Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree [duplicate]

Solution 1:

Extended hints as requested

Assume a factorization $p(x)=q(x)r(x)$.

1: $p(x)=(x-1)(x-2)\cdots (x-n)-1$

As you observed, in this first case you get $q(i)=-r(i)=\pm1$ for all $i=1,2,\ldots,n$, so $q(x)+r(x)$ has at least $n$ zeros. This is a problem, because the leading coefficients of $q(x)$ and $r(x)$ have the same sign.

2: $p(x)=(x-1)(x-2)\cdots (x-n)+1$

In the second case $q(i)=r(i)=\pm1$ for all $i=1,2,\ldots,n$, so we get that $q(x)-r(x)$ has at least $n$ zeros. As observed in the comments this leaves only the possibility $q(x)=r(x)$. Indeed, when $n=4$ we have $$ p(x)=(x^2-5x+5)^2. $$ Anyway, the remaining case is that $n=2k$, $q(x)=r(x)$, $k=\deg q(x)$. The factorization $p(x)=q(x)^2$ shows that $p(x)\ge0$ for all real numbers $x$. Estimate $p(2k-\dfrac12)$ when $k>2$.