(2k + 1) if k > 0. In the same way, for k = 0, we obtain card(S0) ≤ 9: ∑ n∈Sk 1 |an|3 ≤ card(Sk) k3 ≤ 8(2k + 1) k3 ≤ 24 k2 , ∞∑ n=1 1 |an|3 = ∞∑ k=0 ∑ n∈Sk 1 |an|3 ≤ ∑ n∈S0 1 |an|3 + ∞∑ k=1 24 k2 < ∞, and the series converges absolutely. Problem 4.12. Consider the sequence Sn given by Sn = n + 12n+1 n∑ i=1 2i i . Find limn→∞ Sn. Solution 4.12. We have Sn+1 = n + 22n+2 n+1∑ i=1 2i i = n + 2 2(n + 1) (Sn + 1), so that Sn+2 − Sn+1 = (n + 2) 2(Sn+1 − Sn) − Sn+1 − 1 2(n + 1)(n + 2) . Since Sn ≥ 0, we obtain Sn+2−Sn+1 ≤ 0, and the sequence Sn is bounded decreasing, thus convergent. If we set s for its limit, then s satisfies s = lim n→∞ n + 2 2(n + 1) (Sn + 1) = 1 2 (s + 1) and thus s = 1. Comments 80 By integrating the geometric series formula, we obtain lim n→∞ n an n∑ i=1 ai−1 i = 1 a − 1 . For a = 2, we recover the previous problem. Problem 4.13. Prove that whenever a, b > 0, we have ∫ 1 0 ta−1 1 + tb dt = 1 a − 1 a + b + 1 a + 2b − 1 a + 3b + · · · . 192 8 Analysis Solutions Solution 4.13. Consider the identity ta−1 1 + tb = t a−1 ( n−1∑ k=0 (−1)ktkb + (−1) ntbn 1 + tb ) . Integrating between 0 and 1 and using the fact that lim n→∞ ∣∣ ∣∣ ∫ 1 0 tbn 1 + tb dt ∣ ∣∣∣ = 0, we obtain the equality from the statement. Problem 4.14. Let a, b, c, d ∈ Z∗+, and r = 1 − ab − cd . If r > 0 and a + c ≤ 1982, then r > 119833 . Solution 4.14. We have three cases to analyze: 1. If b, d ≥ 1983, then r ≥ 1 − a+c1983 ≥ 1 − 19821983 > 119833 . 2. If b, d ≤ 1983, then r = bd−ad−bc bd > 0, so that bd − ad − bc ≥ 1. Thus r ≥ 1/bd > 1/19833. 3. Suppose now that b < 1983 < d. If d > 19832 and r < 119833 , then c d < 1982 19832 < 1 1983 ; thus 1− ab < 11983 . This implies that b > 1983(b−a) > 1983 because a ≤ b + 1, which is absurd. If d ≤ 19832, then, as in the case 2 above, we have r ≥ 1 bd > 119833 . Problem 4.15. Let ai ∈ R. Prove that nmin(ai) ≤ n∑ i=1 ai − S ≤ n∑ i=1 ai + S ≤ nmax(ai), where (n − 1)S2 =∑1≤i x, then we apply the first inequality above, and use the periodicity and the fact that ξ x ≤ n in order to get the claim. If ξ ≤ x, then we have two cases: 1. If x ≤ 1 − 1 n , then 1−ξ1−x ≤ 11/n = n, and we apply the second inequality above. 2. If x = 1− 1 n +α for some 0 < α < 1 n , then nx = n−1+nα, and so {nx} = nα. Since 1−ξ1−x = 1−nα1 n −α = n, we use the second inequality again and the claim follows. Comments 81 Whenf (x) = | sin x|, we obtain the well-known inequality | sin nx| ≤ n| sin x|. Problem 4.17. Let (an) be a sequence of positive numbers such that lim n→∞ (a1 + · · · + an) n < ∞, lim n→∞ an n = 0. Does this imply that limn→∞ a21+···+a2n n2 = 0? Solution 4.17. The answer is yes. If the sequence an is bounded, it is obvious. If the sequence is not bounded, consider the subsequence ank defined by the properties that n1 is the smallest integer such that an1 > a1, and nk the smallest integer nk > nk−1 such that ank > ank−1 . If nk ≤ n < nk+1, then we have 0 ≤ a 2 1 + · · · + a2n n2 < ( a1 + · · · + an n ) ank nk . Passing to the limit, we find that the claim holds. Problem 4.18. Show that for a fixed m ≥ 2, the following series converges for a single value of x: S(x) = 1 + ... + 1 m − 1 − x m + 1 m + 1 + · · · + 1 2m − 1 − x 2m + 1 2m + 1 + · · · + 1 3m − 1 − x 3m + · · · . Solution 4.18. Let us denote by Sn(x) = ∑nk=1( 1(k−1)m+1 + · · · + 1km−1 − xkm) the truncation of S(x). Assume that S(x) and S(y) are finite for two distinct values x and y. 194 8 Analysis Solutions We now observe that Sn(x) − Sn(y) = y−xm ∑n k=1 1k , and therefore lim n→∞ |Sn(x) − Sn(y)| = ∞, which contradicts our assumptions. Comments 82 If xm is the value for which S(x) converges, then S(xm) = logm. Problem 4.19. Prove that if zi ∈ C, 0 < |zi | ≤ 1, then |z1 − z2| ≤ | log z1 − log z2|. Solution 4.19. From Lagrange’s mean value theorem, we have | log r1 − log r2| = 1 ξ |r1 + r2| for any 0 < r1, r2 < 1 and ξ ∈ [0, 1]. Thus | log r1 − log r2| ≤ |r1 − r2|. Write zj = rj exp(iθj ). We have (z1 − z2)2 = r21 + r22 − 2r1r2 cos(θ1 − θ2) = (r1 − r2)2 + 2r1r2(1 − cos(θ1 − θ2)) = (r1 − r2)2 + 4r1r2 sin2 12 (θ1 − θ2) ≤ (r1 − r2) 2 + r1r2(θ1 − θ2)2 ≤ (r1 − r2)2 + (θ1 − θ2)2 ≤ (log r1 − log r2)2 + (θ1 − θ2)2 = | log z1 − log z2|2. We have used above the inequality sin θ < |θ |, and its consequence 4 sin2 12 (θ1 − θ2) ≤ (θ1 − θ2)2. Problem 4.20. The roots of the function of a complex variable ζ4(s) = 1 + 2s + 3s + 4s are simple. Solution 4.20. Suppose that ζ4(s) and ζ ′4(s) have common zeros. Let us set x = 2s . Then we have 1 + x + x2 = −3s and x log 2 + x2 log 4 = −3s log 3, so that x2(log 3/2) + x(log 3/2) + log 3 = 0. The discriminant of this quadratic equation is positive; hence the roots are real: 2s = x ∈ R, 3s = −(1 + x + x2) ∈ R. It is known that s cannot be real, since ζ4 has no real zeros. However, the reality condition above shows that the (nonzero) imaginary part of s must be an integer multiple of both πilog 2 and πi log 3 . This is impossible, because log 3/ log 2 �∈ Q. Comments 83 A related function that generalizes the finite sum ζ4(s) is the Riemann zeta function ζ defined by ζ(s) = ∑ k≥1 k−s . 8 Analysis Solutions 195 It is not hard to see that ζ(s) is defined for all complex s with real part �(s) > 1 and can be continued analytically in s. The function ζ(s) has zeros at the negative even integers −2,−4, . . . and one usually refers to them as the trivial zeros. One of the most important problems in mathematics today, known as the Riemann hypothesis, is to decide whether all nontrivial zeros of ζ(s) in the complex plane lie on the “critical line” consisting of points of real part 12 . The nontrivial zeros correspond precisely to the zeros of the following even entire function: ξ(t) = π −s/2s(s − 1) 2 � ( s 2 ) ζ(s), where s = 12 + √−1t and � is the usual gamma function. This is known to hold true for the first 1013 zeros and for 99 percent of the zeros. The validity of the conjecture is equivalent to saying that π(n) − ∫ n 0 dt log t = O(√n log n), where π(n) counts the number of primes less than n. The Riemann hypothesis was included as one of the seven Millennium Problems (for which the Clay Mathematics Institute will award one million dollar for a solution). The interested reader might consult: • B. Conrey: The Riemann hypothesis, Notices Amer. Math. Soc. March, 2003, 341-353. • E. Bombieri: Problems of the millennium: The Riemann Hypothesis, The Clay Mathematics Institute, 2000. http://www.claymath.org/millennium/Riemann_Hypothesis/ Problem 4.21. Compute I = ∫ π/2 0 log sin x dx. Solution 4.21. We have I = ∫ π/2 0 log cos x dx = ∫ π π/2 log sin x dx = 1 2 ∫ π 0 log sin x dx and thus 2I = 2 ∫ π/2 0 log sin 2x dx = 2 (∫ π/2 0 log 2 dx + ∫ π/2 0 log sin x dx + ∫ π/2 0 log cos x dx ) = 4I + π log 2 and thus I = −π2 log 2. 196 8 Analysis Solutions Problem 4.22. Letf a be positive, monotonically decreasing function on [0, 1]. Prove the following inequality: ∫ 1 0 xf 2(x)dx ∫ 1 0 xf (x)dx ≤ ∫ 1 0 f 2(x)dx ∫ 1 0 f (x)dx . Solution 4.22. The above inequality is equivalent to ∫ 1 0 f 2(x) dx ∫ 1 0 yf (f ) dy − ∫ 1 0 yf 2(y) dy ∫ 1 0 f (x) dx ≥ 0, which can be rewritten as follows: I = ∫ 1 0 ∫ 1 0 f (x)f (y)y(f (x) − f (y)) dx dy ≥ 0. Interchanging the variables and the integration order, we obtain 2I = ∫ 1 0 ∫ 1 0 f (x)f (y)(y − x)(f (x) − f (y)) dx dy ≥ 0. Since f is decreasing, we have (y − x)(f (x) − f (y)) ≥ 0 and thus 2I ≥ 0, as claimed. Comments 84 It can be proven by the same method that ∫ 1 0 f (x)g(x)w(x) dx ∫ 1 0 w(x) dx ≤ ∫ 1 0 f (x)w(x) dx ∫ 1 0 g(x)w(x) dx, where f is decreasing, g is increasing, and w is nonnegative, with strict inequality if f and g are not constant on the support of w. Problem 4.23. Prove that every real number x such that 0 < x ≤ 1 can be represented as an infinite sum x = ∞∑ k=1 1 nk , where the nk are natural numbers such that nk+1nk ∈ {2, 3, 4}. Solution 4.23. We have two cases to consider, 13 ≤ x ≤ 1 and 0 < x < 13 . In the first case, let us put n0 = 1 and x0 = x. We define the sequence xk = nk nk−1 xk−1 − 1, where nk = ⎧ ⎪⎨ ⎪⎩ 4nk−1, if 13 ≤ xk−1 < 12 , 3nk−1, if 12 ≤ xk−1 < 23 , 2nk−1, if 23 ≤ xk−1 < 1. 8 Analysis Solutions 197 If 13 ≤ xk−1 ≤ 1, then it is immediate that 13 ≤ xk ≤ 1 and so, 13 ≤ xk ≤ 1. Further, the sequence {nk} is unbounded, which implies that lim k→∞ xk nk = 0. The recurrence relation reads xk nk = xk−1 nk−1 − 1 nk , and hence xk nk = x0 n0 − k∑ i=1 1 ni = x − k∑ i=1 1 ni . Thus by letting k go to infinity, x =∑∞i=1 1ni . In the second case, 0 < x < 13 , we choose a natural number k ≥ 4 such that 1/k ≤ x < 1/(k − 1). Set N = 2(k − 1). Then the number x1 = Nx − 1 satisfies 1 3 < x1 < 1. From the above, x1 = ∞∑ k=1 1 nk , where nk+1 nk ∈ {2, 3, 4}. Then we write x = 1 N + x1 N = 1 N + ∞∑ k=1 1 Nnk = ∞∑ k=1 1 mk , where mk = Nnk−1, for k ≥ 2, and m1 = N . Comments 85 In general, the representation is not unique, since for example, 1 2 = ∞∑ k=2 1 3k = ∞∑ k=1 1 2k . Moreover, the result of the problem holds as well for those x that belong to the interval[ 4 3 , 2 ] . Problem 4.24. Let Sn denote the set of polynomials p(z) = zn + an−1zn−1 + · · · + a1z + 1, with ai ∈ C. Find Mn = min p∈Sn ( max|z|=1 |p(z)| ) . Solution 4.24. Let us show that Mn = 2 if n ≥ 2. We have first Mn ≤ max|z|=1 |zn + 1| = 2. Suppose now that Mn < 2, for some n, so that there exists a polynomial p(z) = zn + an−1zn−1 + · · · + a1z + 1 with the property that max|z|=1 |p(z)| < 2. This implies that 198 8 Analysis Solutions −2 < �(p(z)) < 2, for z ∈ C, with |z| = 1. Let ξj , j = 1, . . . , n be the nth roots of unity. Sum the inequalities above for z = ξj , and get −2n < � n∑ j=1 p(ξj ) < 2n. On the other hand, an immediate computation shows that n∑ j=1 p(ξj ) = 2n for any p as above, which is a contradiction. Thus our claim follows. Problem 4.25. Find an explicit formula for the value of Fn that is given by the recur- rence Fn = (n + 2)Fn−1 − (n − 1)Fn−2 and initial conditions F1 = a, F2 = b. Solution 4.25. We will use the so-called Laplace method. Letϑ : R → R be a smooth function such that Fn = ∫ β α tn−1ϑ(t) dt. The recurrence satisfied by Fn is then a consequence of the following differential equation: dϑ ϑ = 1 − 3t + t 2 t − t2 dt. This differential equation can be integrated, and one obtains the solutionϑ = e−t t (t− 1). Further, the integral’s limits α and β are among the roots of the equation tn+1(t − 1)e−t = 0, yielding t1 = 0, t2 = ∞, t3 = 1. The general solution is therefore Fn = C1 ∫ ∞ 0 tn(t − 1)e−t dt + C2 ∫ 1 0 tn(t − 1) e−tdt = C1n!n + C2 ( n!n ( 1 − 1 e n∑ k=0 1 k! ) − 1 e ) = (3b − 11a)n!n + (4a − b) ( 1 + n!n n∑ k=0 1 k! ) . Problem 4.26. Find the positive functions f (x, y) and g(x, y) satisfying the follow- ing inequalities: ( n∑ i=1 aibi )2 ≤ ( n∑ i=1 f (ai, bi) )( n∑ i=1 g(ai, bi) ) ≤ ( n∑ i=1 a2i )( n∑ i=1 b2i ) for all ai, bi ∈ R and n ∈ Z+. 8 Analysis Solutions 199 Solution 4.26. (ab)2 ≤ f (a, b)g(a, b) ≤ a2b2 and thus (1) f (a, b)g(a, b) = a2b2. For n = 2, use the hypothesis 2a1b1a2b2 ≤ f (a1, b1)g(a2, b2) + f (a2, b2)g(a1, b1) ≤ a21b22 + a22b21 and the identity (1) to find that (2) 2 ≤ f (a1, b1) f (a2, b2) a2b2 a1b1 + f (a2, b2) f (a1, b1) a1b1 a2b2 ≤ a1b2 a2b1 + a2b1 a1b2 . We replace (a1, b1) = (a, b) and (a2, b2) = (λa, λb); then 2 ≤ f (a, b) f (λa, λb) λ2 + f (λa, λb) f (a, b) λ−2 ≤ 2, from which we derive f (λa, λb) = λ2f (a, b). Set f (a) = f (a, 1), and so f (a, b) = b2f (a/b). Then (2) becomes (3) 2 ≤ f (a)/a f (b)/b + f (b)/b f (a)/a ≤ a b + b a . This implies that when a ≥ b, we have f (a)b f (b)a ≤ a b and f (b)a f (a)b ≤ a b . This yields (4) f (b) ≤ f (a), f (a) a2 ≤ f (b) b2 , (a ≥ b). Conversely, if f, g satisfy (1) and (4), then by applying the inequality (2) for all pairs of indices i �= j ≤ n, we find that 2 ∑ aibiaj bj ≤ ∑ i,j [f (ai, bi)g(aj , bj )+f (aj , bj )g(ai, bi)] ≤ ∑ i,j ( a2i b 2 j +a2j b2i ) , and the inequality from the statement follows. Comments 86 If f (x, y) = x2 + y2, g(x, y) = x2y2/(x2 + y2), then we obtain an inequality due to Hardy, Littlewood, and Pólya. By choosing f (x, y) = x1+αy1−α , g(x, y) = x1−αy1+α , we obtain the Calderbank inequality. Problem 4.27. Let g ∈ C1(R) be a smooth function such that g(0) = 0 and |g′(x)| ≤ |g(x)|. Prove that g(x) = 0. 200 8 Analysis Solutions Solution 4.27. Let x ≥ 0. We have |g(x)| ≤ ∣∣ ∣∣ ∫ x 0 g′(t) dt ∣∣ ∣∣ ≤ ∫ x 0 |g′(t)| dt ≤ ∫ x 0 |g(t)| dt. Let a > 0 be arbitrary. Since g ∈ C1(R), g is bounded on any bounded interval and thus there exists k, such that |g(x)| ≤ k, for 0 ≤ x ≤ a. If 0 ≤ t ≤ x ≤ a, then |g(t)| ≤ k, and so |g(x)| ≤ ∫ x 0 k dt = kx for every x with 0 ≤ x ≤ a. Now the condition |g(t)| ≤ kt , for 0 ≤ t ≤ x ≤ a, implies that |g(x)| ≤ ∫ x 0 kt dt = 1 2 kx2, for 0 ≤ x ≤ a. Continuing this way, we prove by induction that |g(x)| ≤ 1 n! kxn, provided that 0 ≤ x ≤ a. Putting x = a and passing to the limit n → ∞, we obtain |g(a)| ≤ lim n→∞ 1 n! kan = 0. Since a was arbitrary, we have g(x) = 0 for all x ∈ [0,∞). A similar argument settles the case x ∈ (−∞, 0]. Problem 4.28. Let a1, b1, c1 ∈ R+ such that a1 + b1 + c1 = 1 and define an+1 = a2n + 2bncn, bn+1 = b2n + 2ancn, cn+1 = c2n + 2anbn. Prove that the sequences (an), (bn), (cn) have the same limit and find that limit. Solution 4.28. Let un = an + bn + cn, vn = an + ωbn + ω2cn, wn = an + ω2bn + ωcn, where ω3 = 1 is a third root of unity. The given recurrence for an, bn, cn induces the following recurrence relations: un+1 = u2n = 1, vn+1 = w2n, wn+1 = v2n. Thus wn+1 equals either v2n1 or w2n1 , depending on the parity of n (and similarly for vn+1). Since v1 and w1 are convex combinations with positive coefficients of the complex points 1, ω, ω2, it follows that |v1|, |w1| < 1. But then we derive that limn→∞ vn = limn→∞ wn = 0. Since un is constant, we obtain 8 Analysis Solutions 201 lim n→∞ an = limn→∞ 1 3 (un + vn + wn) = 13 , lim n→∞ bn = limn→∞ 1 3 (un + ω2vn + ωwn) = 13 , lim n→∞ cn = limn→∞ 1 3 (un + ωvn + ω2wn) = 13 . Problem 4.29. Consider the sequence (an), given by a1 = 0, a2n+1 = a2n = n− an. Prove that an = n3 , for infinitely many values of n. Does there exist an n such that|an − n3 | > 2005? Also, prove that lim n→ an n = 1 3 . Solution 4.29. 1. We have an = [ n 2 ]− a[ n 2 ] , and thus an = [n 2 ] − [n 4 ] + [n 8 ] − [ n 10 ] + · · · . If n = 3 · 2k , then we have a3·2k = 3 · 2k 2 − 3 · 2 k 4 + · · · + 3 · 2 k 2k − [ 3 2 ] = 3(2k−1 + (−2k−2) + 2k−3 + (−2k−4) · · · ) − 1 = 3 · 2 k + 1 3 − 1 = 2k, and therefore an = n3 . 2. Set nk = 4k−13 . Then nk is odd and satisfies the recurrence nk − 1 = 4nk−1. This implies that ank = ank−1 = a4nk−1 = 2nk−1 − a2nk−1 = 2nk−1 − nk−1 + ank−1 = nk−1 + ank−1 and thus ank = nk−1 + nk−2 + · · · + n1 + a1 = 1 3 ( 4k − 1 3 − k ) . Thus ∣∣ank − nk3 ∣∣ = k3 , and so ∣∣ank − nk3 ∣∣ can be arbitrarily large. 3. Use the fact that n2k − 1 ≤ [ n 2k ] ≤ n2k and find that ⎛ ⎝ [log n]∑ k=1 (−1)k n 2k ⎞ ⎠− [log n] ≤ an ≤ ⎛ ⎝ [log n]∑ k=1 (−1)k n 2k ⎞ ⎠+ [log n] + 1, and after dividing by n and letting n go to infinity, we obtain the claim. 202 8 Analysis Solutions Problem 4.30. Compute the integral f (a) = ∫ 1 0 log(x2 − 2x cos a + 1) x dx. Solution 4.30. f (a 2 ) + f ( π − a 2 ) = ∫ 1 0 log(x2 − 2x cos( a2 ) + 1)(x2 + 2x cos(π − a2 ) + 1) x dx = ∫ 1 0 log(x4 − 2x2 cos a + 1) x dx. Make the variable change x = √t and find that the last integral equals 1 2 ∫ 1 0 (log t2 − 2t cos a + 1) t dt = f (a) 2 , for a ∈ [0, 2π ]. Therefore, f satisfies the identity f (a 2 ) + f ( π − a 2 ) = f (a) 2 . By differentiating twice (we have f is C2), we obtain f ′′ (a 2 ) + f ′′ ( π − a 2 ) = 2f ′′(a). Suppose that f ′′(a) has the maximum M and the minimum m, and let a0 be such that f ′′(a0) = M, a0 ∈ [0, 2π ]. Putting a = a0 in the identity above we get f ′′ (a0 2 ) + f ′′ ( π − a0 2 ) = 2f ′′(a0) = 2M, from which we obtain f ′′(a0/2) = M . Iterating this procedure, we find that f ′′(a0) = f ′′ ( a0 2n ) = M , for any M . From the continuity of f ′′, we find that lim n→0 f ′′ (a0 2n ) = f ′′(0) = M. But now a similar argument dealing with the minimum shows that f ′′(0) = m and therefore M = m; hence f ′′ is constant. This implies that f (a) = αa2/2 + βa + γ . Substituting in the identity above, we obtain the following identities in the coefficients: −πα = β, π2α/2 + βπ + 2γ = γ /2. Using now f ′(π2 ) = π2 , it follows that πα/2 + β = π/2, therefore α = −1, β = π, γ = −π2/3. Thus we obtain f (a) = −a − 2 2 + πa − π 2 3 . 8 Analysis Solutions 203 Problem 4.31. Let −1 < a0 < 1, and define an = ( 1 2 (1 + an−1) )1/2 for n ≥ 1. Find the limits A,B, and C of the sequences An = 4n(1 − an), Bn = a1 · · · an, Cn = 4n(B − a1a2 · · · an). Solution 4.31. Let a0 = cos θ . By making use of the formula cos2 θ/2 = (1 + cos θ)/2 and a recurrence on n, we derive that an = cos(θ/2n). Then An = 4n(1−an) = 4n ( θ2 2 · 4n − θ4 24 · 42n + O(4 −3n) ) = θ 2 2 − θ 4 24 · 4n +O ( 4−2n ) . Thus A = θ22 . Further, sin(θ/2n)Bn = cos(θ/2) cos(θ/22) · · · cos(θ/2n) sin(θ/2n) and thus Bn = sin θ2n sin(θ/2n) = sin θ θ + sin θ 6 · 4n + O(4 −2n) and thus B = sin θ θ . Finally, Cn = −θ sin θ6 + O ( 4−n ) , and so C = −θ sin θ6 . Comments 87 The same method works when a0 > 1, by making use of the hyperbolic trigonometric functions. If we put θ = log(a0 + √ a20 − 1 ) = arccosh a0, then A = −θ2/2, B = sinh θ6 , C = −6 sinh θ6 . Problem 4.32. Consider the sequence given by the recurrence a1 = a, an = a2n−1 − 2. Determine those a ∈ R for which (an) is convergent. Solution 4.32. For |a| > 2, we have |an| > 2 and consequently an+1 > an for all n ≥ 2. Since the sequence is increasing, it is either convergent or tends to ∞. If we have a finite limit an → λ, then it satisfies λ = λ2 − 2 and thus λ ∈ {−1, 2}, contradicting our assumptions. Therefore, we need |a| ≤ 2 for the convergence. Let α ∈ (−π2 , π2 ) be such that 2 cosα = a. By induction on n, we obtain an = 2 cos 2nα. Assume that α is not commensurable with π , that is, α π /∈ Q. Observe that the function ϕ(x) = 2xα is continuous. Thus, the map ϕ : R/πZ → R/πZ sends the dense subset {nα (modπ), n ∈ Z} into a dense subset, which is {2nα (modπ), n ∈ Z}. Therefore, the image under cos, namely {cos 2nα; n ∈ Z}, is dense in [−1, 1], and thus the sequence (an) cannot be convergent. If α/π = p/q, where p, q ∈ Z, then the sequence (an) is periodic for n large enough. Thus, in order for the sequence to be convergent it is necessary and sufficient that it be constant for n large enough. This condition reads 204 8 Analysis Solutions cos ( 2nα ) = cos(2n+1α). Thus 2nα = ±2n+1α + kπ, where k ∈ Z, and so α ∈ { m 2n3 π, where m, n ∈ Z } . One sees that an is convergent and constant for n ≥ m. Problem 4.33. Let 0 < a < 1 and I = (0, a). Find all functions f : I → R satisfying at least one of the conditions below: 1. f is continuous and f (xy) = xf (y) + yf (x). 2. f (xy) = xf (x) + yf (y). Solution 4.33. 1. Let g(x) = f (x) x . Then g(exp(x)) is continuous and additive on I . This implies that g(exp(x)) is linear and hence g(x) = C log x, which implies f (x) = Cx log x for some constant C. 2. Give y the values x, x2, x3, and x4. We find therefore that f ( x2 ) = 2xf (x), f ( x3 ) = xf (x) + x2f (x2) = (x + 2x3)f (x), f ( x4 ) = xf (x) + x3f (x3) = (x + x4 + 2x6)f (x), f ( x4 ) = x2f (x) + x2f (x2) = 4x3f (x). From the last two lines, one derives that f (x) = 0 for all but those points x for which x + x4 + 2x6 = 4x3. This means that f vanishes for all but finitely many (actually 6) values of x. If t ∈ I is such that f (t) �= 0, then the first identity shows that f (t2) �= 0. By induction on n we obtain f (t2n) �= 0. This means that f does not vanish for infinitely many values of the argument, which contradicts our former result. Thus f is identically zero. Problem 4.34. If a, b, c, d ∈ C, ac �= 0, prove that max(|ac|, |ad + bc|, |bd|) max(|a|, |b|)max(|c|, |d|) ≥ −1 + √5 2 . Solution 4.34. Set ab−1 = r, dc−1 = s, −1+ √ 5 2 = k, k2 = 1 − k, 0 < k < 1. The inequality from the statement is equivalent to μ = max(1, |r + s|, |rs|) ≥ k max(1, |r|)max(1, |s|) = ν. (i) If |r| ≥ 1, |s| ≥ 1, then μ ≥ |rs| > k|s||r| = ν, 8 Analysis Solutions 205 proving our claim. (ii) If |r| ≤ 1, |s| ≥ 1, then our inequality is equivalent to max(1, |r + s|, |rs|) ≥ k|s|. Moreover, if k|s| ≤ 1 or |r + s| ≥ k|s|, then this is obviously true. Further, let us assume that k|s| > 1 and |r+s| < k|s|. We have then |r|+|r+s| ≥ |s| and consequently, |r| ≥ |s| − |s + r| ≥ |s| − k|s| = (1 − k)|s| = k2|s|. Furthermore, |rs| ≥ k2|s|2 > k|s| > 1; thus max(1, |r + s|, |rs|) ≥ k|s| holds. (iii) If |r| ≥ 1, |s| ≤ 1, then the inequality is proved as in case (ii), by using the obvious symmetry. Problem 4.35. Let ∑∞ i=1 xi be a convergent series with decreasing terms x1 ≥ x2 ≥· · · ≥ xn ≥ · · · > 0 and let P be the set of numbers which can be written in the form∑ i∈J xi for some subset J ∈ Z+. Prove that P is an interval if and only if xn ≤ ∞∑ i=n+1 xi for every n ∈ Z+. Solution 4.35. 1. Suppose that there exists p such that xp > ∑∞ i=p+1 xi and consider α such that ∑ i>p xi < α < xp. Set S(J ) = ∑ i∈J xi . We claim that there does not exist any J ⊂ Z+ such that S(J ) = α. In fact, if J ∩ {1, 2, . . . , p} �= ∅, then∑ i∈J xi ≥ xp > α, becausexk is decreasing. On the other hand, ifJ∩{1, 2, . . . , p} =∅, then ∑i∈J xi ≤ ∑ i>p xi < α. Finally, P contains numbers smaller than α, for instance ∑ i>p xi , as well as numbers bigger than α, such as xp. Thus P is not an interval. 2. Assume now that the hypothesis of the problem is satisfied and choose any element y from the interval (0, S], where S = ∑∞i=1 xi . We will show that there exists L such that S(L) = y. Let n1 be the smallest index such that xn1 < y. There exists such a one because limk→∞ xk = 0. By induction, we define nk to be the smallest integer with the property that xn1 + · · · + xnk < y. Let L = {n1, . . . , nk, . . . }. It is clear that S(L) ≤ y. If p ∈ Z+ \ L, then choose the smallest k with the property that nk > p. Since p does not belong to L, we have xn1 + · · · + xnk−1 + xp ≥ y. This implies that ( ∞∑ i∈L xi ) + xp ≥ y. 206 8 Analysis Solutions (i) Assume that the set Z+ − L is finite. This set is empty only when y = S, which obviously belongs to P . Now take p to be the maximal element of Z+ − L. This implies that all elements bigger than p already belong to L and hence L = {n1, . . . , nk−1} ∪ {p + 1, p + 2, . . . }. Then ∑ i∈L xi = k−1∑ i=1 xn1 + ∞∑ j=p+1 xj ≥ k−1∑ i=1 xni + xp ≥ y, from which we obtain that ∑ i∈L xi = y. (ii) If Z+ − L is infinite, then for any ε > 0, one can choose some pε ∈ Z+ − L such that xpε < ε. The inequality above for pε implies that � + ∑ i∈L xi ≥ y. This is true for any positive ε, and thus ∑ i∈L xi = y. Problem 4.36. Does there exist a continuous function f : (0,∞) → R such that f (x) = 0 if and only if f (2x) �= 0? What if we require only that f be continuous at infinitely many points? Solution 4.36. 1. There exists some real number α such that f (α) = 0, and thus f (2α) �= 0. Let γ ∈ [α, 2α] be given by γ = sup{x ∈ [α, 2α] such that f (x) = 0}. Since f is continuous, we find that f (γ ) = 0, and thus γ < 2α. Moreover, if xn ∈ R is a sequence such that limn xn = γ , then f (xn) = 0 for n large enough. In fact, otherwise there exists a subsequence f (xnk ) �= 0, which implies that f (2xnk ) = 0, and thus limk f (2xnk ) = 0 �= f (2γ ). Finally, recall that γ is the supremum of those x ∈ [α, 2α] for which f (x) = 0 and γ < 2α. Thus for n large such that γ + 1 n < 2α, there exists xn with γ < xn < γ + 1n for which f (xn) �= 0. This contradicts the previous claim. Thus there does not exist a continuous f as in the statement. 2. The function f (x) = { 0, for 22k < x ≤ 22k+1,where k ∈ Z, 1, for 22k−1 < x < 22k,where k ∈ Z, satisfies the claim and has a countable number of discontinuities. Problem 4.37. Find the smallest number a such that for every real polynomial f (x) of degree two with the property that |f (x)| ≤ 1 for all x ∈ [0, 1], we have |f ′(1)| ≤ a. Find the analogous number b such that |f ′(0)| ≤ b. Solution 4.37. Let f (x) = ux2 + vx + w. We have |f ′(1)| = |2u + v| = |3f (1) − 4f (1/2) + f (0)| ≤ 3|f (1)| + 4|f (1/2)| + |f (0)| ≤ 3 + 4 + 1 = 8. 8 Analysis Solutions 207 Further, a = 8 because we have equality above for f (x) = 8x2 − 8x + 1. For the second case, we observe that |f ′(0)| = |v| = |4f (1/2) − 3f (0) − f (1)| ≤ 8 and thus b = 8 by taking f (x) = 8x2 − 8x + 1. Problem 4.38. Let f : R → R be a function for which there exists some constant M > 0 satisfying |f (x + y) − f (x) − f (y)| ≤ M, for all x, y ∈ R. Prove that there exists a unique additive function g : R → R such that |f (x) − g(x)| ≤ M, for all x ∈ R. Moreover, if f is continuous, then g is linear. Solution 4.38. Let us show first the existence of the upper limit ϕ(x) = lim supn→∞ f (nx) n . In fact, using the hypothesis, one finds that f (nx) ≤ n(f (x) + M), and hence the sequence f (nx) n is bounded from above, and thus ϕ(x) exists. Furthermore, ∣∣∣∣ f (n(x + y)) n − f (nx) n − f (ny) n ∣∣∣∣ ≤ M n , which yields ϕ(x) + ϕ(y) = ϕ(x + y); thus ϕ is additive. Next, observe that f (nx) ≤ n(f (x) + M) implies that ∣∣nx n − f (x)∣∣ ≤ M , and we can take g(x) = ϕ(x). If f is continuous, then it is easy to see that the lower limit, lim infn→∞ f (nx)n , coincides with ϕ(x), and moreover, ϕ(x) is continuous. An additive continuous func- tion is therefore linear. In fact, we have f (nx) = nf (x) for n ∈ Z, which yields f (rx) = rf (x) for any r ∈ Q. Since f is continuous and every real number r can be approximated by a sequence of rational numbers, we find that f (rx) = rf (x) for any r ∈ R, and hence the claim. Let us prove the uniqueness of such additive functions. Assume that there exists another additive function h satisfying |f (x) − h(x)| ≤ M, for all x ∈ R. This implies that |g(x) − h(x)| ≤ M, for all x ∈ R. However, an additive function is either zero or unbounded. Therefore, the additive function g − h must vanish. 208 8 Analysis Solutions Comments 88 Functions f satisfying the condition from the statement are called quasimorphisms. Notice that there exist additive functions ϕ : R → R that are not linear (not contin- uous of course). Examples can be constructed by choosing a basis of R as a vector space over Q and defining arbitrarily the values of the function on each vector of that basis. Problem 4.39. Show that if f is differentiable and if lim t→∞ ( f (t) + f ′(t)) = 1, then lim t→∞ f (t) = 1. Solution 4.39. Let us show first that if limt→∞(f ′(t) + αf (t)) = 0, where a = �(α) > 0, then limt→∞ f (t) = 0. Let ε > 0 and c < ∞ be such that |f ′(t) + αf (t)| ≤ aε whenever t ≥ c. Then ∣∣∣∣ d dt ( eαtf (t) ) ∣∣∣∣ = ∣∣eαt (f ′(t) + αf (t))∣∣ ≤ aεeat for t ≥ c. Therefore, using the mean value theorem we have |eαtf (t) − eacf (c)| ≤ ε (eαt − eac) , for t > c. This yields |f (t)| ≤ ea(c−t) · |f (c)| + ε∣∣1 − ea(c−t)∣∣. In particular, for sufficiently large t , we have |f (t)| ≤ 2ε, as claimed. Now the statement follows by applying this result to f − 1. Comments 89 Let P(D) be a polynomial in the derivation operator D = ddt . Let us write P(D) = (D − λ1)(D − λ2) · · · (D − λn). The above result can be generalized as follows. Assume that�λi < 0, where� denotes the real part; if the function f satisfies limt→∞ P(D)f (t) = 0, then limt→∞ f (t) = 0. This is proved by recurrence on the degree of P . The recurrence step follows from the argument given above. In particular, the claim holds for P(D) = D2 + D + 1. The condition �λi < 0 is necessary. In fact, we can consider f (t) = eλi t , which satisfies limt→∞ P(D)f → 0. Also, one notices that P(D) = D0 + D1 + ... + Dn does not fulfill this condition for n ≥ 3. Problem 4.40. Let c be a real number and letf : R → R be a smooth function of class C3 such that limx→∞ f (x) = c and limx→∞ f ′′′(x) = 0. Show that limn→∞ f ′(x) = limx→∞ f ′′(x) = 0. 8 Analysis Solutions 209 Solution 4.40. According to Taylor’s formula, there exist ξx, ηx ∈ (0, 1) such that f (x + 1) = f (x) + f ′(x) + 1 2 f ′′(x) + 1 6 f ′′′(x + ξx), f (x − 1) = f (x) − f ′(x) + 1 2 f ′′(x) − 1 6 f ′′(x + ηx). Suitable linear combinations of these yield f ′′(x) = f (x + 1) − 2f (x) + f (x − 1) − 1 6 f ′′′(x + ξx) + 16f ′′′(x − ηx), 2f ′(x) = f (x + 1) − f (x − 1) − 1 6 f ′′′(x + ξx) − 16f ′′′(x − ηx). Since x + ξx → ∞ and x −ηx → ∞ as x → ∞, by passing to the limit, one obtains lim x→∞ f ′′(x) = c − 2c + c − 1 6 · 0 + 1 6 · 0 = 0, lim x→∞ f ′(x) = 1 2 ( c − c − 1 6 · 0 − 1 6 · 0 ) = 0. Comments 90 It can be proved that if limx→∞ f (x) exists and if f (n) is bounded, then limx→∞ f (k)(x) = 0 for any 1 ≤ k < n. See for instance: • J. Littlewood: The converse of Abel’s theorem, Proc. London Math. Soc. (2) 9 (1910–11), 434–448. Problem 4.41. Prove that the following integral equation has at most one continuous solution on [0, 1] × [0, 1]: f (x, y) = 1 + ∫ x 0 ∫ y 0 f (u, v) du dv. Solution 4.41. If f1, f2 are two solutions, then their difference g satisfies g(x, y) = ∫ x 0 ∫ y 0 g(u, v) du dv. Since g is continuous, its absolute value is bounded by some constant M on the square [0, 1] × [0, 1]. Thus |g(x, y)| ≤ ∫ x 0 ∫ y 0 M du dv = Mxy. By induction, we show that |g(x, y)| ≤ M xnyn (n!)2 . The induction step follows from |g(x, y)| ≤ ∫ x 0 ∫ y 0 Munvn (n!)2 du dv = M(xy) n+1 ((n + 1)!)2 . If we fix x, y, then 0 ≤ |g(x, y)| ≤ lim n→ Mxnyn n!n! = 0, and thus g(x, y) ≡ 0. 210 8 Analysis Solutions Comments 91 There exists a continuous solution of this equation, given by f (x, y) = 1 + xy + x 2y2 2!2! + x 3y3 3!3! + · · · = J0 (√−4xy) = J0 ( 2i √ xy ) , where J0 is the Bessel function of order zero. Problem 4.42. Find those λ ∈ R for which the functional equation ∫ 1 0 min(x, y)f (y) dy = λf (x) has a solution f that is nonzero and continuous on the interval [0, 1]. Find these solutions. Solution 4.42. The equation can be written in the form λf (x) = ∫ x 0 yf (y) dy + x ∫ 1 x f (y) dy. If λ �= 0, then f is differentiable, and after derivation, we obtain λf ′(x) = xf (x) − xf (x) + ∫ 1 x f (y)dy = ∫ 1 x f (y)dy. Thus f ′ is differentiable and λf ′′(x) = −f (x). This implies that f (x) = A cosμx + B sin μx, where μ = λ−1/2 if λ > 0, and f (x) = A cosh νx + B sinh νx, where ν = (−λ)−1/2 if λ < 0. According to the hypothesis, lim n→0 f (x) = 0, and hence A = 0 for any choice of λ. From the above, we also get lim n→0 f ′(x) = 0, and thus Bμ cosμ = 0 and Bν cosh ν = 0. The last equation yields B = 0, and so we do not have nonzero solutions. For λ > 0, if B �= 0, then cosμ = 0 and thus μ = (2k + 1)π2 . Thus λ = μ−2 4(2k+1)2π2 , where k ∈ Z, and f (x) = B cos(2k + 1)π 2 x, B ∈ R. If λ = 0, the same method gives us f (x) = 0. 8 Analysis Solutions 211 Comments 92 We proved that the integral operator T : C0(0, 1) → C0(0, 1) defined on the space C0(0, 1) of continuous functions on (0, 1) and given by the formula (Tf )(x) = ∫ 1 0 min(x, y)f (y)dy has the eigenvalues 4 (2k+1)2π2 and the eigenvectors cos(2k + 1)π2 x. Problem 4.43. Let X be an unbounded subset of the real numbers R. Prove that the set AX = {t ∈ R; tX is dense modulo 1} is dense in R. Solution 4.43. Let {On}n∈Z+ be a countable base of (0, 1), On �= ∅. Let p be the canonical projection p : R → R/Z. Define An = {r ∈ R;p(rX) ∩ On �= ∅}. This amounts to An = ⋃ x∈X−{0} 1 x p−1({On}), and thus An is open. If F is an interval of length f > 0, then there exists x ∈ X such that xf > 1, which implies that p(xF) = R/Z and thus there exists r ∈ F with p(rx) ∈ On. This proves that An is dense in R. Since AX = ∞⋂ n=0 An, we obtain that AX is also dense in R, as the intersection of open dense subsets. Problem 4.44. Consider P(z) = zn+a1zn−1 +· · ·+an, where ai ∈ C. If |P(z)| = 1 for all z satisfying |z| = 1, then a1 = · · · = an = 0. Solution 4.44. Consider the polynomial Q(z) = 1 + a1z+ · · · + anzn, which can be written as Q(z) = P(z), when |z| = 1. This implies that |Q(z)| = 1 for all z of unit modulus. Moreover, Q(0) = 1, and so Q has a value in the interior of the disk whose modulus is equal to its maximum on the boundary circle. According to the maximum principle for holomorphic functions, Q has to be constant, Q(z) = 1, whence the claim. Comments 93 W. Blaschke has studied the complex analytic (i.e., holomorphic) func- tions f : D → C on the unit disk D that are continuous on D and have |f (z)| = 1 for all z satisfying |z| = 1. He has shown that such a function has the form f = σ n∏ k=1 z − bk 1 − bkz , where σ is a constant of modulus one and bk ∈ C are such that |bk| < 1. 212 8 Analysis Solutions Problem 4.45. Let I ⊂ R be an interval and u, v : I → R smooth functions satisfy- ing the equations u′′(x) + A(x)u(x) = 0, v′′(x) + B(x)v(x) = 0, where A,B are continuous on I and A(x) ≥ B(x) for all x ∈ I . Assume that v is not identically zero. If α < β are roots of v, then there exists a root of u that lies within the interval (α, β), unless A(x) = B(x), in which case u and v are proportional for α ≤ x ≤ β. Solution 4.45. The roots of v are isolated and we can suppose that v|(α, β) > 0. Assume that u|(α, β), so that after a possible sign change, u|(α, β) > 0. Also, we have v′(α) ≥ 0, v(β) ≤ 0, and since v is not identically zero, then v and v′ do not have common roots; thus v′(α) > 0 and v′(β) < 0. We consider w(x) = u(x)v′(x) − u′(x)v(x). We have w′(x) = u(x)v′′(x) − u′′(x)v(x) = (A(x) − B(x))uv ≥ 0, and therefore w is increasing; so w(α) ≤ w(β) and thus u(α)v′(α) ≤ u(β)v′(β). Since u(α) ≥ 0, u(β) ≥ 0, v′(α) > 0, v′(β) < 0, we obtain u(α) = u(β) = 0, and so w(α) = w(β) = 0. Now w is increasing and hence w ≡ 0 and thus w′ ≡ 0, which yields A(x) = B(x). Moreover, for α < x < β we have w(x) v2(x) = ( u v )′, and hence u v is constant. Comments 94 This result is known as Sturm’s comparison theorem in differential equations. Problem 4.46. Let V be a finite-dimensional real vector space and f : V → R a continuous mapping. For any basis B = {b1, b2, . . . , bn} of V , consider the set EB = {z1b1 + · · · + znbn, where zi ∈ Z}. Show that if f is bounded on EB for any choice of the basis B, then f is bounded on V . Solution 4.46. We solve first the case n = 1. Given any pair of positive numbers 0 < a < b, there exists some r = r(a, b) (depending on a, b) such that for any x > r , one can find m ∈ Z, with ma ≤ x ≤ mb. In fact, the intervals (ma,mb), for integral m, will cover all of R but a compact set. Thus, for any x > r , one can find an interval (c, d) containing x, and some m ∈ Z such that a < c m < d m < b. Assume that f is unbounded. Let us consider an interval I1 = [a1, b1] with the property that f (x) > 1, for x ∈ I1. Since f is unbounded, there exist x1 > r(a1, b1) arbitrarily large such that f (x1) > 4 and m1 ∈ Z such that m1a1 < x1 < m1b1. Moreover, f is continuous and so f (x) > 2 for any x ∈ [c1, d1], for some small interval around x1, which can be chosen to be contained in (m1a1,m1b1). Set I2 = 8 Analysis Solutions 213 {x ∈ I1;m1x ∈ [c1, d1]}. Thus I2 ⊂ I1 is a compact nonempty interval with the property that f (m1x) > 2 for any x ∈ I2. We define in this way a nested sequence of compact intervals I1 ⊂ I2 ⊂ I3 ⊂ · · · Ik ⊂ · · · , and a sequence of integers m1 < m2 < m3 < · · · with the property that f (x) > 2k for x ∈ mkIk . This nested sequence of nontrivial compact intervals must have nontrivial intersection. Let a ∈ ∩∞k=1Ik . Then the restriction of f to aZ is unbounded, since for any k, there exists some mk with f (mka) > 2k , by making use of the fact that a ∈ Ik . Let now deal with the general case. Let zk ∈ V be a sequence for which f (zk) is unbounded, e.g., f (zk) > 2k+1. Then zk is unbounded, because f is continuous. Let B = {b1, . . . , bn} be a basis of B and write zk = ∑ni=1 cikbi in the basis B. There exists at least one i0 ∈ {1, 2, . . . , n} such that the sequence ci0k is unbounded; otherwise, zk would be bounded. One can slightly modify the basis B so that this condition is satisfied for all i. In fact, assume that cik is unbounded for 1 ≤ i ≤ m < n and cik is bounded for m + 1 ≤ i ≤ n. Take then the new basis vectors b′1 = b1 − n∑ i=m+1 bi, b ′ i = bi, when i ≥ 2. In the new basis, one can express the vector zk as zk = m∑ i=1 cikb ′ i + n∑ i=m+1 (cik + c1k)b′i , and all components are now unbounded. We consider now the intervals Jik with the property that Uk = {x; x =∑n i=1 αikbi, with αik ∈ Jik} is a neighborhood of zk and f (x) > 2k , for all x ∈ Uk . The argument for n = 1 shows that there exists a sequence of integers (mik) going to infinity such that we have a nested sequence of nontrivial compact intervals · · · ⊂ 1 mik+1 Jik+1 ⊂ 1 mik Jik ⊂ · · · ⊂ Ji1. Take γi ∈ ∞⋂ k=1 1 mik Jik and construct a new basis B˜ = {γibi, i ∈ {1, 2, . . . , n}}. It follows thatf is unbounded on EB˜ . Problem 4.47. It is known that if f, g : C → C are entire functions without common zeros then there exist entire functionsa, b : C → C such thata(z)f (z)+b(z)g(z) = 1 for all z ∈ C. 1. Prove that we can choose a(z) without any zeros. 2. Is it possible to choose both a and b without zeros? 214 8 Analysis Solutions Solution 4.47. 1. We know that Af + Bg = 1 for some entire functions A and B. Therefore, for any entire function λ, we have also (A + λg)f + (B − λf )g = 1. We now choose a = A+λg such thatA+λg = eh. If z0 is a zero of g, we need to have eh(z0) = A(z0) (with the same multiplicity). This is possible from the interpolation theorem for entire functions, because A(z0)f (z0) = 1. 2. Let f, g ∈ C[z] be nonconstant polynomials of different degrees. We will show that there are no entire functions a, b such that af + bg = 1 and a, b have no zeros. There are no integer nonconstant functions. First, there exist nonzero polynomials A and B such that Af + Bg = 1. As above, for any entire function λ : C → C, one can construct other solutions, namely a = A − λg and b = B + λf . Let us now show that any pair a, b can be obtained in this way. We have (A − a)f + (B − b)g = 0. Thus the set of zeros of (A − a) contains the set of zeros of g (since f and g have no common zeros). In particular, there exists an entire function λ such that A − a = λg. In a similar way, there exists an entire function μ such that B − b = μf . The condition above shows that λ = μ, as claimed. If a, b do not have zeros, then the meromorphic function F = 1 λ must be distinct everywhere from the rational functions a1, a2, a3 given below: a1(z) = 0, a2(z) = g(z)/A(z), a3(z) = −f (z)/B(z). Consider the meromorphic function G = F − a1 F − a2 · a3 − a2 a3 − a1 . The points where G(z) ∈ {0, 1,∞} are among those satisfying one of the conditions a1(z) = a2(z), a1(z) = a3(z), a2(z) = a3(z), and therefore they are finitely many. According to Picard’s theorem, G does not have an essential singularity at ∞ and thus G is rational. This implies that F is rational; hence λf is a polynomial and thus a and b are polynomials. But polynomials without zeros are constant, and this is impossible since f and g have different degrees. Problem 4.48. Consider a compact set X ⊂ R. Show that a necessary and sufficient condition for the existence of a monic nonconstant polynomial with real coefficients h ∈ R[x] such that |h(x)| < 1 for all x ∈ X is the existence of monic nonconstant polynomial g(x) ∈ R[x] such that |g(x)| < 2 for all x ∈ X. Prove that 2 is the maximal number with this property. Solution 4.48. Let P(X) be the Banach space of real polynomials endowed with the norm ‖θ‖ = supx∈X |θ(x)|. We define the nonlinear operator A : P(x) → P(x) given by 8 Analysis Solutions 215 (Af (x)) = f 2(x) − 1 2 ‖f ‖2, for f ∈ P(X). If f is monic, then Af is also monic and we have ‖Af ‖ = 12‖f ‖2. An easy induction implies that ‖Anf ‖ = 2‖f ‖ 2n 22n . Moreover, if ‖f ‖ < 2, then there exists some n for which ‖Anf ‖ < 1. This proves the first part. Let X = [−2, 2]. The norm of p(x) = x on X is ‖p‖ = 2. We will prove that any monic polynomial on X has norm at least 2. Assume the contrary, namely that there exists f ∈ P(x), monic, with ‖f ‖ < 2. We have two intermediate results: 1. There exists an operator T : P(X) → P(X) that takes monic polynomials of degree 2k to monic polynomials of degree k such that ‖T P ‖ < ‖P ‖. 2. Assuming the existence of a monic polynomial of norm ‖f ‖ < 2, there exists n ∈ Z+ and a monic polynomial g, of degree 2n, with ‖g‖ < 2. By points 1 and 2 above there exists a monic polynomial g of degree 2n with ‖g‖ < 2 on [−2, 2], which implies that there exists a monic polynomial of degree 1 such that ‖g‖ < 2 on [−2, 2], which is obviously false. Proof of claim 1. If h ∈ P(x), let us consider h˜(x) = 12 (h(x) + h(−x)), which is even and thus can be written as a polynomial in x2. Thus h˜(x) = Q(x2), where now Q : [0, 4] → R. Define T (h)(x) = Q(x + 2). It is clear that T has the desired properties. Proof of claim 2. If f has degree 2mp, where p is odd, then t = T mf is monic of degreep, ‖t‖ < 2. There existsn ∈ Z+ such thatp divides 2n−1, and so 2n−1 = pq. We define g ∈ P(x) by means of g = t (x)qx. It is immediate that ‖g‖ < 2, and the proof is complete. 9 Glossary 9.1 Compendium of Triangle Basic Terminology and Formulas 9.1.1 Lengths in a Triangle We used the standard notation for the important features of a triangle, as follows: • a, b, c the sides lengths • A,B,C the angles • ma,mb,mc the medians • ha, hb, hc the altitudes • wa,wb,wc the angle bisectors • R the radius of the circumcircle • r the radius of the incircle • ra, rb, rc the radii of the extrinsic circles • S the area • p the semiperimeter We collect below some useful identities between these quantities: • S = 12bc sin A = (p(p − a)(p − b)(p − c))1/2 = 4Rr cos A2 cos B2 cos C2 , • S = rp, • 4RS = abc, • ra(p − a) = rp = S, ra = p tan A2 , • 4R + r = ra + rb + rc, • aha = 2S, • hahbhc = 8S3abc = 2S 2 R , • 1 ha + 1 hb + 1 hc = 1 ra + 1 rb + 1 rc = 1 r , • m2a + m2b + m2c = 34 ( a2 + b2 + c2), • wa = 2 √ bc b+c √ p(p − a), • ra = √ p(p−b)(p−c) p−a , 218 9 Glossary • a2 + b2 + c2 = 8R2(1 + cosA cosB cosC), • sin A2 = √ (p−b)(p−c) bc , • cosA + cosB + cosC = 1 + 4 sin A2 sin B2 sin C2 = 1 + rR , • tan A2 = √ (p−b)(p−c) p(p−a) , • R = a2 sin A = b2 sin B = c2 sin C , • 4m2a = 2b2 + 2c2 − a2 . 9.1.2 Important Points and Lines in a Triangle • The circumcenter O is the center of the triangle’s circumcircle. It can be found as the intersection of the perpendicular bisectors of the sides. • The incenter I is the center of the incircle and thus the intersection of the angle bisectors. • The orthocenter H is the intersection of the three altitudes of the triangle. • The centroid G is the intersection of the three medians. • The excenter JA is the center of the excircle lying in the angle B̂AC, outside the triangle ABC and tangent to the lines AB, AC and to the segment BC. There are three excenters, JA, JB, JC , using similar definitions. • If MA,MB,MC are the midpoints of BC,CA,AB respectively, then the three lines MAJA,MBJB,MCJC are concurrent, and their intersection point is called the mittenpunkt. • If SA, SB, SC are the contact points of the excircles with BC,CA,AB respec- tively, then ASA,BSB,CSC are concurrent and their intersection is called the Nagel point Na. • Let TA, TB, TC be the contact points of the incircle with sides BC,CA,AB re- spectively. Then the lines ATA,BTB,CTC are concurrent, and their intersection point is called the Gergonne point Ge. • The isogonal conjugate of the point X is the intersection of the three concurrent lines that are the reflections of AX,BX,CX about the angle bisectors of A,B,C respectively. • The intersection point of the symmedians is called the symmedian point (or the Lemoine point). It is the isogonal conjugate of the centroid G. • Two lines passing through the vertex A of a triangle ABC are called isotomic if they cut the side BC in points symmetric with respect to the midpoint of BC. Given a point P , the isotomic lines of the Cevians AP , BP , and CP meet at a point P ′ called the isotomic point of P . • The Fermat point F of an acute triangle is the point that minimizes the sum of the distances to the vertices. If we draw equilateral triangles BCA′, ABC′, ACB ′ on the outside of the triangleABC, then the three linesAA′, BB ′, CC′ are concurrent at F . • The Euler line: the points O,G,H are collinear and |HG| : |GO| = 2 : 1. 9.1 Compendium of Triangle Basic Terminology and Formulas 219 • The Gergonne point Ge, the triangle centroid G, and the mittenpunkt M are collinear, with |Ge G| : |GM| = 2 : 1. • The Nagel line: the points I,G,Na are collinear and |Na G| : |GI | = 2 : 1. Kimberling tabulated and enumerated properties of more than 1477 centers in a trian- gle (later Brisse extended the list to 2001 items). The interested reader might consult his updated web page: • C. Kimberling: see http://faculty.evansville.edu/ck6/encyclopedia/ETC.html 9.1.3 Coordinates in a Triangle We are given a reference triangle ABC. Barycentric coordinates. The barycentric coordinates of the point P are given by an ordered triple FP = (tA, tB, tC) (up to nonzero scalar multiplication) such that P is the centroid of the system consisting of the three masses tA, tB, tC located at the respective vertices A,B,C of the triangle, i.e., the mass tA is located at A, etc. These coordinates were introduced by Möbius in 1827. The coordinates tA are proportional to the areas area(PBC). If they are actually equal to these areas, they are called homogeneous barycentric coordinates. Here are the coordinates of the principal points in a triangle. • FG = (1, 1, 1) • FO = ( a2 ( b2 + c2 − a2), b2(a2 + c2 − b2), c2(a2 + b2 − c2)) • FJA = (−a, b, c), FI = (a, b, c) • FGe = ((p − b)(p − c), (p − c)(p − a), (p − a)(p − b)) • FNa = (p − a, p − b, p − c) • FH = (( a2 + c2 − b2)(a2 + b2 − c2), (a2 + b2 − c2)(b2 + c2 − a2),( a2 + c2 − b2)(b2 + c2 − a2)) • FK = ( a2, b2, c2 ) The line determined by the points (s1, s2, s2) and (t1, t2, t3) has the equation det ⎛ ⎝ s1 s2 s3 t1 t2 t3 x1 x2 x3 ⎞ ⎠ = 0. Trilinear coordinates. The trilinear coordinates of a point P are given by an ordered triple of numbers proportional to the directed distance from P to the sides, up to multiplication by a nonzero scalar. If we normalize them so that they give the distances to the sides, then these are called the exact trilinear coordinates. Trilinear coordinates are also known as homogeneous coordinates. They were introduced by Plücker in 1835. For instance, the vertices have trilinear coordinates (1 : 0 : 0), (0 : 1 : 0), and (0 : 0 : 1). 220 9 Glossary For most important points in a triangle, the trilinear coordinates are given by triples of the form (f (a, b, c) : f (b, c, a) : f (c, a, b)), for some function f on the sides a, b, c. For instance, we have fH (a, b, c) = cosB cosC, fG(a, b, c) = 1a , fO(a, b, c) = cosA, fI (a, b, c) = 1. For the Fermat point, we can take fF (a, b, c) = sin(A + π3 ). The homogeneous barycentric coordinates corresponding to (x : y : z) are (ax, by, cz). The trilinear coordinates of the isogonal conjugate of (x : y : z) are( 1 x , 1 y , 1 z ) . 9.2 Appendix: Pell’s Equation Pell’s equation is the Diophantine equation x2 − dy2 = 1. The first treatment of this equation was actually given by Lord William Brouncker in 1657, but his solution was erroneously ascribed to John Pell by Euler. Lagrange developed the theory of continued fractions, which gives the actual method of finding the minimal solution, and published the first proof in 1766. However, some methods (the cyclic method) were already known to the Indian mathematicians Brahmagupta and Bhaskara in the twelfth century. General solutions. Pell’s equation has infinitely many solutions if d is not a perfect square and none otherwise. If (x0, y0) denotes its minimal solution (called also the fundamental solution) different from (1, 0), then all solutions (xn, yn) are obtained from it by means of the recurrence relations xn+1 = x0xn + dy0yn, yn+1 = y0xn + x0yn, x1 = x0, y1 = y0. The fundamental solution. Thus the main point is finding the fundamental solution. This can be achieved using continued fractions. According to Lagrange, the continued fraction of a quadratic irrational number (i.e., a real number satisfying a quadratic equation with rational coefficients that is not rational itself) is periodic after some point; in our case, even more can be obtained, namely √ d = [ a0, a1, . . . , am, 2a0 ] . The numbersaj can be calculated recursively, as follows. Seta0 = [√ d ] and construct the sequences P0 = 0, Q0 = 1, P1 = a0, Q1 = d − a20, Pn = an−1Qn−1 − Pn−1, Qn = d−P 2 n Qn−1 . Then an = [ a0 + Pn Qn ] . 9.2 Appendix: Pell’s Equation 221 Furthermore, consider the sequences p0 = a0, q0 = 1, p1 = a0a1 + 1, q1 = a1, pn = anpn−1 + pn−2, qn = anqn−1 + qn−2. Then we have the identities p2n − dq2n = (−1)n+1Qn+1, which permit one to construct solutions to the Pell-type equations. If am+1 = 2a0, as happens for a quadratic irrational, thenQm+1 = 1. In particular, we find that the fundamental solution is determined by the parity of m, as follows: x0 = pm, y0 = qm, if m is odd, x0 = p2m+1, y0 = q2m+1, if m is even. Observe that pr/qr = [a0, a1, . . . , ar ]. The general solution (xn, yn) is obtained by means of the trick x2 − dy2 = (x20 − dy20 )n = 1, and setting x + √dy = (x0 + √ dy0 )n , x − √dy = (x0 − √ dy0 )n , we obtain the family of solutions xn = ( x0 + √ dy0 )n + (x0 − √ dy0 )n 2 , yn = ( x0 + √ dy0 )n − (x0 − √ dy0 )n 2 . The negative Pell equation. The Pell-like equation x2 − dy2 = −1 can also be solved, but it does not always have solutions. When it has one, then it has infinitely many. A necessary condition for this equation to be solvable is that all odd prime factors of d be of the form 4k + 1, and that d �≡ 0 (mod 4). However, these conditions are not sufficient, as can be seen from the equation for d = 34, which has no solutions. The method of continued fractions works for these equations as well. In fact, for this Pell-type equation, we have the fundamental solution x0 = pm, y0 = qm, if m is even. The equation does not have solutions if m is odd. Furthermore, the general solution is given again by xn = ( x0 + √ dy0 )n + (x0 − √ dy0 )n 2 , yn = ( x0 + √ dy0 )n − (x0 − √ dy0 )n 2 , 222 9 Glossary but only for odd n. Pell-like equations. The Pell-like equation x2 − dy2 = c can be solved using the same ideas. If |c| < √d, then the equation has solutions if and only if c ∈ { Q0,−Q1, . . . , (−1)m+1Qm+1 } . Moreover, if c > √ d , then the procedure is significantly more complicated. See for instance the comments of Dickson on this subject. It is clear that using the solutions (xn, yn) to the companion Pell equation x2 −dy2 = 1 and a particular solution (u, v) of the general Pell-type equation from above, we are able to find infinitely many solutions by means of the recurrence un = xnu ± dynv, vn = xnv ± ynu. However, even if we start with the minimal solution, we cannot always obtain all solutions of the equation using this recurrence. The reason is that we might well have several fundamental solutions. For instance, if d = 10 and c = 9, then we have the fundamental solutions (7, 2), (13, 4) and (57, 18). The general equation ax2 − by2 = c can be reduced to the former equation by setting d = ab, x′ = ax, c′ = ac and looking for those solutions where x′ is divisible by a. The degree-two equation in two unknowns. Moreover, the general degree-two equa- tion ax2 + bxy + cy2 + dx + cy + f = 0 where a, b, c, d, e, f ∈ Z, can be reduced to either Pell-type equations or elliptic equations. 1. The case � = b2 − 4ac < 0. The equation represents an ellipse in the plane, and therefore it has only finitely many integer solutions. These equations were solved by Gauss and earlier considered by Euler. According to Dickson one can find all its solutions using the solutions to Pell’s equations. 2. If � = 0, the parabolic-type equation becomes 1 4a ( (2ax + by + d)2 + 2(ae − bd)y + 4af − d2 ) = 0. If 2ae− bd = 0, then the equation is equivalent, through a linear transformation, to (2ax + by + d)2 = d2 − 4af ; hence the parabola degenerates into two lines. If d2 − 4af is a perfect square, then we have two infinite families of solutions 2ax + by + d = ±√d2 − 4af ; otherwise, there are no integer solutions. 9.2 Appendix: Pell’s Equation 223 If 2ae−bd �= 0, then we have to solve the equation 2(ae−bd)y = −t2+d2−4af , for an arbitrary parameter t ∈ Z. This depends only on computing the residues modulo 2(ae − bd) of the perfect squares, and yields infinitely many solutions if one exists. Furthermore, 2ax = t − d − by determines x. 3. If � > 0, then the hyperbolic-type equation becomes: a(�x − 2cd + bc)2 + b(�x − 2cd + bc)(�y + 2ac + bd) + c(�y − 2ac + bd)2 = −�(ac2 + cd2 + ef 2 − bdc − 4acf ), which, through a transformation u = �x − 2cd + bc, v = �y − 2ac+ bd, becomes au2 + buv + cv2 = h. This equation can be reduced to a Pell equation. It will have then either zero or infinitely many solutions. The Ankeny–Artin–Chowla Conjecture. Although the Pell equation seems to be well understood, there are many subtle questions concerning its solutions. Here is a problem that has resisted to all efforts until now. 1. If p is a prime p ≡ 1 (mod 4) and y0 is the smallest positive value of y such that x2 − py2 = −4, then y0 �≡ 0 (mod p). 2. If p is a prime p ≡ 3 (mod 4) and y0 is the smallest positive value of y such that x2 − py2 = 1, then y0 �≡ 0 (mod p). The first conjecture is due to Ankeny, Artin, and Chowla, and the second to Mordell. The Thue theorem for higher degree. The situation of equations of higher degree is completely different. Let f (x, y) = a0xn + a1xn−1y + a2xn−2y2 + · · · + anyn be a homogeneous form of degree n. If n ≥ 3 and f is irreducible over Q, then the equation f (x, y) = g(x, y), where g(x, y) is an arbitrary form of degree n − 2 (not necessarily homogeneous), has only finitely many solutions. This is a celebrated theorem due to Thue and based on results of Siegel and Roth. Later A. Baker gave explicit bounds on the size of the solutions for a wide class of such f . • L.E. Dickson: Pell equation: ax2 + by2 + c made square, Chapter 12 in History of the Theory of Numbers, Vol. 2: Diophantine Analysis, New York, Chelsea, 341–400, 1952. • H.W. Lenstra, Jr.: Solving the Pell equation, Notices Amer. Math. Soc. 49 (2002), 2, 182–192. Index of Mathematical Results ˇCebotarev density theorem, 83 Ankeny–Artin–Chowla conjecture, 223 Birch–Swinnerton-Dyer conjecture, 81 Blaschke theorem, 211 Bonnesen’s isoperimetric inequality, 170 Bonsé–Pósa inequality, 46 Borsuk problem, 153, 156, 175 Borsuk–Lusternik–Shnirelman theorem, 175 Catalan’s conjecture, 56 Cauchy’s integral formula, 167 Cauchy’s rigidity theorem, 171, 173 Cauchy–Schwartz inequality, 167 Cauchy–Schwarz inequality, 175 Ceva’s theorem, 139 Chebyshev theorem, 40, 45, 46, 58 Chinese Remainder theorem, 82 Chinese remainder theorem, 50, 78 congruent numbers problem, 80 De Moivre’s identity, 60 Dirac theorem, 97 Dirichlet theorem on arithmetic progressions, 50, 82, 97 Eisenstein criterion, 38 Erdo˝s conjecture, 114 Erdo˝s theorem, 126 Erdo˝s–Straus conjecture, 63 Euler conjecture, 42 Euler criterion, 72, 73 Euler theorem in a triangle, 182 Fermat equation, 42, 67 four color theorem, 104 Frobenius coin problem, 50 fundamental theorem of symmetric polynomials, 85, 89 Gowers theorem, 123 Green–Tao theorem, 123 Hadamard inequality, 186 Hahn–Banach lemma, 131 Heawood conjecture, 103 Helly’s theorem, 159 intermediate value theorem, 188 isoperimetric inequality, 147 Jensen inequality, 179 Jung’s theorem, 159 Kahn–Kallai theorem, 156 Kaprekar constant, 95 Lagrange four squares theorem, 74 Lagrange mean value theorem, 194 Lagrange multiplier method, 189 Lagrange theorem on periodic continued fractions, 220 Laplace method, 198 Lehmer conjecture, 57 Ljunggren equation, 69 Maclaurin series, 189 226 Index of Mathematical Results maximum principle for holomorphic functions, 211 Means inequality, 161, 179, 180, 189 Millennium problems, 81, 195 Minkowski’s theorem, 165 odd perfect number conjecture, 59 Pell equation, 43 Pell’s equation, 220 Picard theorem, 214 pigeonhole principle, 86, 92, 96, 97, 108, 112 quadratic reciprocity theorem, 50 Rayleigh–Beatty theorem, 101 Riemann hypothesis, 195 Roché inequality, 28, 182 Sturm comparison theorem, 212 superperfect numbers conjecture, 60 Sylvester’s problem, 172 Szemerédi theorem, 122 Taylor formula, 209 Thales’ theorem, 140 Thue’s theorem, 223 Ulam problem, 126 Van der Waerden theorem, 122 Waring problem, 52, 74, 76 Wiles theorem, 42 Wilson theorem, 44, 53 Index of Mathematical Terms abundant number, 9, 81 altitude, 22, 25, 27, 139, 160, 162, 175, 217 annulus, 190 area, 8, 19, 23, 25–27, 70, 80, 124, 140–142, 145, 154, 157, 159, 161, 165–167, 170, 173, 176, 177, 181, 182, 190, 217 arithmetic progression, 4, 7, 12, 40, 50, 56, 82, 87, 97, 122 asymptotic, 16, 78, 106, 114, 128 ball, 25, 27, 143, 156–159, 174, 175 Banach space, 214 barycentric coordinates, 142, 219 Bernoulli numbers, 54 Bessel function, 210 binomial binomial coefficient, 14, 41, 62, 97, 98, 105, 109 binomial expansion, 38, 62, 104 binomial coefficient, 3, 37 bisector, 21, 25, 27, 133, 134, 162, 175, 217 body, 25, 145, 156–158, 166 boundary, 18, 25, 150, 151, 156–158, 161, 165, 166, 174, 211 cardinal of a set, 15, 17, 50, 78, 88, 92, 100–102, 104, 108, 111, 112, 117, 128, 191 cardinality of a set, 15 centroid, 218 Cevian, 139 choice number, 3, 37 chord, 21, 22, 24, 135–137, 152, 157, 159, 160 circular permutation, 62, 190 circumcenter, 218 circumscribable, 23, 140 collinear points, 23, 126, 138, 140–142, 172, 182, 218, 219 commensurable, 126, 203 compact, 34, 151, 157, 212–214 complex number, 13, 24, 30, 89, 92, 126, 135, 136, 147, 148, 190, 200 complex plane, 55 computer search, 43, 63, 65 computer-assisted proof, 104 concave, 149 congruence modulo p, 3, 9, 39, 40, 44, 50, 52, 61, 63, 66, 72, 73, 78–80, 82, 83 congruence transformation, 24, 151 congruent number, 80 continued fraction, 220 convergent, 9, 30–33, 77, 78, 190, 191, 193, 194, 203, 205 convex, 16, 22–26, 106, 136, 139, 143, 144, 147, 149, 156–158, 167, 168, 170, 171 convex hull, 136, 156, 158 covering, 27, 173 critical, extremal point, 180, 190 deficient number, 9, 81 dense, 34, 126, 131, 142, 157, 203, 211 density, 83, 109, 122, 128 determinant, 11, 86, 87, 109, 138, 141, 186 diameter, 24, 25, 145, 147, 151, 152, 154, 156, 158, 167, 174 differential equation, 198 digits, 4, 5, 13, 40, 47, 95 228 Index of Mathematical Terms Diophantine equation, 42, 53, 56, 64, 67, 69, 70, 80, 115, 220 discriminant, 48, 83, 194 distance, 16, 19, 20, 22–24, 26, 107, 122, 126, 128, 130, 131, 136, 142, 143, 145, 146, 148, 151, 152, 165, 167, 169, 219 domain, 26, 170, 171, 174 eigenvalue, 90, 91, 211 Euler inequality, 28, 177 Euler line, 22, 138, 218 Euler totient function, 3, 6, 7, 39, 51, 57, 61, 92 Euler–Poincaré characteristic, 103, 150, 171, 172 Eulerian circuit, 19, 123 excenter, 218 factorial, 3, 37 Fermat point, 22, 138, 218 Fibonacci sequence, 118, 185 fractional part, 193 function, 12, 23, 31, 33, 88, 144, 165, 194, 204, 207 additive function, 208 bijective function, 15, 102 concave (or convex) function, 179 continuous function, 33, 34, 71, 153, 157, 206, 209, 210 decreasing function, 31, 139, 196 entire function, 34, 213 function, 104 generating function, 16, 106 holomorphic function, 211 hyperbolic trigonometric function, 203 increasing function, 15, 64, 101, 110, 189 meromorphic function, 214 periodic function, 31, 193 positive function, 32, 198 rational function, 13, 42, 92 smooth function, 32–34, 87, 162, 198, 199, 208, 212 Gauss map, 156 Gergonne point, 23, 141, 218, 219 graph, 15, 19, 97, 102, 103, 123, 147, 150 Hadwiger number, 163 Hamiltonian cycle, 97 homographic division, 138 homothety, 26, 106, 126, 136, 170, 171 incenter, 218 index of mathematical terms, 37 infinite descent method, 56, 67 inscribable, 8, 22, 23, 26, 70, 137, 140, 141, 169 integer part, 4, 39 integral equation, 34, 209 integral operator, 211 intrinsic perimeter, 164 isosceles, 18, 21, 25, 122, 134, 135, 138, 159, 160 isotomic point, 218 Jacobian matrix, 190 lattice, 26, 122, 129, 165 Legendre symbol, 72, 82 Lemoine (symmedian) point, 218 length, 19, 21–27, 29, 94, 96, 124, 125, 135, 137, 140, 143, 145, 146, 151, 152, 154–156, 159, 160, 162, 166–169, 171, 175, 182, 187, 211, 217 linear linear combination, 88, 93, 209 linear equation, 113 linear function, 33, 86, 204, 207 linear functional, 131 linear independence, 13, 25, 91, 92, 162 linear map, 131 matrix, 26, 86, 87, 91, 109, 111, 141, 142, 162, 164, 186, 190 median, 27, 175, 217 Mersenne prime number, 55 Mersenne primes, 59 Minkowski norm, 164 mittenpunkt, 218 Nagel line, 219 Nagel point, 23, 141, 218 nilpotent, 90, 93 non-collinear points, 19 noncollinear points, 27, 124, 126, 172 nonlinear operator, 214 number of combinations, 3, 37 orthocenter, 218 Index of Mathematical Terms 229 orthogonal, 130, 136, 137, 143, 145, 153–157 packing, 19, 125, 129 parameterization (of a curve), 41 partition, 15, 17, 26, 99, 103, 113, 114, 151, 153, 156, 158, 168, 169, 171, 174 pentagonal number, 8, 69 perfect number, 59 periodic, 31, 193, 203, 225 poder, 181 polygon, 16, 23, 26, 29, 106, 121, 143, 144, 146, 147, 150, 158, 168, 171, 187 polyhedron, 27, 153, 157, 171–173 polynomial characteristic polynomial, 91 irreducible polynomial, 3, 38, 83, 92, 223 monic polynomial, 34, 214 polynomial with arbitrary coefficients, 11, 13, 86, 91, 93, 211, 214 polynomial with integer coefficients, 6, 9, 12, 40, 51, 61, 82, 89 real polynomial, 12, 29, 30, 33, 86, 87, 185, 188, 206, 208 set of polynomials, 32, 197 symmetric polynomial, 85, 89 projective duality, 172 projective transformation, 138 Pythagorean triple, 5, 42, 44, 66, 67, 69 quadratic irrational number, 220 quadratic residue, 50, 68, 72, 73 �, 29, 185 relatively prime, 39, 40, 46, 65, 78, 86, 93 Riemann zeta function, 54, 194 root, 34, 61, 83, 89, 91–93, 160, 200, 212 sequence, 4, 6, 9, 12, 14, 17, 18, 30–32, 40, 43, 48, 50, 61, 74, 76, 77, 81, 87, 95, 96, 98, 100, 110, 112, 114, 116, 117, 119, 127, 128, 147, 191, 193, 196, 201, 203, 206, 207, 213 series, 77 simplex, 25, 156, 158, 159, 171 stereographic projection, 41 sum-free set, 17, 114 superperfect number, 60 surface, 26, 103, 154, 166, 174 symmedian, 181 symmetric, 137, 154, 155, 161, 166, 174, 175 symmetry, 51, 88, 97, 106, 133, 136, 137, 160, 166, 170, 172, 205 tiling, 121, 149 torus, 13, 93 transcendental number, 55, 128 triangular number, 8, 69 trilinear coordinates, 138, 141, 219 vector space, 90, 130, 208 width, 23, 145, 152, 153, 155, 166, 167 Index of Topics and Methods Classical arithmetic functions and ap- plications: 1.4, 1.29, 1.34, 1.36, 1.42, 1.43, 1.44, 1.45, 1.46, 1.69 Congruences and divisibility: 1.1, 1.3, 1.8, 1.11, 1.15, 1.17, 1.22, 1.24, 1.27, 1.38, 1.48, 1.50, 1.51, 1.67, 1.68, 2.7, 2.45 Rational and algebraic numbers: 1.47, 2.14, 2.15 Polynomials: 1.2, 1.7, 1.30, 2.2, 2.4, 2.9, 2.12, 2.13, 2.24 Estimates of arithmetically defined functions: 1.5, 1.6, 1.23, 1.28, 1.37, 1.55, 1.65, 1.66, 2.25, 2.35, 2.47 Symmetric functions: 2.1, 2.8 General Diophantine equations (Pyth- agorean, exponential): 1.12, 1.14, 1.16, 1.20, 1.25, 1.33, 1.35, 1.39, 1.41, 1.49, 1.52, 1.56, 1.57, 1.60 Pell’s equation and its applications: 1.13, 1.58, 1.59 Continued fractions: 1.13 Prime numbers, Chebyshev theorem: 1.9, 1.18, 1.19, 1.21, 1.32, 1.54, 2.46 Arithmetic progressions: 1.10, 1.26, 1.40, 2.5, 2.21 Representing integers by algebraic functions, Waring problem: 1.31, 1.53, 1.61, 1.62, 1.63, 1.64 Algebraic number theory: 1.70 Groups: 2.10 Combinatorial identities: 2.33, 2.34 Algebraic equations: 2.3, 2.16 Linear algebra, matrices, determi- nants: 2.6, 2.11, 2.17, 2.18, 2.40, 2.43 Combinatorial properties of sets of in- tegers: 2.26, 2.27, 2.28, 2.29, 2.41, 2.44 Sequences of integers: 2.19, 2.42, 2.52, 2.53 Counting problems: 2.23, 3.21 Extremal problems concerning sets of integers: 2.36, 2.37, 2.48, 2.49, 2.50, 2.51 232 Index of Topics and Methods Ramsey-type problems: 2.20, 2.30, 2.32, 2.38, 2.39, 2.56, 2.57, 2.59, 2.62 Graphs and applications: 2.22, 2.31, 2.58 Tiling, packing and covering: 2.55, 2.60, 2.64, 3.46, 3.47, 3.53 Area and volume: 3.42, 3.45, 3.51, 3.56, 3.60 Distances in combinatorial geometry: 2.61, 2.63, 3.23, 3.48, 3.54 Geometry of the triangle: 3.1, 3.11, 3.12, 3.14, 3.16 Complex numbers and applications: 3.5, 3.6, 3.8, 4.1, 4.19, 4.34 Geometric constructions, geometric loci: 3.2, 3.3, 3.4, 3.7, 3.9, 3.10, 3.15 Trigonometry: 3.17, 3.30 Analytic methods in geometry: 3.18 Higher dimensional geometry: 3.13, 3.57 Convex geometry, integral formulas, isoperimetric inequalities: 3.24, 3.25, 3.29, 3.49, 3.50, 3.55 Extremal problems concerning finite sets of points: 2.54, 3.20, 3.27, 3.28, 3.31, 3.37, 3.40, 3.41, 3.44, 3.52 General geometric inequalities: 3.22, 3.26, 3.43, 4.5 Geometric inequalities in the triangle: 3.62, 3.63, 3.64, 3.65, 3.66, 3.67, 3.68, 3.69, 3.70, 3.71, 3.72, 3.73 Topological methods in geometry: 3.32, 3.33, 3.34, 3.35, 3.36, 3.38, 3.39, 3.58, 3.59, 3.61 General inequalities: 4.2, 4.3, 4.4, 4.6, 4.14, 4.15, 4.22, 4.26 Convexity of real functions: 4.16 Minima and maxima of real functions, mean values: 4.7, 4.8, 4.9, 4.10, 4.37 Functions of a complex variable, holo- morphic functions: 4.20, 4.24, 4.44, 4.47 Sums of series and convergence of se- quences: 4.11, 4.12, 4.17, 4.18, 4.23, 4.28, 4.29, 4.31 Dynamical systems, density of sets of reals: 3.19, 4.32, 4.35, 4.43 Integrals: 4.13, 4.21, 4.30 Functional equations and inequations: 4.27, 4.33, 4.36, 4.38 Integral and differential equations: 4.25, 4.39, 4.40, 4.41, 4.42, 4.45 Linear functionals and operators: 2.65, 4.46, 4.48 cover-image-large front-matter Part I fulltext_1 fulltext_2 fulltext_3 fulltext_4 Part II fulltext_5 fulltext_6 fulltext_7 fulltext_8 fulltext_9 back-matter