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Unformatted text preview: Gausss Quadratic Reciprocity Theorem : NumThy Jonathan L.F. King University of Florida, Gainesville FL 32611-2082, USA email@example.com Webpage http://www.math.ufl.edu/ squash/ 30 November, 2011 (at 01:30 ) 1 : Nomenclature. For odd D , use H D to mean D- 1 2 . ( The H is to suggest Half. ) In the sequel, p is an odd prime and S p is the stridelength ; we will walk around the circumference= p circle using strides of length S . Use H := H p and hh x ii := hh x ii p for the symmetric residue of integer x modulo p ; so hh x ii is in [ H ..H ] . Let mean p . Let G = G p ( S ) be the set of indices [ 1 ..H ] such that hh S ii p is neGative. Letting P be the indices with hh S ii Positive, we have that ( disjointly ) G t P = [ 1 ..H ] . Finally, use N = N p ( S ) for the number of negative indices; N := # G . 2 : Propn. Fix an S p , with notation from (1) . Then the mapping ( absolute-value of symm-residue ) 7 hh S ii , is a permutation of [ 1 ..H ] . We say that the map- ping 7 hh S ii is a permutation up to sign of [ 1 ..H ] . Proof. Given indices 1 6 6 k 6 H , we want that either equality hh S ii = hh k S ii forces = k . For either choice of sign in , note that hh S ii = hh k S ii IFF [ k ] S IFF k , since S p . Thus 6 k 6 2 H < p . Together with k 0, this forces k to actually be zero. Thus the is a minus sign, and k = . 3 : Gauss Lemma. Fix an odd prime p and integer S p . Then the Legendre symbol ( S p ) satisfies ( S p ) = [ 1] N . Proof of Gauss Lemma. Necessarily H Y =1 hh S ii H Y =1 S = H ! S H H ! S p ! , 4 : with the last step following from LSThm. Observe that hh S ii equals hh S ii as is-not/is in G . Propn 2, consequently, tells us that LhS(4) can be written as H ! times [ 1] N . Thus RhS(4) equals H ! ( S p ) H ! [ 1] N . The H !, being co-prime to p , cancels mod- p to hand us congruence ( S p ) [ 1] N . An important application is the following. 5 : Two-is-QR Lemma. Consider an oddprime p . Then 2 is a p-QR IFF p 8 1 . Abbrev. An odd integer D is 8Near if D 8 1; it is 8Far if D 8 3. ( The names come from being, mod 8, near/far from zero. ) Proof. Call p good if 2 is a p-QR. As usual, let H := p- 1 2 . It is easy to check that....
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