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?