# Number Theory (Legendre symbols, quadratic residues/nonresidues)

1. Nov 4, 2007

### Ignea_unda

1. The problem statement, all variables and given/known data
"Tell whether the statement is true and give a counterexample if it is false"
"Let m > 0 and (m, ab) = 1. If neither x^2 congruent to a (mod m) nor y^2 congruent to b (mod m) is solvable, then z^2 congruent to ab (mod m) is solveable.

2. Relevant equations
Legendre symbols

3. The attempt at a solution
I know this is true, using Legendre symbols. But I want to be able to prove it. I started multiplying the congruences and came up with:
x^2y^2 congruent ab (mod m)
But I'm not sure if I'm even heading in the right direction.

Okay, so maybe I didn't show my work correctly enough.

My thoughts that this is true is that:

(a / p) (b / p) = (ab / p) Being the fact that both are not congruent, this means that there is an odd number of divisors, such that x^2 has no solutions. Therefore, when you multiply the divisors together you get an even number of "nonresidues" and multiplying them all together (because of multiplicity) get's you that there is an answer to z^2 = ab. I am more curious if there is a better way to prove this.

Last edited: Nov 5, 2007