Square Root Proof Problem

**cryptocrow** · June 23rd 2008, 07:23 PM

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...

**ThePerfectHacker** · June 23rd 2008, 07:56 PM

Hint: $x^2\equiv y^2 ~ (p^2) \implies (x-y)(x+y) \equiv 0 ~ (p^2)$

**CaptainBlack** · June 23rd 2008, 10:41 PM

Originally Posted by cryptocrow

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...

Suppose $x>0$ , and $x \not\equiv 0 \mod p$ , then there exist $k \ge 0$ and $p>r>0$ such that:

$x=kp+r$

Also:

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

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

RonL

**cryptocrow** · June 24th 2008, 06:10 AM

Thanks a lot, i see where i went wrong.

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