Show that if $d$ is even then $(p− 1) \pmod p$ is a solution to
$$ x^d − 1 \equiv 0 \pmod p $$
where $p$ is prime.
Let's remind ourselves of Proposition (6.19).
Let $p$ be prime and $d \mid (p-1)$. The congruence $x^d \equiv 1 \pmod p$ has exactly $d$ incongruent solutions.
Since $d$ is even, we can write it as $d=2k$ for some integer $k$. And so
$$ x^{2k} \equiv 1 \pmod p $$
We have $2k \mid p-1$ because $p-1$ is even, and so by Proposition (6.19) the congruence has $2k$ incongruent solutions.
Using $p-1 \equiv -1 \pmod p$, we have
$$ (p-1)^2 \equiv 1 \pmod p $$
Furthermore
$$ \begin{align} ((p-1)^2)^k & \equiv 1^k \pmod p \\ \\ (p-1)^{2k} & \equiv 1 \pmod p \end{align}$$
That is, $(p-1)$ is a solution to $x^d -1 \equiv 0 \pmod p$.