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 $g^{\frac{p-1}{2}} \equiv -1 \pmod{p}.$

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, 
$$g^{\frac{p-1}{2}\frac{p+1}{2}}=\left(g^{\frac{p-1}{2}}\right)^{\frac{p+1}{2}}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$ 
