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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 ) = p-1 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 .)...
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