Number Theory (Legendre symbols, quadratic residues/nonresidues)

[Ignea_unda]

Homework Statement

"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.

Homework Equations

Legendre symbols
Quadratic Congruences

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.

 

[mighty2000]

Thank you for your post. I am a scientist and I will provide a response to your question.

The statement you provided is actually false. A counterexample to this statement is when m = 6, a = 3, b = 2. In this case, (m, ab) = (6, 6) = 1, and neither x^2 congruent to 3 (mod 6) nor y^2 congruent to 2 (mod 6) is solvable. However, z^2 congruent to 6 (mod 6) is not solvable either.

The reason for this is that the Legendre symbol only gives information about the solvability of a quadratic congruence when the modulus is a prime number. In this case, m = 6 is not a prime number, so the Legendre symbol cannot be used to determine the solvability of z^2 congruent to ab (mod m).

In general, the statement you provided is not true for composite numbers as the modulus. A better way to prove this statement would be to use the Chinese Remainder Theorem, which states that if two congruences x^2 congruent to a (mod m) and y^2 congruent to b (mod m) are solvable, then the congruence z^2 congruent to ab (mod m) is also solvable. However, this theorem does not hold for non-prime moduli, so the statement is not true in general.

I hope this clarifies things for you. Keep investigating and asking questions, that's what being a scientist is all about. Good luck with your studies!