I know that someone has already posted this problem, but I wanted to see if my proof is also valid. Here is the problem:
Let $r$ be a primitive root of the odd prime $p$. Prove that if $p\equiv 3\pmod4$, then $-r$ has order $(p-1)/2$ modulo $p$.
Proof
Note that since $r$ is a primitive root of $p$, we have $r^{(p-1)/2}\equiv -1\pmod{p}$.
Suppose $o_p(-r)=m$, some $0<m<\frac{p-1}{2}$.
First suppose $m$ is even. Then, $(-r)^m=r^m$, so $(-r^m)\equiv 1\pmod{p}\implies r^m\equiv 1\pmod{p}$, which contradicts $r$ is a primitive root of $p$.
Now, suppose $m$ is odd. Then $(-r)^m=-r^m$, so $(-r^m)\equiv 1\pmod{p}\implies -r^m\equiv 1\pmod{p}\implies r^m\equiv -1\pmod{p}$.
But since $r$ is a primitive root of $p$, we have $r^{(p-1)/2}\equiv -1\pmod{p}$ (we had shown this in a previous problem). So $r^m\equiv -1\pmod{p}\implies m=\frac{p-1}{2}$, contradicting $m<\frac{p-1}{2}$.
Therefore, $o_p(-r)$ is not less than $\frac{p-1}{2}$.
I'm not sure where to go from here...