Let $n$ be an integer, prove that every prime factor of $12x^2+1$ has the form $6m+1$, where $m$ is an integer.
I've gone as far as knowing that all primes $p>3$ are either $6m+1$ or $6m-1$, how do I use this, not sure whether to even use that.
I know about the Legendre symbol and quadratic reciprocity
2 and 3 can be easily excluded, so let $p$ be a prime which is at least 5.
Then $p$ is either congruent to 1 or 5 modulo 6. Suppose $p \ | \ 12n^2 + 1$ and $p \equiv 5 \pmod 6$. Then we have $12n^2 \equiv -1 \pmod p$ and hence
$$ \Big(\frac{-3}{p}\Big) = 1, $$
where $()$ is the Legendre symbol. By the rules of Legendre symbol and the quadratic reciprocity, we have
$$ \Big(\frac{-3}{p}\Big) = \Big(\frac{-1}{p}\Big)\Big(\frac{3}{p}\Big) = (-1)^{\frac{p-1}{2}} (-1)^{\frac{2\times(p-1)}{4}} \Big(\frac{p}{3}\Big) = \Big(\frac{p}{3}\Big) = -1, $$
which leads to a contradiction.