Prove Corollary (7.18).
Let's remind ourselves of Corollary (7.18).
Let $p$ be an odd prime then
$$ (\frac{2}{p}) = (-1)^{\frac{p^2-1}{8}} $$
We start with Proposition (7.15).
Let $p$ be an odd prime then
$$ (\frac{2}{p}) = \begin{cases} 1 & \text{ if } p \equiv \pm 1 \pmod 8 \\ -1 & \text{ if } p \equiv \pm 3 \pmod 8 \end{cases} $$
We consider both cases for $p$, that is, $p \equiv \pm 1 \pmod 8$ and $p \equiv \pm 3 \pmod 8$.
Case $p \equiv \pm 1 \pmod 8$
Here $p = 8k \pm 1$ for some integer $k$. This means
$$ \frac{p^2 -1}{8} = \frac{8^2k^2 + 1 -1 \pm 16k}{8} = 8k^2 \pm 2k = 2(4k^2 \pm k) $$
In this case $\frac{p^2 -1}{8} = 2(4k^2 \pm k)$ is even, and so $(-1)^{\frac{p^2 -1}{8}} = 1$. And so this case satisfies Corollary (7.18).
Case $p \equiv \pm 3 \pmod 8$
Here $p = 8k \pm 3$ for some integer $k$. This means
$$ \frac{p^2 -1}{8} = \frac{8^2k^2 + 9 -1 \pm 48k}{8} = 8k^2 + 1 \pm 6k = 2(4k^2 \pm 3k) + 1$$
In this case $\frac{p^2 -1}{8} = 2(4k^2 \pm 3k) + 1$ is odd, and so $(-1)^{\frac{p^2 -1}{8}} = -1$. And so this case satisfies Corollary (7.18).
For both possible cases of $p$ we have shown the Corollary holds, and so we have proven Corollary (7.18).