Gauss' well-known law of quadratic reciprocity states the following:
Main Theorem. Given two disttinct primes, $p$ and $q$, if at least one of $p-1$ and $q-1$ is divisable by $4$, then the congruency $x^2 \equiv q ~\operatorname{mod} p$ is solvable iff $x^2 \equiv p ~\operatorname{mod} q$ is. But if $p\equiv q \equiv 3 \operatorname{mod} 4$, then $x^2 \equiv q ~\operatorname{mod} p$ is solvable iff $x^2 \equiv -p ~\operatorname{mod} q$ is. Using the Legandre symbol, it can be stated simply as $\left({\frac {p}{q}}\right)\left({\frac {q}{p}}\right)=(-1)^{{\frac {p-1}{2}}{\frac {q-1}{2}}}$.
Wikipedia provides a short proof, originally by B. Veklych [2019]. They first prove some sort of lemma:
Lemma. Let $p$ be an odd prime. Thus $2$ is a quadratic residue iff the number of solutions to the equation $x^2 +y^2 = 2$ in $\Bbb{F}_p$ is divisable by $8$.
Afterwards, they prove that the number of solutions to this equation can by expressed by ${\displaystyle p-(-1)^{\frac {p-1}{2}}}$. They finish the proof by stating:
"Therefore, $2$ is a residue modulo $p$ if and only if $8$ divides ${\displaystyle p-(-1)^{\frac {p-1}{2}}}$. This is a reformulation of the condition stated above."
Question. Why is this a reformulation of the main threom's statement?