Show that $4x^2+3$ has at least a prime divisor of the form $12n+7$ If $x$ is not divisible by $3$, how to prove that $4x^2+3$ has at least a prime divisor of the form $12n+7$?
Thanks.

Since $4x^2+3$ is of the form $4k+3$, it has at least one prime divisor $p$ of the form $4k+3$. 
and therefore of the form $12n+3$, $12n+7$, or $12n+11$. The possibility $12n+3$ is ruled out by the fact $x$ is not divisible by $3$. We now proceed to rule out the form $12n+11$. 
Note that since $4x^2\equiv -3\pmod{p}$, it follows that $-3$ is a QR of $p$. Thus $(-3/p)=1$. Since $p$ is of the form $4k+3$, we have $(-1/p)=-1$, so $(p/30=-1$. Thus by Quadratic Reciprocity, we have $(p/3)=1$. It follows that $p\equiv 1\pmod{3}$, ruling out $p$ of the form $12n+11$.
