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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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    $\begingroup$ 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. $\endgroup$
    – mercio
    Commented Nov 27, 2011 at 11:54
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    $\begingroup$ The question reminded me of $n^2-n+41$. $\endgroup$
    – Martin Sleziak
    Commented Nov 27, 2011 at 11:58

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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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  • $\begingroup$ If anyone is wondering why there exists $m$ with $f(m)\neq \pm 1$, this is the subject of this question. $\endgroup$
    – Arnaud D.
    Commented Oct 10, 2018 at 11:59
  • $\begingroup$ But $f(m+pk)\equiv f(m) (mod p)$ valids when $a_n>0$. @Gerry Myerson $\endgroup$
    – TaylorR
    Commented Aug 26, 2020 at 13:01
  • $\begingroup$ See this answer more generally for a UFD with finitely many units. $\endgroup$
    – Bill Dubuque
    Commented Jul 7, 2023 at 21:45
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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.

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  • $\begingroup$ What if $\,f(x)=x^3+x^2+x+1\,$. $\endgroup$
    – dxiv
    Commented Jul 27, 2018 at 20:25

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