Does $f(x)\in\mathbb P$ for polynomial $f$ and some $x>0$ imply infinitely many prime $f(x)$? Let $$f(x):=\sum_{n=0}^{k} a_n x^n$$ be some non-constant polynomial where all coefficients $a_n$ are nonnegative integers except for the constant term $a_0$, which must be positive.

Assuming the Bunyakovsky conjecture, if $f(c)$ is prime for some integer $c&gt;0$, would it follow that $f(x)$ yields infinitely many primes?
Yes.

$f(x)$ must be strictly increasing given the constraints. If $f(c)$ is prime and $c&gt;0$, then since $f(0)&lt;f(c)$, we know $f(c) \nmid f(0)$ and $f(0)$ must be relatively prime to $f(c)$, thus fulfilling the requirements of the Bunyakovsky conjecture.
Yes.

$f(x)$ must be strictly increasing given the constraints. If $f(c)$ is prime and $c&gt;0$, then since $f(0)&lt;f(c)$, we know $f(c) \nmid f(0)$ and $f(0)$ must be relatively prime to $f(c)$, thus fulfilling the requirements of the Bunyakovsky conjecture.