Solvable form of the polynomial equation n n−1 x + an−1x + ... + a1x + a0 = 0,(n = 2k + 1) Nguyen Van Quang Hue - Vietnam, 04-2021
Abstract It is known, there is no solution in radicals to general polynomial equation of degree five or higher with arbitrary coefficients [?]. In this article, we give a form of the polynomial equations with odd degree can be solved in radicals. From there, we come up some solvable equations with one or more zero coefficients,especially for the quintic and septic equations.
3 2 x + a2x + a1x + a0 = 0 It is known, the cubic equation is solvable.
2.Quintic equation 5 4 3 2 x + a4x + a3x + a2x + a1x + a0 = 0 We have found the coefficients below to give a solvable form.
(a − a2)(2a + a2) a = 3 4 3 4 1 9 (a − a2)(a3 − a a + 3a ) a = 3 4 4 4 3 2 0 9 In other words , the quintic equation:
(q − p2)(2q + p2) (q − p2)(p3 − pq + 3r) x5 + px4 + qx3 + rx2 + x + = 0 9 9 is solvable.
Some special cases: p = 0:
2q2 qr x5 + qx3 + rx2 + x + = 0 9 3 q = 0:
p4 p2(p3 + 3r) x5 + px4 + rx2 − x − = 0 9 9 r = 0:
(q − p2)(2q + p2) p(q − p2)2 x5 + px4 + qx3 + x − = 0 9 9 1 And for the case q = 0, r = 0, we obtain:
p4 p5 p4 x5 + px4 − x − = (x + p)(x4 − ) = 0 9 9 9
3.Septic equation 7 6 5 4 3 2 x + a6x + a5x + a4x + a3x + a2x + a1x + a0 = 0 We have found the coefficients below to give a solvable form.
(a − a2)a a = 5 6 5 3 3 (a − a2)a a = 5 6 4 2 3 (a − a2)2(2a + a2) a = 5 6 5 6 1 27 (a − a2)2(a3 − a a + 3a ) a = 5 6 6 6 5 4 0 27 In other words , the septic equation: (q − p2)q (q − p2)r (q − p2)2(2q + p2) (q − p2)2(p3 − pq + 3r) x7 +px6 +qx5 +rx4 + x3 + x2 + x+ = 0 3 3 27 27 is solvable.
Some special cases: p = 0:
q2 qr 2q3 q2r x7 + qx5 + rx4 + x3 + x2 + x + = 0 3 3 27 9 q = 0:
p2r p6 p4(p3 + 3r) x7 + px6 + rx4 − x2 + x + = 0 3 27 27 r = 0:
(q − p2)q (q − p2)2(2q + p2) p(q − p2)3 x7 + px6 + qx5 + x3 + x − = 0 3 27 27
And for the case q = 0, r = 0, we obtain:
p6 p7 p6 x7 + px6 + x + = (x + p)(x6 + ) = 0 27 27 27
2 Generally, for the equation :
n n−1 n−2 n−3 n−4 x + an−1x + an−2x + an−3x + an−4x ... + a1x + a0 = 0 , n is odd
There are always coefficients ai: (i = 0;1...n-4 ) depend on coefficients an−1; an−2; an−3 to have a solvable from of the equations. Here an−1; an−2; an−3 are three arbitrary coefficients.
There is an algorithm for the above theorem.
References
[1] Quintic function - Wikipedia
[2] Septic equation - Wikipedia
[3] Abel - Ruffini theorem - Wikipedia
[4] Quang N V, Export a sequence of prime numbers Vixra: 1601.0048 (CO),studyres.com
[5] Quang N V, A proof of the four color theorem by induction Vixra: 1601.0247 (CO), Seman- ticscholar.org :124682326
[6] Quang N V, A new conjecture on sequence of consecutive natural number Vixra:1603.0136 (NT), Semanticscholar.org :124779148
[7] Quang N V, Dirichlet ’s proof of Fermat’s Last theorem for n = 5 is flawed Vixra:1607.0400 (NT)
[8] Quang N V, Upper bound of prime gaps, Lengendre’s conjecture was verified Vixra:1707.0237(NT)
[9] Quang N V, A parametric equation of the equation a5 + b5 = 2c5 Vixra:1901.0116(NT)
[10] Quang N V, Part 1 : My theorem Vixra:1910.0563(NT)
[11] Quang N V, Part 2 : Another version of my theorem and infinite ascent Vixra:2004.0690(NT)
[12] Quang N V, Proof of the Beal conjecture and Fermat - Catalan conjecture (summary) Vixra:2005.0237(NT),docplayer.net,Semanticscholar.org :225929433 PQ PR P 3 [13] Quang N V, Solvable sextic equation x6 + P x4 + Qx3 + Rx2 + x + − 2 = 0 3 3 27 Vixra:2010.0234(AL)
[14] Quang N V, A new Solvable quintic equation of the shape x5+ax2+b = 0 Vixra:2011.0165(AL), Semanticscholar.org :229488183
Email: [email protected] [email protected]
3