If $a \in \mathbb{Z}_5$ and $a \equiv \pm1 \text{ }(\text{mod }5)$, does there exist $x \in \mathbb{Z}_5$ where $x^2 = a$? 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?
 A: It should be clear that it’s enough to prove that if $a\equiv1\pmod5$, then there is a square root of $a$ in $\Bbb Z_5$. You can use the Binomial Series for $(1+x)^{1/2}$, in which the only denominators are powers of $2$, and hence $5$-adic integers. This means that when you substitute $a-1$ for $x$, you get a $5$-adically convergent series, since $5|(a-1)$.
Alternatively, you may want to use a direct appeal to Strong Hensel. He says that if $f\in\Bbb Z_p[x]$ and the corresponding polynomial factors over $\Bbb F_p$ into relatively prime factors, then these lift to characteristic zero. More precisely, calling $\tilde f$ the reduced polynomial in $\Bbb F_p[x]$, if $\tilde f(x)=\gamma(x)\eta(x)$, product of two $\Bbb F_p$ polynomials, and if $\gcd(\gamma,\eta)=1$, then there are polynomials $g(x),h(x)\in\Bbb Z_p[x]$ with $\tilde g=\gamma$, $\tilde h=\eta$, $f=gh$, and $\deg(g)=\deg(\gamma)$.
(By the way, for my money, the Wikipedia article on Hensel’s Lemma is worthless, ’cause it doesn’t say anything about the Strong version.)
