A hint on why if $c$ is not a square in $\mathbf{F}_p$, then $c^{(p - 1)/2} \equiv -1 \mod p$ Let $\mathbf{F}_p$ be a finite field and let $c \in (\mathbf{Z}/p)^\times$. If $x^2 = c$ does not have a solution in $\mathbf{F}_p$, then $c^\frac{p - 1}{2} \equiv -1 \mod p$.
I will try to prove the contrapositive: Suppose that $c^\frac{p - 1}{2} \not\equiv -1 \mod p$. We show that $x^2 = c$ has a solution in $\mathbf{F}_p$. By Fermat's Theorem, $c^{p - 1} \equiv 1 \mod p$. Then $c^{p - 1} - 1 \equiv 0 \mod p$. Then $(c^\frac{p - 1}{2} + 1)(c^\frac{p - 1}{2}  - 1) \equiv 0 \mod p$. This implies that either $c^\frac{p - 1}{2} \equiv -1 \mod p$ or $c^\frac{p - 1}{2} \equiv 1 \mod p$. 
Hence it must be that $c^\frac{p - 1}{2} \equiv 1 \mod p$. 
I'm not sure how to derive an $a \in \mathbf{F}_p$ such that $a^2 = c$.

Let $\alpha$ be a generator of $\mathbb F_p^{\times}$, $p$ a prime greater than $2$. Then, each nonzero element of $\mathbb F_p$ can be expressed as $\alpha^i$ where $i$ ranges from $0$ to $p-2$.  The elements of the form
$\alpha^{2k}$ are quadratic residues and have square roots $\pm \alpha^i$ in the field, while the elements of the form
$\alpha^{2k+1}$ are quadratic nonresidues.   Thus,
the equation $x^2 = c$ (where $c \in \mathbb F_p$) has no roots in $\mathbb F_p$ if and only
if $c= \alpha^{2k+1}$ for some $k$.
 Now, $\alpha^{p-1} = 1$ and $\alpha^{(p-1)/2} = -1$.  If $c = \alpha^{2k+1}$,
then
$$c^{(p-1)/2} = \left(\alpha^{2k+1}\right)^{\frac{(p-1)}{2}}
= \alpha^{k(p-1) + (p-1)/2} = \alpha^{(p-1)/2} = -1.$$
