# Dirichlet's theorem/Bunyakovsky conjecture for infinite composites/a single prime

From https://en.wikipedia.org/wiki/Bunyakovsky_conjecture, the Bunyakovsky Conjecture is an open problem that states that $$f(x)$$ has infinitely many primes in sequence $$f(1),f(2),...$$ if

1) The leading coefficient is positive.

2) The polynomial is irreducible over the integers.

3) The coefficients of $$f(x)$$ are relatively prime.

It is conjectured as a generalization of Dirichlet's theorem on arithmetic progressions, which states that $$a+xd$$ contains infinitely many primes for $$a$$ and $$d$$ relatively prime.

I am curious whether it is known that $$f(x)$$ must contain at least one prime if the three conditions are met, or whether we know that $$f(x)$$ must have infinitely many composites for certain conditions, even just in the case of first degree polynomials.

• to be an odd prime, either the constant term is odd and the index is likely even, or an odd number of odd coefficients happen and it's likely to be of odd index. at least for univariate polynomials. There are other possible constaints, just not usefully. – Roddy MacPhee Jun 11 at 12:02
• The condition 3) is that the $f(n)$ are relatively primes, it is much stronger than just the coefficients are coprime, see $n^2+n+2$ is always even. – reuns Jun 11 at 21:24

Let $$g$$ be a positive integer. For any f be an irreducible polynomial of degree $$g$$, define $$N_f = \text{gcd}\left\{ f(n) : n \in \{1, \ldots, g - 1\}\right\}$$. If for each irreducible polynomial $$f$$ of degree $$g$$, there exists one integer $$m$$ for which $$f(m)/N_f$$ is prime, then for each irreducible polynomial $$f$$ of degree $$g$$, there exists infinitely many integers $$m$$ for which $$f(m)/N_f$$ is prime.
• The proof is the way $N_f$ behaves under $f(x) \mapsto f(px+m)$ – reuns Jun 11 at 21:34