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Theory and Modern Applications

On the fourth power mean of the general k th Kloosterman sums

Abstract

Let q>2 be an integer, and let χ be a Dirichlet character modulo q. For integers m, n and k, the general k th Kloosterman sum S(m,n,k,χ;q) is defined by S(m,n,k,χ;q)= a = 1 q χ(a)e( m a k + n a ¯ k q ), where ∑ denotes the summation over all a with (a,q)=1, e(y)= e 2 π i y , and a ¯ is the inverse of a modulo q such that a a ¯ 1modq and 1 a ¯ q. In this paper we further study the fourth power mean χ mod q m = 1 q |S(m,n,k,χ;q) | 4 , and we give some identities.

MSC:11F20.

1 Introduction

Let q>2 be an integer, and let χ be a Dirichlet character modulo q. For arbitrary integers m and n, the general Kloosterman sum S(m,n,χ;q) is defined by

S(m,n,χ;q)= a = 1 q χ(a)e ( m a + n a ¯ q ) ,

where ∑ denotes the summation over all a with (a,q)=1, e(y)= e 2 π i y , and a ¯ is the inverse of a modulo q such that a a ¯ 1modq and 1 a ¯ q. For q=p a prime, Chowla [1] and Malyshev [2] proved an upper bound:

|S(m,n,χ;p)| ( m , n , p ) 1 / 2 p 1 / 2 + ϵ ,

where (m,n,p) is the great common divisor of m, n and p.

For general integer q>2, we do not know how large |S(m,n,χ;q)| is. However, |S(m,n,χ;q)| enjoys good value distribution properties. For fixed integer n with (n,q)=1, Zhang [3] showed the identity

χ mod q m = 1 q |S(m,n,χ;q) | 4 = ϕ 2 (q) q 2 d(q) p α q ( 1 2 α + 1 p α 1 1 p α ( p 1 ) + α 4 p α 1 ( α + 1 ) p α ) ,

where d(q) is the divisor function, ϕ(q) is the Euler function, and p α q denotes the product over all prime divisors p of q with p α q and p α + 1 q.

For integers m, n and k, the general k th Kloosterman sum S(m,n,k,χ;q) is defined by

S(m,n,k,χ;q)= a = 1 q χ(a)e ( m a k + n a ¯ k q ) .

Suppose that (nk,q)=1. Liu and Zhang [4] proved the identity:

χ mod q m = 1 q | S ( m , n , k , χ ; q ) | 4 = ϕ 2 ( q ) q 2 p α q ( k , p 1 ) 2 ( α 1 + 2 ( p 1 ) ( k , p 1 ) p + α p α 2 ( p α 1 ) p α ( p 1 ) ) .

In this paper we further consider the situation (k,q)>1. Our result is a generalization of [4].

Theorem 1.1 Let p be an odd prime, n be any integer. Let α and k be positive integers with d=(k,p1) and (k, p α 1 )= p δ . Then we have

χ mod p α m = 1 p α | S ( m , n , k , χ ; p α ) | 4 = { 2 d ϕ 3 ( p α ) p α + δ ( 1 d 2 ϕ ( p ) ) + d 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p α δ 1 1 ) , α 2 2 δ α 1 ; 2 d ϕ 3 ( p α ) p α + δ ( 1 d 2 ϕ ( p ) ) + d 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p δ + 1 1 ) + d 2 ϕ 2 ( p α ) p 2 α + 2 δ 1 ( p 1 ) ( α 2 δ 2 ) , 0 δ α 3 2 .

From Theorem 1.1 we immediately get the following corollary.

Corollary 1.1 Let p be an odd prime, n be any integer. Let α and k be positive integers. Assume that one of the following conditions holds:

  1. (1)

    α=1;

  2. (2)

    α=2 and (k, p α 1 )>1;

  3. (3)

    α2 and (k, p α 1 )= p α 1 .

Then we have

χ mod p α m = 1 p α |S ( m , n , k , χ ; p α ) | 4 =2 ϕ 3 ( p α ) p α ( k , ϕ ( p α ) ) ( 1 d 2 ϕ ( p ) ) .

Theorem 1.2 Let q>2 be an odd number, n be an integer with (n,q)=1, and let k be a positive integer. Then we have

χ mod q m = 1 q | S ( m , n , k , χ ; q ) | 4 = p α q ( k , p α 1 ) = p δ α 2 2 δ α 1 ( 2 ( k , p 1 ) ϕ 3 ( p α ) p α + δ ( 1 ( k , p 1 ) 2 ϕ ( p ) ) + ( k , p 1 ) 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p α δ 1 1 ) ) × p α q ( k , p α 1 ) = p δ 0 δ α 3 2 ( 2 ( k , p 1 ) ϕ 3 ( p α ) p α + δ ( 1 ( k , p 1 ) 2 ϕ ( p ) ) + ( k , p 1 ) 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p δ + 1 1 ) + ( k , p 1 ) 2 ϕ 2 ( p α ) p 2 α + 2 δ 1 ( p 1 ) ( α 2 δ 2 ) ) .

From Theorem 1.2 we can get the following corollary.

Corollary 1.2 Let q>2 be an odd number, n be an integer with (n,q)=1, and let k be a positive integer. Assume that one of the following conditions holds:

  1. (1)

    q is square-free;

  2. (2)

    q= p 1 2 p r 2 and p 1 p r k;

  3. (3)

    q= i = 1 α i 2 r p i α i and qk i = 1 r p i .

Then we have

χ mod q m = 1 q |S(m,n,k,χ;q) | 4 = 2 ω ( q ) ϕ 3 (q)q ( k , ϕ ( q ) ) p α q ( 1 ( k , p 1 ) 2 ϕ ( p ) ) .

2 Some lemmas

To complete the proof of theorems, we need the following lemmas.

Lemma 2.1 Let p be an odd prime, and let α and k be positive integers with (k, p α 1 )= p δ . Write d=(k,p1). Then we have

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1= { 2 d ϕ ( p α ) p δ ( 1 d 2 ϕ ( p ) ) + d 2 p α + δ 1 ( p α δ 1 1 ) , α 2 2 δ α 1 ; 2 d ϕ ( p α ) p δ ( 1 d 2 ϕ ( p ) ) + d 2 p α + δ 1 ( p δ + 1 1 ) + d 2 p α + 2 δ 1 ( p 1 ) ( α 2 δ 2 ) , 0 δ α 3 2 .

Proof For α=1, we get

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1= a = 1 p 1 b = 1 p 1 p ( a k 1 ) ( b k 1 ) 1=2d(p1) d 2 =2dϕ(p) ( 1 d 2 ϕ ( p ) ) .
(2.1)

Now we assume that α2. It is not hard to show that

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1 = β = 0 α γ = 0 α β + γ α a = 1 p α p β a k 1 b = 1 p α p γ b k 1 1 = β = 0 α a = 1 p α p β a k 1 γ = α β α b = 1 p α p γ b k 1 1 = β = 0 α a = 1 p α p β a k 1 b = 1 p α p α β b k 1 1 = a = 1 p α p a k 1 b = 1 p α p α b k 1 1 + β = 1 α 1 a = 1 p α p β a k 1 b = 1 p α p α β b k 1 1 + a = 1 p α p α a k 1 b = 1 p α 1 = ( ϕ ( p α ) p α 1 ( k , ϕ ( p ) ) ) ( k , ϕ ( p α ) ) + ( k , ϕ ( p α ) ) ϕ ( p α ) + β = 1 α 1 ( p α β ( k , ϕ ( p β ) ) p α β 1 ( k , ϕ ( p β + 1 ) ) ) p β ( k , ϕ ( p α β ) ) = p α 1 ( 2 ϕ ( p ) ( k , ϕ ( p ) ) ) ( k , p α 1 ) ( k , ϕ ( p ) ) + p α 1 ( k , ϕ ( p ) ) 2 β = 1 α 1 ( p ( k , p β 1 ) ( k , p β ) ) ( k , p α β 1 ) = 2 d ϕ ( p α ) ( k , p α 1 ) ( 1 d 2 ϕ ( p ) ) + d 2 p α 1 β = 1 α 1 ( p ( k , p β 1 ) ( k , p β ) ) ( k , p α β 1 ) .
(2.2)

If (k, p α 1 )=1, by (2.2) we have

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1=2dϕ ( p α ) ( 1 d 2 ϕ ( p ) ) + d 2 ϕ ( p α ) (α1).
(2.3)

If (k, p α 1 )= p α 1 , then

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1=2dϕ ( p α ) p α 1 ( 1 d 2 ϕ ( p ) ) .
(2.4)

On the other hand, for α=2 and (k, p α 1 )>1, we have

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1=2dϕ ( p α ) ( k , p α 1 ) ( 1 d 2 ϕ ( p ) ) .
(2.5)

Next we suppose that α3 and (k, p α 1 )= p δ with 1δα2. Then

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1 = 2 d ϕ ( p α ) ( k , p α 1 ) ( 1 d 2 ϕ ( p ) ) + d 2 p α 1 β = 1 α 1 ( p ( k , p β 1 ) ( k , p β ) ) ( k , p α β 1 ) = 2 d ϕ ( p α ) p δ ( 1 d 2 ϕ ( p ) ) + d 2 ϕ ( p α ) p δ β = δ + 1 α 1 ( k , p α β 1 ) .

Noting that

β = δ + 1 α 1 ( k , p α β 1 ) = u = 0 α δ 2 ( k , p u ) = { u = 0 α δ 2 p u , α δ 2 δ , u = 0 δ p u + u = δ + 1 α δ 2 p δ , α δ 2 δ + 1 = { p α δ 1 1 p 1 , δ α 2 2 , p δ + 1 1 p 1 + ( α 2 δ 2 ) p δ , δ α 3 2 .

Therefore

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1 = { 2 d ϕ ( p α ) p δ ( 1 d 2 ϕ ( p ) ) + d 2 p α + δ 1 ( p α δ 1 1 ) , α 2 2 δ α 2 ; 2 d ϕ ( p α ) p δ ( 1 d 2 ϕ ( p ) ) + d 2 p α + δ 1 ( p δ + 1 1 ) + d 2 p α + 2 δ 1 ( p 1 ) ( α 2 δ 2 ) , 1 δ α 3 2 .
(2.6)

Now combining (2.1), (2.3)-(2.6) we immediately get

a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1 = { 2 d ϕ ( p α ) p δ ( 1 d 2 ϕ ( p ) ) + d 2 p α + δ 1 ( p α δ 1 1 ) , α 2 2 δ α 1 ; 2 d ϕ ( p α ) p δ ( 1 d 2 ϕ ( p ) ) + d 2 p α + δ 1 ( p δ + 1 1 ) + d 2 p α + 2 δ 1 ( p 1 ) ( α 2 δ 2 ) , 0 δ α 3 2 ,

for α1. □

Lemma 2.2 Let n, k, k 1 , k 2 be integers with (n, k 1 k 2 )=( k 1 , k 2 )=1. Then for any character χmod k 1 k 2 , there exist integers n 1 and n 2 with ( n 1 , k 1 )=( n 2 , k 2 )=1 such that

n n 1 k 2 2 + n 2 k 1 2 (mod k 1 k 2 ),

and for these integers we have

|S(m,n,k,χ; k 1 k 2 )|=|S(m k ¯ 2 , n 1 k 2 ,k, χ 1 ; k 1 )||S(m k ¯ 1 , n 2 k 1 ,k, χ 2 ; k 2 )|,

where χ= χ 1 χ 2 with χ 1 mod k 1 and χ 2 mod k 2 .

Proof This is Lemma 2.2 of [4]. □

3 Proof of the theorems

First we prove Theorem 1.1. Let p be an odd prime, and let α and k be positive integers. Assume that n is any integer. We have

| S ( m , n , k , χ ; p α ) | 2 = a = 1 p α χ ( a ) e ( m a k + n a ¯ k p α ) b = 1 p α χ ¯ ( b ) e ( m b k + n b ¯ k p α ) = a = 1 p α b = 1 p α χ ( a ) e ( m b k ( a k 1 ) + n b ¯ k ( a ¯ k 1 ) p α ) .

Then from the orthogonality relation for characters and the trigonometric identity we get

χ mod p α m = 1 p α | S ( m , n , k , χ ; p α ) | 4 = χ mod p α m = 1 p α a = 1 p α b = 1 p α χ ( a ) e ( m b k ( a k 1 ) + n b ¯ k ( a ¯ k 1 ) p α ) × c = 1 p α d = 1 p α χ ¯ ( c ) e ( m d k ( c k 1 ) + n d ¯ k ( c ¯ k 1 ) p α ) = ϕ ( p α ) a = 1 p α b = 1 p α d = 1 p α m = 1 p α e ( m ( a k 1 ) ( b k d k ) + n ( a ¯ k 1 ) ( b ¯ k d ¯ k ) p α ) = ϕ ( p α ) a = 1 p α b = 1 p α d = 1 p α m = 1 p α e ( m d k ( a k 1 ) ( b k 1 ) + n d ¯ k ( a ¯ k 1 ) ( b ¯ k 1 ) p α ) = ϕ ( p α ) a = 1 p α b = 1 p α d = 1 p α m = 1 p α e ( m d k ( a k 1 ) ( b k 1 ) + n d ¯ k a ¯ k b ¯ k ( a k 1 ) ( b k 1 ) p α ) = ϕ 2 ( p α ) p α a = 1 p α b = 1 p α p α ( a k 1 ) ( b k 1 ) 1 .

Write d=(k,p1) and (k, p α 1 )= p δ . By Lemma 2.1 we immediately have

χ mod p α m = 1 p α | S ( m , n , k , χ ; p α ) | 4 = { 2 d ϕ 3 ( p α ) p α + δ ( 1 d 2 ϕ ( p ) ) + d 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p α δ 1 1 ) , α 2 2 δ α 1 ; 2 d ϕ 3 ( p α ) p α + δ ( 1 d 2 ϕ ( p ) ) + d 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p δ + 1 1 ) + d 2 ϕ 2 ( p α ) p 2 α + 2 δ 1 ( p 1 ) ( α 2 δ 2 ) , 0 δ α 3 2 .

This completes the proof of Theorem 1.1.

Now we prove Theorem 1.2. Let q>2 be an odd number, n be an integer with (n,q)=1, and let k be a positive integer. Write

q= i = 1 r p i α i andm= i = 1 r m i q p i α i .

By Lemma 2.2 and Theorem 1.1 we have

χ mod q m = 1 q | S ( m , n , k , χ ; q ) | 4 = i = 1 r ( χ i mod p i α i m i = 1 p i α i | S ( m i q p i α i ( q p i α i ) ¯ , n i q p i α i , k , χ i ; p i α i ) | 4 ) = i = 1 r ( χ i mod p i α i m i = 1 p i α i | S ( m i , n i q p i α i , k , χ i ; p i α i ) | 4 ) = p α q ( k , p α 1 ) = p δ α 2 2 δ α 1 ( 2 ( k , p 1 ) ϕ 3 ( p α ) p α + δ ( 1 ( k , p 1 ) 2 ϕ ( p ) ) + ( k , p 1 ) 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p α δ 1 1 ) ) × p α q ( k , p α 1 ) = p δ 0 δ α 3 2 ( 2 ( k , p 1 ) ϕ 3 ( p α ) p α + δ ( 1 ( k , p 1 ) 2 ϕ ( p ) ) + ( k , p 1 ) 2 ϕ 2 ( p α ) p 2 α + δ 1 ( p δ + 1 1 ) + ( k , p 1 ) 2 ϕ 2 ( p α ) p 2 α + 2 δ 1 ( p 1 ) ( α 2 δ 2 ) ) .

This proves Theorem 1.2.

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Acknowledgements

This work is supported by the P.E.D. (2013JK0561) and N.S.F. (11371291) of P.R. China.

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Guo, X., Geng, G. & Pan, X. On the fourth power mean of the general k th Kloosterman sums. Adv Differ Equ 2014, 103 (2014). https://doi.org/10.1186/1687-1847-2014-103

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Keywords

  • general k th Kloosterman sum
  • fourth power mean
  • identity