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 $a^2 = c$.
 A: We assume $p$ is odd, and use an argument that yields additional information. 
There are two possibilities, $p$ is of the form $4k-1$, and $p$ is of the form $4k+1$. 
Let $p$ be of the form $4k-1$. If $c^{(p-1)/2}\equiv 1\pmod{p}$, then $c^{(p+1)/2}\equiv c\pmod{p}$. But $\frac{p+1}{2}=2k$, and therefore 
$$(c^k)^2\equiv c\pmod{p}.$$ 
To complete things, we show that if $p$ is of the form $4k+1$, then the congruence $x^2\equiv -1\pmod{p}$ has a solution. The argument goes back at least to Dirichlet. 
Suppose that $x^2\equiv -1\pmod{p}$ has no solution. Consider the numbers $1,2,\dots,p-1$. For any $a$ in this collection, there is a $b$ such that $ab\equiv -1\pmod{p}$. Pair numbers $a$ and $b$ if $ab\equiv -1\pmod{p}$.   Since the congruence $x^2\equiv -1\pmod{p}$ has no solution, no number is paired with itself. The product of all the pairs is $(p-1)!$, and it is also congruent to  $(-1)^{(p-1)/2}$ modulo $p$. Since $\frac{p-1}{2}$ is even, it follows that $(p-1)!\equiv 1\pmod{p}$, which contradicts Wilson's Theorem.   
A: 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
$$c^{(p-1)/2} = \left(\alpha^{2k+1}\right)^{\frac{(p-1)}{2}}
= \alpha^{k(p-1) + (p-1)/2} = \alpha^{(p-1)/2} = -1.$$
A: Hints (fill in the missing details):


*

*If $c=x^2$, $x\neq0$, then $c^{(p-1)/2}=x^{p-1}=1$ by Little Fermat. 

*If $x^2=y^2$, $x\neq0$, then $(x-y)(x+y)=0$, so $y=\pm x$. Therefore there are $(p-1)/2$ distinct non-zero squares $q$, and all those satisfy $q^{(p-1)/2}=1$.

*Because the equation $x^{(p-1)/2}=1$ is of degree $(p-1)/2$, it can have at most $(p-1)/2$ solutions. And we have found all of them.

*Put this together with what you have to conclude that if $c$ is a non-square, then $c^{(p-1)/2}=-1$.

