Prime divisors of $n^2-5$

Question

Q: Show that if $p$ prime divides $n^2-5$, then $p\equiv\pm1$(mod 5), with $p\neq2,5$

I have shown that the only common quadratic residues between $p$ and $5$ are $1$ and $4$. I note that $4\equiv-1$(mod 5). I'm not quite sure where to go from here, I'm quite lost.

Answer 1

Note that $p\mid n^2-5$ implies that $n^2\equiv 5\mod{p}$, so that $5$ is a square modulo $p$. Can you finish from here? (Hint: Quadratic reciprocity.)

Answer 2

The way I like to say it this: given odd prime $$ q \equiv \pm 2 \pmod 5 \; , \; $$ so that Legendre symbol $$ (20| q) = (5|q) = -1. $$

With integers $x,y.$

IF $$ x^2 - 5 y^2 \equiv 0 \pmod q \; , \; $$ then both $$ x,y \equiv 0 \pmod q \; , \; $$ so that $$ x^2 - 5 y^2 \equiv 0 \pmod {q^2} \; , \; $$

Your problem says my $y=1.$ This is impossible for my prime $q$