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I would like to ask the following question:

Is it possible to find two non-constant polynomials $p(x), q(x)$ with integer coefficients, such that $\gcd(p(n), q(m))=1$ for every $(n, m)\in \mathbb{N}^2$?

If such $p(x), q(x)$ exist, we will call them "completely" coprime, since all of their values will be coprime.
Obviously $p(x), q(x)$ must not have the same root, but this does not seem to help. The problem seems to be quite simple and I suspect that the answer is no, but I was unable to prove this.

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Fedor Petrov has pointed out that there is a flaw in the published solution, and neither of us sees an easy way to fix the argument. Since MathOverflow will not allow me to delete the solution, I am replacing it with this text, and will change my answer to Community wiki to remove the credit.

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    $\begingroup$ can not we simply say that if $\alpha_1,\ldots,\alpha_k$ are roots of $f(x)$, then $\operatorname{Res}_x\bigl(f(x),g(x+y)\bigr)=C\prod_{i=1}^k g(y+\alpha_i)$ is a polynomial in $y$ of degree $\deg g\cdot \deg f$, thus not a constant? $\endgroup$
    – Fedor Petrov
    Commented Jan 7 at 22:51
  • $\begingroup$ @FedorPetrov Sure, that's an even more efficient way to finish the proof, using the definition of the resultant instead of Bezout's identity. Thanks. $\endgroup$
    – Joe Silverman
    Commented Jan 8 at 3:50
  • $\begingroup$ Wait, but why if $p$ divides a resultant there is a common root mod $p$? There is a common factor, not necessarily linear. $\endgroup$
    – Fedor Petrov
    Commented Jan 9 at 8:52
  • $\begingroup$ @FedorPetrov You're right, that's an error in the proof. I thought a little bit about how to fix it and haven't come up with a fix. I'll continue to think about it. Hmmm... If it's not fixable, should I delete the answer, or just edit it to point out the error? $\endgroup$
    – Joe Silverman
    Commented Jan 10 at 13:22
  • $\begingroup$ At least it should be clarified in a disclaimer. Possibly somebody may see how to fix it. But I am sceptical on this:( $\endgroup$
    – Fedor Petrov
    Commented Jan 10 at 13:46
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Let me rename to $f$ and $g$ the two polynomials (which you called $p$ and $q$ in the question) as this will make notations less confusing.

Clearly we can assume that $f$ and $g$ are irreducible and distinct. Let $K$ be the decomposition field of $fg$ over $\mathbb{Q}$. Possibly excluding finitely many primes that divide the leading coefficient of $f$ or $g$ or at which $f$ or $g$ ramifies, to say that a prime $p$ splits completely in $K$ means that $fg$ has $\deg f + \deg g$ roots in $\mathbb{Z}/p\mathbb{Z}$, and certainly implies that $f$ and $g$ have roots there. Now by Čebotarëv's density theorem (or something weaker), there is a positive density of such primes, so we can find $p$ such that both $f$ and $g$ have roots, say $\bar m,\bar n$ in $\mathbb{Z}/p\mathbb{Z}$. Lifting them arbitrarily to integers, we see that $p$ divides both $f(m)$ and $g(n)$, which are therefore not relatively prime.

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    $\begingroup$ Regarding "something weaker", I asked this very question a while ago and got some excellent answers: mathoverflow.net/questions/15220/… $\endgroup$
    – François G. Dorais
    Commented Mar 21, 2023 at 11:19
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    $\begingroup$ @FrançoisG.Dorais When you say that you asked "this very question", I assumed you meant you'd asked the OP's question, and then wondered why the OP's question wasn't then flagged as a duplicate. So you might want to clarify that what you asked was for a more elementary proof that infinitely many primes split completely, which is also a great question, of course. $\endgroup$
    – Joe Silverman
    Commented Mar 21, 2023 at 20:05
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    $\begingroup$ Also "a while ago" is a nice understatement for a question from the first half year of this site's existence! $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 21, 2023 at 21:19

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