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As far as I know there is no polynomial function $f:\mathbb N\to\mathbb P$ with just prime number results, but how about polynomials $g:\mathbb N\to\mathbb N$ such that for every prime $p$ there is a $n\in\mathbb N$ such that $p=g(n)$? For the odd primes there is the trivial polynomial $g(n)=2n+1$, but besides from that?

Lehs
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    No polynomial of degree greater than $1$ can have this property; their values thin out too quickly, and it would contradict eg the prime number theorem. – Qiaochu Yuan Jun 18 '17 at 19:28
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    Do we assume that $g \in \mathbb{Z}[X]$? In this case, such a polynomial $f$ would have to be irreducible. Then it is even impossible for $f$ to have a root modulo $p$ for every prime $p$, see this answer. – Lukas Heger Jun 18 '17 at 19:48

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