For primes $p≡3\pmod 4$, prove that $[(p−1)/2]!≡±1\pmod p$. I know to use Wilson's Theorem and that each element in the second half is congruent to the negative of the first half, but I'm not sure how to construct a proof for it.
 A: $p-r\equiv -r\pmod p\implies r\equiv-(p-r)$
For uniqueness, $r\le p-r$ or $2r\le p\implies r\le\frac p 2$
So, $1\le r\le \frac{p-1}2$ as $p$ is odd
Putting $r=1,2,3,\cdots,\frac{p-3}2,\frac{p-1}2$ we get, 
$1\equiv-(p-1)$
$2\equiv-(p-2)$
...
$\frac{p-3}2\equiv-(p-\frac{p-3}2)=\frac{p+3}2$
$\frac{p-1}2\equiv-(p-\frac{p-1}2)=\frac{p+1}2$
So, there are $\frac{p-1}2$ pairs so, 
$(p-1)!=(-1)^{\frac{p-1}2}\left((\frac{p-1}2)!\right)^2$ 
Using Wilson's theorem, $(-1)^{\frac{p-1}2}\left((\frac{p-1}2)!\right)^2\equiv-1\pmod p$ 
If $p\equiv3\pmod 4,p=4t+3$ for some integer $t$,
So, $\frac{p-1}2=2t+1$ which is odd, so $(-1)^{\frac{p-1}2}=-1$
$\implies \left((\frac{p-1}2)!\right)^2\equiv1\pmod p$
$\implies \left(\frac{p-1}2 \right)!\equiv\pm1\pmod p$
A: Recall that $\binom{p-1}{k}\equiv (-1)^k \pmod{p}$ (apply the fact that $p\mid \binom{p}{k}$ for $1\leq k\leq p-1$, and the recursive formula for obtaining binomial coefficients).
Now we have $\frac{(p-1)!}{(\frac{p-1}{2})!(\frac{p-1}{2})!}\equiv\binom{p-1}{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}} \equiv -1 \pmod{p}$, and thus $-1\equiv(p-1)!\equiv (-1)((\frac{p-1}{2})!)^2 \pmod{p}$. So $(\frac{p-1}{2})!$ is a square root of 1 modulo $p$, hence is $\pm1$.
