Primitive root modulo p Let $p$ be an odd prime with a primitive root $g$. Prove that $$\prod_{x=1}^{\frac{p-1}{2}}x^2 \equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$
Remark: I intend to use the relationship $g^{\frac{p-1}{2}} \equiv -1 \pmod{p}.$
 A: If you really want to use a primitive root, note that your product is equal to the product of all the non-zero squares (the quadratic residues) modulo $p$. But these are congruent to the even powers of our primitive root $g$. So your product is congruent to
$$g^2 g^4 g^6\cdots g^{2\frac{p-1}{2}}.$$
This is equal to
$$g^{2\left(1+2+3+\cdots +\frac{p-1}{2}\right)}.$$
The arithmetic progression which is the exponent of $g$ has sum
$$\frac{p-1}{2}\frac{p+1}{2}.$$
Finally, 
$$g^{\frac{p-1}{2}\frac{p+1}{2}}=\left(g^{\frac{p-1}{2}}\right)^{\frac{p+1}{2}}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$ 
A: $\prod_1^{p-1}x=\prod_1^{(p-1)/2}x\prod_{(p+1)/2}^{p-1}x\equiv\prod_1^{(p-1)/2}x\prod_1^{(p-1)/2}(-x)=(-1)^{(p-1)/2}\prod_1^{(p-1)/2}x^2$ but also by Wilson's Theorem $\prod_1^{p-1}x\equiv-1\pmod p$ and there you are. 
A: $\prod_{1\le x\le p-1}x\equiv\prod_{1\le y\le p-1}g^y$
$=g^{1+2+\cdots +p-1}=g^{\frac{p(p-1)}2}=(g^\frac{p-1}2)^p\equiv (-1)^p=-1$
$\prod_{1\le x\le p-1}x=\prod_{1\le x\le \frac{p-1}2}x(p-x)$ as to avoid omission and repetition of terms $x\le p-x\implies x\le \frac p 2$ i.e., $x\le\frac{p-1}2$ as $p$ is odd
So, $\prod_{1\le x\le p-1}x\equiv(-1)^\frac{p-1}2\prod_{1\le x\le \frac{p-1}2}x^2$ 
So, $(-1)^\frac{p-1}2\prod_{1\le x\le \frac{p-1}2}x^2\equiv-1$
Multiplying either sides by $(-1)^{p-1},$ (which is legal as $(-1)^{p-1}=1$ as $p$ is odd)
$\prod_{1\le x\le \frac{p-1}2}x^2\equiv (-1)(-1)^\frac{p-1}2=(-1)^\frac{p+1}2$
