781hw9sol

781hw9sol - 18.781 Fall 2007 Problem Set 9 Solutions to...

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Unformatted text preview: 18.781, Fall 2007 Problem Set 9 Solutions to Selected Problems Problem 1 When x = y , it is clear that p- 1 p- 1 X t =0 ζ t ( x- y ) p = p- 1 p- 1 X t =0 ζ p = p- 1 p- 1 X t =0 1 = 1 . Now recall the fact that if α is not 0 in modulo p , then { , α, 2 α, ··· , ( p- 1) α } is a complete residue system modulo p . Also a ≡ b (mod p ) implies that ζ a p = ζ b p . Thus, when x 6≡ y (mod p ), we have p- 1 X t =0 ζ t ( x- y ) p = p- 1 X t =0 ζ t p = 0 , where the last equality can be verified easily. Therefore, we have p- 1 p- 1 X t =0 ζ t ( x- y ) p = 0 . This gives the desired conclusion. 2 Problem 2 We can easily find that χ ( a ) g ( a, χ ) = p- 1 X t =0 χ ( at ) ζ at p = p- 1 X x =0 χ ( x ) ζ x p = g (1 , χ ) because if a is not 0 in modulo p , then { , a, 2 a, ··· , ( p- 1) a } is a complete residue system modulo p . Since χ ( a ) χ ( a- 1 ) = 1, we have g ( a, χ ) = χ ( a- 1 ) g (1 , χ ) . 2 1 Problem 3 (a) The sum is going through ( a, b ) = (0 , p ) , (1 , 0) , (2 , p- 1) , (3 , p- 2) , ··· , ( p- 1 , 2)....
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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.

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781hw9sol - 18.781 Fall 2007 Problem Set 9 Solutions to...

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