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Let $p$ be an odd prime and let $g$ be a primitive root $pmod{p}$.
Show that $−g$ is also a primitive root of $p$ if $p equiv 1 pmod{4}$, but that $ord_p(−g) = frac{p−1}{2}$ if $p equiv 3 pmod{4}$.

So far, I have shown as $p equiv1 pmod{4}$ I can use Fermat's Little Theorem.
$$g equiv g^{p} equiv -(-g)^{p} pmod{p}$$

Since $p equiv 1 pmod{4}$, $x^2 equiv -1 pmod{p}$. ($-1$ is a QR of $p$)
There $exists k in mathbb{Z}$ such that

$$-1 equiv g^{2k} equiv (-g)^{2k} pmod{p}$$

Thus, $g equiv (-g)^{2k}(-g)^{p} pmod{p}$.
As $g$ is congruent to $-g^{p}$, $-g$ is a primitive root of $p$.

Is this enough to show the first part of the question, also how do I begin to show the 2nd part?

First of all, consider this fact:

If $a$ is of order $h$ $pmod n$, then $a^k$ is of order $frac{h}{gcd(h,k)} quad quad quadquad quad (1)$

The Proof:

Since $g$ is a primitive root, $-1 equiv g^{frac{p-1}{2}} pmod p$. Therefore, $-g equiv (-1)(g) equiv g^{frac{p-1}{2}}g equiv g^{frac{p+1}{2}} pmod p$. Now, the order of $g^{frac{p+1}{2}} pmod p$ according to $(1)$ is $frac{p-1}{gcd(frac{p+1}{2},p-1)}$. If $pequiv 1 pmod 4$, then $frac{p+1}{2}$ is odd and ${gcd(frac{p+1}{2},p-1)}$ is 1 making the order of $-g$ to be $p-1$. i.e. a primitive root. Otherwise, the term $frac{p+1}{2}$ is even and ${gcd(frac{p+1}{2},p-1)} = frac{p-1}{2} > 1$. Therefore, the order of $-g$ is not $p-1$. i.e. not a primitive root.

I am quoting my answer on my question here with some minor changes.