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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?

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  • $\begingroup$ Where did you find this conjecture? $\endgroup$
    – SmileyCraft
    Jan 10, 2019 at 21:37
  • $\begingroup$ @SmileyCraft: as I wrote, I've done some experimenting. With my own software Bigz. $\endgroup$
    – Lehs
    Jan 10, 2019 at 21:39
  • $\begingroup$ 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 $n\in \mathbb Z$. $\endgroup$
    – lulu
    Jan 10, 2019 at 21:47
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    $\begingroup$ 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 $p$. $\endgroup$
    – lulu
    Jan 10, 2019 at 21:54
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    $\begingroup$ And little Fermat implies that $p | x^p-x$ for all primes $p$ $\endgroup$
    – Mark Bennet
    Jan 10, 2019 at 21:55

1 Answer 1

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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.

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