<s>[INST] Is $x^2 \equiv 1 \pmod{p^k} \iff x \equiv \pm 1 \pmod {p^k}$ for odd prime $p$? Problem: As the title says, is $x^2 \equiv 1 \pmod {p^k} \iff x \equiv \pm 1 \pmod {p^k}$ for odd prime $p$? If not, is it at least true that there are two solutions to the congruence equation?
Attempt: I am a beginner in number theory, and I am not sure if the statement in question is true. The following is my attempt. Assume $x^2 \equiv 1 \pmod {p^k}$. Then $p^k | (x-1)(x+1) \Rightarrow p^k|(x-1)$ or $p^k|(x+1)$, because if $p^k$ can divide both, then $p^k|2$ which is false. This shows the forward direction.
The converse is easy.
For any $y^2 \equiv 1 \pmod {p^k}$, $y^2 \equiv x^2 \pmod {p^k}$ which yields the conclusion before, so $\pm 1$ are the only solutions $\pmod {p^k}$.
 [/INST] Assume $x^2 \equiv 1 \pmod {p^k}$. Then $x^2-1 \equiv 0 \pmod {p^k}$, meaning $p^k | x^2-1=(x-1)(x+1)$. Since $p$ is prime, we can conclude that  $p^k\vert(x-1)$ or $p^k\vert(x+1)$. Yet, $p^k\vert(x-1) \Rightarrow x-1 \equiv 0 \pmod {p^k} \Rightarrow x \equiv +1 \pmod {p^k}$ and $p^k\vert(x+1) \Rightarrow x+1 \equiv 0 \pmod {p^k} \Rightarrow x \equiv -1 \pmod {p^k}$. 
Thus, there are two solutions: If $x^2 \equiv 1 \pmod {p^k}$, then, $x \equiv -1 \pmod {p^k}$ and $x \equiv +1 \pmod {p^k}$.
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