Infinitude of prime number of the form $x^2+14y^2$ In the book "Primes of the form $x^2+ny^2$", David Cox had shown that:
$$p=x^2+14y^2 \Longleftrightarrow (-14/p)=1 \;\text{and}\; (x^2+1)^2=8 \mod p \; \text{has an integer solution.} $$
Is this imply that there are infinitely many primes of the form $x^2+14y^2$? It is easy to see that there are infinitely many primes that the equation $(x^2+1)^2=8 \mod p$ has an integer solution but I don't have any clue to check if some of them can take $-14$ as their quadratic residue.
 A: According to a theorem due to Dirichlet (proved later by various mathematicians such as A. Meyer and H. Weber), every primitive binary quadratic form represents infinitely many prime numbers.
In the case of $x^2 + 14y^2$, this follows from the observation above in addition with the fact that there are infinitely many primes that split completely in a given number field (a very special case of the standard density theorems in algebraic number theory). In your case, the number field is the Hilbert class field of $k = {\mathbb Q}(\sqrt{-14})$ defined by the roots of $(x^2+1)^2 = 8$ over $k$.
A: COMMENT.- I bet there is an infinity of primes and I rely for this on the following:
for instance,the four identities $$(a+3)^2+14(3b\pm2)^2=6(6a^2+6a+21b^2\pm28+11)-1\\(6a+1)^2+14(3b)^2=6(6a^2+2a+21b^2)+1\\(6a+5)^2+14(3b)^2=6(6a^2+10a+21b^2+4)+1$$ There are an infinity of values for the factors of $6$ in the three $RHS$ of the four identities which give solutions of the equation $$x^2+14y^2=6z\pm1$$ and I find very plausible suppose that there are in a lot of them primes (all prime is of the form $6z\pm1$).
