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

I am working through Ireland/Rosen at the moment, and I cannot solve a (probably) simple exercise. Any nudges you can give me in the right direction are appreciated.

How does one show that $x^2 \equiv 1 \pmod{p^{\ell}}$ has only two solutions (namely $\pm 1$)? Here, $\ell$ is a positive integer (which we can take to be at least $2$ as $\mathbb{F}_p$ is a field and must therefore only have two solutions).

I am aware of how to prove this statement in general using Hensel's Lemma, but there must be an elementary (maybe two line) proof as it is in the first few chapters of Ireland/Rosen. The book also discusses Euler's theorem around this point:

$$ x^{\varphi(n)} \equiv 1 \pmod{n}, $$

where $\varphi(n)$ is the totient function, and I suspect this plays a role.

How does one show that $x^2 = 1 \pmod{p^{\ell}}$ has only two solutions, without invoking Hensel's lemma? (Please provide hints only!!)

Solution 1:

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.)

Solution 2:

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 $\ (p,\,a_i,\,a_j) = 1$.