$r$ primitive root of prime $p$, where $p \equiv 1 \mod 4$: prove $-r$ is also a primitive root 
Let $p$ be a prime with $p \equiv  1 \mod 4$, and $r$ be a primitive root of $p$.  Prove that $-r$ is also a primitive root of $p$.

I have shown that $-r^{\phi(p)} \equiv 1 \mod p$.  What I am having trouble showing, however, is that the order of $-r$ modulo $p$ is not some number (dividing $\phi(p)$ that is LESS than ($\phi(p)$).
By way of contradiction, I've shown that the order of $-r$ cannot be an EVEN number less than $\phi(p)$.  But my methodology does not work for the hypothetical possibility of an ODD order that is less than $\phi(p)$.
Any and all help appreciated.  Happy to show methodology for any of the parts I have managed to do, if requested.

Result: Let $r$ be a primitive root $\pmod{p}$. Then the order of $r^k \pmod{p}$ is $\frac{p-1}{\gcd(k, p-1)}$.
Proof: Let $m$ be the order of $r^k \pmod{p}$. Then $1 \equiv (r^k)^m \equiv r^{km} \pmod{p}$, so $p-1 \mid km$ as $r$ is a primitive root. Thus $\frac{p-1}{\gcd(k, p-1)} \mid \frac{k}{\gcd(k, p-1)}m$ and $\gcd(\frac{p-1}{\gcd(k, p-1)},\frac{k}{\gcd(k, p-1)})=1$ so $\frac{p-1}{\gcd(k, p-1)} \mid m$. On the other hand $(r^k)^{\frac{p-1}{\gcd(k, p-1)}} \equiv (r^{\frac{k}{\gcd(k, p-1)}})^{p-1} \equiv 1 \pmod{p}$ so $m \mid \frac{p-1}{\gcd(k, p-1)}$. Thus $m=\frac{p-1}{\gcd(k, p-1)}$, as desired.

Now note $r^{\frac{p+1}{2}} \equiv r^{\frac{p-1}{2}}r \equiv -r \pmod{p}$.
