TITLE: Polynomials representing primes
QUESTION [10 upvotes]: Suppose over $\mathbb{Z}$ we are given an irreducible polynomial $p(x)$.
Can we say that at the very least that $p(x)$ represents a prime as $x$ runs through integers?
Thanks in advance.

REPLY [8 votes]: Let $P(x)$ be a polynomial of degree $\ge 1$, with integer coefficients, such that no $d \gt 1$ divides all the coefficients. 
If $P(x)$ has degree $1$, then $P$ represents at least one prime. This is a consequence of Dirichlet's Theorem on primes in arithmetic progression (and easily implies that Theorem). 
As has been pointed out, for degree $\ge 2$, irreducibility is not enough to ensure that a polynomial represents a prime. For some irreducible polynomials $P(x)$, there exists a $d \gt 1$ such that $d$ divides $P(n)$ for every integer $n$. 
However, that can only happen for relatively simple congruential reasons. So let us focus attention on polynomials $P(x)$ for which there is no such universal $d$. Unfortunately, it is an open problem whether such a polynomial must necessarily represent at least one prime.  
Example: There is a good deal of evidence that there are infinitely many primes of the form $x^2+1$. However, whether or not there are infinitely many is a long-standing open problem, often called the Hardy-Littlewood Conjecture. If we could show that for all $a\ne 0$, (or even infinitely many $a$) there exists $x$ such that $(2ax)^2+1$ is prime, that would settle the Hardy-Littlewood Conjecture.  (Conversely, the Hardy-Littlewood Conjecture implies that there are infinitely many such $a$.) 
So the question you raised seems to be extremely difficult even for polynomials of degree $2$!