Let us consider a polynomial $p\in \mathbb{Q}[x]$ with $p(\mathbb{Z})\subseteq \mathbb{Z}$, such that for each $a\in \mathbb{N},\,p(n)\equiv 0 \,(\text{ mod } a\,) $ has a solution for some $n\in \mathbb{Z}.$

Does this imply $p(n)=0$ for some $n\in \mathbb{Z}$ ?

Thanks in advance for any help.


I claim that $f(x) = (x^3-3)(x^2-2)(x^2+2)(x^2+1)$ is a counterexample: for any nonzero integer $a$, we have $a \mid f(n)$ for some $n \in \mathbb{Z}$. However, it is clear that $f(n) \ne 0$ for all $n \in \mathbb{Z}$.

First, by the Chinese Remainder Theorem, it suffices to prove this in cases where $a$ is a prime power. Now, if $a = p$ is an odd prime, then at least one of $-1, 2, -2$ must be a quadratic residue $\pmod{p}$ since $-2 = (-1) \cdot 2$. Then, for example, if $n^2 \equiv -2 \pmod{p}$, then $p \mid n^2+2$ so $p \mid f(n)$; and similarly for the other cases. Now for example if $n^2 + 2 \equiv 0 \pmod{p}$, then in particular since $p$ is odd we have $2n \not\equiv 0 \pmod{p}$; therefore, by Hensel's lemma, for each $m > 0$ that implies that there exists $n' \equiv n \pmod{p}$ such that $(n')^2 + 2 \equiv 0 \pmod{p^m}$. This proves what we wanted in cases where $a$ is an odd prime power.

All that remains is to show the case $a = 2^m$. However, here the $x^3-3$ factor comes into play: set $g(x) = x^3 - 3$. Then since $g(1) = -2 \equiv 0 \pmod{2}$ and $g'(1) = 3 \not\equiv 0 \pmod{2}$, again by Hensel's lemma we get that for each $m$, there exists $n' \equiv 1 \pmod{2}$ such that $g(n') \equiv 0 \pmod{2^m}$, finishing the proof.

  • 1
    $\begingroup$ Come to think of it, $(2x-1)(3x-1)$ should also work: if $a$ is a prime power, then at least one of 2 or 3 has a multiplicative inverse $\pmod{a}$. The initial example has the advantages of being monic and of not even having any rational roots, though. $\endgroup$ – Daniel Schepler Jul 27 '18 at 17:01
  • $\begingroup$ Hi, Daniel. Sorry to be late. Thanks for yours answer. But, I didn't get one part, i,e; how does it suffices to prove the result in case where $a$ is a prime power. Please allow me to explain. If $a \in \mathbb{N}$, then by UFT, $a=a_1 a_2 ...a_m$ for some $m \in \mathbb{N}$, where each $a_i$'s are prime power and coprime to each other. Now by CRT, if we have $f(n) \equiv 0 (\text{ mod } a_i)$ has solution for some $n \in \mathbb{N}$, then there exists some integer $x$ such that $x \equiv 0 (\text{ mod } a)$. But my confusion is that how can we say that $x$ is of the form $f(n)$? $\endgroup$ – Surajit Aug 1 '18 at 19:06
  • 1
    $\begingroup$ Choose $n_1$ such that $f(n_1) \equiv 0 \pmod{a_1}$, and choose $n_2$ such that $f(n_2) \equiv 0 \pmod{a_2}$, etc. Now, choose $n$ such that $n \equiv n_1 \pmod{a_1}$, $n \equiv n_2 \pmod{a_2}$, etc. Then $f(n) \equiv f(n_1) \equiv 0 \pmod{a_1}$, $f(n) \equiv f(n_2) \equiv 0 \pmod{a_2}$, etc. $\endgroup$ – Daniel Schepler Aug 1 '18 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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