Prove that if $a$ has order $2k$ modulo prime $p$ where $p$ is odd then
$$ a^k \equiv -1 \pmod p $$
We're given $a$ has order $2k$ modulo odd prime $p$
$$ a^{2k} \equiv 1 \pmod p $$
That is
$$ (a^{k})^2 \equiv 1 \pmod p $$
By Lemma 4.3, this means $a^k \equiv -1 \pmod p$ or $a^k \equiv 1 \pmod p$.
Since $2k$ is the order of $a$, it is the least integer $j$ such that $2^j \equiv 1 \pmod p$, and so $a^k \not \equiv 1 \pmod p$, leaving only $a^k \equiv -1 \pmod p$.