Show that $x^2 \equiv 1 \pmod {n} \;\not\!\!\!\implies x \equiv \pm 1 \pmod {n}$.
We show this with a counter-example.
Consider $x \equiv 3 \pmod 8$.
$$ 3^2 \equiv 9 \equiv 1 \pmod 8 $$
Here $x^2 \equiv 1 \pmod n$, but $x \not \equiv \pm 1 \pmod n$.
Note that $x^2 \equiv 1 \pmod {n} \implies x \equiv \pm 1 \pmod {n}$ only if $n$ is prime, as we have shown in a previous exercise.