If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$ 
If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$.

I don't know how to prove this statement.
$p=4m+3$, so $(2m+1)! \equiv \pm1\pmod p$
This is all I did.
 A: $\left( \frac{p-1}{2}! \right)^2 = \prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}n=\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}-n \equiv\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=\frac{p+1}{2}}^{p-1}n=(p-1)! \equiv 1$
You can change $n$ for $-n$ because there is an even number of factors from $1$ to $\frac{n-1}{2}$ 
Therefore, $ \frac{p-1}{2}!$ is congruent to either $1$ or $-1$
A: 
Wilson's theorem
$$(p-1)!\equiv_p -1$$
Proof: Every element $x\in\Bbb Z_p^\times$ has a unique inverse $y\in\Bbb Z_p^\times$ such that $xy\equiv_p 1$, except $-1,1$ since they are their own inverses, i.e., they satisfy $x^2\equiv_p 1$.
This inverse is unique, since if both $y,y'$ were inverses of $x$, then $$xy\equiv_p xy'\equiv_p 1\Rightarrow y\equiv_p y'$$ Now in the product $(p-1)!$ every element is multiplied with it's corresponding inverse, except $-1$ and $1$, which means $$(p-1)!\equiv_p 1\cdot 1\cdot\ldots\cdot 1\cdot (-1)\equiv_p -1$$

$(p-1)!\equiv_p 1\cdot (-1)\cdot 2\cdot (-2)\cdot \ldots\cdot \frac{p-1}{2}\cdot (-\frac{p-1}{2})\equiv_p (-1)^{\frac{p-1}{2}}\left (\frac{p-1}{2}\right )!^2$
By Wilson's theorem $-1\equiv_p(-1)^{\frac{p-1}{2}}\left (\frac{p-1}{2}\right )!^2$
Using that $p\equiv_4 3$ we get $$\left (\frac{p-1}{2}\right )!\equiv_p \pm1$$
