A Tale of Three Solvable Octics

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.

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One response to this post.

  1. […] 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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