Show -1 and +1 are only solutions to x^2 = 1 (mod p^2) for odd prime p.

[Vespero]

Homework Statement

Let p be an odd prime and n be an integer. Show that -1 and +1 and the only solutions to x^2 \equiv 1\ mod\ p^n. Hint: What does a \equiv b\ mod\ m mean, then think a bit.

Homework Equations

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

The Attempt at a Solution

I've tried using the fact that x^2 \equiv 1\ mod\ p^n implies that x is its own inverse and thus that x \equiv x\ mod\ p^n and that x^2 \equiv 1\ mod\ p^n implies x^2 \equiv 1\ mod\ p^{n-1}, 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 x^2 = p^nk + 1 for p > 2), 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.

 

[micromass]

If x^2=1 (mod p^n), then p^n divides x^2-1.

Rewrite x^2-1.

 

[Vespero]

So, x^2 \equiv 1\ mod\ p^n implies p^n|x^2 - 1.
Factoring x^2 - 1, we have that p^n|(x+1)(x-1).
Since (x + 1) - (x - 1) = 2, p^n|(x+1) and p^n|(x-1) only if p|2, which would imply that p=2. However, since we are dealing with odd primes, Euclid's Lemma requires that either p^n|(x+1) or p^n|(x-1). For x\ mod\ p^n, the only solution to the first relation is x\ =\ p^n - 1 \equiv -1\ (mod p^n), and the only solution to the second relation is x\ =\ 1 \equiv 1\ (mod p^n). Thus, x \equiv \pm 1\ (mod\ p^n) are the only solutions to x^2 \equiv 1\ (mod\ p^n) for p < 2.

Is this correct?

 

[micromass]

Seems ok! Congratulations!