# A hybrid mean value involving a new sum and Kloosterman sums

## Abstract

In this paper, we introduce a new sum, analogous to Cochrane sums, and use elementary and analytic methods to study the hybrid mean value problem involving this sum and Kloosterman sums, and we give an interesting asymptotic formula for it.

MSC:11L40, 11F20.

## 1 Introduction

Let q be a natural number and h an integer with $(h,q)=1$. The Cochrane sum $C(h,q)$ is defined by

$C(h,q)= ∑ ′ a = 1 q ( ( a ¯ q ) ) ( ( a h q ) ) ,$

where

$a ¯$ is defined by $a a ¯ ≡1modq$, and $∑ ′ a = 1 q$ denotes the summation over all $1≤a≤q$ such that $(a,q)=1$.

This sum was introduced by Professor Todd Cochrane, and it has been studied by many authors, and one obtained many interesting results. Related works can be found in  and . For example, Zhang  studied the hybrid mean value properties of Cochrane sums and Kloosterman sums and proved that for any prime $p>3$, we have the asymptotic formula

$∑ h = 1 p − 1 K(h,1;p)C(h,p)= − 1 2 π 2 p 2 +O ( p ⋅ exp ( 3 ln p ln ln p ) ) ,$

where $exp(y)= e y$,

$K(m,n;q)= ∑ ′ a = 1 q e ( m a + n a ¯ q )$

denotes the Kloosterman sum, and $e(y)= e 2 π i y$.

Now for any odd prime $p>1$ and integer h with $(h,q)=1$, we define another sum analogous to the Cochrane sum as follows:

$C 1 (h,p)= ∑ a = 1 p − 1 2 ( ( a ¯ 2 p ) ) ( ( a 2 h p ) ) .$

The main purpose of this paper is using elementary and analytic methods to study the hybrid mean value problem involving $C 1 (h,p)$ and the Kloosterman sum and give an asymptotic formula for it. That is, we shall prove the following.

Theorem Let p be an odd prime. Then we have the asymptotic formula

$∑ h = 1 p − 1 K(h,1;p) C 1 (h,p)= − 1 4 π 2 ⋅ p 2 +O ( p 3 / 2 ⋅ exp ( 3 ln p ln ln p ) ) .$

For general odd number $q>2$, whether there exists an asymptotic formula for

$∑ ′ h = 1 q K(h,1;q) C 1 (h,q)$

is still an open problem.

## 2 Several lemmas

In this section, we shall give several lemmas, which are necessary in the proof of our theorem. Hereinafter, we shall use many properties of Gauss sums; all of these can be found in reference , so they will not be repeated here. First we have the following.

Lemma 1 Let p be an odd prime. Then for any integer a with $(a,p)=1$, we have the identity

$C(a,p)= − 1 π 2 ( p − 1 ) ∑ χ mod p χ ( − 1 ) = − 1 χ ¯ (a) τ 2 (χ) L 2 (1, χ ¯ ),$

where χ runs through the Dirichlet characters modp with $χ(−1)=−1$, and $τ(χ)= ∑ a = 1 p − 1 χ(a)e( a p )$ denotes the classical Gauss sum corresponding to χ.

Proof See reference . □

Lemma 2 Let p be an odd prime and h an integer with $(h,p)=1$. We define $C 2 (h,p)$ as follows:

$C 2 (h,p)= ∑ a = 1 p − 1 ( a p ) ( ( a ¯ p ) ) ( ( a h p ) ) .$

Then we have the identity

where $χ 2 =( ∗ p )$ denotes the Legendre symbol.

Proof From the orthogonality relation for characters modp we have

$C 2 ( h , p ) = 1 p − 1 ∑ a = 1 p − 1 ∑ b = 1 p − 1 ( a p ) ∑ χ mod p χ ( a b ) ( ( a p ) ) ( ( h b p ) ) = 1 p − 1 ∑ χ mod p ( ∑ a = 1 p − 1 ( a p ) χ ( a ) ( ( a p ) ) ) ( ∑ b = 1 p − 1 χ ( b ) ( ( h b p ) ) ) .$
(1)

If $χ= χ 0$ is the principal character modp, then we have

$∑ b = 1 p − 1 χ(b) ( ( h b p ) ) = ∑ b = 1 p − 1 ( ( b p ) ) = ∑ b = 1 p − 1 ( b p − 1 2 ) =0.$
(2)

If χ is an even character modp (that is, χ is a non-principal character and $χ(−1)=1$), then we have

$∑ b = 1 p − 1 χ ( b ) ( ( h b p ) ) = χ ¯ ( h ) ∑ b = 1 p − 1 χ ( b ) ( b p − 1 2 ) = χ ¯ ( h ) 1 p ( ∑ b = 1 p − 1 χ ( b ) b − 1 2 ∑ b = 1 p − 1 χ ( b ) ) = χ ¯ ( h ) 1 p ∑ b = 1 p − 1 χ ( b ) b .$
(3)

Since

$∑ b = 1 p − 1 χ ( b ) b = ∑ b = 1 p − 1 χ ( p − b ) ( p − b ) = ∑ b = 1 p − 1 χ ( b ) ( p − b ) = p ∑ b = 1 p − 1 χ ( b ) − ∑ b = 1 p − 1 χ ( b ) b = − ∑ b = 1 p − 1 χ ( b ) b ,$

so that

$∑ b = 1 p − 1 χ(b)b=0.$

From this identity, (2), and (3) we know that if χ is an even character modp, then

$∑ b = 1 p − 1 χ(b) ( ( h b p ) ) =0.$
(4)

If χ is an odd character modp and $p≡3mod4$. Let $χ 2 =( ∗ p )$ denote the Legendre symbol, then $χ χ 2$ must be an even character modp, so we have

$∑ a = 1 p − 1 ( a p ) χ(a) ( ( a p ) ) =0.$
(5)

If χ is an odd character modp and $p≡1mod4$, then $χ χ 2$ must be an odd character modp, so we have

$∑ a = 1 p − 1 ( a p ) χ ( a ) ( ( a p ) ) = 1 p ∑ a = 1 p − 1 ( a p ) χ ( a ) a = i π τ ( χ χ 2 ) L ( 1 , χ ¯ χ 2 )$
(6)

and

$∑ b = 1 p − 1 χ(b) ( ( h b p ) ) = χ ¯ ( h ) p ∑ b = 1 p − 1 χ(b)b= i π χ ¯ (h)τ(χ)L(1, χ ¯ ).$
(7)

Combining (1), (2), (4)-(7) we may immediately deduce the identity

This proves Lemma 2. □

Lemma 3 Let $p>3$ be an odd prime, then we have the asymptotic formula

$∑ χ mod p χ ( − 1 ) = − 1 L 2 (1,χ)= 1 2 p+O ( exp ( 3 ln p ln ln p ) ) .$

Proof For any non-principal character $χmodp$, applying Abel’s identity (see Theorem 4.2 of ) we have

$L 2 (1,χ)= ∑ n = 1 p 3 χ ( n ) d ( n ) n + ∫ p 3 ∞ A ( y , χ ) y 2 dy,$
(8)

where $d(n)$ denotes the divisor function, and $A(y,χ)= ∑ p 3 < n ≤ y χ(n)d(n)$.

From  we know that for any real number $y> p 3$, we have the estimate

$∑ χ mod p χ ( − 1 ) = − 1 |A(y,χ) | 2 ≪y ϕ 2 (p).$
(9)

From (8) and (9) we can deduce that

$∑ χ mod p χ ( − 1 ) = − 1 L 2 ( 1 , χ ) = 1 2 ∑ n = 1 p 3 d ( n ) n ∑ χ mod p χ ( − 1 ) = − 1 χ ( n ) + O ( ∑ χ mod p χ ( − 1 ) = − 1 ∫ p 3 ∞ | A ( χ , y ) | y 2 d y ) = p − 1 2 ∑ n = 1 p 3 n ≡ 1 mod p d ( n ) n − p − 1 2 ∑ n = 1 p 3 n ≡ − 1 mod p d ( n ) n + O ( 1 ) = 1 2 p + O ( ∑ l = 1 p 2 d ( p l ± 1 ) l ) = 1 2 p + O ( exp ( 3 ln p ln ln p ) ) ,$

where we have used the estimate $d(n)≪exp( ( 1 + ϵ ) ln n ln ln n )$, $ϵ>0$ is any fixed real number. This proves Lemma 3. □

Lemma 4 Let $p>2$ be a prime with $p≡1mod4$. Then we have the estimates

$∑ a = 1 p − 1 | ∑ χ mod p χ ( − 1 ) = − 1 χ(a)L(1, χ ¯ )L(1, χ ¯ χ 2 )|=O ( p ⋅ exp ( 3 ln p ln ln p ) ) .$

Proof From the method of proving Lemma 3 we have

$∑ χ mod p χ ( − 1 ) = − 1 χ ( a ) L ( 1 , χ ¯ ) L ( 1 , χ ¯ χ 2 ) = 1 2 ∑ n = 1 p 3 ∑ d | n χ 2 ( d ) n ∑ χ mod p ( χ ( a n ¯ ) − χ ( − a n ¯ ) ) + O ( 1 ) = O ( d ( a ) a ⋅ p ) + O ( exp ( 3 ln p ln ln p ) ) .$
(10)

So applying (10) we may immediately deduce the estimate

$∑ a = 1 p − 1 | ∑ χ mod p χ ( − 1 ) = − 1 χ ( a ) L ( 1 , χ ¯ ) L ( 1 , χ ¯ χ 2 ) | = O ( ∑ a = 1 p − 1 d ( a ) a p ) + O ( ∑ a = 1 p − 1 exp ( 3 ln p ln ln p ) ) = O ( p ⋅ exp ( 3 ln p ln ln p ) ) .$

This proves Lemma 4. □

## 3 Proof of the theorem

In this section, we shall use the lemmas to complete the proof of our theorem. For any prime $p>2$, note that the identities

$∑ h = 1 p − 1 χ ¯ (h)K(h,1;p)= ∑ b = 1 p − 1 e ( b ¯ p ) ∑ h = 1 p − 1 χ ¯ (h)e ( h b p ) =τ( χ ¯ ) ∑ b = 1 p χ(b)e ( b ¯ p ) = τ 2 ( χ ¯ )$

and

$C 1 (h,p)= 1 2 ∑ a = 1 p − 1 ( 1 + ( a p ) ) ( ( a ¯ p ) ) ( ( a h p ) ) = 1 2 C(h,p)+ 1 2 C 2 (h,p).$

If $p≡1mod4$, then from Lemma 1, Lemma 2, and the properties of the Gauss sum $τ(χ)$ we have

$∑ h = 1 p − 1 K ( h , 1 ; p ) C 1 ( h , p ) = − 1 2 π 2 ( p − 1 ) ∑ χ mod p χ ( − 1 ) = − 1 ∑ h = 1 p − 1 χ ¯ ( h ) K ( h , 1 ; p ) τ 2 ( χ ) L 2 ( 1 , χ ¯ ) + − 1 2 π 2 ( p − 1 ) ∑ χ mod p χ ( − 1 ) = − 1 ∑ h = 1 p − 1 χ ¯ ( h ) K ( h , 1 ; p ) τ ( χ ) τ ( χ χ 2 ) ( 1 , χ ¯ ) ( 1 , χ ¯ χ 2 ) = − 1 2 π 2 ( p − 1 ) ∑ χ mod p χ ( − 1 ) = − 1 τ 2 ( χ ) τ 2 ( χ ¯ ) L 2 ( 1 , χ ¯ ) + − 1 2 π 2 ( p − 1 ) ∑ χ mod p χ ( − 1 ) = − 1 τ 2 ( χ ¯ ) τ ( χ ) τ ( χ χ 2 ) L ( 1 , χ ¯ ) L ( 1 , χ ¯ χ 2 ) = p 2 π 2 ( p − 1 ) ∑ χ mod p χ ( − 1 ) = − 1 τ ( χ ¯ ) τ ( χ χ 2 ) L ( 1 , χ ¯ ) L ( 1 , χ ¯ χ 2 ) − p 2 2 π 2 ( p − 1 ) ∑ χ mod p χ ( − 1 ) = − 1 L 2 ( 1 , χ ¯ ) .$
(11)

Note that $|τ( χ 2 )|= p$ and that we have the identity

$τ ( χ ¯ ) τ ( χ χ 2 ) = ∑ a = 1 p − 1 ∑ b = 1 p − 1 χ ¯ ( a ) χ ( b ) χ 2 ( b ) e ( a + b p ) = ∑ a = 1 p − 1 χ ¯ ( a ) ∑ b = 1 p − 1 χ 2 ( b ) e ( a b + b p ) = τ ( χ 2 ) ∑ a = 1 p − 1 χ ¯ ( a ) χ 2 ( a + 1 ) = τ ( χ 2 ) ∑ a = 1 p − 1 χ ( a ) χ 2 ( a ¯ + 1 ) ,$
(12)

so from Lemma 4 we have

$| ∑ χ mod p χ ( − 1 ) = − 1 τ ( χ ¯ ) τ ( χ χ 2 ) L ( 1 , χ ¯ ) L ( 1 , χ ¯ χ 2 ) | = p ⋅ | ∑ χ mod p χ ( − 1 ) = − 1 ∑ a = 1 p − 1 χ ( a ) χ 2 ( a ¯ + 1 ) L ( 1 , χ ¯ ) L ( 1 , χ ¯ χ 2 ) | ≤ p ⋅ ∑ a = 1 p − 1 | ∑ χ mod p χ ( − 1 ) = − 1 χ ( a ) L ( 1 , χ ¯ ) L ( 1 , χ ¯ χ 2 ) | = O ( p 3 2 ⋅ exp ( 3 ln p ln ln p ) ) .$
(13)

Combining (11), (13), and Lemma 3 we can deduce the asymptotic formula

$∑ h = 1 p − 1 K(h,1;p) C 1 (h,p)=− 1 4 π 2 ⋅ p 2 +O ( p 3 2 ⋅ exp ( 3 ln p ln ln p ) ) .$
(14)

If $p≡3mod4$, then from Lemma 2 we know that $C 1 (h,p)= 1 2 C(h,p)$, so from Lemma 3 and the method of proving (14) we also have

$∑ h = 1 p − 1 K ( h , 1 ; p ) C 1 ( h , p ) = − 1 2 π 2 p 2 p − 1 ∑ χ mod p χ ( − 1 ) = − 1 L 2 ( 1 , χ ¯ ) = − 1 4 π 2 ⋅ p 2 + O ( p 3 2 ⋅ exp ( 3 ln p ln ln p ) ) .$
(15)

Now the theorem follows from the asymptotic formulas (14) and (15).

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## Acknowledgements

The authors would like to thank the referees for their 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.

## Author information

Authors

### Corresponding author

Correspondence to Xiaoxue Li.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

XW carried out the proofs of the lemmas, XL carried out the part of Introduction, XW and XL carried out the theorem’s proof. All authors read and approved the final manuscript.

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Wang, X., Li, X. A hybrid mean value involving a new sum and Kloosterman sums. J Inequal Appl 2014, 44 (2014). https://doi.org/10.1186/1029-242X-2014-44 