Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question

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).

share|cite|improve this answer

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 $d=\gcd\{f(a)\mid a\in\mathbb Z\}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.