Let $p$ be prime. Show that $a$ modulo $p$ has its own inverse if and only if $a ≡ ±1 \pmod p$.
($\implies$)
If $a$ has its own inverse then
$$ \begin{align} a \times a & \equiv 1 \pmod p \\ \\ a^2 -1 & \equiv 0 \pmod p \\ \\ (a+1) \times (a-1) & \equiv 0 \pmod p \end{align} $$
Proposition (3.14)(a) tells us that either $a -1 \equiv 0 \pmod p$ or $a+1 \equiv 0 \pmod p$. That is $a \equiv \pm 1 \pmod p$.
($\impliedby$)
We start with $a \equiv \pm 1 \pmod p$. This means either $a -1 \equiv 0 \pmod p$ or $a+1 \equiv 0 \pmod p$. This gives us
$$ \begin{align} (a-1) \times (a+1) \equiv 0 \pmod p \\ \\ a^2 - 1 \equiv 0 \pmod p \\ \\ a \times a \equiv 1 \pmod p \end{align} $$
That is, $a$ has its own inverse.
By showing the implication in both directions, we have shown that for prime $p$, $a$ modulo $p$ has its own inverse if and only if $a ≡ ±1 \pmod p$.