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Generalized quadratic Gauss sums and their 2mth power mean

  • Dewang Cui and Wenpeng Zhang EMAIL logo
Published/Copyright: November 2, 2024
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Open Mathematics
From the journal Open Mathematics Volume 22 Issue 1

Abstract

The main purpose of this article is to study the problem of calculating the 2mth power mean of the generalized quadratic Gauss sums, and using the analytic method and an interesting combinatorial identity to give a sharp asymptotic formula for the 2mth power mean. Thus, a new simple proof of this existing result [N. Bag, A. Rojas-León, and W. P. Zhang, On some conjectures on generalized quadratic Gauss sums and related problems, Finite Fields Appl. 86 (2023), 10213] is provided.

Keywords: generalized quadratic Gauss sums; 2mth power mean; analytic method; asymptotic formula
MSC 2010: 11L03; 11L05

1 Introduction

Let q ≥ 2 be an integer. For integer n , and nontrivial Dirichlet character λ modulo q , the generalized 2th Gauss sums G ( n , 2 , λ ; q ) are defined as follows:

G ( n , 2 , λ ; q ) = ∑ s = 1 q λ ( s ) e q ( n s 2 ) ,

where e q ( x ) = exp ( 2 π i x ⁄ q ) with i the imaginary unit (i.e. i 2 = − 1 ).

The objective of the article is to study the generalized quadratic Gauss sums G ( n , 2 , λ ; q ) when k = 2 . Such sums have been studied for long, and numerous results were provided by many mathematicians. Gauss sums and related exponential sums have important applications in number theory, based on the estimation of individual sums and their weighted sums. In [1], Weil proved that if p ≥ 3 is a prime, then

∣ G ( n , 2 , λ ; p ) ∣ ≤ 2 p .

In fact, Cochrane and Zheng [2] generalized this result to any integer. That is, for any integer n with ( n , q ) = 1 , we have

∣ G ( n , 2 , λ ; q ) ∣ ≤ 2 ω ( q ) q ,

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

There are numerous results on power mean values of G ( n , 2 , λ ; q ) . For example, W. P. Zhang [1] proved the identities

1 p − 1 ∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 4 = 3 p 2 − 6 p − 1 + 4 λ 2 ( n ) if 4 ∣ ϕ ( p ) , 3 p 2 − 6 p − 1 if 4 ∤ ϕ ( p )

and

1 p − 1 ∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 6 = 10 p 3 − 25 p 2 − 4 p − 1 , if 4 ∤ ϕ ( p ) ,

where p is an odd prime and n is any integer with ( n , p ) = 1 .

In 2020, Bag and Barman [3] proved that for odd prime p and any n ∈ Z with ( n , p ) = 1 . They have the asymptotic formulae

∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 6 = 10 p 4 + O p 7 2

and

∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 8 = 35 p 5 + O p 9 2 .

Over the next 2 years, Bag et al. [4,5] proved not only the asymptotic formula for the 10th power mean of G ( n , 2 , λ ; p ) , but also the asymptotic formula for any 2kth power mean of G ( n , 2 , λ ; p ) . In particular, they proved the following:

∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 10 = 126 p 6 + O p 11 2

and

(1) ∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 2 m = 2 m − 1 m p m + 1 + O p 2 m + 1 2 .

Of course, there are a number of other forms of the higher power mean of G ( n , k , λ ; q ) , interested readers can refer to [6–9].

In this article, as a note of [5], we are using the elementary and analytic methods, and an interesting combinatorial identity to provide a new and simple proof for the asymptotic formula (1). That is, we prove the following conclusion:

Theorem 1

Let p be an odd prime. Then for any positive integer m, we have the asymptotic formula:

∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 2 m = 2 m − 1 m p m + 1 + O p m + 1 2 .

2 Several lemmas

In order to make the structure of the article concise and complete, we need to introduce three simple lemmas in this section. The proof of these lemmas requires some knowledge of elementary and analytic number theory, all of which can be found in [10] and [11], so we will not repeat them here.

Lemma 1

Let p be an odd prime. Then for any integers c and k ≥ 1 , we have the estimates

∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ( τ ( λ ) τ ( λ λ 2 ) ¯ ) k ≪ p k + 1 2 ,

where τ ( λ ) denotes the classical Gauss sums, and λ 2 is the second-order Dirichlet character modulo p.

Proof

Without loss of generality we assume that ( c , p ) = 1 . Note that ∣ τ ( λ 2 ) ∣ = p . From the definition of the classical Gauss sums τ ( λ ) and the orthogonality of the characters modulo p (see Apostol [10], Theorem 6.16)

∑ λ mod p λ ( − 1 ) = − 1 λ ( s ) = p − 1 2 if s ≡ 1 mod p , − p − 1 2 if s ≡ − 1 mod p , 0 otherwise

we have

(2) ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) τ ( λ ) τ ( λ λ 2 ) ¯ = ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ∑ s = 1 p − 1 λ ( s ) ∑ t = 1 p − 1 λ ¯ ( t ) λ 2 ( t ) e p ( s − t ) = ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ∑ s = 1 p − 1 λ ( s ) ∑ t = 1 p − 1 λ 2 ( t ) e p ( t ( s − 1 ) ) = τ ( λ 2 ) ∑ s = 1 p − 1 ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) λ ( s ) λ 2 ( s − 1 ) = τ ( λ 2 ) ∑ s = 1 p − 1 c s ≡ 1 mod p λ 2 ( s − 1 ) p − 1 2 − τ ( λ 2 ) ∑ s = 1 p − 1 c s ≡ − 1 mod p λ 2 ( s − 1 ) p − 1 2 = p − 1 2 p ∣ λ 2 ( c ¯ − 1 ) − λ 2 ( − c ¯ − 1 ) ∣ ≪ p 3 2 .

If k = 2 , then we have

(3) ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ( τ ( λ ) τ ( λ λ 2 ) ¯ ) 2 = ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ∑ s = 1 p − 1 λ ( s ) ∑ t = 1 p − 1 λ ¯ ( t ) λ 2 ( t ) e p ( s − t ) 2 = τ 2 ( λ 2 ) ∑ s = 1 p − 1 ∑ t = 1 p − 1 ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) λ ( s ) λ ( t ) λ 2 ( s − 1 ) λ 2 ( t − 1 ) = p p − 1 2 ∑ s = 1 p − 1 ∑ t = 1 p − 1 s t c ≡ 1 mod p λ 2 ( s − 1 ) λ 2 ( t − 1 ) − ∑ s = 1 p − 1 ∑ t = 1 p − 1 s t c ≡ − 1 mod p λ 2 ( s − 1 ) λ 2 ( t − 1 ) = p 2 − p 2 ∑ s = 1 p − 1 λ 2 ( s − 1 ) λ 2 ( s c ¯ − 1 ) − ∑ s = 1 p − 1 λ 2 ( s − 1 ) λ 2 ( − s c ¯ − 1 ) ≪ p 2 + 1 2 .

Assuming that the estimate holds for 2 ≤ r ≤ k . That is,

(4) ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ( τ ( λ ) τ ( λ λ 2 ) ¯ ) r ≪ p r + 1 2 .

Then for r = k + 1 , from the method of proving (3) and the induction hypothesis (4) we have the estimate

(5) ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ( τ ( λ ) τ ( λ λ 2 ) ¯ ) k + 1 = ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) ∑ s = 1 p − 1 λ ( s ) ∑ t = 1 p − 1 λ ¯ ( t ) λ 2 ( t ) e p ( s − t ) k + 1 = τ k + 1 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k + 1 = 1 p − 1 ∑ λ mod p λ ( − 1 ) = − 1 λ ( c ) λ ( s 1 ) … λ ( s k + 1 ) λ 2 ( s 1 − 1 ) … λ 2 ( s k + 1 − 1 ) = p − 1 2 τ k + 1 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k + 1 = 1 p − 1 c s 1 … s k + 1 ≡ 1 mod p λ 2 ( s 1 − 1 ) … λ 2 ( s k + 1 − 1 ) − τ k + 1 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k + 1 = 1 p − 1 c s 1 … s k + 1 ≡ − 1 mod p λ 2 ( s 1 − 1 ) … λ 2 ( s k + 1 − 1 ) = p − 1 2 τ k + 1 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k = 1 p − 1 λ 2 ( s 1 − 1 ) … λ 2 ( s k − 1 ) λ 2 ( s 1 … s k c ¯ − 1 ) − τ k + 1 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k = 1 p − 1 λ 2 ( s 1 − 1 ) … λ 2 ( s k − 1 ) λ 2 ( − s 1 … s k c ¯ − 1 ) = p − 1 2 τ k + 1 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k = 1 p − 1 λ 2 ( s 1 − 1 ) … λ 2 ( s k − 2 − 1 ) λ 2 ( s k − 1 − 1 ) × λ 2 ( s k s k − 1 s k − 2 ¯ − 1 ) λ 2 ( s 1 … s k − 3 s k c ¯ − 1 ) − τ k + 1 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k = 1 p − 1 λ 2 ( s 1 − 1 ) … λ 2 ( s k − 2 − 1 ) λ 2 ( s k − 1 − 1 ) × λ 2 ( s k s k − 1 s k − 2 ¯ − 1 ) λ 2 ( − s 1 … s k − 3 s k c ¯ − 1 ) ∣ ≤ p − 1 2 τ 3 ( λ 2 ) ∑ s k − 1 = 1 p − 1 λ 2 ( s k − 1 − 1 ) ∑ s k − 2 = 1 p − 1 λ 2 ( s k − 2 − 1 ) ∑ s k = 1 p − 1 λ 2 ( s k s k − 1 s k − 2 ¯ − 1 ) × τ k − 2 ( λ 2 ) ∑ s 1 = 1 p − 1 … ∑ s k − 3 = 1 p − 1 λ 2 ( s 1 − 1 ) … λ 2 ( s k − 3 − 1 ) λ 2 ( s 1 … s k − 3 s k c ¯ − 1 ) − ∑ s 1 = 1 p − 1 … ∑ s k − 3 = 1 p − 1 λ 2 ( s 1 − 1 ) … λ 2 ( s k − 3 − 1 ) λ 2 ( − s 1 … s k − 3 s k c ¯ − 1 ) ≤ p 3 2 ∑ s k − 1 = 1 p − 1 λ 2 ( s k − 1 − 1 ) ∑ s k − 2 = 1 p − 1 λ 2 ( s k − 2 − 1 ) ∑ s k = 1 p − 1 λ 2 ( s k s k − 1 s k − 2 ¯ − 1 ) × ∑ λ mod p λ ( − 1 ) = − 1 λ ( c s k ) ( τ ( λ ) τ ( λ λ 2 ) ¯ ) k − 2 ≪ p 3 2 + 3 2 + k − 2 + 1 2 = p k + 1 + 1 2 ,

where we have used the fact that for any integral coefficient polynomial f ( x ) , if it is not the square of any polynomial, then we have the estimate (see Weil [1])

∑ s = 1 p − 1 λ 2 ( f ( s ) ) ≪ p .

Now Lemma 1 follows from (2), (3), (5) and the induction.□

Lemma 2

For any positive integer m, we have the identity

2 m − 1 m = 1 2 ∑ r = 0 m 2 ∣ ( m − r ) 2 r m r m − r m − r 2 .

Proof

First for any positive l and real number x , from Gould [12] (Combinatorial identities (3.63)) we have

(6) ∑ k = 0 l 2 ( − 1 ) k x k 2 x − 2 k l − 2 k = x l 2 l .

Applying (6) we may immediately deduce that

∑ r = 0 m 2 ∣ ( m − r ) 2 r m r m − r m − r 2 = ∑ m − 2 k = 0 m 2 m − 2 k m m − 2 k m − ( m − 2 k ) m − ( m − 2 k ) 2 = ∑ k = 0 m 2 m m − 2 k 2 m − 2 k 2 k k = 2 m ∑ k = 0 m 2 m m − 2 k ( − 1 ) k − 1 2 k = ( − 2 ) m ∑ k = 0 m 2 − 1 2 k ( − 1 ) k − 1 − 2 k m − 2 k = ( − 2 ) m − 1 2 m 2 m = 2 m m = 2 2 m − 1 m ,

then

2 m − 1 m = 1 2 ∑ r = 0 m 2 ∣ ( m − r ) 2 r m r m − r m − r 2 .

This proves Lemma 2.□

Lemma 3

Let p be an odd prime. Then for any positive integer m, we have

∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) + λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m = p − 1 2 m m ⁄ 2 p m + O p m + 1 2 i f 2 ∣ m , O p m + 1 2 i f 2 ∤ m .

Proof

Note that τ ( λ ) ¯ τ ( λ λ 2 ) τ ( λ ) τ ( λ λ 2 ) ¯ = p 2 , unless λ = λ 2 , in this case, its value is p . If m is an odd number, then from Lemma 1 we have

(7) ∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) + λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m = ∑ r = 0 m m r ∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) ) r ( λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m − r ≪ ∑ r = 0 m 2 m r p 2 r ∣ λ 2 ( n ) ∣ m − 2 r ∑ λ mod p λ ( − 1 ) = − 1 ( τ ( λ ) τ ( λ λ 2 ) ¯ ) m − 2 r ≪ ∑ r = 0 m 2 m r p 2 r p m − 2 r + 1 2 ≪ p m + 1 2 .

If m is an even number, then note that the special case r = m 2 , from Lemma 1 and the method of proving (7) we have

(8) ∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) + λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m = ∑ r = 0 m m r ∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) ) r ( λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m − r = p − 1 2 m m 2 p m + ∑ r = 0 m r ≠ m 2 m r ∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) ) r ( λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m − r = p − 1 2 m m 2 p m + O p m + 1 2 .

Now Lemma 3 follows from (7) and (8).□

3 Proof of the theorem

It is clear that if λ is an odd character modulo p , then we have

∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) = ∑ s = 1 p − 1 λ ( − s ) e p ( n ( − s ) 2 ) = − ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) = 0 .

So we only consider all even characters modulo p . Note that

(9) ∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 2 m = ∑ λ mod p λ ( − 1 ) = − 1 ∑ s = 1 p − 1 λ 2 ( s ) e p ( n s 2 ) 2 m .

Then from (9), Lemma 2, Lemma 3 and the properties of the classical Gauss sums we have the asymptotic formula

∑ λ mod p ∑ s = 1 p − 1 λ ( s ) e p ( n s 2 ) 2 m = ∑ λ mod p λ ( − 1 ) = − 1 ∑ s = 1 p − 1 λ 2 ( s ) e p ( n s 2 ) 2 m = ∑ λ mod p λ ( − 1 ) = − 1 ∑ s = 1 p − 1 λ ( s ) ( 1 + λ 2 ( s ) ) e p ( n s ) 2 m = ∑ λ mod p λ ( − 1 ) = − 1 ∑ s = 1 p − 1 λ ( s ) e p ( n s ) + ∑ t = 1 p − 1 λ ( t ) λ 2 ( t ) e p ( n t ) 2 m = ∑ λ mod p λ ( − 1 ) = − 1 ∣ λ ( n ) ¯ τ ( λ ) + λ λ 2 ( n ) ¯ τ ( λ λ 2 ) ∣ 2 m = ∑ λ mod p λ ( − 1 ) = − 1 ( ( λ ( n ) ¯ τ ( λ ) + λ λ 2 ( n ) ¯ τ ( λ λ 2 ) ) ( λ ( n ) τ ( λ ) ¯ + λ λ 2 ( n ) τ ( λ λ 2 ) ¯ ) ) m = ∑ λ mod p λ ( − 1 ) = − 1 ( 2 p + λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) + λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m + O ( p m ) = ∑ r = 0 m m r ( 2 p ) r ∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) + λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m − r + O ( p m ) = ∑ r = 0 m 2 ∣ ( m − r ) m r ( 2 p ) r ∑ λ mod p λ ( − 1 ) = − 1 ( λ 2 ( n ) ¯ τ ( λ ) ¯ τ ( λ λ 2 ) + λ 2 ( n ) τ ( λ ) τ ( λ λ 2 ) ¯ ) m − r + O p m + 1 2 = p − 1 2 p m ∑ r = 0 m 2 ∣ ( m − r ) m r 2 r m − r m − r 2 + O p m + 1 2 = p m + 1 2 m − 1 m + O p m + 1 2 ,

where O ( p m ) may come from p ≡ 3 mod 4 and λ = λ 2 .

This proves our theorem.

Acknowledgments

The authors would like to thank the referees for their very helpful and detailed comments.

  1. Funding information: This work was supported by the Natural Science Foundation of China (12126357).

  2. Author contributions: All authors have equally contributed to this work. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors declare that there are no conflicts of interest regarding the publication of this article.

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Received: 2024-04-03
Revised: 2024-09-12
Accepted: 2024-09-17
Published Online: 2024-11-02

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  114. On the existence of tripartite graphs and n-partite graphs
  115. Classifying pentavalent symmetric graphs of order 12pq
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  129. Erratum
  130. Erratum to “Infinitesimals via Cauchy sequences: Refining the classical equivalence”
  131. Corrigendum
  132. Corrigendum to “Matrix stretching”
  133. Corrigendum to “A comprehensive review of the recent numerical methods for solving FPDEs”
https://doi.org/10.1515/math-2024-0074
Keywords for this article
generalized quadratic Gauss sums; 2mth power mean; analytic method; asymptotic formula
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