$p$ an odd prime, $p \equiv 3 \pmod 8$. Show that $2^{(\frac{p-1}{2})}*(p-1)! \equiv 1 \pmod p$ $p$ an odd prime, $p \equiv 3 \pmod 8$. Show that $2^{\left(\frac{p-1}{2}\right)}\cdot(p-1)! \equiv 1 \pmod p$
From Wilson's thm: $(p-1)!= -1 \pmod p$. 
hence, need to show that $2^{\left(\frac{p-1}{2}\right)} \equiv -1 \pmod p. $
we know that $2^{p-1} \equiv 1 \pmod p.$ 
Hence:  $2^{\left(\frac{p-1}{2}\right)} \equiv \pm 1 \pmod p. $
How do I show that this must be the negative option? 

Euler's Criterion
Need an $x \in \Bbb N$ such that $2 \equiv x^2 \pmod p$. However no such $x$ exists in $\Bbb Z_p$. 
For example, in $\Bbb Z_3$, $1^2=2^2=1 \neq 2$. In $\Bbb Z_{11}$, no square number is $\equiv 2 \pmod {11}$. Leave that proof to you. If you get stuck examine $x^2-2 \equiv 0 \pmod p$.
$\Rightarrow$   $2^{(p-1)/2} \equiv -1 \pmod p$
