For prime $p$, I understand that each $x \in \mathbb{Z}^{*}_{p}$ has exactly 2 square roots (as there are at least 2, and the quadratic residues are exactly half of the group (the group is cyclic so we take even exponents of a generator g) so there can't be more than 2 by counting (otherwise there is an element that is a root of 2 different elements). I also understand what happens at $x \in \mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes as we can use chinese remainder theorem and compute roots separately and get 4 roots).
I however don't understand what happens at $x \in \mathbb{Z}^{*}_{p^2}$ as I am unsure if its cyclic so I think my previous argument doesn't work.
thanks.