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.



Polynomials splitting into linear factors modulo certain primes
**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


