# Congruence Problem

• Jun 18th 2011, 06:15 PM
VinceW
Congruence Problem
This is from chapter on linear congruences:

Show that if p is an odd prime and a is a positive integer not divisible by p, then the congruence $x^2 \equiv a \pmod {p}$ has either no solution or exactly two incongruent solutions.

I can see If $p \mid x$, then there are no solutions. I'm really not sure how to approach the rest of this...
• Jun 18th 2011, 09:23 PM
chisigma
Re: Congruence Problem
$x^{2} \equiv a (\text{mod}\ p)$ (1)
$a + p\ y = x^{2}$ (2)
Now if $x_{0}$ satisfies (2) for some y, then $x_{1} = p-x_{0} \ne x_{0}$ satisfies also (2) for some other y...
$\chi$ $\sigma$