# Conjecture about fixed prime divisors of polynomials with integer coefficients

While experimenting with random polynomials I've found this conjecture:

A polynomial $$f\in\mathbb Z[X]$$ of degree $$n$$ with co-prime coefficients have no fixed prime divisor $$p> n$$.

A fixed prime divisor is a prime $$p$$ such that $$p|f(m)$$ for all $$m\in\mathbb Z$$.

Is this known? Proved? Or are there counterexamples?

• Where did you find this conjecture? Jan 10, 2019 at 21:37
• @SmileyCraft: as I wrote, I've done some experimenting. With my own software Bigz.
– Lehs
Jan 10, 2019 at 21:39
• do you mean pairwise co-prime? If you mean that there should be no prime which divides all the coefficients then it seems to be false. For instance, take $f(x)=x(x+1)(x+2)=x^3+3x^2+2x$. Clearly $2\,|\,f(n)$ for all $n\in \mathbb Z$.
– lulu
Jan 10, 2019 at 21:47
• Ok, but the $p>n$ case is trivial. No polynomial of degree $n<p$ can vanish for all the residues $\pmod p$ and your polynomial can't reduce to $0\pmod p$ since that would imply that every coefficient was divisible by $p$.
– lulu
Jan 10, 2019 at 21:54
• And little Fermat implies that $p | x^p-x$ for all primes $p$ Jan 10, 2019 at 21:55

Since $$p$$ does not divide each of the coefficients of $$f(x)$$, $$f(x)$$ reduces to a non-trivial polynomial $$\overline f(x)$$ of degree $$≤n$$ $$\pmod p$$. But if $$p>n$$ then $$\overline f(x)$$ can have at most $$n$$ roots $$\pmod p$$, hence it is non-zero on at least one residue $$a \pmod p$$. But then $$f(a)\not \equiv 0 \pmod p$$ and we are done.