# Difference between revisions of "Talk:Quadratic residues"

m |
|||

Line 9: | Line 9: | ||

Thanks for clarifying -Cosinator | Thanks for clarifying -Cosinator | ||

+ | |||

+ | ''Whereas the above are properties of the Legendre symbol, they still hold for any odd integers p and q when using the Jacobi symbol defined below''. | ||

+ | Hmmm... The quadratic reciprocity law clearly fails (at least in the form as written for primes) if <math>\gcd(m,n)>1</math>. So some correction is needed. My knowledge of number theory really needs some refreshment, so could someone else write the correct statement here? --[[User:Fedja|Fedja]] 14:34, 28 June 2006 (EDT) |

## Revision as of 13:34, 28 June 2006

I'm sure someone wants to write out all the fun properties of Legendre symbols. It just happens not to be me right now. -- ComplexZeta

Is it any number n, or any integer n? --- cosinator

Where? --ComplexZeta 11:07, 27 June 2006 (EDT)

In the introduction it says 'We say that a is a quadratic residue modulo m if there is some number n so that n^2 − a is divisible by m.' If it were any number, I would think that any a could be a quadratic residue modulo m

Thanks for clarifying -Cosinator

*Whereas the above are properties of the Legendre symbol, they still hold for any odd integers p and q when using the Jacobi symbol defined below*.
Hmmm... The quadratic reciprocity law clearly fails (at least in the form as written for primes) if . So some correction is needed. My knowledge of number theory really needs some refreshment, so could someone else write the correct statement here? --Fedja 14:34, 28 June 2006 (EDT)