Prove that the polynomial $f(X)=X^{2017}+X^{2016}+...+X^3+X^2+2017$ is irreducible in $\mathbb{Z}[X]$.

Question

Prove that the polynomial $f(X)=X^{2017}+X^{2016}+...+X^3+X^2+2017$ is irreducible in $\mathbb{Z}[X]$.
Here is what I have tried : $f(X)=X^{2017}+X^{2016}+...+X^3+X^2+X+1-X+2016=\frac{X^{2018}-1}{X-1}-X+2016$.
Then I wanted to prove that $f(X+1)$ is irreducible in $\mathbb{Z}[X]$ (this would imply that $f(X)$ is irreducible in $\mathbb{Z}[X]$).
$f(X+1)=\frac{(X+1)^{2018}-1}{X}-X+2016=\frac{X^{2018}+\binom{2018}{1}X^{2017}+...+\binom{2018}{2016}X^2+\binom{2018}{2017}X}{X}-X+2016=$ $=X^{2017}+\binom{2018}{1}X^{2016}+...+\left[\binom{2018}{2016}-1\right]X+4034$.
Now, what I had in mind was applying Eisenstein's criterion, but it doesn't seem to work.

Answer 1

The following argument works. I'm fairly sure I have seen it earlier on the site, but fifteen minutes with Approach0 searches like this left me with nothing.

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Let $\alpha$ be a complex zero of your polynomial. Because there are 2016 powers of $x$ in the polynomial, the triangle inequality forces us to conclude that $|\alpha|>1$.

This leads to a contradiction as follows. Assume that there is a factorization $f(X)=g(X)h(X)$. By Gauss's lemma we can assume that both factors $g(X),h(X)$ are monic and in $\Bbb{Z}[X]$. But, up to a sign, $g(0)$ and $h(0)$ are the products of the respective zeros of the said factor. By the above observation $g(0)$ and $h(0)$ thus both have absolute values $>1$.

But $2017=f(0)=g(0)h(0)$ is a prime, so this is impossible.