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I have the following problem:

Let $p$ be prime, $\alpha \in \mathbb{N}$ and $x \in \mathbb{Z}_{p^\alpha}$. What is the number of solutions to the equation $x^2 \equiv 1 \pmod{p^\alpha}$?

What I've got so far is $p^\alpha|(x - 1)(x + 1)$ but I don't know what to do next.

Bartłomiej Sługocki
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    For $p=2$ see here, otherwise see this duplicate, where for $m=p^n$ primitive roots exist. – Dietrich Burde Nov 08 '16 at 20:17
  • For $p$ odd, observe that you can have either $p^{\alpha} \mid (x-1)$ or $p^{\alpha} \mid (x+1),$ but nothing else (that is, it cannot happen that $p \mid (x-1)$ and $p \mid (x+1)$ at the same time;, thus, something like $p^{\alpha-1} \mid (x-1), p \mid (x+1)$ is not posible (assuming $\alpha >1$)). – Pavel Čoupek Nov 08 '16 at 20:19

2 Answers2

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We know that when $p$ is odd $\Bbb Z/{p^\alpha}^*$ is isomorphic to $\Bbb Z/p-1\times\Bbb Z/p^{\alpha-1}$ which means we have a cyclic group, so there are $2$ solutions. (Hensel's lemma also verifies exactly $2$ solutions in this case).

For $p=2$, when $\alpha =1, 2$ there is clearly one solution, otherwise we have $\Bbb Z/{2^\alpha}^*\cong \Bbb Z/2^{\alpha-2}\times\Bbb Z/2$ which has $4$ elements of order $2$. (Again, Hensel's lemma verifies this without any mess)

Adam Hughes
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HINT: For an odd prime $p^a|(x-1)(x+1)$ implies $p|x-1$ or $p|x+1$. In the special case of $p=2$, at most one of the factors can be divisible by $2^2$, so that factor must be divisible by $2^{a-1}$, and then the other will automatically be divisible by $2$.

sbares
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