Proving $\frac{p-1}{2}$ is a primitive root modulo $p$ if and only if $2(-1)^{(p-1)/2}$ is a primitive root modulo $p$

Question

Let $p$ be an odd prime. Prove that $\frac{p-1}{2}$ is a primitive root modulo $p$ if and only if $2(-1)^{(p-1)/2}$ is a primitive root modulo $p$.

I was thinking that since $\frac{p-1}{2}$ is a primitive root modulo $p$ that its order is $p-1$.

Answer 1

If $g$ is a primitive root modulo $p$ then it's order is $p-1$ :

$$\left(\frac{p-1}{2}\right)^n\equiv 1\mod p\ \ \ \Leftrightarrow (-1)^n \equiv 2^n\mod p\ \Leftrightarrow 1 \equiv (-2)^n\mod p$$

hence $\frac{p-1}{2}$ is a primitive root if and only if $-2$ is a primitive root modulo $p$.

• If $\frac{p-1}{2}$ is odd this is what we need.

• If $\frac{p-1}{2}$ is even prove that $-2$ is a primitive root implies that $2$ is a primitive root, let $t$ be the order of $-2$ then $(-2)^t=1$ hence $2^{2t}=1$ but the order of $2$ is $p-1$ so $p-1$ divides $2t$ hence $2t\geq p-1$ and because the order of any element is less then $p-1$ we have: $$ \frac{p-1}{2} \leq t \leq p-1$$ but if $t=\frac{p-1}{2}$ then $(-2)^t=2^t=1$ which is absurd because the order of $2$ is $p-1$ hence $t=p-1$