Hi. I'm having trouble with these problems.

p is and odd prime for both.

Prove x = 0 mod p if x = -x mod p.

Prove x = +/-y mod p^2 if x^2 = y^2 mod p^2, neither x nor y are 0 mod p.

Thanks...

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- Jun 23rd 2008, 07:23 PMcryptocrowSquare Root Proof Problem
Hi. I'm having trouble with these problems.

p is and odd prime for both.

Prove x = 0 mod p if x = -x mod p.

Prove x = +/-y mod p^2 if x^2 = y^2 mod p^2, neither x nor y are 0 mod p.

Thanks... - Jun 23rd 2008, 07:56 PMThePerfectHacker
Hint: $\displaystyle x^2\equiv y^2 ~ (p^2) \implies (x-y)(x+y) \equiv 0 ~ (p^2)$

- Jun 23rd 2008, 10:41 PMCaptainBlack
Suppose $\displaystyle x>0$, and $\displaystyle x \not\equiv 0 \mod p$, then there exist $\displaystyle k \ge 0$ and $\displaystyle p>r>0$ such that:

$\displaystyle x=kp+r$

Also:

$\displaystyle -x=(-k)p-r=(-1-k)p+(p-r)$

So if $\displaystyle x \equiv -x \mod p$ then $\displaystyle p-r=r$, or $\displaystyle 2r=p$, but $\displaystyle p$ is an odd prime which is a contradiction, so our premis fails and $\displaystyle x \equiv 0 \mod p$.

RonL - Jun 24th 2008, 06:10 AMcryptocrow
Thanks a lot, i see where i went wrong.