Prove that $X^n+aX^{n-1}+\cdots+aX-1$ is irreducible in $\mathbb{Z}[X]$.

Question

Prove that $X^n+aX^{n-1}+\cdots+aX-1$ is irreducible in $\mathbb{Z}[X]$, where $n\ge 2$ and $a\in \mathbb{Z}$.

Let $f=X^n+aX^{n-1}+...+aX-1$. Through direct computations I could show that $f=(X-1+a)(X^{n-1}+X^{n-2}+...+X+1)-a$ and then I tried to assume that it isn't irreducible i.e. $\exists g, h \in \mathbb{Z}[X]$ such that $f=g\cdot h$ and $\deg g, \deg h<n$.
This wasn't useful since I could only get that $g(0)=1$ and $h(0)=-1$ or $g(0)=-1$ and $h(0)=1$ (which could be obtained from $f$'s initial form). I also tried using the fact that $f(1-a)=-a$, but to no avail.

Answer 1

Below criterion by Bauer is useful (you can find its proof in Polynomials by Victor V. Prasolov, Theorem 2.2.6):

Bauer's criterion. Let $a_1 \geq a_2 \geq \dots \geq a_n$ be positive integers and $n \geq 2$. Then the polynomial $p(x)=x^n-a_1x^{n-1}-a_2x^{n-2}-\dots-a_n$ is irreducible over $\mathbb{Z}$.

Denote your polynomial $f(x)=x^n+ax^{n-1}+\dots+ax-1$. Notice that for $a<0$ the $f(x)$ satisfies conditions of Bauer criterion, and hence is irreducible. For $a>0$, note that $f(x)$ will be irreducible if and only if $-x^nf(1/x)$ is irreducible (see Prove that the polynomial $x^nf(1/x)$ with reverted coefficients is also irreducible polynomial over $\mathbb{Q}$). However polynomial $$ -x^nf(1/x)=x^n-ax^{n-1}-\dots-ax-1 $$ satisfies conditions of Bauer's criterion, so it must be irreducible. Consequently, $f(x)$ is irreducible.

Alternative way idea (incomplete proof).

If you could someshow prove that $f(x)$ has all but one root lying outside the complex unit circle, then irreducibility follows. Indeed, assume $f(x)=g(x)h(x)$ is a factorization, then absolute values of constant coefficients of $g,h$ are $|g(0)|=|h(0)|=1$. However, one of the two polynomials must have all its roots outside the unit circle, say $g(x)$. So $g(x)=\prod(x-\alpha_i)$ with $|\alpha_i|>1$, and write $$ 1=|g(0)|=|(-1)^k \prod \alpha_i|=\prod|\alpha_i|>1, $$ a contradiction. Such factorization is not possible and $f(x)$ is irreducible.

The difficulty seems to be showing that those all but one roots lie in the unit circle, maybe fact that $$(x-1)f(x)=x^{n+1}+(a-1)x^n-(a+1)x+1$$ could be somehow used, but I don't see how.