# Irreducible polynomial $\frac{x^{n}+x^{m}-2}{x^{\gcd(n,m)}-1}$ over $\mathbb{Q}$.

I have to show that the polynomial $$f(x)=\frac{x^{n}+x^{m}-2}{x^{\gcd(n,m)}-1}$$ is irreducible over $\mathbb{Q}$, for all $n,m \in \mathbb{N}$. Any idea as to how I can show this.

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– Julian Kuelshammer Oct 24 '12 at 5:54

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Here is a generalization: mathproblems123.wordpress.com/2009/11/02/… And here is a useful lemma: mathproblems123.wordpress.com/2009/11/09/position-of-roots – Beni Bogosel Oct 24 '12 at 7:00
@JulianKuelshammer: Since this is an old Miklos Schweitzer problem, I don't think it was given as homework. Although, the mention of the OP's work would make the question more valuable. – Beni Bogosel Oct 24 '12 at 10:43

## 1 Answer

Hint: Say that $gcd(n,m)=d$, and write $f(x) = \dfrac{x^n+x^m-2}{x^d-1}$ as

$$f(x) = x^{(c-1)d}+x^{(c-2)d} + \cdots x^{bd}+2x^{(b-1)d)}+ \cdots +2x^d +2$$ where $n = cd$, $m=bd$. Consider the polynomial $g(x) = x^{c-1} + \cdots + x^b +2x^{b-1}+ \cdots +2x + 2$. Show that this polynomial only has roots of absolute value greater than 1. Use this to show that the roots of f satisfies a similar property, and derive a contradiction.

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