# Do infinitely many prime numbers occur in $P(x) \in \mathbb{N}[x]$ when the coefficients are relative prime?

I have read about Dirichlet's theorem recently, that is, for relative prime positive integers $a,b$, there exists infinitely primes with the form $ax+b$.

What I want to ask is the situation when the $ax+b$ is changed as any irreducible polynomial with relative prime positive integer coefficients. Is there still infinitely many primes?

• This is entirely unknown, see this. We don't even know a single polynomial (of degree $>1$) that can be shown to take infinitely many prime values. – lulu Sep 7 '18 at 11:47
• Note: it's clear that what you wrote is too broad. The polynomial $x^2+2x+1$ passes your tests, but clearly can't take prime values. Similarly, every value of $x^2+x$ is even. – lulu Sep 7 '18 at 11:50
• Talking about polynomials in two variables, there is this : www.michaelnielsen.org/polymath1/index.php?title=Friedlander-Iwaniec_theorem – Teresa Lisbon Sep 7 '18 at 11:53
• Note; "irreducible" does not solve the problem. $x^2+x+2$ is also always even. – lulu Sep 7 '18 at 11:55
• So far the problem appears to be astonishingly intractable. Even a concrete example, like $x^2+1$ seems to be beyond existing methods. – lulu Sep 7 '18 at 11:58

No, $x^2+2x+1=(x+1)^2$ obviously is not prime for natural $x>0$.
• Irreducible is not sufficient: $x^2+x+2$ is always even, so almost never prime. – xarles Sep 7 '18 at 13:04