This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: x 3 2 + 3 (11) is dicult to solve in R since there are 121 residue classes mod 11. By cubic reciprocity 2 + 3 11 = 11 2 + 3 . Use this fact, together with your knowledge of the congruence x 3 11 (7) to prove that the initial congruence mod 11 in R is not solvable. 6. Prove g (1 , p ) = t (mod p ) t 2 p by evaluating the sum X t (mod p ) [1 + t p ] t p (Recall that p denotes a p th root of unity e 2 i/p and the Gauss sum g ( a, p ) = X t (mod p ) t p at p 7. Let f be a function from Z C , the complex numbers. Suppose that for p prime, f ( n + p ) = f ( n ) for all integers n . Dene f ( a ) = p1 X t (mod p ) f ( t ) at p . Prove that f ( t ) = X a (mod p ) f ( a ) at p (Note the similarity with Fourier analysis for periodic functions on R .)...
View
Full
Document
This note was uploaded on 04/28/2009 for the course MATH 18.781 taught by Professor Brubaker during the Spring '09 term at MIT.
 Spring '09
 BRUBAKER
 Algebra, Number Theory

Click to edit the document details