Assuming that $r$ is a primitive root of the odd prime $p$ prove that $ r^{(p-1)/2}\equiv -1 \pmod p $ holds I know if $r$ is primitive root $r^{^{n}}\equiv a\pmod n$ 
from the set of residue $\{1,2,3....(n-1)\}$ but if change to $r^{(p-1)/2}\equiv -1 \pmod p$.
My assumption: It's no longer be primitive root of $p$ since the residue is $-1$
but don't know where $-1$ is come from ?
I think Euler's Criterion might related it ,that can prove $-1$ can exit.
But I can't do next step. Anyone can give me a hint? 

Note that we have that $r^{p-1} \equiv 1 \pmod p$ by Fermat's Little Theorem. Then we have that: $p \mid r^{p-1} - 1 = (r^{\frac{p-1}{2}} - 1)(r^{\frac{p-1}{2}} + 1)$. Now obviously $p$ must divide one of the factors and it can't be $r^{\frac{p-1}{2}} - 1$, as then $r^{\frac{p-1}{2}} \equiv 1 \pmod p$, meaning that $r$ isn't a primitive root modulo $p$. 
On the other side the fact that such a primitive root modulo $p$ exists comes from the fact that the multiplicative group modulo $p$ is cyclic.
