Let $p$ be a prime number with $p \equiv 1 \ (\mathrm{mod} \ 4)$ and $a \in \mathbb{N}$ with $p \nmid a$.
How to show that
$a$ is a primitive root modulo $p \Leftrightarrow -a$ is a primitive root modulo $p$?
I tried to:
$\Rightarrow:$ Let $a$ be a primitive root modulo $p$.
Then it's $\mathrm{ord_p(a)}=\varphi(p)$
So $a^{\varphi(p)-1} \neq a^{\varphi(p)}=1$
Since $p \nmid a$, it follow that $a \neq mp$ for $m \in \mathbb{Z}$
$p \equiv 1 \ (\mathrm{mod} \ 4) \Rightarrow 4 \vert (p-1)$
Here I don't know how to continue.
How to show this equivalence?
Best Answer
You only need to show the $\Rightarrow$ implication, since $-(-a)=a$.
What you know is that $a^i\ne 1$ for all $1\le i <p-1$, and want to know the same for $-a$.
But $$(-a)^i=(-1)^ia^i=\begin{cases} a^i & \text{ if $i$ is even} \\ -(a^i) & \text{ if $i$ is odd} \end{cases}$$ so you only need to be sure that $a^i\ne -1$ for odd $i$. But $a^i=-1$ implies that $a^i$ has order $2$, so $i=\frac{p-1}2$, that it is even if $p\equiv 1 \pmod 4$.