Prove that every odd prime divisor of $n^2+100$ is of the form $12k+1$ or $12k+5$

Question

Prove that every odd prime divisor of $n^2+100$ is of the form $12k+1$ or $12k+5$.

I'm not sure how to do this.

Answer 1

If $p|(n^2+100), n^2\equiv-100\pmod p\iff (n\cdot (10)^{-1})^2\equiv-1\pmod p$

Using this, $p\equiv1\pmod 4$

$\implies p\equiv1,5,9\pmod{12}$ but $p$ needs to be prime

Observe that $p\equiv9\pmod {12}\equiv0\pmod3$