Let $p$ be a prime. Prove that
$a^p \equiv b^p \pmod {p} \implies a \equiv b \pmod {p}$
If $p$ is prime, then by Corollary 4.2 we have
$$ a^p \equiv a \pmod p $$
$$ b^p \equiv b \pmod p $$
We start by assuming the antecedent of the implication is true
$$ a^p \equiv b^p \pmod p $$
Using $ a^p \equiv a \pmod p $ we can substitute for $a^p$
$$ a \equiv b^p \pmod p $$
Using $ b^p \equiv b \pmod p $ we can substitute for $b^p$
$$ a \equiv b \pmod p $$
This conclusion tells us
$$ a^p \equiv b^p \pmod {p} \implies a \equiv b \pmod {p} $$
Note: the official solution says that for the algebra of real numbers $a^p = b^p$ does not imply $a=b$ in general. The textbook suggests the implication is only true for $a=b=1$ and $a=b=0$. This is not right, as we can see with an example $a=b=2, p=2$. Another example is $a=-2, b=-2, p=2$.