Number Theory: $m^2\equiv 1\pmod{p^n}$

Question

I have this problem assigned for homework, and I think I've completed the majority of the proof, but I'm stuck at one part.

Problem

If $p>2$ is a prime, $n\in \mathbb{N}$, and $m\in \mathbb{Z}$, show that $m^2 \equiv 1\pmod{p^n} \implies$ either $m\equiv 1\pmod{p^n}$ or $m\equiv -1\pmod{p^n}$.

Proof

Let $m^2\equiv 1\pmod{p^n}$.

Then, $m^2 = qp^n + 1$, some $q\in \mathbb{Z}$

$\implies m^2 - 1=qp^n$

$\implies (m+1)(m-1)=qp^n$.

$\implies m+1=\frac{qp^n}{m-1}$.

Note that $\frac{q}{m-1}=x$, some $x\in\mathbb{Z}$. (This is the step I'm unable to prove.)

Thus, we have $m+1=xp^n\implies m=xp^n-1\implies m\equiv -1\pmod{p^n}$.

It also follows from $(m+1)(m-1)=qp^n$ that $m-1=\frac{qp^n}{m+1}$.

Note that $\frac{q}{m+1}=y$, some $y\in\mathbb{Z}$. (Again, I'm unable to prove this.)

Thus, we have $m-1=yp^n\implies m=yp^n+1\implies m\equiv 1\pmod{p^n}$.

Therefore, $m^2 \equiv 1\pmod{p^n} \implies$ either $m\equiv 1\pmod{p^n}$ or $m\equiv -1\pmod{p^n}$. $\blacksquare$

Thanks in advance!

Answer 1

What you need to show is either $\frac{q}{m-1}=x$ or $\frac{q}{m+1}=y$ must be true.

This can be shown by using the fact $p$ is prime so $p^n$ must be coprime with either $m+1$ or $m-1$. ($m=2$ will be a special case and can be easily shown as $p$ must be $3$ for the premise to be true)

If $p^n$ is coprime with $m-1$, then $\frac{q}{m-1}=x$ must be true because $\frac{qp^n}{m-1}=m+1$ is an integer. Similar for $m+1$ case.

Answer 2

HINT:

If $p^n|(m-1)(m+1)$

let $p^a|(m-1), p^{n-a}|(m+1)$ where $0\le a\le n$

$\implies2=m+1-(m-1)$ will be divisible by $p^{\text{min}(a,n-a)}$

$\implies$min$(a,n-a)=0$