Let $f(n)=a\_3n^3+a\_2n^2+a\_1n+a\_0$, with $a\_i\in\mathbb{Z}$, $a\_3>0, a\_0\neq 0$ such that $f(n)>0$ for all positive integers $n$.


Given a prime $p$, when is $f(p)$ again prime?


For example, let $f(n)=7n^3-50n+30$. Then,
$$f(7)=2081\quad {\rm (prime)},$$
$$f(11)=19\cdot463,$$
$$f(13)=14759\quad {\rm (prime)}.$$


Are there conditions on the $a\_i$'s that guarantee that $f(p)$ is prime for all primes $p$?



Cubic polynomial mapping primes to primes
There is no non-constant polynomial sending primes to primes, aside from $f(x)=x$. Indeed, it suffices to consider the case where $f$ is irreducible, as if $f(x)$ factors as $g(x)h(x)$, $f(p)$ is clearly composite for large $p$.


Now if $f(x)\not=x$, choose some large prime $p$ such that $f$ has a non-zero root $a$ in $\mathbb{F}\_p$. By Dirichlet's theorem on primes in arithmetic progressions, there exist infinitely many primes $q\_i$ with $q\_i\equiv a\bmod p$. But for such $q\_i$, $f(q\_i)$ is divisible by $p$, and thus taking $q\_i$ large, we have that $f(q\_i)$ is composite.


I've been a bit glib about why one can choose $p$ as claimed; just look for primes which split in the splitting field of $f$ over $\mathbb{Q}$.


EDIT: More simply, use Lemma 1 [here](http://math.stanford.edu/~dlitt/exposnotes/primes1mod4.pdf) to pick $p$.


