TITLE: $r$ primitive root of prime $p$, where $p \equiv 1 \mod 4$: prove $-r$ is also a primitive root
QUESTION [8 upvotes]: 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.

REPLY [7 votes]: Let $p=4k+1$. Since $r$ is a primitive root of $p$, we have $r^{2k}\equiv -1\pmod{p}$. Thus $(-r)^{2k}\equiv -1\pmod{p}$, and therefore
$$(-r)^{2k+1}\equiv (-1)(-r)\equiv r\pmod{p}.$$
Since $r$ is congruent to a power of $-r$, and $r$ is a primitive root of $p$, it follows that $-r$ is a primitive root of $p$.