WATSON’S METHOD OF SOLVING A
QUINTIC EQUATION
Melisa J. Lavallee, Blair K. Spearman, and Kenneth S. Williams
Abstract. Watson’s method for determining the roots of a solvable quin-
tic equation in radical form is examined in complete detail. New methods
in the spirit of Watson are constructed to cover those exceptional cases to
which Watson’s original method does not apply, thereby making Watson’s
method completely general. Examples illustrating the various cases that
arise are presented.
1. Introduction. In the 1930’s the English mathematician George Neville
Watson (1886-1965) devoted considerable effort to the evaluation of singular
moduli and class invariants arising in the theory of elliptic functions [6]-[11].
These evaluations were given in terms of the roots of polynomial equations
whose roots are expressible in terms of radicals. In order to solve those equa-
tions of degree 5, Watson developed a method of finding the roots of a solvable
quintic equation in radical form. He described his method in a lecture given
at Cambridge University in 1948. A commentary on this lecture was given
recently by Berndt, Spearman and Williams [1]. This commentary included a
general description of Watson’s method. However it was not noted by Wat-
son (nor in [1]) that there are solvable quintic equations to which his method
does not apply. In this paper we describe Watson’s method in complete detail
treating the exceptional cases separately, thus making Watson’s method appli-
cable to any solvable quintic equation. Several examples illustrating Watson’s
method are given. Another method of solving the quintic has been given by
Dummit [4].
———————–
2000 Mathematics Subject Classification: 12E10, 12E12, 12F10.
Key words and phrases: solvable quintic equations, Watson’s method.
The second and third authors were supported by research grants from the Natural Sciences
and Engineering Research Council of Canada.
1
2 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
2. Watson’s method. Let f(x) be a monic solvable irreducible quintic
polynomial in Q[x]. By means of a linear change of variable we may suppose
that the coefficient of x4 is 0 so that
(2.1) f(x) = x5 + 10Cx3 + 10Dx2 + 5Ex+ F
for some C , D, E, F ∈ Q. Let x1, x2, x3, x4, x5 ∈ C be the five roots of f(x).
The discriminant δ of f(x) is the quantity
(2.2) δ =
∏
1≤j 0.
We set
(2.5) K = E + 3C2,
(2.6) L = −2DF + 3E2 − 2C2E + 8CD2 + 15C4,
(2.7) M = CF 2 − 2DEF + E3 − 2C2DF − 11C2E2
+28CD2E − 16D4 + 35C4E − 40C3D2 − 25C6.
Let x1, x2, x3, x4, x5 ∈ C be the five roots of f(x). Cayley [2] has shown that
(2.8)
φ1 = x1x2 + x2x3 + x3x4 + x4x5 + x5x1 − x1x3 − x3x5 − x5x2 − x2x4 − x4x1,
φ2 = x1x3 + x3x4 + x4x2 + x2x5 + x5x1 − x1x4 − x4x5 − x5x3 − x3x2 − x2x1,
φ3 = x1x4 + x4x2 + x2x3 + x3x5 + x5x1 − x1x2 − x2x5 − x5x4 − x4x3 − x3x1,
φ4 = x1x2 + x2x5 + x5x3 + x3x4 + x4x1 − x1x5 − x5x4 − x4x2 − x2x3 − x3x1,
φ5 = x1x3 + x3x5 + x5x4 + x4x2 + x2x1 − x1x5 − x5x2 − x2x3 − x3x4 − x4x1,
φ6 = x1x4 + x4x5 + x5x2 + x2x3 + x3x1 − x1x5 − x5x3 − x3x4 − x4x2 − x2x1,
are the roots of
(2.9) g(x) = x6 − 100Kx4 + 2000Lx2 − 32
√
δx+ 40000M ∈ Q[x].
WATSON’S METHOD 3
Watson [1] has observed as f(x) is solvable and irreducible that g(x) has a
root of the form φ = ρ
√
δ, where ρ ∈ Q, so that φ ∈ Q(
√
δ). We set
(2.10) θ =
φ
√
5
50
∈ Q(
√
5δ) ⊆ R.
Clearly θ is a root of
(2.11) h(x) = x6 − K
5
x4 +
L
125
x2 −
√
5δ
390625
x+
M
3125
.
The following simple lemma enables us to determine the solutions of a
quintic equation.
Lemma. Let C , D, E, F ∈ Q. If u1, u2, u3, u4 ∈ C are such that
(2.12) u1u4 + u2u3 = −2C,
(2.13) u1u
2
2
+ u2u
2
4
+ u3u
2
1
+ u4u
2
3
= −2D,
(2.14) u2
1
u2
4
+ u2
2
u2
3
− u3
1
u2 − u32u4 − u33u1 − u34u3 − u1u2u3u4 = E,
(2.15) u5
1
+ u5
2
+ u5
3
+ u5
4
− 5(u1u4 − u2u3)(u21u3 − u22u1 − u23u4 + u24u2) = −F,
then the five roots of f(x) = 0 are
(2.16) x = ωu1 + ω
2u2 + ω
3u3 + ω
4u4,
where ω runs through the fifth roots of unity.
Proof. This follows from the identity(
ωu1 + ω
2u2 + ω
3u3 + ω
4u4
)5 − 5U (ωu1 + ω2u2 + ω3u3 + ω4u4)3
− 5V (ωu1 + ω2u2 + ω3u3 + ω4u4)2 + 5W (ωu1 + ω2u2 + ω3u3 + ω4u4)
+ 5(X − Y )− Z = 0,
where
U = u1u4 + u2u3,
V = u1u
2
2
+ u2u
2
4
+ u3u
2
1
+ u4u
2
3
,
W = u2
1
u2
4
+ u2
2
u2
3
− u3
1
u2 − u32u4 − u33u1 − u43u3 − u1u2u3u4,
X = u3
1
u3u4 + u
3
2
u1u3 + u
3
3
u2u4 + u
3
4
u1u2,
Y = u1u
2
3
u2
4
+ u2u
2
1
u2
3
+ u3u
2
2
u2
4
+ u4u
2
1
u2
2
,
Z = u5
1
+ u5
2
+ u5
3
+ u5
4
,
see for example [5, p. 987].
4 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
If θ 6= 0,±C Watson’s method of determining the roots of f(x) = 0 in
radical form is given in the next theorem.
Theorem 1. Let f(x) be the solvable irreducible quintic polynomial (2.1).
Suppose that θ 6= 0, ±C . Set
p(T ) = T 4 + (−14Cθ2 − 2D2 + 2CE − 2C3)T 2 + 16Dθ3T(2.17)
+ (−25θ6 + (35C2 + 6E)θ4 + (−11C4 − 2CD2 − 4C2E −E2)θ2
+ (C6 + 2C3D2 − 2CD2E − 2C4E + C2E2 +D4)
and
q(T ) = −CT 3 −DθT 2 + (25θ4 − (10C2 + E)θ2 + C4 + CD2 − C2E)T(2.18)
+ (−Fθ3 + (−2CDE + C2F +D3)θ).
Then the pair of equations
(2.19) p(T ) = q(T ) = 0
has at least one solution T ∈ C, which is expressible by radicals. Set
(2.20) R1 =
√
(D − T )2 + 4(C − θ)2(C + θ)
and
(2.21) R2 =
C(D2 − T 2) + (C2 − θ2)(C2 + 3θ2 −E)
R1θ
, if R1 6= 0
√
(D + T )2 + 4(C + θ)2(C − θ), if R1 = 0.
If R1 6= 0 we have
(2.22) R2 = ±
√
(D + T )2 + 4(C + θ)2(C − θ).
Set
(2.23) X =
1
2
(−D + T +R1), X = 1
2
(−D + T − R1),
(2.24) Y =
1
2
(−D − T +R2), Y = 1
2
(−D− T − R2),
(2.25) Z = −C − θ 6= 0, Z = −C + θ 6= 0.
Let u1 be any fifth root of X
2Y/Z2. Set
(2.26) u2 =
X
Z
2
u2
1
, u3 =
XY
ZZ
3
u3
1
, u4 =
X
2
Y
Z2Z
4
u4
1
.
Then the five roots of f(x) = 0 are given in radical form by (2.16).
WATSON’S METHOD 5
The proof of Theorem 1 is given in Section 3. Theorem 1 does not apply
when θ = 0 since in this case R2 is not always defined; when θ = C since Z
= 0 and u2, u3, u4 are not defined; and when θ = −C since Z = 0 and u1,
u3, u4 are not defined. These excluded cases were not covered by Watson [1]
and are given in Theorems 2, 3, 4. Theorem 2 covers θ = ±C 6= 0, Theorem
3 covers θ = C = 0, and Theorem 4 covers θ = 0, C 6= 0.
Theorem 2. Let f(x) be the solvable irreducible quintic polynomial (2.1).
Suppose that θ = ±C 6= 0. Set
(2.27) r(T ) = T 3 +DT 2 + (−16C3 + 2CE −D2)T + (2CDE −D3)
and
s(T ) = T 6 + (88C3 − 3D2)T 4 + (112C3D − 4C2F )T 3(2.28)
+ (−64C6 − 32C3D2 − 4C2DF + 3D4)T 2
+ (128C6D − 48C3D3 + 4C2D2F )T
+ (−64C6D2 + 8C3D4 + 4C2D3F −D6).
Then the pair of equations
(2.29) r(T ) = s(T ) = 0
has at least one solution T ∈ C expressible by radicals.
(a) If D 6= ±T we let u1 be any fifth root of
(2.30)
−(D− T )2(D + T )
4C2
, if θ = C,
−16C4(D − T )
(D + T )2
, if θ = −C,
and set
(2.31) u2 =
2C
(D − T )u
2
1
, if θ = C,
−(D + T )
4C2
u2
1
, if θ = −C,
(2.32) u3 =
4C2
D2 − T 2u
3
1
, if θ = C,
0, if θ = −C,
6 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
(2.33) u4 =
0, if θ = C,
(D + T )2
8(D − T )C3u
4
1
, if θ = −C.
(b) If D = ±T then
(2.34) D = 0
and either
(2.35) (i) E = 4C2 or (ii) E =
16C3 − Z2
2C
, F =
64C6 − 88C3Z2 − Z4
4C2Z
for some Z ∈ Q, Z 6= 0. Let u1 be any fifth root of
(2.36)
−F +√128C5 + F 2
2
, (i),
2CZ, (ii).
and set
(2.37) u2 = 0,
(2.38) u3 =
{
0, (i),
1
2C
u3
1
, (ii),
(2.39) u4 =
1
16C4
(
−F −√128C5 + F 2
2
)
u4
1
, (i),
− 1
Z
u4
1
, (ii).
Then in both cases (a) and (b) the five roots of f(x) = 0 are given by (2.16).
Theorem 3. Let f(x) be the solvable irreducible quintic polynomial (2.1).
Suppose that θ = C = 0. In this case
(2.40) (i) D = E = 0 or (ii) D 6= 0, E 6= 0.
In case (ii) we have
(2.41) F =
E3 − 16D4
2DE
.
Let u1 be any fifth root of
(2.42)
{ −F, (i),
8D3
E
, (ii).
WATSON’S METHOD 7
Set
(2.43) u2 = u4 = 0
and
(2.44) u3 =
{
0, (i),
− E
4D2
u3
1
, (ii).
Then the five roots of f(x) = 0 are given by (2.16).
Theorem 4. Let f(x) be the solvable irreducible quintic polynomial (2.1).
Suppose that θ = 0, C 6= 0. Set
(2.45) T =
√
C3 −CE +D2.
Then
(2.46) T 2((D− T )2 + 4C3)((D + T )2 + 4C3)
= (2C3D − C2F +D3 −DT 2)2.
Let R1, R2 ∈ C be such that
(2.47) R2
1
= (D − T )2 + 4C3,
(2.48) R2
2
= (D + T )2 + 4C3,
(2.49) TR1R2 = 2C
3D − C2F +D3 −DT 2.
Set
(2.50) X =
1
2
(−D + T +R1), X = 1
2
(−D + T − R1),
(2.51) Y =
1
2
(−D − T +R2), Y = 1
2
(−D− T − R2).
Let u1 be any fifth root of
X2Y
C2
. Set
(2.52) u2 =
X
C2
u2
1
, u3 =
XY
C4
u3
1
, u4 =
X
2
Y
C6
u4
1
.
Then the five roots of f(x) = 0 are given by (2.16).
3. Proof of Theorem 1. If C 6= 0 or D 6= 0 or 25θ4 − (10C2 + E)θ2 +
(C4+CD2−C2E) 6= 0 the polynomial q(T ) is non-constant and the resultant
R(p, q) of p and q is
(3.1) R(p, q) = 510θ4(C2 − θ2)3h(θ)h(−θ).
As h(θ) = 0, we have R(p, q) = 0. Thus (2.19) has at least one solution T ∈ C,
which is expressible by radicals as it is a root of the quartic polynomial p(T ).
8 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
On the other hand if C = D = 25θ4−(10C2+E)θ2+(C4+CD2−C2E) = 0
we show that q(T ) is identically zero and the assertion remains valid. In this
case 25θ4 − Eθ2 = 0. As θ 6= 0 we have θ2 = E/25. Thus (2.3), (2.5), (2.6)
and (2.7) give
δ = 2855E5 + 55F 4, K = E, L = 3E2, M = E3.
Hence
h(x) = x6 − E
5
x4 +
3E2
53
x2 − (2
8E5 + F 4)
1/2
55
x+
E3
55
.
Thus
0 = h(θ) =
24E3
56
± (2
8E6 + EF 4)
1/2
56
.
As E 6= 0 we deduce that F = 0 proving that q(T ) ≡ 0.
Multiplying p(T ) = 0 by C2 − θ2 ( 6= 0) and rearranging, we obtain
(C(D2 − T 2) + (C2 − θ2)(C2 + 3θ2 −E))2(3.2)
= θ2((D − T )2 + 4(C − θ)2(C + θ))((D + T )2 + 4(C + θ)2(C − θ)).
Define R1 and R2 as in (2.20) and (2.21). If R1 6= 0, as θ 6= 0 we deduce from
(3.2) that
R2
2
= (D + T )2 + 4(C + θ)2(C − θ),
which is (2.22). Define X, X, Y , Y , Z, Z as in (2.23), (2.24) and (2.25).
Clearly
(3.3) X +X + Y + Y = −2D, X +X − Y − Y = 2T.
From (2.20), (2.23) and (2.25) we deduce that
(3.4) XX = ZZ
2
.
From (2.21), (2.22), (2.24) and (2.25) we deduce that
(3.5) Y Y = Z2Z.
From (2.23), (2.24) and (2.25) we obtain
Z2 − ZZ + Z2 −
(
XY
Z
+
XY
Z
+
XY
Z
+
XY
Z
)
=
−C4 + 3θ4 − 2C2θ2 +R1R2θ − CD2 + CT 2
−(C2 − θ2) .
Appealing to (2.20) and (3.2) if R1 = 0 and to (2.21) if R1 6= 0, we deduce
that
(3.6) Z2 − ZZ + Z2 −
(
XY
Z
+
XY
Z
+
XY
Z
+
XY
Z
)
= E.
WATSON’S METHOD 9
From (2.20)-(2.25) we obtain
X2Y
Z2
+
XY 2
Z
2
+
XY
2
Z
2
+
X2Y
Z2
− 20Tθ(3.7)
=
q(T ) + Fθ3 − C2Fθ
(C2 − θ2)θ = −F.
Now define u1, u2, u3, u4 by (2.26). As Z, Z 6= 0 we deduce from (3.4) and
(3.5) that X, X , Y , Y 6= 0. Further, by (3.4) and (3.5), we have
XY
ZZ
3
=
Z2
XY
,
X
2
Y
Z2Z
4
=
Y
X2
,
so that
u3 =
Z2
XY
u3
1
, u4 =
Y
X2
u4
1
.
Then
(3.8) u1u4 =
Y Y
Z2
= Z, u2u3 =
XX
Z
2
= Z
and
(3.9) u5
2
=
XY 2
Z
2
, u5
3
=
XY
2
Z
2
, u5
4
=
X2Y
Z2
.
Hence
u1u4 + u2u3 = Z + Z = −2C,
which is (2.12). Next
u2
1
u3 ± u22u1 ± u23u4 + u24u2
= u5
1
Z2
XY
± u5
1
X
2
Z
4
± u10
1
Z4Y
X4Y 2
+ u10
1
Y
2
X
X4Z
2
= X ± Y ± Y +X,
that is
(3.10) u2
1
u3 ± u22u1 ± u23u4 + u24u2 =
{ −2D, with + signs,
2T, with − signs.
The first of these is (2.13).
10 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
Further
u2
1
u2
4
+ u2
2
u2
3
− u1u2u3u4 − u31u2 − u32u4 − u33u1 − u43u3
= u10
1
(
Y
2
X4
+
Z6
X4Y 2
− Z
3Y
X4Y
)
−
(
u5
1
Z
X
+ u10
1
Z3Y
X5
+ u10
1
Z6
X3Y 3
+ u15
1
Y
3
Z2
X7Y
)
= Z
2
+ Z2 − ZZ −
(
XY
Z
+
XY
Z
+
XY
Z
+
XY
Z
)
= E,
which is (2.14).
Finally, from (3.7), (3.8), (3.9) and (3.10), we obtain
u5
1
+ u5
2
+ u5
3
+ u5
4
− 5(u1u4 − u2u3)(u21u2 − u22u1 − u23u4 + u24u2)
=
X2Y
Z2
+
XY 2
Z
2
+
XY
2
Z
2
+
X2Y
Z2
− 20Tθ
= −F,
which is (2.15). By the Lemma the roots of f(x) = 0 are given by (2.16). As
θ and T are expressible by radicals so are R1, R2, X, X , Y , Y , Z, Z. Hence
u1, u2, u3, u4 are expressible by radicals. Thus the roots x1, x2, x3, x4, x5 of
f(x) = 0 are expressible by radicals.
4. Proof of Theorem 2. Using MAPLE we find that
(4.1) R(r, s) = −5
15
C2
h(C)h(−C) = 0,
as θ = ±C , so that there is at least one solution T ∈ C of (2.29). As T is a
root of a cubic equation, T is expressible in terms of radicals.
If D 6= ±T we define u1, u2, u3, u4 as in (2.30)-(2.33). Then
u1u4 + u2u3 =
8C3u5
1
(D − T )2(D + T ), if θ = C,
(D + T )2u5
1
8C3(D − T ) , if θ = −C,
= −2C,
which is (2.12).
WATSON’S METHOD 11
Next
u1u
2
2
+ u2u
2
4
+ u3u
2
1
+ u4u
2
3
=
8C2Du5
1
(D − T )2(D + T ), if θ = C,
4C2u5
1
(D − T )2 +
(D + T )4u10
1
32C5(D − T )3 , if θ = −C,
= −2D,
which is (2.13).
Further, for both θ = C and θ = −C, we have
u2
1
u2
4
+ u2
2
u2
3
− u3
1
u2 − u32u4 − u33u1 − u43u3 − u1u2u3u4
= 4C2 +
(D2 − T 2)
2C
− 4C2 (D − T )
(D + T )
= E,
by (2.27) and (2.29), which is (2.14).
Finally, using (2.30)-(2.33), we obtain
u5
1
+ u5
2
+ u5
3
+ u5
4
− 5(u1u4 − u2u3)(u21u3 − u22u1 − u23u4 + u24u2)
=
u5
1
+ u5
2
+ u5
3
− 20CT, if θ = C,
u5
1
+ u5
2
+ u5
4
− 20CT, if θ = −C,
=
−(D− T )
2(D + T )
4C2
+ 2C
(D + T )2
(D − T ) −
16C4(D − T )
(D + T )2
− 20CT, if θ = C,
−16C
4(D− T )
(D + T )2
− (D + T )(D − T )
2
4C2
+
2C(D + T )2
(D − T ) − 20CT, if θ = −C,
= −F,
by (2.28) and (2.29), which is (2.15).
If D = ±T , from r(T ) = r(±D) = 0, we obtain{
128C3D4 = 0, if T = D,
−256C6D2 = 0, if T = −D.
As C 6= 0 we deduce that D = 0. Then (2.5)-(2.7) become
K = 3C2 + E,
L = 15C4 − 2C2E + 3E2,
M = −25C6 + 35C4E − 11C2E2 + CF 2 + E3.
From
h(C)h(−C) = h(θ)h(−θ) = 0
12 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
and (2.3) with D = 0, we obtain(
4C2 − E)2 (−160000C8 + 48000C6E − 4400C4E2
+ 16C3F 2 + 120C2E3 − 2CEF 2 − E4) = 0.
If E = 8C2 this equation becomes −212C12 = 0, contradicting C 6= 0. Thus
E 6= 8C2. Hence either
(i) E = 4C2
or
(ii) F 2 =
(400C4 − 60C2E + E2)2
2C(8C2 −E) .
Since f(x) is irreducible, F 6= 0, and in case (ii) we have
2C(8C2 − E) = Z2
for some Z ∈ Q with Z 6= 0. Then
E =
16C3 − Z2
2C
and
F =
400C4 − 60C2E + E2
Z
=
64C6 − 88C3Z2 − Z4
4C2Z
.
Now define u1, u2, u3, u4 as in (2.36)-(2.39). Then
u1u4 + u2u3 = u1u4 = −2C,
which is (2.12).
Next
u1u
2
2
+ u2u
2
4
+ u3u
2
1
+ u4u
2
3
= u3u
2
1
+ u4u
2
3
= 0 = −2D,
which is (2.13).
Further
u2
1
u2
4
+ u2
2
u2
3
− u3
1
u2 − u32u4 − u33u1 − u43u3 − u1u2u3u4
= u2
1
u2
4
− u3
3
u1 − u34u3
=
{
4C2, (i),
16C3 − Z2
2C
, (ii).
= E,
which is (2.14).
Finally
u5
1
+ u5
2
+ u5
3
+ u5
4
− 5(u1u4 − u2u3)(u21u2 − u22u1 − u23u4 + u24u2)
WATSON’S METHOD 13
=
u5
1
+ u5
4
, (i)
u5
1
+ u5
3
+ u5
4
+ 20CZ, (ii)
=
−F, (i)
22CZ +
Z3
4C2
− 16C
4
Z
, (ii)
= −F,
which is (2.15).
In both cases (i) and (ii), by the Lemma the roots of f(x) = 0 are given by
(2.16). As T is expressible in terms of radicals, so are u1, u2, u3, u4, and thus
x1, x2, x3, x4, x5 are expressible in radicals.
5. Proof of Theorem 3. As θ = C = 0 we deduce from h(θ) = 0 that
M = −16D4 − 2DEF + E3 = 0.
If E = 0 then D = 0 and conversely. Thus
(i) D = E = 0 or (ii) D 6= 0, E 6= 0.
In case (ii) we have
F =
E3 − 16D4
2DE
.
Define u1, u2, u3, u4 as in (2.42)-(2.44). Then
u1u4 + u2u3 = 0 = −2C,
which is (2.12). Also
u1u
2
2
+ u2u
2
4
+ u3u
2
1
+ u4u
2
3
= u1u
2
3
=
{
0, (i)
−2D, (ii)
}
= −2D,
which is (2.13). Further
u2
1
u2
4
+ u2
2
u2
3
− u3
1
u2 − u32u4 − u33u1 − u43u3 − u1u2u3u4
= −u1u33 =
{
0, (i)
E, (ii)
}
= E,
which is (2.14). Finally
u5
1
+ u5
2
+ u5
3
+ u5
4
− 5(u1u4 − u2u3)(u21u3 − u22u1 − u23u4 + u24u2)
= u5
1
+ u5
3
=
{ −F, (i)
16D4 −E3
2DE
, (ii)
}
= −F,
which is (2.15). Hence by the lemma the roots of f(x) = 0 are given by (2.16).
Clearly u1, u2, u3, u4 can be expressed in radical form so that x1, x2, x3, x4,
14 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
x5 are expressible by radicals.
6. Proof of Theorem 4. We define T by (2.45). As θ = 0 we have
h(θ) = h(0) =
M
3125
= 0 so that
(6.1)
−25C6 + 35C4E − 40C3D2 − 2C2DF − 11C2E2
+28CDE + CF 2 − 16D4 − 2DEF + E3 = 0.
Replacing E by
(C3 +D2 − T 2)
C
in (6.1), we obtain (2.46). Define R1 and R2
as in (2.47)-(2.49). Define X, X, Y , Y as in (2.50) and (2.51). Clearly
X, X, Y, Y 6= 0
and
(6.2) XX = Y Y = −C3, X +X + Y+Y = −2D.
Next
(6.3) C2 +
(X +X)(Y + Y )
C
=
C3 − T 2 +D2
C
= E.
Also
X2Y +XY 2 +XY
2
+X
2
Y
C2
(6.4)
=
TR1R2 − 1
4
DR2
1
− 1
4
TR2
1
− 1
4
DR2
2
+
1
4
TR2
2
− 1
2
D3 +
1
2
DT 2
C2
= −F.
Now define u1, u2, u3, u4 by (2.52). Then
u1u4 = u2u3 = −C.
Also
u5
2
=
XY 2
C2
, u5
3
=
XY
2
C2
, u5
4
=
X2Y
C2
.
Then
u1u4 + u2u3 = −2C,
which is (2.12). Also
u1u
2
2
+ u2u
2
4
+ u3u
2
1
+ u4u
2
3
= Y +X + Y +X = −2D,
WATSON’S METHOD 15
which is (2.13). Further
u2
1
u2
4
+ u2
2
u2
3
− u1u2u3u4 − (u31u2 + u32u4 + u33u1 + u43u3)
= C2 +
(
XY
C
+
XY
C
+
XY
C
+
XY
C
)
= E,
by (6.3), which is (2.14). Finally
u5
1
+ u5
2
+ u5
3
+ u5
4
− 5(u1u4 − u2u3)(u21u2 − u22u1 − u23u4 + u24u2)
= u5
1
+ u5
2
+ u5
3
+ u5
4
=
X2Y
C2
+
XY 2
C2
+
XY 2
C2
+
X2Y
C2
= −F,
by (6.4), which is (2.15). By the Lemma the roots of f(x) = 0 are given by
(2.16). As T , R1, R2, X, X, Y , Y are expressible by radicals, so are u1, u2,
u3, u4, and thus x1, x2, x3, x4, x5 are expressible by radicals.
7. Examples. We present eight examples.
Example 1. This is Example 3 from [1] with typos corrected
f(x) = x5 − 25x3 + 50x2 − 25, Gal(f) = Z/5Z [MAPLE]
C = −5
2
, D = 5, E = 0, F = −25
K =
75
4
, L =
5375
16
, M =
−30625
64
δ = 51272
h(x) = x6 − 15
4
x4 +
43
16
x2 − 7
25
√
5x− 49
320
=
(
x+
√
5
2
)(
x5 −
√
5
2
x4 − 5
2
x3 +
5
√
5
4
x2 − 7
16
x− 49
√
5
800
)
θ = −
√
5
2
, T = 0
R1 =
√
−25 + 10
√
5, R2 = −
√
−25− 10
√
5
16 MELISA J. LAVALLEE, BLAIR K. SPEARMAN, AND KENNETH S. WILLIAMS
X =
−5 +
√
−25 + 10√5
2
, Y =
−5−
√
−25− 10√5
2
, Z =
5 +
√
5
2
u1 =
(
X2Y
Z2
)1/5
=
(
25 + 15
√
5 + 5
√
−130− 58√5
4
)1/5
u2 =
(
XY 2
Z
2
)1/5
=
(
25− 15
√
5 + 5
√
−130 + 58
√
5
4
)1/5
u3 =
(
XY
2
Z
2
)1/5
=
(
25− 15
√
5− 5
√
−130 + 58
√
5
4
)1/5
u4 =
(
X
2
Y
Z2
)1/5
=
(
25 + 15
√
5− 5
√
−130− 58
√
5
4
)1/5
.
Example 2. f(x) = x5 + 10x3 + 10x2 + 10x + 78, Gal(f) = F20 [MAPLE]
C = 1, D = 1, E = 2, F = 78
K = 5, L = −125, M = 5625, δ = 24513
h(x) = x6 − x4 − x2 − 4
5
x+
9
5
= (x− 1)(x5 + x4 − x− 9
5
)
θ = 1, T = 3
Theorem 2(a) (θ = C) gives
u5
1
= −4, u2 = −u21, u3 = −
1
2
u3
1
, u4 = 0
x = −ω22/5 − ω224/5 + ω321/5
Example 3. f(x) = x5 + 10x3 + 20x+ 1, Gal(f) = F20 [MAPLE]
C = 1, D = 0, E = 4, F = 1
K = 7, L = 55, M = 4, δ = 3255432
h(x) = x6 − 7
5
x4 +
11
25
x2 − 129
3125
x+
4
3125
= (x− 1)(x5 + x4 − 2
5
x3 − 2
5
x2 +
1
25
x− 4
3125
)
WATSON’S METHOD 17
θ = 1, T = 0
Theorem 2 (b)(i) gives
u1 =
(
−1 +√129
2
)1/5
, u2 = u3 = 0, u4 =
(
−1−√129
2
)1/5
x = ω
(
−1 +√129
2
)1/5
+ ω4
(
−1−√129
2
)1/5
Example 4. f(x) = x5 + 10x3 + 30x− 38, Gal(f) = F20 [MAPLE]
C = 1, D = 0, E = 6, F = −38
K = 9, L = 111, M = 1449, δ = 24554312
h(x) = x6 − 9
5
x4 +
111
125
x2 − 1724
3125
x+
1449
3125
= (x− 1)(x5 + x4 − 4
5
x3 − 4
5
x2 +
11
125
x− 1449
3125
)
θ = 1, T = 0
Theorem 2(b)(ii) gives
Z = 2
u1 = 2
2/5, u2 = 0, u3 = 2
1/5 u4 = −23/5
x = ω22/5 + ω321/5 − ω423/5
Example 5. f(x) = x5 − 20x3 + 180x − 236, Gal(f) = F20 [MAPLE]
C = −2, D = 0, E = 36, F = −236
K = 48, L = 3840, M = −103200,
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