We need to show that if $x$ is an integer satisfying $x^2 \equiv 1 \pmod{p^n}$, then it must be congruent to $\pm 1 \pmod{p^n}$. That is, we need to prove:

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

Step 1: Rewrite the Given Congruence

We can rewrite the given congruence as:

$x^2 - 1 \equiv 0 \pmod{p^n}$

which can be factored as:

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

This means that $p^n$ divides the product $(x-1)(x+1)$.

Step 2: Consideration of Prime Factors

Since $p^n$ is a power of the prime $p$, we analyze the divisibility conditions. Observe that $x - 1$ and $x + 1$ are two consecutive integers, meaning that they are coprime unless one of them is divisible by $p$.

Since they are coprime, all of the prime power $p^n$ must be entirely contained in one of these factors. This leads to two cases:

1. $x - 1 \equiv 0 \pmod{p^n}$, which implies $x \equiv 1 \pmod{p^n}$.
2. $x + 1 \equiv 0 \pmod{p^n}$, which implies $x \equiv -1 \pmod{p^n}$.

Step 3: Conclusion

Since these are the only two possibilities, we conclude:

$x \equiv \pm 1 \pmod{p^n}.$

Thus, we have proven that the only solutions to $x^2 \equiv 1 \pmod{p^n}$ are $x \equiv \pm 1 \pmod{p^n}$, completing the proof.

---

Do you want more details or have any questions?

Here are five related questions to expand on this topic:

1. How does this result generalize to other moduli that are not prime powers?
2. What happens if we replace $p^n$ with an arbitrary composite number?
3. Can we extend this result to rings other than $\mathbb{Z}$?
4. How does this relate to the structure of the multiplicative group modulo $p^n$?
5. What are the solutions to $x^2 \equiv 1$ modulo a prime $p$ alone?

Tip: The argument here relies on the fact that two consecutive integers are coprime. This property is crucial in many number theory proofs!