The Bunyakosvsky-conjecture states the following :
Let $f(x)$ be a primitive polynomial over the integers. If $f(x)$ is irreducible over $\mathbb Q[x]$ and there is no prime $p$, such that $p|f(n)$ holds for every integer $n$, then $f(n)$ is a prime number for infinite many integers $n$.
The condition "for no prime we have $p|f(n)$ for all integers $n$" menas $\gcd${$f(n)|n\in\mathbb Z$}$=1$. Denote $g:=\gcd${$f(n)|n\in\mathbb Z$}. Then, I wonder whether the following statement is likely to be true (or perhaps follows from the Bunyakovsky conjecture.
Let $f(x)$ be a primitive polynomial over the integers, irreducible over $\mathbb Q[x]$. Suppose, $g$ is defined as above. Then, there are infinite many primes of the form $\frac{f(n)}{g}$ with $n\in\mathbb Z$.
For example : are there infinite many primes of the form $$\frac{n^2+n+2}{2}$$ with $n\in\mathbb Z$ ?