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Let $f(x)$ and $g(x)$ be polynomials with integer coefficients such that $f(x)>0$ and $g(x)>0$ for all real values of $x$. Suppose that for every integer $n$, if $f(x)=0$ has a solution mod $n$, then $g(x)=0$ has a solution mod $n$. What can generally be said about $f(x)$ and $g(x)$?

My guess is that $g(x)$ must have a factor $h(x) \in \mathbb{Z}[x]$ such that $f(x)=h\circ k(x)$ for some $k(x) \in \mathbb{Z}[x]$.

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    $\begingroup$ Without the positivity condition, non-isomorphic number fields with the same Dedekind zeta function should give counterexamples. I don't know if the positivity condition would rule these out but I suspect not. $\endgroup$ – Felipe Voloch Apr 28 '18 at 21:30
  • $\begingroup$ Yes, without the positivity conditions, one can take $f(x)=x$ and $g(x)$ to be any intersective polynomial. $\endgroup$ – Marco Apr 28 '18 at 21:53
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The question becomes more tractable if we assume that $f$ and $g$ are monic irreducible polynomials, and we restrict to $n$'s coprime to the discriminants of $f$ and $g$. In this case, by Hensel's lemma and standard facts on discriminants, the modified question remains unchanged if we restrict to primes $n$ coprime to the discriminants of $f$ and $g$.

Let us consider the corresponding number fields $K=\mathbb{Q}[x]/(f(x))$ and $L=\mathbb{Q}[x]/(g(x))$ whose discrimants divide the discriminants of $f$ and $g$, respectively. Then, in the special case when $L$ is Galois (i.e. $L$ contains all the roots of $g$), the modified condition holds for $f$ and $g$ if and only if $K$ contains $L$. This follows from Prop. 15 in Ch. VIII-5 of Weil: Basic number theory.

Example. Let $p$ be a prime congruent to $1$ mod $4$. Let $f(x)=x^{p-1}+\dots+1$ be the $p$-th cyclotomic polynomial, and let $g(x)=x^2-p$. Consider an arbitrary prime $n\nmid 2p$. Then, $f(x)$ has a root modulo $n$ if and only if $n$ is congruent to $1$ mod $p$, and $g(x)$ has a root modulo $n$ if and only if $n$ is a quadratic residue mod $p$. Indeed, in this case, $K=\mathbb{Q}(e^{2\pi i/p})$ is the $p$-th cyclotomic field and $L=\mathbb{Q}(\sqrt{p})$ is its unique quadratic subfield, and we just recorded the splitting primes in $K$ and $L$.

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  • $\begingroup$ It seems the positivity condition did not matter in this argument. Can one use the Inclusion theorem to derive the same conclusion for general irreducible f(x) and g(x)? I am in particular interested in the case $f(x)=x^2+1$. $\endgroup$ – Marco Apr 28 '18 at 23:13
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    $\begingroup$ @Marco: The inclusion theorem cannot be extended to general monic irreducible $f(x)$ and $g(x)$, for the reason explained by Felipe Voloch's comment (under your post). But $f(x)=x^2+1$ is a special case, for which I don't know the answer. The question is interesting! $\endgroup$ – GH from MO Apr 29 '18 at 0:27

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