- #1

- 28

- 0

## Homework Statement

Let p be an odd prime and n be an integer. Show that -1 and +1 and the only solutions to [itex]x^2 \equiv 1\ mod\ p^n.[/itex]

*Hint: What does [itex]a \equiv b\ mod\ m[/itex] mean, then think a bit.*

## Homework Equations

[itex]x^2 \equiv 1\ mod\ p^n \rightarrow x^2 = 1 + p^nk[/itex] for k an integer.

## The Attempt at a Solution

I've tried using the fact that [itex]x^2 \equiv 1\ mod\ p^n[/itex] implies that x is its own inverse and thus that [itex]x \equiv x\ mod\ p^n[/itex] and that [itex]x^2 \equiv 1\ mod\ p^n[/itex] implies [itex]x^2 \equiv 1\ mod\ p^{n-1}[/itex], attempting to find a way to combine the equations to force x to be either 1 or -1, but neither of these got me anywhere. The hint the teacher left makes me think the answer is something fairly simple (perhaps perfect squares cannot have the form [itex]x^2 = p^nk + 1[/itex] for [itex]p > 2[/itex]), but I just can't seem to see it. I have a feeling that just the tiniest push in the right direction would lead me to the answer.