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.

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*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 $\equiv 1 \pmod{12}$.
