elementary number theory - Show that if $p$ is prime and $p equiv 3 pmod 4$, then $frac{p-1}{2} notequiv pm1 pmod p$

Show that if $p$ is prime and $p \equiv 3 \pmod 4$ then $\frac{p-1}{2} \not\equiv \pm1 \pmod p$.

Edit:

Could I say that a given is $x^2 \equiv 1 \pmod p \iff x \equiv \pm 1 \pmod p$ and then substitute in $x = \frac{p-1}{2}$ and show $(p-1) \not\equiv 1 \mod p$ giving me the final answer? If so how would I show this?

Answer

Let $q=\frac{p-1}{2}$. Note that the numbers $q+1,q+2, \dots, 2q$ are congruent modulo $p$, in reverse order, to $-1,-2,\dots,-q$. It follows that
$$(p-1)!\equiv (-1)^q(q!)^2\pmod{p}.$$
But by Wilson's Theorem, we have $$(p-1)!\equiv -1\pmod{p}.$$
It follows that

$$(q!)^2\equiv (-1)^{q-1}\pmod{p}.$$
If $p=4k+3$, then $q-1=2k$, which is even. Thus
$$(q!)^2\equiv 1\pmod{p}.$$
The result now follows.