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Solvable quintic trinomials of form, $$x^5+ax+b=0$$ have been completely parameterized. A nice example using complex numbers is, $$x^5+10ix+8i=0$$ discussed in this post. Finding $6$th-deg versions is relatively easy to do such as, $$x^6+3x+3=0$$ which factors over $\sqrt{-3}$. No $7$th-deg are known, but surprisingly there are octic ones, such as the simple, $$x^8-5x-5=0$$ which factors over $\sqrt{5}$. And the not-so-simple ones,

$$x^8-11(4x+3)=0\\x^8+16(4x+7)=0\\x^8 + 5\cdot23^2(12 x+43) =0$$

which factors over a quartic extension (and needs the cube root of unity).

Q: Any other octic examples, if possible parametric?

$\color{green}{Update:}$

Klajok in his answer below has found a family for the class of octic trinomials that factor over a quadratic extension. However, another class needs a quartic extension. For example, $$x^8-44x-33=0\tag1$$ 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.

Tito Piezas III
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  • @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 03:10
  • 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 03:12
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    @AhmedS.Attaalla: Yes, special privileges once you get above a certain rep. :) – Tito Piezas III Jan 13 '17 at 03:30
  • The reference for the quintic is: https://www.jstor.org/stable/2975165 – Kenta S Dec 14 '22 at 04:04
  • @KentaS Thanks, yes, I've read their paper. The solution of the trinomial quintic $x^5-5x+12$ involves the Fibonacci constant (golden ratio). The solution of the trinomial octic $x^8-44x-33$ involves the Tribonacci constant. Kindly see this new MSE post. – Tito Piezas III Dec 14 '22 at 04:13

2 Answers2

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

which factors over $\sqrt{-3}$ and $\sqrt{-15}$, respectively.

See https://sites.google.com/site/klajok/polynomials/x8-ax-b0

$\color{green}{Added:}$

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-ax-b0 .

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:

$C_2 \wr A_4$: $$x^8+64x+112=x^8+16(4x+7)$$

$C_2 \wr S_4$: $$x^8-44x-33=x^8-11(4x+3)$$ $$x^8+768x+1344=x^8+3\cdot8^2(4x+7)$$ $$x^8+31740x+113735=x^8+5\cdot23^2(12x+43)$$ $$x^8-856251x-2023866=x^8-3^4\cdot31^2(11x+26)$$ $$x^8-5992704x-304129728=x^8-6^4\cdot34^2(4x+203)$$

user402556
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    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 05:21
  • 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. – user402556 Jan 16 '17 at 23:12
  • Checking out your page I found that you are interested in Klotski. I warmly recommend that you take a look at another game evolved from Klotski. As is possible when the tiles consist of bits as opposed to plywood, certain features become available. Over the years Andreas's game has come up with several tile/floor features that make the game more challenging and fun. – Jyrki Lahtonen Dec 14 '22 at 17:37
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A result of Harris [1] is that every monic palindromic polynomial of degree-8 can be factored into two monic palindromic polynomials of degree-4.

$$ \begin{align} f(x) & = x^8 + ax^7 + bx^6 + cx^5 + dx^4 + cx^3 + bx^2 + ax + 1 \\ & = (x^4 + px^3 + qx^2 + px + 1)(x^4 + rx^3 + sx^2 + rx + 1) \\ & = x^8+x^7 (p+r)+x^6 (pr+q+s)+x^5 (p(s+1)+qr+r)+x^4 (2pr+qs+2)+x^3 (p(s+1)+qr+r)+ x^2 (pr+q+s)+x (p+r)+1 \end{align} $$

Equating the coefficients:

$$ \begin{align} a &= p+r \\ b &= pr + q + s \\ c &= p(s+1) + qr + r \\ d &= 2pr + qs + 2 \end{align} $$

For the subset of degree-8 monic palindromic polynomials of the form

$$ \begin{align} f(x) & = x^8 + 0x^7 + 0x^6 + 0x^5 + dx^4 + 0x^3 + 0x^2 + 0x + 1 \\ & = x^8 + dx^4 + 1 \\ \text{we have } 0 &= p+r \\ 0 &= pr + q + s \\ 0 &= p(s+1) + qr + r \\ d &= 2pr + qs + 2 \end{align} $$

We have the parametric solutions:

$$ r = 0 ∧ s = -q ∧ p = 0 ∧ d = 2 - q^2, \\ r = \sqrt{2} \sqrt{q} ∧ s = q ∧ p = -\sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2, \\ r = -\sqrt{2} \sqrt{q} ∧ s = q ∧ p = \sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2. \\ $$

Note that all of the above are parametrized on $d$.

On similar lines, we can factor the monic quartic palindromic polynomials into two monic quadratic palindromic polynomials each and then use the quadratic formula to get the roots.

References

[1]: J. R. Harris, "96.31 Palindromic Polynomials," The Mathematical Gazette, vol. 96, no. 536, p. 266–69, 2012. https://doi.org/10.1017/S0025557200004526

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