$1 + \prod_{i = 1}^{n}(x - i)$ is irreducible in $\mathbb{Q}[x]$

Question

How to show that $1 + \prod_{i = 1}^{n}(x - i)$ is irreducible in $\mathbb{Q}[x]$ for all $n \ne 4$? If I suppose by contradiction that it is reducible, as $p(x)q(x)$, then we can assume without loss of generality that $p(x)$ and $q(x)$ are both monic. We also have that $p(i)q(i) = 1$ for all $1 \leq i \leq n$. Can I draw a contradiction from this? Or should I try something else

Answer 1

Let $r(x)=1+\prod_{k=1}^n(x-k)$. Assume that $r=pq$.

At least one of the polynomials $p$ or $q$ has degree $\le n/2$. Assume that it is $p$. Note that $p(i)^2=1$ for $i=1,2,\ldots,n$. Then $p^2$ and $r$ has the same values at $n$ points, and they are monic, so $$p(x)^2=r(x)$$

But $r(n+1)=n!+1$, that is a perfect square for $n=4,5,7$. Paul Erdős conjectured that there are no more integer numbers $n$ such that $n!+1$ is a square. See Brocard's problem.

Nonetheless, for even $n\ge 6$, $$r(1.5)=1-\frac{3\cdot5\cdot7\cdot\ldots\cdot(2n-3)}{2^n}<1-\frac{945}{64}<0$$

Answer 2

Hints: note that for $1\leq i\leq n$, $p(i)=q(i)$ (why?). Hence $p(x)-q(x)$ has $n$ roots, but has degree smaller than $n$, so $p(x)=q(x)$. Therefore your polynomial is a square. This is impossible for odd $n$, clearly. For even $n\geq 6$, evaluate at $x=1.5$ to derive a contradiction.