Let $a,m$ an integers s.t $(a,m)=1$. Let $K$ a quadratic field, I would like to calculate the natural density of the set


$$\{p \;\; \text{rational prime}\; /\; p\;\text{inert in}\; K,\; p\equiv a\pmod m\}$$


I think that is equal to $1/2\phi(m)$, but I couldn't prove that.



Inert primes in arithmetic progression
This is not true in general. Take $K = \mathbb{Q}(i), a =3$ and $m = 4$. Then the density is just $1/2$ in this case. This is because a prime $p$ is inert in $K$ if and only if $p \equiv 3 \bmod 4$. So the congruence condition implies already that $p$ is inert.


In general, saying that a prime is inert in $K$ can be written in terms of congruence conditions modulo $|d\_K|$, where $d\_K$ is the discriminant of $K$. So if $\gcd(d\_K,m) = 1$, then the density is indeed $1/2\varphi(m)$. But otherwise you can get different densities occuring.


