proof of quadratic reciprocity rule
The quadratic reciprocity law is:
( is the Legendre symbol.)
Proof: Let be the subset of . Let be the interval
of . By the Chinese remainder theorem, there exists a unique bijection such that, for any , if we write , then and . Let be the subset of consisting of the values of on . contains, say, elements of the form such that , and elements of the form with . Intending to apply Gauss’s lemma, we seek some kind of comparison between and .
We define three subsets of by
and we let be the cardinal of for each .
has elements in the region , namely for all of the form with and . Thus
Swapping and , we have likewise
Furthermore, for any , if then . It follows that for any other than , either or is in , but not both. Therefore
Adding (1), (2), and (3) gives us
which, in view of Gauss’s lemma, is the desired conclusion.
For a bibliography of the more than 200 known proofs of the QRL, see http://www.rzuser.uni-heidelberg.de/ hb3/fchrono.htmlLemmermeyer .
|Title||proof of quadratic reciprocity rule|
|Date of creation||2013-03-22 13:16:15|
|Last modified on||2013-03-22 13:16:15|
|Last modified by||mathcam (2727)|