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Let $ a $ be a positive integer and $ p $ be an odd prime. Show that if $ x^2 \equiv 1 \mod{p^a}, $ then $ x \equiv \pm 1 \mod{p^a}. $

From $ x^2 \equiv 1 \mod{p^a} $ we have that $ p^a \vert (x^2 - 1) = (x + 1)(x - 1). $ It's easy to see that the proof is true when $ a = 1 $ from Euclid's lemma.

How can I show this is true for $ a = 2. $ I am trying to build towards an induction proof.

Hints only, please.

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    $\begingroup$ If it weren't true then $p$ would divide both $x + 1$ and $x - 1$ and hence also their difference, so $p$ divides $2$. $\endgroup$ – Tob Ernack Feb 5 '18 at 4:25
  • $\begingroup$ Is it not possible for $p$ to not be able to divide either $x+1$ and $x-1$? $\endgroup$ – Zed1 Feb 5 '18 at 4:27
  • $\begingroup$ I don't think I am explaining my question well enough. I'm asking about three cases; $p$ divides both, $p$ divides $1$ factor, or $p$ divides neither factor. The first case is not possible, that I understand. What about the third case, why is that not possible? $\endgroup$ – Zed1 Feb 5 '18 at 4:39
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    $\begingroup$ Alright, so this is ensured by Euclid's lemma: if $p$ is prime and $p | ab$, then either $p | a$ or $p | b$. It is not possible that neither $a$ nor $b$ are divisible by $p$. $\endgroup$ – Tob Ernack Feb 5 '18 at 5:33
  • $\begingroup$ Would about for $p^a$ with $a>1$? $\endgroup$ – Zed1 Feb 5 '18 at 18:47

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