Squares modulo an odd prime

Question

In Theorem 2.6 of Examples Of Mordell's Equation, Conrad writes:

..., so $y^{2} \equiv -18 \,\mathrm{mod}\,p$. Hence $-18 \equiv \square\,\mathrm{mod}\,p$, so $-2 \equiv \square\,\mathrm{mod}\,p$.

Note that $p$ is an odd prime. Why does the second implication hold? If $-18$ is some quadratic residue, then why does it follow that $-2$ is also the quadratic residue?

Answer 1

For the Legendre symbol we have $$ 1=\left(\frac{-18}{p}\right)=\left(\frac{-3^2\cdot 2}{p}\right)=\left(\frac{-2}{p}\right). $$ So if $-18$ is a quadratic residue modulo $p$, then also $-2$.