International Journal of Pure and Applied Mathematics ————————————————————————– Volume 18 No

International Journal of Pure and Applied Mathematics ————————————————————————– Volume 18 No. 2 2005, 213-222

NEW WAY FOR A TWO-PARAMETER CANONICAL FORM OF SEXTIC EQUATIONS AND ITS SOLVABLE CASES

Yoshihiro Mochimaru Department of International Development Engineering Graduate School of Science and Engineering Tokyo Institute of Technology 2-12-1, O-okayama, Meguru-ku, Tokyo 152-8550, JAPAN e-mail: [email protected]

Abstract: A new way for a two-parameter canonical form of sextic equations is given, through a Tschirnhausian transform by multiple steps, supplemented with some solvable cases at most in terms of hypergeometric functions.

AMS Subject Classiﬁcation: 11D41 Key Words: sextic equation, Tschirnhausian transform

1. Introduction

6 6 6−n Actually the general sextic equation x + anx = 0 can be solved in nX=1 terms of Kamp´ede F´eriet functions, and it can be reduced to a three-parameter

Received: October 21, 2004 c 2005, Academic Publications Ltd. 214 Y. Mochimaru

6 4 2 resolvent of the form y + ay + by + cy + c = 0 by Joubert [3]. Such a type of the sextic equation was treated by Felix Klein and a part was given in [4]. Other contributions to treatment on sextic equations were given e.g. by Cole [2] and Coble [1]. In this paper, a new way leading to a two-parameter canonical form of sextic equations is given.

2. Analysis

2.1. General

Let the original given sextic equation be expressed as

6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ ∗ x + a x + b x + c x + d x + e x + f = 0. (1)

3 Here the quantity 5a∗ − 18a∗b∗ + 27c∗ corresponding to the polynomial of the left hand side of equation (1) is invariant with respect to the parallel shift x → x+constant.

3 2.2. In Case of 5a∗ − 18a∗b∗ + 27c∗ = 0

Let x = y − a∗/6, then equation (1) becomes

6 5 ∗2 ∗ 4 25 ∗4 1 ∗2 ∗ ∗ 2 y + − a + b y + a − a b + d y 12 432 6 1 ∗5 1 ∗3 ∗ 1 ∗ ∗ ∗ + − a + a b − a d + e y 81 27 3 35 ∗6 1 ∗4 ∗ 1 ∗2 ∗ 1 ∗ ∗ ∗ + a − a b + a d − a e + f = 0. (2) 46656 432 36 6

3 2.3. In Case of 5a∗ − 18a∗b∗ + 27c∗ = 0

Now apply a Tschirnhausian transform of the following type:

2 −y = p + qx + x . (3)

2 Eliminating x from equation (3) and equation (1) with p = −(a∗ −2b∗ −a∗q)/6, using Sylvester’s resultant, gives

6 4 3 2 y + b0y + c0y + d0y + e0y + f0 = 0, (4) NEW WAY FOR A TWO-PARAMETER CANONICAL... 215

2 b0 ≡ 15p + 5py5 + y4, (5)

3 2 5 ∗3 2 ∗ ∗ ∗ 3 c0 ≡ 20p + 10p y5 + 4py4 + y3 = − a + a b − c q 27 3 5 ∗4 22 ∗2 ∗ 4 ∗2 ∗ ∗ ∗ 2 + a − a b + b + 3a c − 4d q 9 9 3 5 ∗5 26 ∗3 ∗ 26 ∗ ∗2 10 ∗2 ∗ ∗ ∗ 13 ∗ ∗ ∗ + − a + a b − a b − a c + 3b c + a d − 5e q 9 9 9 3 3 5 ∗6 10 ∗4 ∗ 14 ∗2 ∗2 4 ∗3 4 ∗3 ∗ 8 ∗ ∗ ∗ + a − a b + a b − b + a c − a b c 27 9 9 27 3 3 ∗2 4 ∗2 ∗ 2 ∗ ∗ ∗ ∗ ∗ + c − a d + b d + 2a e − 2f , (6) 3 3

4 3 2 d0 ≡ 15p + 10p y5 + 6p y4 + 3py3 + y2, (7)

5 4 3 2 e0 ≡ 6p + 5p y5 + 4p y4 + 3p y3 + 2py2 + y1, (8)

6 5 4 3 2 f0 ≡ p + p y5 + p y4 + p y3 + p y2 + py1 + y0, (9)

∗ ∗2 ∗ y5 ≡−a q + a − 2b , (10) ∗ 2 ∗ ∗ ∗ ∗ ∗ ∗2 ∗ y4 ≡b q + (−a b + 3c ) q − 2a c + b + 2d , (11) ∗ 3 ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗ ∗ y3 ≡−c q + (a c − 4d ) q + (3a d − b c − 5e ) q ∗ ∗ ∗ ∗ ∗2 ∗ +2a e − 2b d + c − 2f , (12) ∗ 4 ∗ ∗ ∗ 3 ∗ ∗ ∗ ∗ ∗ 2 y2 ≡d q + (−a d + 5e ) q + (−4a e + b d + 9f ) q ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗2 + (−5a f + 3b e − c d ) q + 2b f − 2c e + d , (13) ∗ 5 ∗ ∗ ∗ 4 ∗ ∗ ∗ ∗ 3 y1 ≡−e q + (a e − 6f ) q + (5a f − b e ) q ∗ ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗ ∗ ∗ ∗2 + (−4b f + c e ) q + (3c f − d e ) q − 2d f + e , (14) ∗ 6 ∗ ∗ 5 ∗ ∗ 4 ∗ ∗ 3 ∗ ∗ 2 ∗ ∗ ∗2 y0 ≡f q − a f q + b f q − c f q + d f q − e f q + f . (15)

Thus if q is to be one of the roots of the cubic equation c0 = 0, then equation (1) becomes 6 4 2 y + b0y + d0y + e0y + f0 = 0. (16)

Therefore, now consider a sextic equation of the form:

6 ∗ 4 ∗ 2 ∗ ∗ y + B y + D y + E y + F = 0. (17) 216 Y. Mochimaru

2.4. In Case of B∗ = 0

Apply a Tschirnhausian transform of the type: 2 3 4 −z = p + qy + ry + sy + y . (18)

Eliminating y from equation (18) and equation (17), using Sylvester’s resultant, gives 6 n (p + z) Gn (q,r,s)=0 with G6 = 1, (19) Xn=0 ∗ ∗2 ∗ G5 ≡−2B r + 2B − 4D , (20)

∗ 2 ∗2 ∗ ∗2 ∗ 2 ∗ G4 ≡ B q + −2B + 4D qs + B + 2D r + 5E rs ∗3 ∗ ∗ ∗ 2 ∗ ∗3 ∗ ∗ ∗ + B − 3B D + 3F s + 5E q + −2B + 2B D + 6F r ∗ ∗ ∗4 ∗2 ∗ ∗2 ∗ ∗ − 7B E s + B − 4B D + 6D − 4B F , (21)

∗ 2 ∗ 2 ∗ 2 ∗ ∗ ∗ G3 ≡−4D q r − 5E q s − 5E qr + (4B D − 12F ) qrs ∗ ∗ 2 ∗ ∗ ∗ 3 ∗ ∗ 2 + 7B E qs + (−2B D − 2F ) r − 3B E r s ∗2 ∗ ∗ ∗ ∗2 2 ∗2 ∗ ∗ ∗ 3 + −2B D + 2B F + 4D rs + −3B E + 3D E s ∗ ∗ ∗ 2 ∗ ∗ ∗2 ∗ ∗ + (2B D − 6F ) q + 4B E qr + −8D + 16B F qs ∗2 ∗ ∗2 ∗ ∗ 2 ∗2 ∗ ∗ ∗ + 4B D − 4D − 4B F r + 6B E − 2D E rs ∗2 ∗ ∗2 ∗2 ∗ ∗ ∗ 2 ∗2 ∗ ∗ ∗ + 5E + 2B D − 4B F − 2D F s + B E − 11D E q ∗3 ∗ ∗ ∗2 ∗2 ∗ ∗ ∗ ∗2 + −2B D + 2B D + 10B F − 14D F + 5E r ∗3 ∗ ∗ ∗ ∗ ∗ ∗ + −3B E + 6B D E + 11E F s ∗3 ∗ ∗2 ∗2 ∗ ∗ ∗ ∗ ∗2 ∗3 ∗2 − 4B F + 2B D + 8B D F − 4B E − 4D + 2F . (22)

G2, G1, G0 are polynomials of degree 4, 5, and 6 with respect to the combination of q, r, and s respectively, and are not shown here. Equation (19) can be rearranged as a polynomial of z: 5 6 n z + Hnz = 0, (23) nX=0 NEW WAY FOR A TWO-PARAMETER CANONICAL... 217 where

H5 = 6p + G5, (24) 2 H4 = 15p + 5pG5 + G4, (25) 3 2 H3 = 20p + 10p G5 + 4pG4 + G3, (26) 4 3 2 H2 = 15p + 10p G5 + 6p G4 + 3pG3 + G2, (27) 5 4 3 2 H1 = 6p + 5p G5 + 4p G4 + 3p G3 + 2pG2 + G1, (28) 6 5 4 3 2 H0 = p + p G5 + p G4 + p G3 + p G2 + pG1 + G0. (29)

Since H5 is linear in p,q,r,s, and H4 is of second degree, and H3 is of third degree, one way of determing p,q,r,s to satisfy H3 = H4 = H5 = 0 is as follows: at ﬁrst p,q,r, and s are selected with at least one free parameter so as to satisfy H5 = H4 = 0, then the parameter is determined from the restriction H3 = 0. That is, let q0 be one of the roots of the quadratic equation:

∗ 2 ∗ 2 ∗4 8 ∗2 ∗ ∗ ∗ 2 ∗2 B q0 + 5E q0 − B + B D − 4B F − D = 0. (30) 3 3 3

Using q0, the quantities q1, r1,s1 are deﬁned as

∗ ∗ q1 ≡ 2B q0 + 5E , (31)

4 ∗3 14 ∗ ∗ ∗ r1 ≡ B − B D + 6F , (32) 3 3 ∗2 ∗ ∗ ∗ s1 ≡ −2B + 4D q0 − 7B E . (33) In case of q1 = 0, i.e.

∗2 ∗ 2 ∗4 8 ∗2 ∗ ∗ ∗ 2 ∗2 25E + 4B B − B D + 4B F + D = 0, 3 3 3

q ≡ q0 − (r1r + s1s)/ q1, (34)

r ≡ λϕ0, (35)

s ≡ µϕ0, (36) where λ, µ are eigenvalues corresponding to the homogeneous part (second degree) of H4 = 0. It is necessary only to get the ratio λ/µ or µ/λ through the homogeneous quadratic equation. That is,

2 ∗2 ∗ 16 ∗7 112 ∗5 ∗ 196 ∗3 ∗2 ∗4 ∗ − B + 2D + B − B D + B D + 16B F 3 9 9 9 218 Y. Mochimaru

∗2 ∗ ∗ ∗ ∗2 2 2 − 56B D F + 36B F q1 λ . ∗ 16 ∗6 88 ∗4 ∗ 112 ∗2 ∗2 ∗3 ∗ ∗ ∗ ∗ + 5E + − B + B D − B D −24B F +48B D F 3 3 3 2 56 ∗5 ∗ 196 ∗3 ∗ ∗ ∗2 ∗ ∗ 2 ×q0/ q1 + − B E + B D E − 84B E F q1 3 3 8 ∗5 44 ∗3 ∗ 56 ∗ ∗2 ∗2 ∗ ∗ ∗ + B − B D + B D + 12B F − 24D F q1 λµ 3 3 3 ∗3 ∗ ∗ ∗ ∗5 ∗3 ∗ ∗ ∗2 2 2 + B − 3B D + 3F + 4B − 16B D + 16B D q0 /q1 ∗4 ∗ ∗2 ∗ ∗ 2 ∗3 ∗2 2 + 28B E − 56B D E q0/q1 + 49B E /q1 ∗4 ∗2 ∗ ∗2 + −4B + 16B D − 16D q0/q1 ∗3 ∗ ∗ ∗ ∗ 2 + −14B E + 28B D E /q1 µ = 0 . (37) In case of q1 = 0, i.e.

∗2 ∗ 2 ∗4 8 ∗2 ∗ ∗ ∗ 2 ∗2 25E + 4B B − B D + 4B F + D = 0, 3 3 3

q ≡ q0 + µϕ0, (38)

r ≡−s1λϕ0, (39)

s ≡ r1λϕ0, (40)

∗ 2 ∗2 ∗ ∗3 ∗ ∗ ∗ 2 B µ + −2B + 4D r1µλ + B − 3B D + 3F r1 2 ∗2 ∗ 2 ∗ 2 + − B + 2D s1 − 5E r1s1 λ = 0. (41) 3

ϕ0 can be determined by H3 = 0 (at most a cubic equation), since all the coeﬃcients appearing in H3,H4, and H5 are rational and each component as a polynomial is irreducible. Finally equation (23) is expressed in the form 6 2 z + Dz + Ez + F = 0. (42)

2.5. In Case of B∗ = 0 at equation (17)

Equation (17) itself is already of the form (42). NEW WAY FOR A TWO-PARAMETER CANONICAL... 219

2.6. In Case of E = 0 at equation (42)

Equation (42) itself is a compound cubic equation.

2.7. In Case of F = 0 at equation (42)

Equation(42) itself is substantially a quintic equation (or z = 0), which is solvable at least using a modular function.

2.8. In Case of EF = 0 at equation (42)

Let z ≡ (F/E)w , then equation (42) becomes

4 6 6 E 2 E w + D w + F (w + 1) = 0, (43) F F which is the ﬁnal two-parameter equation of the form: x6 + ax2 + bx + b = 0.

2.9. Solvable Cases of equation (42)

Apply a Tschirnhausian transform

2 3 4 5 −v = p + qz + rz + sz + tz + z . (44)

Eliminating z from equation (44) and equation (42), using Sylvester’s resultant, leads to 5 6 n (v + p) + (v + p) Ln = 0, (45) nX=0 where L5 = −4Dt − 5E, (46)

2 2 2 2 L4 = 4Dqs + 5Eqt + 2Dr + 5Ers + 6Frt + 3Fs + 6D t 2 2 + 6Fq − 4D s + 11DEt − 5DF + 10E , (47)

2 2 2 2 2 L3 = −4Dq r − 5Eq s − 6Fq t − 5Eqr − 12Fqrs − 8D qst 2 3 2 2 2 2 − 11DEqt − 2F r − 4D r t + 4D rs − 2DErst 2 2 3 2 2 + −14DF + 5E rt + 3DEs + −2DF + 5E s t 220 Y. Mochimaru

2 3 2 3 2 + 11EFst + −4D + 2F t + 8D qr − 2DEqs 2 2 2 + −4DF − 15E qt − DEr + 20DF − 15E rs − 8EFrt 2 3 2 2 2 3 2 − 4EFs + 8D + 12F st − 7D Et − 19EFq +−4D + 6F r 2 2 2 3 + 7D Es + 6D F − 9DE t + 15DEF − 10E. (48) L2,L1, and L0 are polynomials of degree 4, 5, 6 respectively and are not shown here. Rearranging equation (45) as a polynomial of v leads to

5 6 n ∗ v + v Hn = 0. (49) nX=0 ∗ ∗ The forms for Hn’s are similar to equations (24)-(29) (replacing H by H , and ∗ ∗ ∗ G by L). Let us try to establish H5 = H3 = H1 = 0. In this case the resulting ∗ equation is a compound cubic equation. From H5 = 0, we get 4Dt + 5E p = . (50) 6

5 5 E2 Let q0 ≡ 3 D − 27 F , and q is deﬁned as

r1r + s1s + t1t q ≡ q0 − , (51) EF where 2 2 4 16 3 2 80 D E 100 DE r1 ≡− D + 6F + − , (52) 9 81 F 729 F 2 3 5 4 2 80 DE 125 E s1 ≡− D E + − , (53) 3 81 F 729 F 2 4 2 82 2 125 E t1 ≡−4D F + DE − . (54) 27 243 F Then we have a one-parameter solution of the form:

r = λϕ1, (55) s = µϕ1, (56) t = νϕ1, (57) where λ, µ, and ν are eigen values to be determined (only the ratios are required) ∗ through the homogeneous parts (second degree and the third degree) of H3 = 0 under the assumptions (50) and (51), and ϕ1 is a parameter to be determined ∗ by H1 = 0. The ratios of eigen values are given, using Sylvester’s resultant, as NEW WAY FOR A TWO-PARAMETER CANONICAL... 221

∗ the roots of at most a sextic equation, (since H3 is a polynominal of degree 3), therefore solvable cases are obtained by assuming the coeﬃcient of the highest (sixth) degree is equal to zero, which is expressed by

k l m CklmD E F = 0, (58) k,l,mX where Cklm stands for a coeﬃcient and eventually 4k + 5l + 6m = 57, 0 ≤ k ≤ 16, 0 ≤ l ≤ 27, k,l,m: integers. A part (but not all) of the coeﬃcients Cijk’s are 573440 1310720 40960 C13,1,0 = − , C16,1,−2 = − , C10,1,2 = − , 81 2187 3 508559360 2884550656 C7,1,4 = 46080, C15,3,−3 = , C12,3,−1 = 177147 59049 In this case the ratios of eigen values, e.g. µ/λ, ν/λ (in case of λ = 0), are determined by at most a quintic equation, and ϕ1 by also a quintic equation ∗ H1 = 0.

2.10. Other Solvable Cases of equation (42)

D = 0: trinomial equation, roots are expressible in terms of a hypergeometric function. E2 − 4DF = 0: extraction of the square root is possible, resulting in a product of cubic polynomials.

3. Conclusion

A new way for a two-parameter canonical form of sextic equations is given, through a Tschirnhausian transform by multiple steps, supplemented with some soluvable cases.

References

[1] A.B. Coble, The reduction of the sextic equation to the Valentier form- problem, Math. Ann, 70 (1911), 337-350.

[2] F.N. Cole, A contribution to the theory of the general equation of the sixth degree, Amer. J. Math., 8 (1886), 265-286. 222 Y. Mochimaru

[3] Father Joubert, Sur l’equation du sixi`eme degr´e, C-R Acad. Sc. Paris, 64 (1876), 1025-1029.

[4] F. Klein, Lectures on the Icosahedron, Second Edition, Translated by G.G. Morrice, New York, Dover Publications (2003).