If $a \in \mathbb{Z}_5$ and $a \equiv \pm1 \text{ }(\text{mod }5)$, does there exist $x \in \mathbb{Z}_5$ where $x^2 = a$? I know we want to use Hensel's Lemma somehow to assess this question, but I'm not sure I see how to do this. Could anybody help?
Edit. OP here with an account. Here is my progress so far. Consider $x \mapsto x^2$ in $(\mathbb{Z}/5^n\mathbb{Z})^\times$. The kernel is $\{\pm1\}$, so half the elements are squares. If $[a]$ is a square $(\mathbb{Z}/5^{n+1}\mathbb{Z})^\times$, it must also be $(\mathbb{Z}/5^n\mathbb{Z})^\times$. Then counting says the only way to get enough squares in $(\mathbb{Z}/5^{n+1}\mathbb{Z})^\times$ is that is if-and-only-if. So we can reduce down to the case $(\mathbb{Z}/5\mathbb{Z})^\times$. I'm stuck here though, could anybody help me finish?