571p12 - N k ( q, h ) = k = ( h ) . 3. Suppose that k | ( p...

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Math 571 Analytic Number Theory I, Spring 2011, Problems 12 Due Tuesday 12th April Given positive integers k and q defne G k ( q,a )= q ° n =1 e ± an k q ² and let N k ( q,h ) denote the number oF solutions oF the congruence x k h (mod q ). p denotes a prime number. 1. (a) Prove that G k ( p, a )= p ° h =1 N k ( p, h ) e ± ah p ² (b) Let l =( k,p 1). Prove that N k ( p, h )= N l ( p, h ) For all h , and hence that G k ( p, a )= G l ( p, a ). (c) Prove that p 1 ° a =1 | G k ( p, a ) | 2 = p p ° h =1 N k ( p, h ) 2 p 2 . (d) Suppose that k | ( p 1). Prove that there are ( p 1) /k non-zero residues h (mod p ) For which N k ( p, h )= k , that N k ( p, 0) = 1, and that N k ( p, h ) = 0 For all other residue classes (mod p ). Prove that the right hand side oF (c) is p ( p 1)( k 1) and hence that in general it is p ( p 1)(( k,p 1) 1). (e) Suppose that p ° a , p ° c , and that b ac k (mod p ). Prove that G k ( p, a )= G k ( p, b ). (F) Suppose that k | ( p 1). Prove that iF p ° a then | G k ( p, a ) | <k p , and hence that in general | G k ( p, a ) | < ( k,p 1) p . 2. Suppose that ( h, q ) = 1. (a) Prove that 1 ϕ ( q ) ° χ χ ( x ) k χ ( h )= ³ 1i F x k h (mod q ) , 0 otherwise. (b) Prove that
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Unformatted text preview: N k ( q, h ) = k = ( h ) . 3. Suppose that k | ( p 1) and let be a character modulo p oF order k , say ( n ) = e ((ind n ) /k ). (a) Prove that For all h , N k ( p, h ) = 1 + k 1 j =1 ( h ) j . (b) Prove that iF p a , then G k ( p, a ) = k 1 j =1 ( a ) j ( j ) . (c) Prove that iF p a , then | G k ( p, a ) | ( k 1) p and hence in general | G k ( p, a ) | (( k, p 1) 1) p . 4. Let M s ( q, n ) denote the number oF solutions in x 1 , . . . , x s oF the congruence x k 1 + + x k s n (mod q ). (a) Prove that M s ( q, n ) = 1 q q a =1 G k ( q, a ) s e ( an/q ). (b) Prove that | M s ( p, n ) p s 1 | (( k, p 1) 1) s p s/ 2 ....
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This note was uploaded on 01/23/2012 for the course MATH 571 taught by Professor Vaughan,r during the Spring '08 term at Pennsylvania State University, University Park.

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