Let $p$ be an odd prime and $x$ be a least positive residue modulo $p$.
Show that $x^{p+1} ≡ 4 \pmod p \implies x ≡ 2$ or $x ≡ −2 \pmod p$.
Since $p$ is prime, Corollary 4.2 gives us
$$ x^p \equiv x \pmod p $$
Multiplying both sides by $x$ gives
$$ x^{p+1} \equiv x^2 \pmod p $$
Comparing with $x^{p+1} ≡ 4 \pmod p$ we have
$$ x^2 \equiv 4 \pmod p $$
This has solutions $x \equiv 2 \pmod p$ and $x \equiv -2 \pmod p$.
Note that the solutions $x \equiv -2 \pmod p$ are not the least positive because for $p=2$ there are no solutions, and for $p=5$ and larger the solutions are larger than $x \equiv 2 \pmod p$.