How to prove $r^{\phi(m)/2} \equiv -1\pmod m$ if $r^{\phi(m)} \equiv 1\pmod m$?

Question

If we have a primitive root $r$ for $m > 2$ (whatever composite or prime), that $r^{\phi(m)} \equiv 1$ (mod $m$), how can we show that \begin{align} r^{\phi(m)/2} \equiv -1\text{ }(\text{mod } m)? \end{align}

Thanks a lot.

Answer 1

If $r$ is a primitive root modulo $m$ then $1,r,r^2,\dots, r^{\phi(m)-1}$ are distinct modulo $m$.

There are a couple of ways to proceed from this observation.

For example: This means that $\mathbb Z_m^{\times}$ is cyclic. In particular, that means there is only two square roots of $1$ modulo $m$, $1$ and $-1$.

Or: This means that $r^k\equiv -1\pmod m$ for some $k$. But then $r^{\phi(m)-k}=-1$ too, which is impossible unless $k=0$ or $k=\phi(m)/2$. But $k=0$ means $1\equiv -1\pmod m$, so $m\mid 2$.