# On solvable octic trinomials like $x^8-5x-5=0$

Solvable quintic trinomials $$x^5+ax+b=0$$ have been completely parameterized. Finding $6$th-deg versions is relatively easy to do such as, $$x^6+3x+3=0$$ No $7$th-deg are known, but surprisingly there are octic ones, such as the simple, $$x^8-5x-5=0\\x^8+\tfrac12x+\tfrac7{16}=0\\x^8-44x-33=0$$ and the not-so-simple, $$x^8 + 23^2(60 x+215) =0$$

Q: Any other octic examples, if possible parametric?

$\color{green}{Update:}$

Klajok has found a family for the class of octic trinomials that factor over a quadratic extension. However, another class needs a quartic extension, such as $$x^8-44x-33=0\tag1$$ which factors into four quadratics, $$x^2 + v x - (2v^3 - 7v^2 + 5v + 33)/13=0$$ and where $v$ is any root of $v^4 + 22v + 22=0$. More generally, eliminating $v$ between $$x^2 + v x + (pv^3 +qv^2 + rv + s)=0$$ $$v^4+av^2+bv+c=0$$ easily done by the resultant function of Mathematica will result in an irreducible but solvable octic and judicious choice of rational coefficients will yield a trinomial. However, it is not known if this second class of trinomials like $(1)$ has a parametric family as well.

• @AhmedS.Attaalla: I just saw your deleted example. It had six non-zero terms. A "trinomial" means it must have only three terms. Specifically, the post looks for $x^8+ax+b = 0$. – Tito Piezas III Jan 13 '17 at 3:10
• Sorry want paying attention to that constraint. @TitoPiezasIII just a question though, you can actually see answers once they are deleted? – Ahmed S. Attaalla Jan 13 '17 at 3:12
• @AhmedS.Attaalla: Yes, special privileges once you get above a certain rep. :) – Tito Piezas III Jan 13 '17 at 3:30

Using brute force approach I have found few others, such as $$x^8+9x+9$$ $$x^8+64x+112$$ $$x^8+75x+150$$

$$\color{green}{Added:}$$

Given pairs of rational numbers $$\left(\alpha,\beta\right)$$ such that $$2\alpha^2+6\alpha+1=\beta^2$$. Define the following parameters:

$$u=\frac{2\alpha+1-\beta}{4},\quad v=\frac{1-\beta}{8},\quad w=\frac{\alpha}{8}\left(3\alpha-2\beta+3\right)$$ $$A=\frac{\alpha u}{2}\left(\alpha+1-4u\right),\quad B=w^2 - \alpha v^2$$

then the following identity is satisfied: $$x^8+Ax+B =\\ \left(x^4+\sqrt{\alpha}x^3+\left(-\frac{\sqrt{\alpha}}{2}+\frac{\alpha}{2}\right)x^2+\left(u\sqrt{\alpha}-\frac{\alpha}{2}\right)x+\left(v\sqrt{\alpha}+w\right)\right)\\ \left(x^4-\sqrt{\alpha}x^3+\left(\frac{\sqrt{\alpha}}{2}+\frac{\alpha}{2}\right)x^2+\left(-u\sqrt{\alpha}-\frac{\alpha}{2}\right)x+\left(-v\sqrt{\alpha}+w\right)\right)$$

Let exclude the pairs $$\left(\alpha, \beta \right) \in \left\{ \left(0, -1\right), \left(0, 1\right), \left(1, 3\right) \right\}$$ for which trinomials degenerate to the simpler form where $$AB=0$$.

Observations: If $$\sqrt{\alpha}$$ is not a rational number then the corresponding octic trinomials are irreducible and solvable. Otherwise the trinomials are still solvable but they are not irreducible.

Notes:

1. All the $$\left(\alpha, \beta \right)$$ pairs can be easily enumerated: $$\left(\alpha, \beta \right) \in \left\{ \left( \frac{2q+6}{q^2-2}, \frac{q^2+6q+2}{q^2-2} \right) : q \in \mathbb Q \setminus \lbrace -3, 4 \rbrace \right\}$$ Excluded values $$q=-3$$ and $$q=4$$ correspond to the degenerate cases $$(0, -1)$$ and $$(1, 3)$$, respectively. The remaining degenerate case $$(0,1)$$ corresponds to the value $$q$$ at infinity (I found this "subparameterization" using the following Philip Gibbs' answer: https://math.stackexchange.com/q/1905148).

2. Examples and preliminary conjectures related to this parameterization is available under the same page: https://sites.google.com/site/klajok/polynomials/x8-plus-ax-plus-b-is-0 .

I searched for more examples of $$C_2 \wr A_4$$ and $$C_2 \wr S_4$$ for $$x^8+ax+b$$ and the only ones for integer $$|a|,|b| \leq 100000000$$ are:

$$C_2 \wr A_4$$: $$x^8+64x+112$$

$$C_2 \wr S_4$$: $$x^8+44x-33$$ $$x^8+768x+1344$$ $$x^8+31740x+113735$$ $$x^8+856251x-2023866$$ $$x^8+5992704x-304129728$$

• The bounty will go to this post in about 24 hours. Since you mention your rep hinders you, then this should increase your rep to 100+ and allow you to participate more in the forum. – Tito Piezas III Jan 16 '17 at 5:21
• I've added an update about a second class of trinomial octics. Maybe you can find more examples for these as well? – Tito Piezas III Jan 16 '17 at 11:50
• Thank you very much Tito for the bounty! Regarding a second class of octics - currently I am aware of just one example with Galois group of its roots $C_2\wr A_4$, and 4 examples for $C_2\wr S_4$, so it might be a problem with guessing. Those examples are shown on my web page, so anyone can use them as a hint. – klajok Jan 16 '17 at 23:12