Show that the only solutions to the congruence $x^2 \equiv 1\pmod {p^n}$ are $x \equiv \pm 1 \pmod {p^n}$

Question

Let $p$ be an odd prime, and $n \in \mathbb{N}$.

Here's what I have so far.

$$x^2 \equiv 1\pmod {p^n}$$

implies

$$p^n|x^2-1$$

$$x^2-1 = (x+1)(x-1)$$

so we have $$(x+1)(x-1) \equiv 0\pmod {p^n}$$

I don't know where to go from here. I know that $\pm 1$ are solutions (obviously) but I don't know how to show that they are the only solutions.

Answer 1

The inequality $(x+1)(x−1)\equiv0(\text{mod }p^n)$shows that the only possible solutions $x$ are such that $x+1\equiv0(\text{mod }p^n)$ or $x-1\equiv0(\text{mod }p^n)$.

So, $x\equiv1(\text{mod }p^n)$ and $x\equiv-1(\text{mod }p^n)$ are the only solutions.

Answer 2

Hint: as $p$ is odd, if $p$ divides $x + 1$, then $p $ does not divide $x - 1$ (and vice versa). We are given that $p^n$ divides $x^2 -1 = (x + 1)(x - 1)$ and so $p$ divides one of $x + 1 $ or $x - 1$. But then $p^n$ must divide whichever of $x + 1$ and $x - 1$ that $p$ divides, i.e., $x \equiv \pm 1 (\mbox{mod}\;p^n)$

Answer 3

The idea is to determine the cardinal of $S=\{x\in (\mathbb{Z}/p^n\mathbb{Z})^*\ \ / \ x^2 =1\}$.

Then, because of the cyclicity of $(\mathbb{Z}/p^n\mathbb{Z})^*$, you have #$S=\gcd(2,\varphi(p^n))=2$ since $p$ is odd.

So there are two solutions and you have identified $\pm1$.

Answer 4

$x^2\equiv 1\pmod{p^n}\Rightarrow x^2\equiv 1\pmod p\iff x^2-1=0$ in the finite field $\mathbb F_p$.

Since $x^2-1=0$ is an equation of second degree with coefficients in a field it can have no more than two roots so $x$ can be only the evident roots of $x^2-1=0$ which are $1$ and $-1$.