Solving $x^2 \equiv 1 \pmod{p^{\ell}}$

Let $p$ be an odd prime. Then $p^{\ell}$ can divide at most one of $x-1$ and $x+1$. (Note that the argument breaks down if $p=2$, and indeed in that case there can be more than 2 solutions.)

Hint $\,\ p\,$ prime, $\, (p,a,b) = 1,\ p^n\mid ab\ \Rightarrow\ p^n\ |\ a\ $ or $\ p^n\ |\ b\ $ by iterating Euclid's Lemma.

Note that $\ (p,x\!-\!1,x\!+\!1) = (p,x\!-\!1,x\!+\!1\!-\!(x\!-\!1)) = (p,x\!-\!1,2) = 1\ $ for odd $\,p.$

i.e. if $\,p^n$ divides a product of pairwise $\,p\,$-coprime elements $\,a_i\,$ then $\,p^n\,$ divides one of the $\,a_i,\, $ for otherwise, by unique factorization, $\ p\,$ divides at least two factors $\,a_i,\ a_j\,$ contra $\ (p,\,a_i,\,a_j) = 1$.