Skip to main content

Retracted Article: On the fourth power mean of the two-term exponential sums

This article was retracted on 03 September 2014

Abstract

Abstract

The main purpose of this paper is using the analytic methods and the properties of Gauss sums to study the computational problem of one kind of fourth power mean of two-term exponential sums, and to give an interesting identity and asymptotic formula for it.

MSC: 11L05.

1 Introduction

Let q3 be a positive integer. For any integers m and n, the two-term exponential sum C(m,n,k;q) is defined as follows:

C(m,n,k;q)= a = 1 q e ( m a k + n a q ) ,

where e(y)= e 2 π i y .

About the various properties of C(m,n,k;q), some authors had studied it, and they obtained a series of results, some related works can be found in references [15]. For example, Gauss’ classical work proved the remarkable formula (see [6])

C(1,0,2;q)= 1 2 q (1+i) ( 1 + e ( q / 4 ) ) = { q , if  q 1 mod 4 , 0 , if  q 2 mod 4 , i q , if  q 3 mod 4 , ( 1 + i ) q , if  q 0 mod 4 ,

where i 2 =1.

Generally, for any odd number q and (2m,q)=1, the exact value of |C(m,n,2;q)| is (e.g. see Berndt, Evans, Williams and Apostol’s related works). Cochrane and Zheng [2] show for the general sum that

|C(m,n,k;q)| k ω ( q ) q 1 2 ,

where ω(q) denotes the number of all distinct prime divisors of q.

In this paper, we study the fourth power mean of the two-term exponential sum C(m,n,k;q) as follows:

m = 1 q |C(m,n,k;q) | 4 ,
(1.1)

where n is any integer with (n,q)=1.

As regards this problem, it seems that none has yet studied it, at least we have not seen any related result before. The problem is interesting, because it can reflect that the mean value of C(m,n,k;q) is well behaved. The main purpose of this paper is to show this point. That is, we shall prove the following conclusion.

Theorem Let p>3 be a prime. Then for any integer n with (n,p)=1, we have the identity

m = 1 p | a = 1 p 1 e ( m a 3 + n a p ) | 4 = { 2 p 3 3 p 2 3 p , if  3 p 1 , 2 p 3 5 p 2 15 p + 4 τ 3 ( χ 1 ) + 4 τ 3 ( χ ¯ 1 ) , if  3 | p 1 ,

where χ 1 is any 3-order character modp.

Note that for any non-principal character χmodp, we have |τ(χ)|= p , so from our theorem we may immediately deduce the following.

Corollary Let p>3 be a prime with 3|p1. Then for any integer n with (n,p)=1, we have the asymptotic formula

m = 1 p | a = 1 p 1 e ( m a 3 + n a p ) | 4 =2 p 3 5 p 2 +O ( p 3 2 ) .

It seems that our method can also be used to deal with (1.1) for all prime p and integer k4. But this time, the computing is very complex.

For any integer h3, whether there exists an exact computational formula for

m = 1 p | a = 1 p 1 e ( m a 3 + n a p ) | 2 h ,

where p is an odd prime and (n,p)=1, is an open problem.

2 Several lemmas

In this section, we will give several lemmas which are necessary in the proof of our theorem. In the proving process of all lemmas, we used many properties of Gauss sums; all these can be found in [6], we will not be repeat them here. First we have the following.

Lemma 1 Let p be an odd prime, χ be any non-principal character modp. Then for any integer n with (n,p)=1, we have the identity

| m = 1 p 1 χ ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 | = { p | a = 1 p 2 χ ¯ ( 3 a 2 + 3 a + 1 ) | , if  χ  is not a  3 -order character  mod p , p | 2 + a = 1 p 1 χ ¯ ( a ( 1 a ) ) | , if  χ  is a  3 -order character  mod p .

Proof Note that χ is a non-principal character modp, so if χ is not a 3-order character modp (that is, χ 3 χ 0 , the principal character modp), then from the properties of Gauss sums we have

m = 1 p 1 χ ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 = a = 1 p 1 b = 1 p 1 m = 1 p 1 χ ( m ) e ( m ( a 3 b 3 ) + n ( a b ) p ) = a = 1 p 1 b = 1 p 1 m = 1 p 1 χ ( m ) e ( m b 3 ( a 3 1 ) + n b ( a 1 ) p ) = τ ( χ ) a = 1 p 1 χ ¯ ( a 3 1 ) b = 1 p 1 χ ¯ 3 ( b ) e ( n b ( a 1 ) p ) = τ ( χ ) τ ( χ ¯ 3 ) a = 1 p 1 χ ¯ ( a 3 1 ) χ 3 ( n ( a 1 ) ) = χ 3 ( n ) τ ( χ ) τ ( χ ¯ 3 ) a = 1 p 2 χ ¯ ( ( a + 1 ) 3 1 ) χ ( a 3 ) = χ 3 ( n ) τ ( χ ) τ ( χ ¯ 3 ) a = 1 p 2 χ ¯ ( ( a ¯ + 1 ) 3 a ¯ 3 ) = χ 3 ( n ) τ ( χ ) τ ( χ ¯ 3 ) a = 1 p 2 χ ¯ ( 3 a 2 + 3 a + 1 ) ,
(2.1)

where τ(χ)= a = 1 p 1 χ(a)e( a p ) denotes the classical Gauss sum.

If χ is a 3-order character modp, then χ 3 = χ 0 ; note that for any integer a with (a,p)=1, we have

χ 2 (a)+χ(a)+1= { 3 , if  a  is a third residue  mod p , 0 , otherwise .

From the method of proving (2.1) we have the identity

m = 1 p 1 χ ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 = τ ( χ ) a = 1 p 1 χ ¯ ( a 3 1 ) b = 1 p 1 χ ¯ 3 ( b ) e ( n b ( a 1 ) p ) = τ ( χ ) a = 1 p 1 χ ¯ ( a 3 1 ) b = 1 p 1 e ( n b ( a 1 ) p ) = τ ( χ ) a = 1 p 1 χ ¯ ( a 3 1 ) = τ ( χ ) a = 1 p 1 ( χ 2 ( a ) + χ ( a ) + 1 ) χ ¯ ( a 1 ) = τ ( χ ) ( a = 1 p 1 χ ¯ ( a 1 ) + a = 1 p 1 χ ¯ ( a ¯ ( 1 a ¯ ) ) + a = 1 p 1 χ ¯ ( 1 a ¯ ) ) = τ ( χ ) ( 2 + a = 1 p 1 χ ¯ ( a ( 1 a ) ) ) .
(2.2)

Now note that |τ(χ)|= p , if χ χ 0 . From (2.1) and (2.2) we may immediately deduce Lemma 1. □

Lemma 2 Let p be an odd prime, χ be any non-real character modp. Then we have the identity

a = 1 p 1 χ ( a ( a 1 ) ) = τ 2 ( χ ) τ ( χ 2 ) .

Therefore,

| a = 1 p 1 χ ( a ( a 1 ) ) |= p .

Proof From the definition and properties of the classical Gauss sums we have

τ 2 ( χ ) = a = 1 p 1 b = 1 p 1 χ ( a ) χ ( b ) e ( a + b p ) = a = 1 p 1 χ ( a ) b = 1 p 1 χ 2 ( b ) e ( b ( a + 1 ) p ) = τ ( χ 2 ) a = 1 p 1 χ ( a ) χ ¯ ( ( a + 1 ) 2 ) = τ ( χ 2 ) a = 2 p χ ( a 1 ) χ ¯ ( a 2 ) = τ ( χ 2 ) a = 2 p 1 χ ( ( 1 a ¯ ) a ¯ ) = τ ( χ 2 ) a = 1 p 1 χ ( a ( 1 a ) )

or

a = 1 p 1 χ ( a ( a 1 ) ) = τ 2 ( χ ) τ ( χ 2 ) .

This proves Lemma 2. □

3 Proof of the theorem

In this section, we shall complete the proof of our theorem. First from the orthogonality of characters modp we have

χ mod p | m = 1 p 1 χ(m)| a = 1 p 1 e ( m a 3 + n a p ) | 2 | 2 =(p1) m = 1 p 1 | a = 1 p 1 e ( m a 3 + n a p ) | 4 .
(3.1)

On the other hand, if 3p1, then any non-principal character χ is not a 3-order character modp. Note that

m = 1 p 1 χ 0 ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 = a = 1 p 1 b = 1 p 1 m = 1 p 1 e ( m b 3 ( a 3 1 ) + n b ( a 1 ) p ) = ( p 1 ) 2 + p 2 = p 2 p 1 .
(3.2)

From (3.2) and Lemma 1 we have

χ mod p | m = 1 p 1 χ ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 | 2 = | m = 1 p 1 χ 0 ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 | 2 + χ mod p χ χ 0 | m = 1 p 1 χ ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 | 2 = ( p 2 p 1 ) 2 + χ mod p χ χ 0 p 2 | a = 1 p 2 χ ¯ ( 3 a 2 + 3 a + 1 ) | 2 = ( p 2 p 1 ) 2 + p 2 χ mod p | a = 1 p 2 χ ¯ ( 3 a 2 + 3 a + 1 ) | 2 p 2 ( a = 1 p 2 χ 0 ( 3 a 2 + 3 a + 1 ) ) 2 = ( p 2 p 1 ) 2 + p 2 ( p 1 ) a = 1 p 2 b = 1 p 2 a 2 + a b 2 + b mod p 1 p 2 ( p 2 ) 2 = ( p 2 p 1 ) 2 + p 2 ( p 1 ) a = 1 p 2 b = 1 p 2 ( a b ) ( a + b + 1 ) 0 mod p 1 p 2 ( p 2 ) 2 = ( p 2 p 1 ) 2 + p 2 ( p 1 ) ( p 2 + p 2 1 ) p 2 ( p 2 ) 2 = ( p 1 ) ( 2 p 3 3 p 2 3 p 1 ) .
(3.3)

If 3p1, then combining (3.1) and (3.3) we may immediately deduce the identity

m = 1 p 1 | a = 1 p 1 e ( m a 3 + n a p ) | 4 =2 p 3 3 p 2 3p1

or

m = 1 p | a = 1 p 1 e ( m a 3 + n a p ) | 4 =2 p 3 3 p 2 3p.
(3.4)

If 3|p1, let χ 1 χ 0 be a 3-order character modp, then χ ¯ 1 = χ 1 2 is also a 3-order character modp; this time note that

m = 1 p 1 χ 0 (m)| a = 1 p 1 e ( m a 3 + n a p ) | 2 =p(p1)2p1= p 2 3p1

and

a = 1 p 2 χ 1 ( ( a + 1 ) 3 a 3 ) = a = 1 p 2 χ 1 ( ( a ¯ + 1 ) 3 1 ) = a = 1 p 1 χ 1 ( a 3 1 ) = a = 1 p 1 ( χ ¯ 1 2 ( a ) + χ ¯ 1 ( a ) + 1 ) χ 1 ( a 1 ) = 2 + a = 1 p 1 χ 1 ( a ( 1 a ) ) ,

and from Lemma 2 and the method of proving (3.3) we have

χ mod p | m = 1 p 1 χ ( m ) | a = 1 p 1 e ( m a 3 + n a p ) | 2 | 2 = ( p 2 3 p 1 ) 2 + χ mod p χ χ 0 , χ 1 , χ 1 2 p 2 | a = 1 p 2 χ ¯ ( 3 a 2 + 3 a + 1 ) | 2 + 2 p | 2 + a = 1 p 1 χ 1 ( a ( 1 a ) ) | 2 = ( p 2 3 p 1 ) 2 + p 2 ( p 1 ) ( 2 p 5 ) 2 p 2 | a = 1 p 2 χ 1 ( ( a + 1 ) 3 a 3 ) | 2 + 2 p ( p + 4 2 a = 1 p 1 χ 1 ( a ( 1 a ) ) 2 a = 1 p 1 χ ¯ 1 ( a ( 1 a ) ) ) p 2 ( p 4 ) 2 = ( p 2 3 p 1 ) 2 + p 2 ( p 1 ) ( 2 p 5 ) p 2 ( p 4 ) 2 ( 2 p 2 2 p ) ( p + 4 2 a = 1 p 1 χ 1 ( a ( 1 a ) ) 2 a = 1 p 1 χ ¯ 1 ( a ( 1 a ) ) ) = ( p 1 ) ( 2 p 3 5 p 2 15 p 1 ) + 4 ( p 1 ) ( τ 3 ( χ 1 ) + τ 3 ( χ ¯ 1 ) ) .
(3.5)

So if 3|p1, then combining (3.1) and (3.5) we can deduce the asymptotic formula

m = 1 p | a = 1 p 1 e ( m a 3 + n a p ) | 4 = 1 + m = 1 p 1 | a = 1 p 1 e ( m a 3 + n a p ) | 4 = 2 p 3 5 p 2 15 p + 4 ( τ 3 ( χ 1 ) + τ 3 ( χ ¯ 1 ) ) .
(3.6)

Now our theorem follows from (3.4) and (3.6).

References

  1. Cochrane T, Zheng Z: Pure and mixed exponential sums. Acta Arith. 1999, 91: 249–278.

    MathSciNet  MATH  Google Scholar 

  2. Cochrane T, Zheng Z: Bounds for certain exponential sums. Asian J. Math. 2000, 4: 757–774.

    MathSciNet  Article  MATH  Google Scholar 

  3. Cochrane T, Zheng Z: Upper bounds on a two-term exponential sums. Sci. China Ser. A 2001, 44: 1003–1015. 10.1007/BF02878976

    MathSciNet  Article  MATH  Google Scholar 

  4. Cochrane T, Pinner C: A further refinement of Mordell’s bound on exponential sums. Acta Arith. 2005, 116: 35–41. 10.4064/aa116-1-4

    MathSciNet  Article  MATH  Google Scholar 

  5. 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.2005.04.001

    MathSciNet  Article  MATH  Google Scholar 

  6. Apostol TM: Introduction to Analytic Number Theory. Springer, New York; 1976.

    MATH  Google Scholar 

Download references

Acknowledgements

The authors would like to thank the referee for his very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P.S.F. (2013JZ001) and N.S.F. (11371291) of P.R. China, and the SF of Education of Shaanxi province (12JK0877).

Author information

Affiliations

Authors

Corresponding author

Correspondence to Di Han.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The main idea of this paper was proposed by MZ and DH. All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

An erratum to this article can be found at http://dx.doi.org/10.1186/1029-242X-2014-344.

Following publication of this article , it was brought to the attention of the editorial and publishing staff that an identically-titled article by Han Zhang and Wenpeng Zhang was published in January 2014 in The Scientific World Journal.

This is a violation of publication ethics which, in accordance with the Springer Policy on Publishing Integrity, warrants a retraction of the article and a notice to this effect to be published in the journal.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (https://creativecommons.org/licenses/by/4.0), which permits use, duplication, adaptation, distribution, and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Reprints and Permissions

About this article

Cite this article

Zhu, M., Han, D. Retracted Article: On the fourth power mean of the two-term exponential sums. J Inequal Appl 2014, 116 (2014). https://doi.org/10.1186/1029-242X-2014-116

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2014-116

Keywords

  • the two-term exponential sums
  • fourth power mean
  • Gauss sums
  • identity
  • asymptotic formula