Let $f\in \mathbb{Z}[x]$ and suppose $f(x) = a\_nx^n + a\_{n-1}x^{n-1} + \ldots + a\_0$ where $a\_n > 0$ and $a\_0 \in \{-1, +1\}$. Is there for every $n\in \mathbb{N}$ an integer $m> n$ such that $f(m)$ is (positive and) composite?



Does every integer polynomial evaluate to infinitely many composite numbers?
Yes.


Choose $n\_0$ large, so the polynomial is always positive, and let $k = f(n\_0)$. Then choose $n = n\_0 + ik$ for $i$ large; we have that $f(n)$ is congruent to $f(n\_0)$ mod $k$, so as long as $i$ is large enough, $f(n)$ is composite.


