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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 $a_0=1$.

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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
up vote 4 down vote accepted

Choose $m$ such that $f(m)\ne\pm1$, then choose any prime $p$ dividing $f(m)$, and think about $f(m+pk)$ for $k=1,2,\dots$.

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