Proving $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers which are not quadratic squares 
Prove that $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers $0<a<\frac{p}{2}$ which are not quadratic squares $\pmod p$ ($p\equiv3\bmod4$) 

I don't know really from where to start (we know from wilson that $(p-1)!\equiv -1\pmod p$ but that won't help here as far as I see). The lecturer suggested lookigng at $p-n$ where $1\le n\le p-1$.   I'd be glad for any hint. 

$$\left ( \left ( \frac{p-1}{2} \right )! \right )^2= 1^2 \cdot 2^2 \cdot 3^2 \cdots \left (\frac{p-1}{2} \right )^2=1 \cdot 2 \cdot 3 \cdots  \frac{p-1}{2}   \cdot 1 \cdot 2 \cdot 3 \cdots  \frac{p-1}{2} \\\equiv 1 \cdot 2 \cdot 3 \cdots  \frac{p-1}{2}  \cdot (-1) \cdot (p-1) \cdot (-1) \cdot (p-2) \cdots (-1) \cdot \frac{p+1}{2} \overset{Wilson}{ \equiv} \ \ (-1) (-1)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p+1}{2}} \pmod p$$
$$p \mid \left ( \left ( \frac{p-1}{2} \right )! \right )^2-(-1)^{\frac{p+1}{2}} \Rightarrow p \mid  \left( \left ( \frac{p-1}{2} \right )!-(-1)^{\frac{p+1}{2}} \right ) \left( \left (\frac{p-1}{2} \right )!+(-1)^{\frac{p+1}{2}} \right )$$
As $p$ is a prime, $p \mid   \left( \left ( \frac{p-1}{2} \right )!-(-1)^{\frac{p+1}{2}} \right )$ OR  $p \mid \left( \left (\frac{p-1}{2} \right )!+(-1)^{\frac{p+1}{2}} \right )$
Therefore, $ \left(\frac{p-1}{2} \right )! \equiv (-1)^{\frac{p+1}{2}} \pmod p$ OR $ \left(\frac{p-1}{2} \right )! \equiv (-1)^{\frac{p+3}{2}} \pmod p$
Now,it remains to show that the number of integers $0<a<\frac{p}{2}$ which are not quadratic squares $\pmod p$ is either $\frac{p+1}{2}$ or $\frac{p+3}{2}$.
