A Tale of Three Solvable Octics

9 Apr

A Tale of Three Solvable Octics

Posted April 9, 2012 by tpiezas in algebra, equations, mathematics. Tagged: octics, radicals, septics. 1 Comment

The following three cute irreducible octics are solvable,

$\text{(1)}\;\; x^8-5x-5 = 0$

$\text{(2)}\;\; x^8-44x-33 = 0$

$\text{(3)}\;\; x^8-x^7+29x^2+29 = 0$

However, each needs to be solved in a different way: they need a quadratic, quartic, and septic subfield, respectively.

The first is the easiest, it factors over $\sqrt{5}$ into two quartics. The second does not factor over a square root extension, but factors into four quadratics,

$x^2+vx -(2v^3-7v^2+5v+33)/13 = 0$

where the coefficients are determined by the quartic,

$v^4+22v+22 = 0$

The command,

Resultant[x^2+vx -(2v^3-7v^2+5v+33)/13 , v^4+22v+22 ,v]

done in Mathematica or in www.wolframalpha.com will eliminate the variable v and recover [2]. The first two octics are by this author. (Any other example for the second kind with small coefficients?)

The third (by Igor Schein) is the hardest, as it needs a septic subfield. Interestingly though, the solution involves the 29th root of unity. Given,

$x^8-x^7+29x^2+29 = 0$

Then,

$\begin{aligned} x_1 &= (1+(a-b-c-d+e-f-g))/8\\ x_2 &= (1-(a-b-c-d-e+f+g))/8\\ x_3&= (1-(a+b-c+d+e-f+g))/8\\ x_4&= (1+(a+b-c+d-e+f-g))/8\\ x_5&= (1-(a+b+c-d+e+f-g))/8\\ x_6&= (1+(a+b+c-d-e-f+g))/8\\ x_7&= (1-(a-b+c+d-e-f-g))/8\\ x_8&= (1+(a-b+c+d+e+f+g))/8 \end{aligned}$

where the 7 constants {a, b, c, d, e, f, g} is the square root of the appropriate root $z_i$ of the solvable septic,

$z^7-7z^6-2763z^5-19523z^4+1946979z^3+34928043z^2+\\119557031z-3247^2 = 0$

namely,

$\begin{aligned} a &\approx \sqrt{ -26.98}\\ b &\approx \sqrt{ -26.95}\\ c &\approx \sqrt{ -19.71}\\ d &\approx \sqrt{ -4.78}\\ e &\approx \sqrt{ 0.08}\\ f &\approx \sqrt{ 36.91}\\ g &\approx \sqrt{ 48.43}\\ \end{aligned}$

Note that,

$(8x_3-1)+(8x_4-1)+(8x_5-1)+(8x_6-1) = -4e$

hence one can use the roots of this octic to express the roots of its resolvent septic, and vice versa. In terms of radicals, Peter Montgomery expressed the septic roots $z_i$ as,

$\begin{aligned}\tfrac{1}{4}(z-1) &= 2(w^{11}+w^{13}+w^{16}+w^{18})-2(w+w^{12}+w^{17}+w^{28})\\ &+(w^3+w^7+w^{22}+w^{26})-(w^2+w^5+w^{24}+w^{27})\\ &+(w^4+w^{10}+w^{19}+w^{25})-(w^8+w^9+w^{20}+w^{21}) \end{aligned}$

where w is any complex root of unity (excluding w = 1) such that $w^{29}=1$ . For example, $w = \exp(2\pi i\cdot4/29)$ will yield the value for $z_1 \approx -26.98$ , and so on.

One response to this post.

Posted by Quartic equations « tpiezas on April 15, 2012 at 7:19 pm

[…] similar method was used to solve the octic in this post, though now one had to take square roots of its resolvent septic. Share this:TwitterFacebookLike […]

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