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Unformatted text preview: 2 5 1 3 43 5 3 Using cubic reciprocity on the latter symbol composed of two PRIMARY primes, this is = 2 5 1 3 543 3 Now we must reduce 5 mod43 . This is: 543 = 5 (43 )43 (43 ) =2015 2 13 Since an element of Z [ ] close to this fraction is2 2 =1 + , we compute 5(43 )(2 2 ) =22 = 2 2 and hence 3 5 3 = 2 5 1 3 2 243 3 = 2 5 1 3 243 3 43 2 3 = 5 2 3 43 2 3  2 3 = ( 2 )2 ( ) 2 ( ) = 2 so 3 is not a cubic residue mod 5....
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 Spring '09
 BRUBAKER
 Number Theory, Congruence

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