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If $f\in\mathbb Z[X]$ and there are $x_1,\dots, x_n\in\mathbb Z_+$, where $n>\deg f$, such that $f(x_i)\in\mathbb P$, $i=1,\dots n$, then $f$ is irreducible over $\mathbb Z$. Because, if $\,f=g\cdot h$ then either $g(x_i)=1$ or $h(x_i)=1$ which can't happen $n>\deg g+\deg h$ times.

But what about the opposite? Are there non constant, irreducible polynomials $p\in\mathbb Z[X]$ such that

$x_1,\dots, x_n\in\mathbb Z_+$ and $p(x_i)\in\mathbb P$ for all $i=1,\dots,n$ $\:\implies\:$ $n\leq \deg p$?

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    $\begingroup$ I think this question is pleny interesting even without the $n \leq \deg p$ limitation. Just that there are only finitely many inputs that yield primes. $\endgroup$ – Arthur Dec 17 '18 at 9:13
  • $\begingroup$ @Arthur: the limitation would be nice for an eventual computational test method. Else I agree. $\endgroup$ – Lehs Dec 17 '18 at 9:18
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    $\begingroup$ See also math.stackexchange.com/a/2942032/589. $\endgroup$ – lhf Dec 17 '18 at 10:55
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Your claim seems to be negative.

For example, if we take $f(x)=x^3-x+3$. It is an irreducible polynomial of degree $3$, and for each positive integer $n$, $f(n)$ is divided by $3$ ( By Fermat's little theorem). Since $f(n)>3$ when $n>2$, the only $n$ such that $f(n)\in \mathbb{P}$ is just $1$.

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    $\begingroup$ I'm not sure, but it seems that the irreducibility of $f$ can be deduced from the method in en.wikipedia.org/wiki/Irreducible_polynomial#Over_the_integers as $f$ is irreducible over $\mathbb Z_2$. Is that correct? $\endgroup$ – Lehs Dec 17 '18 at 11:17
  • $\begingroup$ Yes, you are correct! $\endgroup$ – Bonbon Dec 17 '18 at 11:24
  • $\begingroup$ Thanks a lot! Now I know how to proceed from your example. $\endgroup$ – Lehs Dec 17 '18 at 11:26

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