TITLE: Bunyakovsky conjecture
QUESTION [5 upvotes]: The Bunyakovsky conjecture states the following :

Let $f$ be an irreducible polynomial and $d$ denote the gcd of the set 
   $f(a)$, where $a$ runs over the integers. Then, $f(a)/d$ is prime for
   infinite many integers $a$.

I found statements about the converse:
If $a$ is large enough, then if $f(a)$ is prime, then $f$ is irreducible.
This "large enough" is precised.
But the case $d>1$ is not considered!
There remain two possibilities :

There is a translation to a function $g$ with $d = 1$, such that $f$ is 
irreducible if and only if $g$ is irreducible.
If $a$ is large enough, then if $f(a)/d$ is prime, then $f$ is irreducible.

Which of the two possibilities can be used to check polynomials 
with $d > 1$? And if the second possibility works, which number
 is large enough ?

REPLY [2 votes]: Both possibilities work. Note that in case 1) you may pass to $f(aX)/a\in\mathbb Z[X]$ for $a=f(0)$ to obtain $d=1$ for the new polynomial. This trick also settles 2).