TITLE: Basic question on primitive roots
QUESTION [7 upvotes]: From Ireland and Rosen's A Classical Introduction to Modern Number Theory, p.48:

Let $p$ be a prime of the form $4t+1$. Show that $a$ is a primitive root $\bmod p$ iff $-a$ is a primitive root $\bmod p$.

I can write (letting $p=4t+1$) 
$$\begin{align} 
a^{p-1} &\equiv 1 \bmod p\quad\quad \text{ because }a\text{ is a primitive root}\\

a^{4t} &\equiv 1 \bmod 4t+1\\

a^{4t} -1 &\equiv 0 \bmod 4t+1
\end{align}$$
I notice that $-a$ satisfies this last equation, but I don't feel comfortable with this because I don't think this is enough to prove that $-a$ is in fact a primitive root.

REPLY [8 votes]: Let $g$ be a primitive root of $p$. Then $(g^{2k})^2 \equiv 1 \pmod{p}$ by Fermat's Theorem,  but $g^{2k}\not\equiv 1 \pmod p$, so $g^{2k}\equiv -1\pmod p$.
It follows that $(-g)^{2k+1}\equiv g \pmod p$. Thus any power of $g$ is congruent to a power of $-g$.  It follows that $-g$ is a primitive root of $p$.
Remark: Let $p$ be an odd prime. Recall that $p^e$ and $2p^e$ have primitive roots. If $e \ge 1$ and $p$ is of the form $4k+1$, and $g$ is a primitive root of $p^e$, then $-g$ is a primitive root of $p^e$. The analogous result holds for $2p^e$.
The proof is the same as for the case $e=1$.  For $\varphi(p^e)=(p-1)p^{e-1}=4kp^{e-1}$. We conclude as above that $g^{2kp^{e-1}}\equiv -1 \pmod{p^e}$, and therefore $(-g)^{2kp^{e-1}+1}\equiv g \pmod{p^e}$.  For $2p^e$,  use the fact that $\varphi(2p^e)=\varphi(p^e)$.