Open Access

A sum analogous to the high-dimensional Kloosterman sums and its upper bound estimate

Journal of Inequalities and Applications20132013:130

https://doi.org/10.1186/1029-242X-2013-130

Received: 6 December 2012

Accepted: 9 March 2013

Published: 27 March 2013

Abstract

The main purpose of this paper is, using the properties of Gauss sums and the estimate for the generalized exponential sums, to study the upper bound estimate problem of one kind sums analogous to the high-dimensional Kloosterman sums and to give some interesting mean value formula and an upper bound estimate for it.

MSC:11L05.

Keywords

a sum analogous to the high-dimensional Kloosterman sums Gauss sums upper bound estimate mean value

1 Introduction

For any integer q 3 , the high-dimensional Kloosterman sums K ( c 1 , c 2 , , c k , m ; q ) are defined as follows:
K ( c 1 , c 2 , , c k , m ; q ) = a 1 = 1 q a k = 1 q e ( c 1 a 1 + + c k a k + m a ¯ 1 a ¯ k q ) ,

where e ( x ) = e 2 π i x , a i = 1 q denotes the summation over all integers 1 a i q such that ( a i , q ) = 1 , c i and m are integers with ( m , q ) = 1 , a ¯ i denotes the solution of the congruent equation x a i 1 mod q ( i = 1 , 2 , , k ).

There are several results on the properties of the Kloosterman sums K ( c 1 , c 2 , , c k , m ; q ) . For example, see [1, 2] and [3]. Related works can also be found in [48] and [9].

In this paper, we consider a sum analogous to the high-dimensional Kloosterman sums as follows:
S ( c 1 , c 2 , , c k , m , χ ; q ) = a 1 = 1 q a k = 1 q χ ( c 1 a 1 + + c k a k + m a ¯ 1 a ¯ k ) ,
(1.1)

where χ is a Dirichlet character modq.

If k = 1 and q = p (an odd prime), then for any integer a with ( a , p ) = 1 , applying the Fermat little theorem, one can deduce a p 2 a ¯ mod p . So, the sum (1.1) becomes
a = 1 p 1 χ ( c a + m a ¯ ) .
It is a special case of the general polynomial character sums
a = N + 1 N + M χ ( f ( a ) ) ,
where M and N are any positive integers and f ( x ) is a polynomial. Let χ be a q th-order character modp. If f ( x ) is not a perfect q th power modp, then from Weil’s classical work (see [10]), we can deduce the estimate
x = N + 1 N + M χ ( f ( x ) ) p 1 2 ln p ,

where ‘’ constant depends only on the degree of f ( x ) . Some related results can also be found in [1113] and [14].

Now we are concerned with the upper bound estimate problem of (1.1). Regarding this contents, it seems that nobody has yet studied it, at least we have not seen any related result before. The problem is interesting because it can reflect some new properties of character sums. The main purpose of this paper is, using the analytic methods and the properties of Gauss sums, to study this problem and give a sharp upper bound estimate for (1.1). That is, we prove the following conclusions.

Theorem 1 Let p be an odd prime, let k be a positive integer with k 2 , and let χ be any non-principal character modp. Then for any integers c 1 , c 2 , , c k and m with ( c 1 c 2 c k m , p ) = 1 , we have the identity
| S ( c 1 , c 2 , , c k , m , χ ; p ) | = { p k 2 if ( k + 1 , p 1 ) = 1 , 0 if ( k + 1 ) | ( p 1 ) and χ p 1 k + 1 χ 0 ,

where χ 0 denotes the principal character modp.

Theorem 2 Let p be an odd prime, let k be a positive integer with k 2 , and let χ be any non-principal character modp. Then for any integers c 1 , c 2 , , c k and m with ( c 1 c 2 c k m , p ) = 1 , we have the estimate
| S ( c 1 , c 2 , , c k , m , χ ; p ) | ( k + 1 ) p k 2 .
Theorem 3 Let p and q be two odd primes, let r be any qth non-residue modp. Then for any integers c 1 , c 2 , , c q 1 with ( c 1 c 2 c q 1 , p ) = 1 , we have the identity
i = 0 q 1 [ a 1 = 1 p 1 a 2 = 1 p 1 a q 1 = 1 p 1 ( c 1 a 1 + c 2 a 2 + + c q 1 a q 1 + r i a ¯ 1 a ¯ 2 a ¯ q 1 p ) ] 2 = { q 2 p q 1 if q | ( p 1 ) , q p q 1 if ( q , p 1 ) = 1 .

If p 1 mod 4 , then the above formula also holds for q = 2 , where ( p ) denotes the Legendre symbol.

Taking p = 4 m + 1 and q = 2 in Theorem 3, note that 2 | ( p 1 ) , we may immediately deduce the following.

Corollary Let p be an odd prime with p 1 mod 4 , then we have the identity
p = ( a = 1 p 1 2 ( a + r a ¯ p ) ) 2 + ( b = 1 p 1 2 ( b + s b ¯ p ) ) 2 ,

where r and s are any two integers such that ( r p ) ( s p ) = 1 .

This gives another proof for a classical work in elementary number theory (i.e., see [15] Theorems 4-11): For any prime p with p 1 mod 4 , there exist two positive integers x and y such that p = x 2 + y 2 .

2 Several lemmas

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

Lemma 1 Let p be an odd prime, let χ be any non-principal character modp, and let k be any positive integer such that ( k , p 1 ) = 1 or k | p 1 . Then for any integer m with ( m , p ) = 1 , we have the identity
a = 1 p 1 χ ( a ) e ( m a k p ) = { χ ¯ r ( m ) τ ( χ r ) if ( k , p 1 ) = 1 , 0 if k | ( p 1 ) and χ p 1 k χ 0 , χ ¯ 1 ( m ) i = 0 k 1 χ ¯ k i ( m ) τ ( χ 1 χ k i ) if k | ( p 1 ) and χ p 1 k = χ 0 ,

where r k 1 mod ( p 1 ) , χ 0 denotes the principal character modp, χ k denotes any k-order character modp and χ 1 k = χ .

Proof If ( k , p 1 ) = 1 , then there exists one integer r with ( r , p 1 ) = 1 such that r k 1 mod ( p 1 ) . This time, for any integer a with ( a , p ) = 1 , we have a r k a mod p . If a passes through a reduced residue system modp, then a r also passes through a reduced residue system modp. Therefore, we have
a = 1 p 1 χ ( a ) e ( m a k p ) = a = 1 p 1 χ ( a r ) e ( m a r k p ) = a = 1 p 1 χ r ( a ) e ( m a p ) = χ ¯ r ( m ) τ ( χ r ) .
(2.1)
If k > 1 and k | ( p 1 ) with χ p 1 k χ 0 , then there must exist an integer n with ( n , p ) = 1 such that χ p 1 k ( n ) 1 . For this n, we have
a = 1 p 1 χ ( a ) e ( m a k p ) = a = 1 p 1 χ ( a n p 1 k ) e ( m ( a n p 1 k ) k p ) = χ ( n p 1 k ) a = 1 p 1 χ ( a ) e ( m a k n p 1 p ) = χ p 1 k ( n ) a = 1 p 1 χ ( a ) e ( m a k p )
or
( 1 χ p 1 k ( n ) ) a = 1 p 1 χ ( a ) e ( m a k p ) = 0 .
Since χ p 1 k ( n ) 1 , from the above identity, we have
a = 1 p 1 χ ( a ) e ( m a k p ) = 0 .
(2.2)
If χ p 1 k = χ 0 , then χ must be a k th character modp, so there exists one character χ 1 mod p such that χ = χ 1 k . Let χ k be a k-order character modp (i.e., χ k k = χ 0 ), then for any integer a with ( a , p ) = 1 , note that
1 + χ k ( a ) + χ k 2 ( a ) + + χ k k 1 ( a ) = { k if  a  is a  k th residue  mod p , 0 , otherwise .
From the properties of Gauss sums, we have
a = 1 p 1 χ ( a ) e ( m a k p ) = a = 1 p 1 χ 1 k ( a ) e ( m a k p ) = a = 1 p 1 χ 1 ( a k ) e ( m a k p ) = a = 1 p 1 χ 1 ( a ) ( 1 + χ k ( a ) + χ k 2 ( a ) + + χ k k 1 ( a ) ) e ( m a p ) = χ ¯ 1 ( m ) i = 0 k 1 χ ¯ k i ( m ) τ ( χ 1 χ k i ) .
(2.3)

Now Lemma 1 follows from (2.1), (2.2) and (2.3). □

Lemma 2 Let p and q be two odd primes with q | ( p 1 ) , and let χ q be any q-order character modp. Then for any integers c 1 , c 2 , , c q 1 and m with ( m c 1 c 2 c q 1 , p ) = 1 , we have the identities
( I ) a 1 = 1 p 1 a 2 = 1 p 1 a q 1 = 1 p 1 ( c 1 a 1 + c 2 a 2 + + c q 1 a q 1 + m a ¯ 1 a ¯ 2 a ¯ q 1 p ) = ( n p ) τ q 1 ( χ 2 ) ( i = 0 q 1 χ q i ( n ) τ q ( χ 2 χ q i ) τ q ( χ 2 ) ) ; ( II ) a = 1 p 1 ( a + m a ¯ p ) = { χ ¯ 1 ( m ) τ ( χ 2 ) ( τ 2 ( χ 1 ) + ( m p ) τ 2 ( χ ¯ 1 ) ) if p 1 mod 4 , 0 if p 3 mod 4 ,

where ( p ) = χ 2 denotes the Legendre symbol, n = m c 1 c 2 c q 1 and χ 1 2 = χ 2 .

Proof If q is an odd prime, then χ 2 q = χ 2 and χ ¯ 2 = χ 2 , so applying (2.3) and the properties of Gauss sums, we have
a 1 = 1 p 1 a 2 = 1 p 1 a q 1 = 1 p 1 ( c 1 a 1 + c 2 a 2 + + c q 1 a q 1 + m a ¯ 1 a ¯ 2 a ¯ q 1 p ) = a 1 = 1 p 1 a 2 = 1 p 1 a q 1 = 1 p 1 ( a 1 + a 2 + + a q 1 + n a ¯ 1 a ¯ 2 a ¯ q 1 p ) = 1 τ ( χ 2 ) a 1 = 1 p 1 a 2 = 1 p 1 a q 1 = 1 p 1 b = 1 p 1 χ 2 ( b ) e ( b ( a 1 + + a q 1 ) + b n a ¯ 1 a ¯ q 1 p ) = 1 τ ( χ 2 ) a 1 = 1 p 1 a 2 = 1 p 1 a q 2 = 1 p 1 b = 1 p 1 χ ( b ) e ( b ( 1 + a 1 + + a q 2 ) p ) × a q 1 = 1 p 1 χ 2 ( a q 1 ) e ( b n a ¯ q 1 q a ¯ 1 a ¯ q 2 p ) = 1 τ ( χ 2 ) a 1 = 1 p 1 a 2 = 1 p 1 a q 2 = 1 p 1 b = 1 p 1 χ ( b ) e ( b ( 1 + a 1 + + a q 2 ) p ) × c = 1 p 1 χ 2 ( c ) ( 1 + χ q ( c ) + + χ q q 1 ( c ) ) e ( b n c a ¯ 1 a ¯ q 2 p ) = ( n p ) τ q 1 ( χ 2 ) ( i = 0 q 1 χ q i ( n ) τ q ( χ 2 χ q i ) τ q ( χ 2 ) ) .

This proves formula (I).

To prove formula (II), note that if p 3 mod 4 , then χ 2 must be an odd character modp (i.e., χ 2 ( 1 ) = 1 ) so that
a = 1 p 1 ( a + m a ¯ p ) = a = 1 p 1 ( p a + m p a ¯ p ) = ( 1 p ) a = 1 p 1 ( a + m a ¯ p ) = a = 1 p 1 ( a + m a ¯ p )
or
a = 1 p 1 ( a + m a ¯ p ) = 0 .
If p 1 mod 4 , then there exists one character χ 1 mod p such that χ 1 2 = χ 2 . Note that χ 1 3 = χ ¯ 1 and χ 2 χ ¯ 1 = χ 1 ; from the properties of Gauss sums, we have
a = 1 p 1 ( a + m a ¯ p ) = 1 τ ( χ 2 ) a = 1 p 1 b = 1 p 1 χ 2 ( b ) e ( b ( a + m a ¯ ) p ) = 1 τ ( χ 2 ) b = 1 p 1 χ 2 ( b ) a = 1 p 1 χ 2 ( a ¯ ) e ( b + m b a ¯ 2 p ) = 1 τ ( χ 2 ) b = 1 p 1 χ 2 ( b ) e ( b p ) a = 1 p 1 χ 1 2 ( a ) e ( m b a 2 p ) = 1 τ ( χ 2 ) b = 1 p 1 χ 2 ( b ) e ( b p ) a = 1 p 1 χ 1 ( a ) ( 1 + χ 2 ( a ) ) e ( m b a p ) = χ ¯ 1 ( m ) τ ( χ 2 ) ( τ 2 ( χ 1 ) + ( m p ) τ 2 ( χ ¯ 1 ) ) .

This proves Lemma 2. □

3 Proof of the theorems

In this section, we complete the proof of our theorems. First we prove Theorems 1 and 2. Let n = m c 1 c 2 c k , ( k + 1 , p 1 ) = d . If d = 1 , we can assume r ( k + 1 ) 1 mod ( p 1 ) , then from Lemma 1, the properties of a reduced residue system modp and Gauss sums, we have
S ( c 1 , c 2 , , c k , m , χ ; p ) = a 1 = 1 p 1 a k = 1 p 1 χ ( a 1 + + a k + m c 1 c 2 c k a ¯ 1 a ¯ k ) = 1 τ ( χ ¯ ) a 1 = 1 p 1 a k = 1 p 1 b = 1 p 1 χ ¯ ( b ) e ( b ( a 1 + + a k + n a ¯ 1 a ¯ k ) p ) = 1 τ ( χ ¯ ) a 2 = 1 p 1 a k = 1 p 1 a 1 = 1 p 1 b = 1 p 1 χ ¯ ( b a ¯ 1 ) e ( b ( 1 + + a k ) + n b a ¯ 1 k + 1 a ¯ 2 a ¯ k p ) = 1 τ ( χ ¯ ) a 2 = 1 p 1 a k = 1 p 1 b = 1 p 1 χ ¯ ( b ) e ( b + b ( a 2 + + a k ) p ) × a 1 = 1 p 1 χ ( a 1 ) e ( n b a ¯ 1 k + 1 a ¯ 2 a ¯ k p ) = 1 τ ( χ ¯ ) a 2 = 1 p 1 a k = 1 p 1 b = 1 p 1 χ ¯ ( b ) e ( b + b ( a 2 + + a k ) p ) × a 1 = 1 p 1 χ ¯ r ( a 1 ) e ( n b a 1 a ¯ 2 a ¯ k p ) = τ ( χ ¯ r ) τ ( χ ¯ ) b = 1 p 1 χ ¯ ( b ) e ( b p ) a 2 = 1 p 1 a k = 1 p 1 χ r ( n b a ¯ 2 a ¯ k ) e ( b ( a 2 + + a k ) p ) = χ r ( n ) τ k + 1 ( χ ¯ r ) τ ( χ ¯ ) = χ r ( m c 1 c 2 c k ) τ k + 1 ( χ ¯ r ) τ ( χ ¯ ) .
(3.1)
If d > 1 and χ p 1 d χ 0 , then from the method of proving (2.2), we have
a 1 = 1 p 1 χ ( a 1 ) e ( n b a ¯ 1 k + 1 a ¯ 2 a ¯ k p ) = 0 .
From this identity and the method of proving (3.1), we may immediately deduce that if χ p 1 d χ 0 , then
S ( c 1 , c 2 , , c k , m , χ ; p ) = 0 .
(3.2)
If d > 1 and χ p 1 d = χ 0 , then χ must be a d th character modp, so there exists a character χ 1 mod p such that χ = χ 1 d . Let χ d be a d-order character modp, then we have
a 1 = 1 p 1 χ ( a 1 ) e ( n b a ¯ 1 k + 1 a ¯ 2 a ¯ k p ) = a = 1 p 1 χ 1 ( a d ) e ( n b a ¯ d k + 1 d a ¯ 2 a ¯ k p ) = a = 1 p 1 χ ¯ 1 ( a ) ( 1 + χ d ( a ) + χ d 2 ( a ) + + χ d d 1 ( a ) ) e ( n b a k + 1 d a ¯ 2 a ¯ k p ) = i = 0 d 1 a = 1 p 1 χ ¯ 1 ( a ) χ d i ( a ) e ( n b a k + 1 d a ¯ 2 a ¯ k p ) .
(3.3)
Let ( k + 1 d , p 1 ) = d 1 , then repeat the process of proving (2.1), (2.2) and (2.3). Combining (3.1), (3.2) and (3.3), we may immediately deduce the estimate
| S ( c 1 , c 2 , , c k , m , χ ; p ) | ( k + 1 ) p k 2 .
(3.4)

Now note that | τ ( χ ) | = | τ ( χ r ) | = p , Theorems 1 and 2 follow from (3.1), (3.2) and (3.4).

Now we prove Theorem 3. If q 3 , we separate q into two cases ( q , p 1 ) = 1 and ( q , p 1 ) = q . If ( q , p 1 ) = q , then note that for any q th non-residue r mod p , we have
h = 0 q 1 χ q l ( r h ) = { q if  q | l , 0 if  ( q , l ) = 1 .
From (I) of Lemma 2, we can deduce that
i = 0 q 1 [ a 1 = 1 p 1 a 2 = 1 p 1 a q 1 = 1 p 1 ( c 1 a 1 + c 2 a 2 + + c q 1 a q 1 + r i a ¯ 1 a ¯ 2 a ¯ q 1 p ) ] 2 = p q 1 i = 0 q 1 j = 0 q 1 h = 0 q 1 χ q i + j ( c 1 c 2 c q 1 r h ) τ q ( χ 2 χ q i ) τ q ( χ 2 ) τ q ( χ 2 χ q j ) τ q ( χ 2 ) = q p q 1 i = 0 q 1 j = 0 q 1 i + j = 0 , q τ q ( χ 2 χ q i ) τ q ( χ 2 ) τ q ( χ ¯ 2 χ ¯ q i ) τ q ( χ 2 ) = q 2 p q 1 .
If ( q , p 1 ) = 1 , then from the method of proving (2.1) and the properties of Gauss sums, we can deduce that
i = 0 q 1 [ a 1 = 1 p 1 a 2 = 1 p 1 a q 1 = 1 p 1 ( c 1 a 1 + c 2 a 2 + + c q 1 a q 1 + r i a ¯ 1 a ¯ 2 a ¯ q 1 p ) ] 2 = i = 0 q 1 p q 1 = q p q 1 .
If q = 2 and p 1 mod 4 , then applying (II) of Lemma 2, we have
( a = 1 p 1 ( a + m a ¯ p ) ) 2 = p | 1 + ( m p ) τ 2 ( χ ¯ 1 ) τ 2 ( χ 1 ) | 2 = 2 p + ( m p ) τ 2 ( χ ¯ 1 ) τ 2 ( χ 1 ) + ( m p ) τ 2 ( χ 1 ) τ 2 ( χ ¯ 1 ) .
(3.5)
Therefore, from (3.5) we can deduce that
( a = 1 p 1 ( a + a ¯ p ) ) 2 + ( a = 1 p 1 ( a + r a ¯ p ) ) 2 = 4 p .

This proves Theorem 3.

To prove the corollary, note that
a = 1 p 1 ( a + m a ¯ p ) = a = 1 p 1 2 ( a + m a ¯ p ) + a = p + 1 2 p 1 ( a + m a ¯ p ) = 2 a = 1 p 1 2 ( a + m a ¯ p ) .
From (3.5) we may immediately deduce the identity
p = ( a = 1 p 1 2 ( a + r a ¯ p ) ) 2 + ( b = 1 p 1 2 ( b + s b ¯ p ) ) 2 ,

where r and s are any two integers such that ( r p ) ( s p ) = 1 .

This completes the proof of our corollary.

Declarations

Acknowledgements

The authors would like to thank the referee for carefully examining this paper and providing a number of important comments. This work is supported by the N.S.F. of P.R. China (11071194, 61202437), and by the Youth Science and Technology Innovation Foundation of Xi’an Shiyou University (2012QN012).

Authors’ Affiliations

(1)
Department of Mathematics, Northwest University
(2)
Department of Mathematics, Xi’an Shiyou University

References

  1. Chowla S: On Kloosterman’s sum. Norske Vid. Selsk. Forh., Trondheim 1967, 40: 70–72.MathSciNetGoogle Scholar
  2. Malyshev AV: A generalization of Kloosterman sums and their estimates. Vestn. Leningr. Univ. 1960, 15: 59–75. (in Russian)Google Scholar
  3. Zhang W: On the fourth power mean of the general Kloosterman sums. Indian J. Pure Appl. Math. 2004, 35: 237–242.MathSciNetGoogle Scholar
  4. Cochrane T, Zheng Z: Bounds for certain exponential sums. Asian J. Math. 2000, 4: 757–774.MathSciNetGoogle Scholar
  5. Cochrane T, Zheng Z: Upper bounds on a two-term exponential sums. Sci. China Ser. A 2001, 44: 1003–1015. 10.1007/BF02878976MathSciNetView ArticleGoogle Scholar
  6. Cochrane T, Zheng Z: Pure and mixed exponential sums. Acta Arith. 1999, 91: 249–278.MathSciNetGoogle Scholar
  7. Cochrane T, Pinner C: A further refinement of Mordell’s bound on exponential sums. Acta Arith. 2005, 116: 35–41. 10.4064/aa116-1-4MathSciNetView ArticleGoogle Scholar
  8. Cochrane T, Pinner C: Using Stepanov’s method for exponential sums involving rational functions. J. Number Theory 2006, 116: 270–292. 10.1016/j.jnt.2005.04.001MathSciNetView ArticleGoogle Scholar
  9. Weil A: On some exponential sums. Proc. Natl. Acad. Sci. USA 1948, 34: 204–207. 10.1073/pnas.34.5.204MathSciNetView ArticleGoogle Scholar
  10. Burgess DA: On Dirichlet characters of polynomials. Proc. Lond. Math. Soc. 1963, 13: 537–548. 10.1112/plms/s3-13.1.537MathSciNetView ArticleGoogle Scholar
  11. Apostol TM: Introduction to Analytic Number Theory. Springer, New York; 1976.Google Scholar
  12. Granville A, Soundararajan K: Large character sums: pretentious characters and the Pólya-Vinogradov theorem. J. Am. Math. Soc. 2007, 20: 357–384. 10.1090/S0894-0347-06-00536-4MathSciNetView ArticleGoogle Scholar
  13. Zhang W, Yao W: A note on the Dirichlet characters of polynomials. Acta Arith. 2004, 115: 225–229. 10.4064/aa115-3-3MathSciNetView ArticleGoogle Scholar
  14. Zhang W, Yi Y: On Dirichlet characters of polynomials. Bull. Lond. Math. Soc. 2002, 34: 469–473. 10.1112/S0024609302001030MathSciNetView ArticleGoogle Scholar
  15. Zhang W, Li H: Elementary Number Theory. Shaanxi Normal University Press, Xi’an; 2008.Google Scholar

Copyright

© Li and Han; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.