Polynomial with roots modulo all primes $p \equiv 3 \pmod 4$ Does there exist an irreducible polynomial $f \in \mathbb{Z}[X]$ of degree $n \geq 2$ with a zero modulo all primes $p \equiv 3 \pmod 4$?
For example, there is such a polynomial $X^2+1$ if we choose primes $p \equiv 1 \pmod 4$. Mostly using quadratic residues, we can come up with many such polynomials for various collections of primes, but I wasn't able to find one for primes $p \equiv 3 \pmod 4$.
 A: Let $$f(x)= x^3-3x+4$$ 
It is irreducible and $$Disc(f) = 4(3)^3-27(-4)^2=-18^2$$
Let $k$ be the splitting field of $f\bmod p$. Factorize $$f(x)=\prod_{j=1}^3 (x-a_j)\in k[x]$$ Note that $$Disc(f)^{1/2}=(a_1-a_2)(a_1-a_3)(a_2-a_3)\in k, \qquad Disc(f)=\prod_{i\ne j} (a_i-a_j)$$

Because of the Frobenius automorphism, if $f\bmod p$ is irreducible then $k=\Bbb{F}_p[x]/(f(x))$ ie. $[k:\Bbb{F}_p]=3$ which implies that $k$ doesn't contain any quadratic subfield ie. $Disc(f)^{1/2}\in \Bbb{F}_p$.

And (exlucding the case $p=3$ where $f=(x+1)^3$ is reducible)  since we know that $Disc(f)^{1/2}\in \Bbb{F}_p$ iff $p\not\equiv 3\bmod 4$ we get that $f$ is never irreducible when $p\equiv 3\bmod 4$, ie. $f\bmod p$ has a root.
A: Again partial answer

Not if $K/\Bbb{Q}$ is Galois where $K=\Bbb{Q}[x]/(f(x))$,

This is because $f\bmod p$ has a root (and $p$ unramified) means that $pO_K$ splits completely, and the density of primes that split completely in a Galois extension of degree $n$ can't be $\ge 1/n$, since this would contradict that $$\zeta_K(s)= F(s)\prod_{p \text{ splits completely}} \frac1{(1-p^{-s})^{\deg(f)}}$$ has a simple pole at $s=1$
(where $ F(s)$ is an Euler product which converges for $ \Re(s) > 1/2$)
A: This is not a complete answer, but under the further assumption that the Galois group of $ f $ is abelian over $ \mathbf Q $, it's easy to show with elementary class field theory that this is impossible.
By Kronecker-Weber the splitting field of $ f $ would be a subfield of some $ \mathbf Q(\zeta_n) $, and the requirement that $ f $ have a root modulo every prime $ 3 $ mod $ 4 $ (in this case, equivalent to $ f $ splitting completely modulo every prime $ 3 $ mod $ 4 $) would imply that there is a nontrivial (not equal to $ \mathbf Q $) subfield of $ \mathbf Q(\zeta_n) $ in which almost every prime $ 3 $ mod $ 4 $ (modulo finitely many exceptions coming from the discriminant of $ f $) split completely. This in turn would imply that the corresponding Frobenius elements in the Galois group $ (\mathbf Z/n \mathbf Z)^{\times} $ all fix this subfield, but it's easy to see that for all $ n $, the subgroup of $ (\mathbf Z/n \mathbf Z)^{\times} $ generated by the sufficiently large primes $ 3 $ mod $ 4 $ is in fact the whole group, implying that the only subfield of $ \mathbf Q(\zeta_n) $ in which almost every prime $ 3 $ mod $ 4 $ is split is the trivial subfield $ \mathbf Q $.
This essentially rules out all constructions based on Artin reciprocity, which in particular covers all irreducible polynomials of degree $ 2 $, since such polynomials all have Galois group $ C_2 $, which is abelian. I don't know if it's possible to cook up an example using nonabelian methods, however.
