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$\def\N{\mathbb{N}}$

What can we say about the closed-form solutions for $x^q - 2x + 1$ for $q \in \N^+$ ? I can solve it for small values of $q$ up to 5. And I wonder what happens next, for $q > 5$ ?

Maybe there are some special functions associated to the solutions of this equation...

kerzol
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    If by "closed-form" solutions, you mean solutions involving radicals, it seems like there won't be any. That polynomial has the obvious root 1 hence is not irreducible, but it looks like $x^q-2x+1/x-1$ is irreducible (maybe Eisenstein's criterion could be used to prove). Then if you know Galois Theory and are able to show the Galois Group is $S_{q-1}$, by Galois' Theorem the roots cannot be expressed in terms of radicals. – sharding4 Apr 28 '21 at 01:41
  • Thanks a lot @sharding4 ! Indeed, dividing $x^q - 2x + 1$ by $(x-1)$ we easily obtain $x^{q-1} + x^{q-2} + \cdots + x^2 + x - 1$. Now I will think how to prove that the Galois Group of this polynomial is $S_{q-1}$. If you have any suggestions or a simple argument don't hesitate to write a comment. – kerzol Apr 28 '21 at 17:16
  • The polynomial is indeed irreducible, see https://math.stackexchange.com/a/4120262/104527 for proof link. – kerzol Apr 28 '21 at 23:33

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