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Prove that if $P$ is a polynomial with integer coefficients, which is not a constant polynomial, then there is a natural number $x$ such that $P(x)$ is a composite number.

My first thought was to conduct a proof by contradication, assuming that $P(x)$ is a prime number for all $x \in N$. Such proof would be pretty easy, as $P(0)=p$ is prime, thus $pāˆ£P(pn)$ and $P(pn)$ is prime, hence $P(pn)=p$ for all $n$. But such assumption is insufficient, because there are also numbers that are neither prime nor composite, so we can have e.g. $P(0) = 1$. How to prove it then?

Bill Dubuque
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xterg
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    Hint: $P(n+d)\equiv P(n)\pmod d.$ Now consider an $n$ such that $P(n)\neq 0,\pm1.$ And consider the cases when $d=kP(n).$ ā€“ Thomas Andrews May 04 '24 at 00:29
  • For example, if $P(x)=x^2+1,$ then $n=2$ gives $P(n)=5\neq 0,\pm1,$ and then $P(2+5k)$ is always divisible by $5.$ Can it always be $\pm5?$ ā€“ Thomas Andrews May 04 '24 at 00:34
  • @ThomasAndrews I don't know if I understood your question correctly, but I think it can't always be $\pm5$ as it depends on the $n$ taken. Let $P(n) = p$, then we can say that $p$ divides $P(kp)$ for every integer $k$, right? However, I don't really understand what this leads to.. ā€“ xterg May 04 '24 at 11:44

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