Simple Congruence proof

Question

Show that there do not exist numbers $(a,b) \in \mathbb{Z}$ such that

$a^2+5b^2 \equiv 2$(mod $5$) or $a^2+5b^2 \equiv 3$(mod $5$)

Can anyone help me where to start or go about this question?

Thanks in advance

Answer 1

Since $5b^2 \equiv 0 \pmod 5$, we need to show $\pm 2$ are not quadratic residues. This easily follows from Euler's Criterion or just testing the five cases modulo $5$.

Answer 2

Hint $\ {\rm mod}\ 5\!:\ \color{#c00}{a^2\equiv \pm2}\,\Rightarrow\, a\not\equiv 0\,\overset{\rm Fermat}\Rightarrow 1\equiv \color{#c00}{a^4\equiv 4}\,\Rightarrow\,5\mid 4\!-\!1\Rightarrow\!\Leftarrow$