Cubic polynomial mapping primes to primes 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 $p$?
 A: 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 to pick $p$.
A: It is, I think, better to investigate the last coefficient a0. It is easy to show that a0 cannot have more than 3 distinct prime factors. because, suppose that p is a prime factor of a0. then, if f(p) is prime, it must be = p, since p divides f(p). similar are the cases for other prime factors of a0. in short, if t is a prime factor of  a0, then f(t)= t. but f being a cubic, it can't have more than three zeroes. this leads to my 1st statement. obviously irreducibility and all that are necessary conditions. i think a little more effort can solve your last problem.
