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

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.

• If it weren't true then $p$ would divide both $x + 1$ and $x - 1$ and hence also their difference, so $p$ divides $2$. – Tob Ernack Feb 5 '18 at 4:25
• Is it not possible for $p$ to not be able to divide either $x+1$ and $x-1$? – Zed1 Feb 5 '18 at 4:27
• 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? – Zed1 Feb 5 '18 at 4:39
• 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 $p$. – Tob Ernack Feb 5 '18 at 5:33
• Would about for $p^a$ with $a>1$? – Zed1 Feb 5 '18 at 18:47