If $p$ prime $>3$ and $a$ primitive root modulo $p$, so $-a$ is that too iff $p\equiv 1\pmod 4$ Let $p>3$ be prime, let $a$ be a primitive root modulo $p$. Show that $-a$ is a primitive root modulo $p$ if and only if $p\equiv 1\pmod 4$.
We have $a^{\phi(p)/2}\equiv -1\pmod p$, so $-a=(-1)a\equiv a^{\phi(p)/2+1}\pmod p$. Then
$$
\operatorname{ord}(-a,p)=\frac{\operatorname{ord}(a,p)}{\gcd\{\phi(p)/2+1,\operatorname{ord}(a,p)\}}=\frac{p-1}{\gcd\{(p+1)/2,p-1\}}.
$$
Want find those $p$ so that $\operatorname{ord}(-a,p)=\phi(p)$. Therefore, we only want to show that
$$
\gcd\{(p+1)/2,p-1\}=1\iff p\equiv 1 \pmod 4.
$$
The direction $\impliedby$ is straightforward. I'm struggling with the other direction.
 A: If $p=4t+3$, then $\dfrac{p+1}{2}=2t+2$ and $p-1=4t+2$, and so $\gcd(\dfrac{p+1}{2},p-1) \ge 2$.
If $p=4t+1$, then $\dfrac{p+1}{2}=2t+1$ and $p-1=4t$, and so $\gcd(\dfrac{p+1}{2},p-1)$ divides $2(2t+1)-4t=2$. But it cannot be $2$ because $2t+1$ is odd. Hence, $\gcd(\dfrac{p+1}{2},p-1) = 1$.
A: Let $\langle g\rangle=\Bbb Z_p^\times$. We have $p\equiv_4 1\iff -1\equiv_p g^{2k}$, where $4k=p-1$. Now consider $\langle -g\rangle$.

*

*If $p\equiv_4 1$ then $\langle -g\rangle\cong \langle g^{2k+1}\rangle$, this group is isomorphic to $\Bbb Z_p^\times$. To see this, consider the homomorphism $$\phi:\Bbb Z_p^\times\rightarrow\Bbb Z_p^\times\\g\mapsto g^{2k+1}$$ Then $\phi(g^m)=g^{m(2k+1)}=e$. Using some algebra: $$g^{2mk}g^m=e\\\Rightarrow g^{2m}=e\tag{squaring both sides}$$ This gives that $g^{m(2k+1)}=g^m=e$, so the kernel is trivial and $\phi$ is an isomorphism.


*If $p\equiv_4 3$ then $\langle -g\rangle\cong\langle g^{2k}\rangle$. This group generates nothing but the squares in $\Bbb Z_p^\times$, so $\langle -g\rangle\subset\Bbb Z_p^\times$.
