Prove that the multiplicative inverse of a quadratic residue of $p$ is also a quadratic residue of $p$.
Euler's Criterion gives us
$$ a^{\frac{p-1}{2}} \equiv 1 \pmod p $$
By definition of multiplicative inverse, we have
$$ a \times a^{-1} \equiv 1 \pmod p $$
Raising this to index $\frac{p-1}{2}$
$$ \begin{align} (a)^{\frac{p-1}{2}} \times (a^{-1})^{\frac{p-1}{2}} & \equiv 1^{\frac{p-1}{2}} \pmod p \\ \\ 1 \times (a^{-1})^{\frac{p-1}{2}} & \equiv 1 \pmod p \\ \\ (a^{-1})^{\frac{p-1}{2}} & \equiv 1 \pmod p \end{align}$$
This is Euler's Criterion that tells us $a^{-1}$ is a quadratic residue of $p$.