I am studying the seemingly official accepted answer to this question here. Here is a screen shot of the answer.
If we assume that $p(x)$ is reducible in $\mathbb{Z}$, then we can assume a factorization, $p(x)=q(x)r(x)$, of $p(x)$ into two non-units. Consequently, the degree of both $q(x)$ and $r(x)$ must be greater than zero.
I understand that $q(i)=-r(i)$ for $i = 1,2,...,n$ implies that $q(x)+r(x)$ has at least $n$ roots. I also understand that since $p(x)$ is monic then so are $r(x)$ and $q(x)$, and $p(x)$ is has a positive coefficient in front of $x^n$, so the coefficients of the largest power of $r(x)$ and $q(x)$ must have the same sign. But I am still not seeing how this implies a contradiction.
To me it seems like the argument could work like this: because $q(x)+r(x)$ has at least $n$ zeros, that means $deg(q(x)+r(x)) \ge n$. This means that either $q(x)$ or $r(x)$ has degree greater than or equal to $n$. But by the properties of multiplication of monic polynomials we must have that the degree of $q(x)$ and $r(x)$ be strictly less than $n$.
