Prove Proposition 3.14 (b).
Proposition 3.14(b) states that if $p$ is prime, then $a^2 ≡ b^2 \pmod p \iff a ≡ ±b \pmod p$.
($\implies$)
We first note that $a^2 ≡ b^2 \pmod p$ means, for some integer $k$
$$ \begin{align} kp & = a^2 - b^2 \\ \\ & = (a+b)(a-b) \end{align}$$
Applying Proposition 2.2 to $p \mid (a+b) \times (a-b)$ means either $p \mid (a+b)$ or $p \mid (a-b)$. Let's consider these two cases in turn.
$p \mid (a+b)$ means $a + b \equiv 0 \pmod p$ which is equivalent to $a \equiv -b \pmod p$.
$p \mid (a-b)$ measn $a - b \equiv 0 \pmod p$ which is equivalent to $a \equiv b \pmod p$.
So we have, for prime $p$, $a^2 ≡ b^2 \pmod p \implies a ≡ ±b \pmod p$.
($\impliedby$)
This time we start with $a \equiv \pm b \pmod p$. This gives us two cases to consider, $a \equiv b \pmod p$ and $a \equiv -b \pmod p$. Let's consider these in turn.
$a \equiv b \pmod p$ means $p \mid (a -b)$
$a \equiv -b \pmod p$ means $p \mid (a +b)$
Since at least one of these cases is true, we have $p \mid (a-b) \times (a+b)$, which gives us $p \mid a^2 - b^2$. That is, $a^2 \equiv b^2 \pmod p$.
By showing both implications, we have proved Proposition 3.14(b).