# How to show the $p$ minus primitive root is also a primitive root for $p \equiv1 \pmod {4}$

Let $p$ be a prime such that $p \equiv1 \pmod {4},$ and let $r$ be a primitive root mod $p$. I wonder how to show that $p-r$ is a primitive root mod $p.$

The only way $-r$ can not be a primitive root is if $(-r)^k \equiv 1 \mod p$ for some nontrivial divisor $k$ of $p-1$. Since $(-r)^k \equiv r^k$ if $k$ is even, $k$ must be odd. Thus $r^k \equiv -1 \mod p$. But then $r^{2k} \equiv 1 \mod p$, and then $2k=p-1$ is not divisible by $4$.
Write $p=4k+1$. Then $r^{2k}\equiv-1\pmod p$, so $p-r\equiv -r\equiv r^{2k+1}\pmod p$. As $2k+1$ is coprime with $4k$ then $r^{2k+1}$ is also a primitive root.