781hw9sol

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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)....
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.

### Page1 / 4

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online