Show that the polynomial $(x-1)(x-2) \cdots (x-n)-1$ is irreducible on $\mathbb{Z}[x]$ for all $n \geq 1$ [duplicate]

Show that the polynomial $h(x)=(x-1)(x-2) \cdots (x-n)-1$ is irreducible in $\mathbb{Z}[x]$ for all $n \geq 1$.

This problem seems to be hard to solve. I thought I could use Eisenstein in developping this polynomial, but it is a bad idea. Another idea would be to suppose the existence of $f(x)$ and $g(x)$, and suppose that $f(x)g(x)=(x-1)(x-2) \cdots (x-n)-1$. In this direction, we could analyse the roots of $h(x)$ I guess.

Is anyone could help me to solve this problem?

Solution 1:

Suppose you had a nontrivial factorization $h=fg$, where $f$ and $g$ are (monic) integral polynomials each of degree strictly less than $n$. Then for the integers $1\leq k\leq n$, $h(k)=f(k)g(k)=-1$, so $f(k)$ and $g(k)$ must be one of the values $\pm 1$, and necessarily $f(k)=-g(k)$ for each $1\leq k\leq n$.

Then the polynomial $p(x)=f(x)+g(x)$ has degree strictly less than $n$, but the integers $1\leq k\leq n$ are all roots of $p(x)$, so $p(x)\equiv 0$, so $f(x)=-g(x)$, a contradiction since their leading coefficients aren't equal.

Solution 2:

David's observation that if $f=gh$, then $g(k)=-h(k)=1$ or both $=-1$ for each of $k=1,2,\ldots , n$ is spot on. So both $g$ and $h$ take at least one of these values at least $n/2$ times. If we now take the polynomial of smaller degree (let's say it's $g$), so $\deg g=m\le n/2$, then the only way to avoid a constant $g$ would be $m=n/2$ (so we're done already if $n$ is odd) and a polynomial of the type $$ g = (x-k_1)(x-k_2) \ldots (x-k_m)+ 1 . $$ But we also have to have $g(j)=-1$ for $1\le j\le n$, $j\not= k_r$, and this clearly isn't working.