Prove that if $p$ is an odd prime and $x$ is an integer such that $x^2\equiv 1~mod~p^k$, then $x=\pm 1~mod~p^k$

Question

I'm trying to prove the statement given in the title. I'm quite confused. I would really appreciate if someone can verify, or suggest any changes to what I've got so far.

Assume $x^2 \equiv 1~~mod~~p^k$. Then $p^k$ must divide $x^2-1$, which is equal to $(x-1)(x+1)$. We know that $p^k$ must divide either $(x-1)$or$(x+1)$. Therefore, either $x \equiv 1~~mod~~p^k$ or $x \equiv -1~~mod~~p^k$.

Thank you.

Answer 1

One possible proof is based on the fact that $U(p^k)$ is cyclic.

Indeed, in a cyclic group, there are exactly $\phi(d)$ elements of order $d$ for each $d$ dividing the order of the group.

In particular, in a cyclic group of even order there are exactly two elements such that $x^2=e$: one is $e$ and the other is the single element of order $2$.

Applying all this to $U(p^k)$, we conclude that since $\pm1$ are solutions of $x^2=1$, they are the only solutions.

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A simpler solution uses induction on $k$.

• $k=1$ is easy.

• $k \to k+1$:
  Suppose $x^2 \equiv 1 \bmod p^{k+1}$. Then $x^2 \equiv 1 \bmod p^{k}$ and by induction we have $x \equiv \pm 1 \bmod p^{k}$. Write $x=\pm1 + tp^k$. Then $1 \equiv x^2 \equiv (\pm1 + tp^k)^2 \equiv 1 \pm2tp^k \bmod p^{k+1}$ implies $\pm2tp^k \equiv 0 \bmod p^{k+1}$ and this implies that $p$ divides $t$ and so $x \equiv \pm 1 \bmod p^{k+1}$.

Answer 2

If $p$ is a prime and $p>2$, then $p^k\mid(x-1)(x+1)$ implies $p^k\mid x-1$ or $p^k\mid x+1$, because these two numbers ($x+1$ and $x-1$) cannot have $p$ as a common factor (in that case $p$ would divide the difference, which is equal $2$ - this is impossible for $p>2$).