Krdytkyu

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.

No, e.g. irreducible $rm f(x), =, x(x+1)+4 $ is even but $rm:f(x) ne pm 2.$

However, such fixed divisors of all values of $rm,f,$ are essentially the only known obstruction to prime values. As motivation, let's start with a converse result. In $1918$ Stackel published the following simple observation:

Theorem If $rm, f(x),$ is a composite integer coefficient polynomial then $rm, f(n), $ is composite for all $rm,|n| > B,, $ for some bound $rm,B.,$ In fact $rm, f(n), $ has at most $rm, 2d, $ prime values, where $rm, d = {rm deg}(f)$.

The simple proof can be found online in Mott & Rose [3], p. 8.
I highly recommend this delightful and stimulating $27$ page paper
which discusses prime-producing polynomials and related topics.

Contrapositively, $rm, f(x), $ is prime (irreducible) if it assumes a prime value
for large enough $rm, |x|, $.
As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which
states that $rm, f(x) in mathbb Z[x],$ is prime if $rm, f(b), $
yields a prime in radix $rm,b,$ representation (so necessarily $rm,0 le f_i < b).$

For example $rm,f(x) = x^4 + 6, x^2 + 1 pmod p,$ factors for all primes $rm,p,,$
yet $rm,f(x),$ is prime since $rm,f(8) = 10601rm$ octal $= 4481$ is prime.
Cohn's test fails if, in radix $rm,b,,$ negative digits are allowed, e.g.
$rm,f(x), =, x^3 - 9 x^2 + x-9, =, (x-9),(x^2 + 1),$ but $rm,f(10) = 101,$ is prime.

Conversely Bouniakowski conjectured $(1857)$
that prime $rm, f(x), $ assume infinitely many prime values (excluding
cases where all the values of $rm,f,$ have fixed common divisors, e.g. $rm, 2: |: x(x+1)+2, ).$ However, except for linear polynomials (Dirichlet's theorem), this conjecture has never been proved for any polynomial of degree $> 1.$

Note that a result yielding the existence of one prime value extends to existence of infinitely many prime values, for any class of polynomials closed under shifts, viz. if $rm:f(n_1):$ is prime, then $rm:g(x) = f(x+ n_1!+1):$ is prime for some $rm:x = n_2inBbb N,:$ etc.

For further detailed discussion of Bouniakowski's conjecture and related results, including heuristic and probabilistic arguments, see Chapter 6 of Ribenboim's The New Book of Prime Number Records.

[1] Bill Dubuque, sci.math 2002-11-12, On prime producing polynomials.

[2] Murty, Ram. Prime numbers and irreducible polynomials.

Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.

[3] Mott, Joe L.; Rose, Kermit. Prime producing cubic polynomials.

Ideal theoretic methods in commutative algebra, 281-317.

Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.

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 $ane 0$, (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 $2$!