If $a$ has exponent $h$ mod $p$, and $h$ is even, then $a^{h/2} \equiv -1 \bmod p$

Question

Prove that if $a$ belongs to the exponent $h$ modulo a prime $p$, and if $h$ is even, then $a^{h/2}$ is congruent to $-1 \pmod p$.

Answer 1

Let $h=2r$

Now $p|(a^{2r}-1)=(a^r+1)(a^r-1)$

$\implies$ either $p|(a^r-1)\implies$ord$_pa|r$ which is impossible as ord$_pa=2r$

$\implies p|(a^r+1)$

Answer 2

This is the same answer as lab bhattachararjee, but I'm phrasing it in a slightly different way.

By definition, $a^h \equiv 1 \pmod p$. Thus $a^{h/2}$ is a solution to the congruence $x^2 \equiv 1 \pmod p$. There are only two solutions to $x^2 \equiv 1 \pmod p$, namely $x \equiv \pm 1$. In this case, we know that $a^{h/2} \not \equiv 1 \pmod p$, since $a$ has exponent $h$. Thus $a^{h/2} \equiv -1$.