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Using the approach described in Smart way to solve octics like $x^8+5992704x-304129728=0$ (the method DecomPoly available in GAP) the decic quadrinomial from this question can be decomposed into two quintics: $$x^{10}+100x^2+160x+64=(x^5+10ix+8i)(x^5-10ix-8i)$$ where $i$ is the imaginary unit. But here I got stuck. How to get explicit formula for roots of those solvable quintic trinomials?

I have tried to use the RadiRoot's method RootsOfPolynomialAsRadicals but without success. It seems Galois group 10T33 of the original quadrinomial is too big, and the method does not accept (quintic) polynomials with complex coefficients.

I am afraid the answer will be too complicated for me, but still have hope there is some tool / built-in method that can run calculations without too deep digging into Galois theory...

user402556
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  • What do you expect from an explicit formula? – Servaes Apr 01 '19 at 21:24
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    The exact solutions that are given by Wolfram make it seem like they cannot be simplified very well. One of the solutions for example is $x = -4/5 _4 F_3(1/5, 2/5, 3/5, 4/5;1/2, 3/4, 5/4;i/2)$ – Peter Foreman Apr 01 '19 at 22:16
  • @Servaes The goal is an explicit expression for roots containing only 4 basic operations ($+$,$-$,$/$,$*$) and extracting n-th roots. I.e. expressing roots by radicals. – user402556 Apr 02 '19 at 20:20
  • @PeterForeman Thanks for interesting answer. However Galois group 10T33 is solvable by radicals. – user402556 Apr 02 '19 at 20:20

3 Answers3

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A big difference to Smart way to solve octics like $x^8+5992704x-304129728=0$ is that in the prior example the field $E$ defined by the polynomial $f$ (of degree 8) had a subfield $S$ (of degree 4) over which it is quadratic. Thus finding the subfield reduced everyting to the degree $\le 4$ formulas.

Here you have two difficulties. The first is that -- while there still is a subfield -- the degree over this subfield is 5. So you cannot use a generic formula, but would need to go into the generic Galois theory machinery with resolvent, roots of unity etc. for degree $\ge 5$.

The second difficulty is that the $E$ as extension of $S$ will have Galois group not cyclic, but of order $20$. You thus will have to do three steps, corresponding to the composition series of the Frobenius group of order $20$.

This can be done (even without need to go to hypergeometric series, as Mathematica seems to do), but the resulting expression in radicals will, in all likeness, be a nightmare, and if you are just looking for the result "out of curiosity" most likely not be worth the effort.

ahulpke
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  • thank you for detailed explanations that unfortunately confirmed that one needs to dig into the mentioned "machinery" to solve the above equation. For the "big science" indeed this task is not important, but for a person that likes such stuff is an opportunity to learn something useful. Very likely your answer should be accepted as the best possible, still I will wait some time for, hopefully, some less demanding steps (publicly available tools, scripts,...) – user402556 Apr 02 '19 at 20:13
  • The solution to solvable and depressed quintics can be succinctly expressed in the form $$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}$$ with appropriate $5$th roots of unity affixed, and where the $z_i$ are the roots of its quartic Lagrange resolvent. (Kindly see other answer.) In this manner, it may be written in "shorthand" and its Lagrange resolvent may be also interesting. For example, turns out for the OP's quintic, one needs the square root of the golden ratio. – Tito Piezas III May 11 '19 at 06:57
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To solve equations of Fermat prime degree $p=2^{n}+1$ with a solvable Galois group, one generally needs a resolvent equation of degree $2^n$ (which, using only nested square roots, can factor into quadratics). Thus to solve the solvable quintic,

$$x^5+10i x+8i=0$$

one employs 4 roots of a quartic (factored into 2 quadratics) which in this case nicely involves the golden ratio $\phi$. Let,

$$\begin{aligned}\color{red}u &= \frac{\sqrt{2\,i}}{\;5}\sqrt{1+2\,i}\\ &= \frac{\sqrt{2\,i}}{\;5}\left(\sqrt{\frac{1+\sqrt5}2}+\sqrt{\frac{1-\sqrt5}2}\right)\\ &= 0.0971+0.4116i\end{aligned}$$

(Note: All numerical values are generated by Mathematica and truncated.)

I. New answer

The advantage of this version is there is only one $z^{1/5}$ root extraction and does not involve the $5$th root of several complex numbers (and consequent ambiguity) as in the old answer.

The five roots $x_k$ are then

$$x_k = u\,T+\frac1T-\frac{a}{T^2}+\frac{b}{T^3}$$

where $$T_k = \zeta^k \Big(\frac{ab}{u}\Big)^{1/5}$$

$$a=\frac1u\big(i-\sqrt{u-1}\big) = -0.05367 + 0.5009 i$$

$$b=\frac1u\big(-c+\sqrt{c^2+1}\big) = 0.24289 + 0.00326 i$$

$$\quad c = \frac{1-5i}2+\frac1u \;=\; 1.04322 - 4.8011i$$

with $\zeta = e^{2\pi i /5}$ and $k = 0,1,2,3,4$.

II. Old answer

Let $\color{red}u$ be as defined above. Then the resolvent quadratics are,

$$z^2+2u(5u+2i)z+u^5 = 0\\z^2+2u(5u-2i)z-u^5 = 0$$

So the roots $z_i$ are,

$$z_1 = -0.0657 - 0.3821 i\\ z_2=0.0191 - 0.0291i\\ z_3=0.0028 + 0.0027i\\ z_4=3.2437 - 1.1914i$$

Taking their $5$th roots and affixing a correct $5$th root of unity $\zeta = e^{2\pi i/5}$,

$$z_1^{1/5}\zeta^2 = -0.3464 + 0.3758i \\z_2^{1/5}\zeta^2 = -0.2934 + 0.1511i \\z_3^{1/5}\zeta^2 = -0.4632 + 0.6855i \\z_4^{1/5}\zeta^4 = 0.3092 - 1.2435i$$

The sum of those $5$th roots,

$$x_1 = z_1^{1/5}\zeta^2+z_2^{1/5}\zeta^2+z_3^{1/5}\zeta^2+z_4^{1/5}\zeta^4 = -0.7938 - 0.03104i$$

is then a root of the quintic,

$$x^5+10i x+8i=0$$

P.S. The other roots can be found by affixing appropriate $\zeta^k$ to the $z_i^{1/5}$.

Tito Piezas III
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  • @Klajok, have you found other irreducible decic quadrinomials like in the post? – Tito Piezas III May 14 '19 at 12:01
  • I have not found any other irreducible decic quadrinomials regardless the above concise form. More generally, I searched for irreducible $x^{n}+(ax^{m}+b)^2$ of degrees 10, 12 and 16, for coprime $n$, $m$, with integer $a$, $b$ up to millions... without success. In case of reducibles, small examples: $x^{10}+(4x+8)^2$, $x^{10}+(38x+35)^2$, and similar: $x^{10}-(33x+34)^2$, $(13x^5)^2-(10x+3)^2$. – user402556 Aug 04 '19 at 12:36
  • How did you get the new answer? – Parcly Taxel Jan 08 '21 at 22:59
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Since both quintic factors are in Bring–Jerrard normal form, Malfatti's method (described here) may be applied. Consider the following mpmath program (so all roots are principal unless otherwise stated):

#!/usr/bin/env python3
from mpmath import *

i = root(-1, 2, s)
a = sqrt(1+2i)
l1 = [1, 4(-2 + i + a - ia)/5, 4(-7a + ia)/3125]
z1 = root(polyroots(l1)[0], 5, n1)
z2 = root(polyroots(l1)[1], 5, n2)
l2 = [1, 4(-2 + i - a + ia)/5, 4(+7a - i*a)/3125]
z3 = root(polyroots(l2)[0], 5, n3)
z4 = root(polyroots(l2)[1], 5, n4)
r = z1 + z2 + z3 + z4
print(polyval([1, 0, 0, 0, 10i, 8i], r))

Then for $s=0,1$ the $n_i$ must be set to these values so that $r$ will end up being a root of the quintic beneath it: $$s=0:\begin{array}{cccc} n_1&n_2&n_3&n_4\\ 0&4&1&0\\ 1&3&4&2\\ 2&2&2&4\\ 3&1&0&1\\ 4&0&3&3\end{array}\qquad s=1:\begin{array}{cccc}n_1&n_2&n_3&n_4\\ 0&1&4&0\\ 1&0&2&2\\ 2&4&0&4\\ 3&3&3&1\\ 4&2&1&3\end{array}$$ Since it is easy to solve a quadratic, this leads to explicit reproducible formulas for all the initial decic's roots.

Parcly Taxel
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