Number theory - euler's phi function

[kimberu]

Homework Statement

Let p = a prime. Show {x}^{2} ≡ a (mod {p}^{2}[/tex]) has 0 solutions if {x}^{2} ≡ a (mod p) has 0 solutions, or 2 solutions if {x}^{2} ≡ a (mod p) has 2.

The Attempt at a Solution

OK, my mistake, I don't think this has anything to do with the phi function. But I don't know what to use to solve this - I thought Euclid's criterion would be somehow useful, but I don't know how.

 

[jbunniii]

If p does not divide x^2 - a, can p^2 do so?

 

[kimberu]

If p does not divide x^2 - a, can p^2 do so?

I'd say no, it can't, because p*p can't divide something that doesn't have at least p as a factor. But what about if p does divide x^2 - a?

 

[jbunniii]

I'd say no, it can't, because p*p can't divide something that doesn't have at least p as a factor.

Right, so that proves the first part.

But what about if p does divide x^2 - a?

OK, suppose that

x^2 \equiv a \mod p

has two distinct solutions, x_1 and x_2. Therefore

x^2 - a \equiv (x - x_1)(x - x_2) \mod p

I claim that this implies

x^2 - a \equiv (x - x_1)(x - x_2) \mod p^2

Can you justify this claim?