Playing around with python, I noticed that for all prime $7\ne p<1000$, $x^2+14y^2$ covers all the residue classes in $(\mathbb{Z}/p\mathbb{Z})^{\times}$. I wondered if it is true for all $p\ne 7$, and came up with the following heuristic. Fix $q\ne 2,7$ and a non-zero residue class mod $q$, $a$. We want to find even a prime $p=a\pmod{q}$ representable by $x^2+14y^2$. We note that $p$ splits if $z^2+14$ has a solution mod $p$, this translates to congruence relations on $p$ mod $56$ (by quadratic reciprocity). Dirichlet's theorem gives us infinitely many primes that satisfy both congruences, thus infinitely many primes $p$ which split and have residue $a\pmod{q}$. But we also want the prime above them to be principal. This is where I'm not rigorous- Chebotarev theorem give equidistribution of the Artin symbol and passing to the Hilbert class field, this gives us detail about the images of primes in the class group. But I don't know if we can say that necessarily we'll have a prime satisfying all the above congruence relations, with the prime above it splits... Can we make this argument true? If not, does anyone have an alternative explanation?
In addition, another interesting phenomenon occurred where I looked at this form mod $N$ not prime, here I got a subgroup of the invertible elements (maybe everything- did not check explicitly). I suspect that this result follows from the first, is it true?
Thanks in advance!