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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 :

  1. There is a translation to a function $g$ with $d = 1$, such that $f$ is irreducible if and only if $g$ is irreducible.

  2. 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 ?

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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).

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If $f\in\mathbb Z[X]$ with $f(0)\ne 0$ then $f(n)=0$ implies $|n|\le|f(0)|$. Consequently, if $f$ has no integer roots (as is readily checked per Gauss), then $|f(n)|\le k$ implies $|n|\le |f(0)|+k$.

Now assume $f$ has no integer roots but is reducible, say $f(X)=g(X)h(X)$. Also assume thet $f(n)=pd$ with $p$ prime for some $n$ with $|n|>|f(0)|+|d|$. Then $|g(0)|,|h(0)|\le |f(0)|$ implies $|g(n)|>|d|$ and $|h(n)|>|d|$, but one of the factors $a,b$ of any factorization $pd=ab$ must be $\le |d|$ in absolute value - contradiction.

Remark: I did not have in the above argument to assume that $d=\gcd\{f(a)\mid a\in\mathbb Z\}$.

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I believe there is something wrong in Hagen's answer: Pick $f(X)=X(X-3)+1$. Then $\lvert f(3)\rvert=1=k$, but $3>2=\lvert f(0)\rvert+k$. – Peter Mueller Jun 14 '13 at 23:02

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