Proving a polynomial $f(x)$ composite for infinitely many $x$

Let $f(x)=a_0+a_1x+ \ldots +a_nx^n$ be a polynomial with integer coefficients, where $a_n>0$ and $n \ge 1$. Prove that $f(x)$ is composite for infinitely many integers $x$.

I can easily show that there are infinitely many composite numbers of the form $a_0+a_1x+ \ldots +a_nx^n$ if $a_0 \ge 2$, we just note that $f(x)$ is composite for every $x$ being a multiple of $a_0$. But I can't find a way to prove this in the case $a_0=1$.

• maybe you can try to translate the polynomial : look at $f(x+b)$, and see if this can make a polynomial with a good constant coefficient. – mercio Nov 27 '11 at 11:54
• The question reminded me of $n^2-n+41$. – Martin Sleziak Nov 27 '11 at 11:58

Choose $m$ such that $f(m)\ne\pm1$, then choose any prime $p$ dividing $f(m)$, and think about $f(m+pk)$ for $k=1,2,\dots$.
• If anyone is wondering why there exists $m$ with $f(m)\neq \pm 1$, this is the subject of this question. – Arnaud D. Oct 10 '18 at 11:59
Let $x_k=(\text{lcm}(a_0,a_1,\dots,a_n))^k$ for some $k$. Then all $x_k$ with $k\in\mathbb N$ are values which produce composite numbers.
• What if $\,f(x)=x^3+x^2+x+1\,$. – dxiv Jul 27 '18 at 20:25