五次方程根式解法

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,