TITLE: Prove that the polynomial $(x-1)(x-2)\cdots(x-n) + 1$, $ n\ge1 $, $ n\ne4 $ is irreducible over $\mathbb Z$
QUESTION [38 upvotes]: I try to solve this problem. I seems to come close to the end but I can't get the conclusion. Can someone help me complete my proof. Thanks


Show that the polynomial $h(x) = (x-1)(x-2)\cdots(x-n) + 1$ is irreducible over $\mathbb Z$ for all $n\ge1$ and $ n\ne4$.


Suppose $h(x) = f(x) g(x)$, then we must have $f(i)g(i) = 1$ for all $i = 1,2,...n$. So both $f(i)$ and $g(i)$ are $1$ or $-1$. In either case, $m(x) = f(x) - g(x)$ has degree smaller than $n$ and have $n$ different roots ($1,2,...,n$). So we must have $m(x) = 0$. Then $h(x) = f(x)^{2}$. So $n$ must be even. Let $n = 2k$. Because $f(x)$ has degree $k$, there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.
This is where I got stuck. Hope some one can help me solve this. Thanks.

REPLY [4 votes]: For $n\geq 7$ the irreducibility follows from following criterion of Pólya (see Polynomials by Prasolov).

(Pólya). Let $f$ be a polynomial of degree $n$ with integer coefficients and define $m=\lfloor\frac{n+1}{2} \rfloor$. Suppose that, for $n$ different integers $a_1,\dots,a_n$ we have $|f(a_i)|<2^{-m}m!$ and the numbers $a_1,\dots,a_n$ are not roots of $f$. Then $f$ is irreducible.

Choose $a_i=i$ and for $n \geq 7$ the claim follows.