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I had discussed the solvable octic trinomial, $$x^8-44x-33=0\tag1$$ way back in this old MSE post, but I revisited this inspired by another solvable octic, $$y^8-y^7+29y^2+29=0\tag2$$ which I also discussed in this MO post. Surprisingly, $(2)$ can be solved by the $29\text{th}$ root of unity, which made me wonder if there was some known constant that could solve $(1)$. The clue was $p=11,$ but it couldn't be the $11\text{th}$ root of unity since that involve quintics.

Then I wondered if the tribonacci constant (which involve $p=11$) would do since quartics can be solved by cubics. And happily that was the case. So a pair of roots to the octic $(1)$ is given by,

$$x = \frac12\left(-\frac{p}2\pm\sqrt{\frac{q}2}\right)$$ $$p = -\sqrt{22u}+\sqrt{-22u+\frac{44}{\sqrt{22u}}}$$ $$q = -\sqrt{22v}+\sqrt{-22v+88+\frac{88}{\sqrt{22v}}}$$ where $u,v$ are, $$u = \frac{1-2t}{t-7},\quad v = \frac{21-11t}{6t+13}$$ and $t$ is the tribonacci constant, the real root of, $$t^3-t^2-t-1 = 0\tag3$$ It's nice to see the octic's explicit solution, and then find multiples of $p=11$ all over the place. In summary, given the irreducible but solvable octic trinomial of form,

$$Ax^8+Bx+C = 0\tag4$$

there are infinitely many that factor over a $\sqrt{d}$ extension, but only six are known (listed in the old MSE's answer) that factor over a quartic extension like $(1)$.

Question: Can you find a 7th solvable trinomial octic that factors over a quartic extension like $(1)$? (It might also involve known constants. P.S. I have spent many, many hours trying, but haven't found another so far.)

Tito Piezas III
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  • Here is a related question. – Tito Piezas III Dec 14 '22 at 03:50
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    https://math.stackexchange.com/questions/2054945/on-solvable-octic-trinomials-like-x8-5x-5-0 – Vadim Chernetsov Dec 14 '22 at 04:43
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    https://sites.google.com/site/klajok/polynomials/x8-ax-b0 – Vadim Chernetsov Dec 14 '22 at 04:46
  • Won't klajok's parametrization give you an infinite family? True, if you plug in a random rational $q$, the coefficients won't be integers. But then you can simply clear the denominators in the style of: an algebraic number is always of the form an algebraic integer divided by a rational integer. – Jyrki Lahtonen Dec 14 '22 at 17:28
  • @JyrkiLahtonen The way he found the parameterization was the octic would factor into 2 quartics (with square roots in the coefficients). This one (and the five others like it), factors into 4 quadratics (with quartic roots in the coefficients). So I can't use his solution, sadly. – Tito Piezas III Dec 14 '22 at 17:34
  • @TitoPiezasIII That may have been due to my bad choice of $q$. – Jyrki Lahtonen Dec 14 '22 at 17:40
  • @JyrkiLahtonen It factors over his $$\sqrt{\alpha} = \sqrt{\frac{2q+6}{q^2-2}}$$ So for your example with $q = -3/2$, then your octic factors over $\sqrt3$. – Tito Piezas III Dec 14 '22 at 17:46
  • @JyrkiLahtonen This has a sextic counterpart. The solvable trinomial $x^6+3x+3=0$ factors over $\sqrt{-3},$ but $x^6-7^2x-7^2=0$ factors over a cubic extension, specifically $\cos(2\pi/7)$. (So, of course, $p=7$ would play a role.) – Tito Piezas III Dec 14 '22 at 18:21
  • Ok, I missed the fact that you were not satisfied with having just solvable octic trinomials, irreducible over $\Bbb{Q}$. – Jyrki Lahtonen Dec 14 '22 at 18:47

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