Difference between revisions of "Talk:Quadratic residues"
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Thanks for clarifying -Cosinator | Thanks for clarifying -Cosinator | ||
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| + | ''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)
