Let p be a prime number with p≡3 (mod 4) and let r be a primitive root modulo p . Prove that $\mathrm{ord}_p(-r) = (p-1)/2.$

Question

I only could write this:

Let p = 4k + 3 where k is an nonnegative integer. Since r is a primitive root modulo p .

$r^{(p-1)/2} ≡ - 1 $ mod p. So

$r^{2k+1}≡ -1$ mod p

$(-r)^{2k+1}=-1*(r)^{2k+1}$

$-1*(r)^{2k+1}≡-1*-1 ≡ 1 $ mod p.

I couldn't find the rest.

Answer 1

We can write: $$(-r)^{\frac{p-1}{2}}\equiv(-1)^{\frac{p-1}{2}}r^{\frac{p-1}{2}}\mod p$$ As $p=4k+3$, $\frac{p-1}{2}$ is odd. Therefore: $$(-1)^{\frac{p-1}{2}}r^{\frac{p-1}{2}}\equiv -r^{\frac{p-1}{2}}\mod p\equiv-(-1)\mod p\equiv1\mod p$$ So we know that $(-r)^{\frac{p-1}{2}}\equiv 1\mod p$. Now if $m=ord(-r)<\frac{p-1}{2}$, it must divide $\frac{p-1}{2}$ and hence is odd. Therefore: $$r^m\equiv(-1)^m(-r)^m\equiv -1\mod p$$ Contradiction to the fact that $r$ is a primitive root (since $m\neq\frac{p-1}{2}$). So the order is exactly $\frac{p-1}{2}$ as desired