Prove that the polynomial $(x-1)(x-2)\cdots(x-n) + 1$, $ n\ge1 $, $ n\ne4 $ is irreducible over $\mathbb Z$ 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.
 A: 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.
A: AS @Calvin Lin suggest, I write down the complete answer here for someone who need.
From my deduction above, we must have $h(x) = f(x)^{2}$. And there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.
Let $I$ is subset of $\{1,2,...,2k\}$ consists of elements at which $f(x)$ is $1$ and $J$ is subset of $\{1,2,...,2k\}$ consists elements at which $f(x)$ is $-1$. It's easy to see that if $n = 2$, then $h(x)$ is irreducible, and $n \neq 4$, so we just consider the case $n \ge 6$, which means that $k \ge 3$.
Suppose $I$ consists of $1$. $J$ has at least $3$ distinct elements greater than $1$, so there exists an elment $u$ in $J$ such that $u - 1 \ge 3$. But we must have $u - 1 | f(u) - f(1) = -2$, contradiction.
