Assume $g$ is a primitive root modulo a prime $p$. Show that $p-g$ is a primitive root if and only if $p \equiv 1 \pmod 4$.
I am studying for a number theory exam that is why I am posting a lot of questions related to primitive roots. I would really appreciate your help because they seem like really nice questions.
We solve the first problem. The case $p=2$ is simple, so let $p$ be an odd prime.
If $p\equiv 3\pmod{4}$, then $p-g$ is not a primitive root of $p$. For $p-g\equiv -g\pmod{p}$. But $-1$ is a quadratic non-residue of $p$, and therefore $-g$ is a QR of $p$, so cannot be a primitive root of $p$.
Conversely, let $p\equiv 1\pmod{4}$. We show that $-g$ is a primitive root of $p$.
By Fermat's Theorem, $g\equiv g^p\equiv -(-g)^p\pmod{p}$. But $-1$ is a QR of $p$, so $-1\equiv g^{2k}\equiv (-g)^{2k}\pmod{p}$ for some integer $k$.
t follows that $g\equiv (-g)^{2k}(-g)^p\pmod{p}$. Thus $g$ is congruent to a power of $-g$. Since the powers of $g$ travel, modulo $p$, through $1,2,\dots, p-1$, so do the powers of $-g$, and therefore $-g$ is a primitive root of $p$.
Remark: For other problems that you may post (and this one also) please indicate what you have tried, any progress you may have made, and where difficulties remain.
$$-g=g(-1)\equiv g^{1+(p-1)/2}\equiv g^{(p+1)/2}$$
We know, ord$_ma=d, $ord$_m(a^k)=\dfrac d{(d,k)}$ (Proof @Page$\#95$)
ord$_p g^{(p+1)/2}=\dfrac{p-1}{\left(p-1,\dfrac{p+1}2\right)}$
Now, if integer $d(>0)$ divides both, $d$ must divide $-(p-1)+2\cdot\dfrac{p+1}2=2$
As for odd prime $p,p-1$ is even
$\left(p-1,\dfrac{p+1}2\right)=2$ if $\dfrac{p+1}2$ is even $=2k$(say) $\iff p=4k-1\equiv-1\pmod4$
$\left(p-1,\dfrac{p+1}2\right)=1$ if $\dfrac{p+1}2$ is odd $=2k+1$(say) $\iff p=4k+1\equiv1\pmod4$
