Let $p$ be an odd prime and let $$P:=\{(x,y)\in\mathbb{N}\times \mathbb{N}: 0<x<\frac{p}{2},0<y<\frac{p}{2},x^2+y^2=1\pmod{p}\}$$ Prove that #$P$ is odd if and only if $p=\pm 1\pmod{8}$.
$\textbf{Solution}$ Let $T$ be the involution defined as $T(x,y):=(y,x)$. Then $T(P)=P$, so that the parity of #$P$ is equal to the parity of the subset $P_0$ of elements of $P$ which are fixed by $T$. These elements are of the form $(x,x)$ with $2x^2=1\pmod{p}$, and $0<x<\frac{p}{2}$. The congruence admits solutions if and only if $p=\pm 1\pmod{8}$, by the quadratic reciprocity law. When a solution $\overline{x}$ exists, there is only one more solution modulo $p: -\overline{x}$. The claim follows because exactly one of the points $(\overline{x},\overline{x})$ and $(-\overline{x},-\overline{x})$ belongs to $P_0$.
What I can't understand:
1) I know that the congruence $x^2=2\pmod{p}$ has a solution iff $\left(\frac{2}{p}\right)=1$ and this happens iff $p=\pm1\pmod{8}$. Why do the same hold for $2x^2=1\pmod{p}$?
2) What does it mean the last part of the Solution? It means that #$P_0=1$?
The elements $(\overline x,\overline x)$ and $(-\overline x,-\overline x)$ are both fixed by $T$, but exactly one of them belongs to $P$ because of the condition $0\lt x\lt p/2$, $0\lt y\lt p/2$. So, yes, $P_0$ has just one element.