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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.

Math Guy
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  • Eisenstein works fine here, right? Just let $p=2$. You just have to show that each of the non-leading coefficients of $f(X+1)$ are all even. – Chaotic Good Oct 27 '19 at 20:10
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    @ChaoticGood $\binom{2018}2$ is odd. – Jyrki Lahtonen Oct 27 '19 at 20:17
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    @ChaoticGood yeah I had already tried that, but it's not true – Math Guy Oct 27 '19 at 20:18
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    It fits Osada's criterion from here https://math.stackexchange.com/questions/1935/methods-to-see-if-a-polynomial-is-irreducible/2761444#2761444 (however idea behind the proof is basically the same as idea in the answer given) – Sil Nov 01 '19 at 18:11

1 Answers1

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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.

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.

Jyrki Lahtonen
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    Since $2017> |X^{2017}+X^{2016}+…+X^3+X^2|$ for all complex $x$ with $|x|=1+\epsilon$, by Rouche's Theorem $f(X)$ has no roots on or inside the unit circle... This is a standard "cute" alternative proof of your first claim. Nice solution. – N. S. Oct 27 '19 at 20:13
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    The same argument proves the irreducibility of $$x^p+x^{p-1}+\cdots+x^3+x^2+p$$ over $\Bbb{Q}$ for any prime $p$. The vague recollection I have is about this family, but my search-fu is weak today :-( – Jyrki Lahtonen Oct 27 '19 at 20:13
  • @JyrkiLahtonen Thank you ! I have two questions, which may be dumb, but I am quite new to polynomials : 1. Why from Gauss's lemma we can assume that $g(X)$ and $h(X)$ are both monic? 2. The fact that $g(0)$ and $h(0)$ have absolute values $>1$ follows from your observation because the complex roots of $f$ are also complex roots of $g$ and $h$, right? – Math Guy Oct 27 '19 at 20:46
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    @MathGuy 1) The leading coefficient of $f$ is $1$, which is also the product of the leading coefficients of $g$ and $h$. Hence either both are $+1$ or both $-1$. In the latter case we can use $-g$ and $-h$ instead. 2) Correct. – Jyrki Lahtonen Oct 27 '19 at 20:49
  • Nice argument +1. – Paramanand Singh Jan 11 '20 at 08:53