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.
 A: *

*Fact $0$: A number of the form $4x^2 + 3$ is odd, hence all its prime divisors must necessarily be odd.

*Fact $1$: If $x$ is not a multiple of $3$, then $x^2 \equiv 1 \pmod{3}$, and therefore $4x^2 + 3$ is not a multiple of $3$.

*Fact $2$: If $p$ is a prime dividing a number of the form $4x^2 + 3$ where $x$ is not a multiple of $3$, then $-3$ is a quadratic residue modulo $p$, that is
$$1 = \left(\frac{-3}{p}\right) = \left(\frac{p}{3}\right),$$
the latter by quadratic reciprocity. That means that $p \equiv 1 \pmod{3}$.
Hence a prime dividing a number $4x^2 + 3$ where $x \not\equiv 0 \pmod{3}$ must be either $\equiv 1 \pmod{12}$ or $\equiv 7 \pmod{12}$.
But the number $4x^2 + 3 \equiv 7 \pmod{12}$, so not all its prime divisors can be $\equiv 1 \pmod{12}$.
A: 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 $12n+11$.
