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Given integers $a,b$, we say a polynomial $f(x) \in \mathbb{Z}[x]$ is an $(a,b)$-filter, if $f(x)$ splits completely into linear factors modulo an odd prime $p$ only if $p=a \pmod b$. For example $x^2+1$ is a $(1,4)$-filter and $(x^2+1)(x^2+2)$ is a $(1,8)$-filter.

Problem: Determine all pairs $(a,b)$ for which an $(a,b)$-filter exists.

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  • $\begingroup$ What makes you think this is at all tractable? $\endgroup$
    – Igor Rivin
    Jul 18, 2017 at 3:47
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    $\begingroup$ $x^2+1$ splits into linear factors modulo $2$, and it is consequently not a $(1,4)$ filter. $\endgroup$
    – js21
    Jul 18, 2017 at 8:30
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    $\begingroup$ $\Phi_n$ splits into linear factors over $\mathbf{F}_p$ iff $p \equiv 1 \bmod n$ or $p \equiv 1 \bmod \frac{n}{p^k}$. @js21 Maybe the OP meant splits completely into $n$ different linear factors $\endgroup$
    – reuns
    Jul 18, 2017 at 8:34

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Claim. An $(a,b)$-filter exists if and only if $a\equiv 1\pmod{b}$.

Proof. Assume that $f(x)\in\mathbb{Z}[x]$ is an $(a,b)$-filter. Let $K$ be a number field containing all the roots of $f(x)$ and all the $b$-th roots of unity. Let $p$ be an odd unramified prime that splits completely in $K$ (there are infinitely many such primes). Then on the one hand we have $p\equiv a\pmod{b}$, while on the other hand we have $p\equiv 1\pmod{b}$. This shows that $a\equiv 1\pmod{b}$. Conversely, the $b$-th cyclotomic polynomial is a $(1,b)$-filter. QED

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