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The fourth power mean of the generalized two-term exponential sums and its upper and lower bound estimates

Abstract

In this paper, we use the analytic method and the properties of Gauss sums to study the computational problem of one kind fourth power mean of the generalized two-term exponential sums, and give an exact computational formula for it.

MSC:11L40, 11F20.

1 Introduction

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

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

where χ denotes any Dirichlet character modq, and e(y)= e 2 π i y .

Regarding the upper bound estimate of C(m,n,k,χ;q), many authors have studied it and obtained a series of important results; related contents can be found in [15] and [6]. For example, from Weil’s classical work [7] one can deduce the estimate

|C(m,0,2,χ;p)|2 p 1 2

for (m,p)=1.

Recently, Wang [8] studied the computational problem of the fourth power mean of C(m,n,k,χ;p), and proved the following conclusion:

Let p be an odd prime with p3a+1. Then, for any integer m with (m,p)=1, we have the identity

n = 1 p | C ( m , n , 3 , χ ; p ) | 4 = { p ( 2 p 2 3 p 3 ) if  χ  is the principal character mod p ; p 2 ( 3 p 7 ) if  χ  is the Legendre symbol mod p ; p 2 ( 2 p 6 ) otherwise.

Wang [9] studied the hybrid power mean of the generalized Kloosterman sums a = 1 p 1 λ(a)e( m a + a ¯ p ) and a = 1 p 1 χ(a+ a ¯ ), where λ denotes a Dirichlet character modp, and gave an interesting asymptotic formula for it. That is, she proved the following result:

Let p be an odd prime. Then, for any non-principal even character χmodp and any character λmodp with λ( p ), we have the asymptotic formula

m = 1 p 1 | a = 1 p 1 λ(a)e ( m a + a ¯ p ) | 2 | b = 1 p 1 χ(mb+ b ¯ ) | 2 =2 p 3 +O ( p 2 ) .

In this paper, as a note of [8] and [9], we found that there exists a close relationship between the fourth power mean of C(m,n,2,χ;p) and | b = 1 p 1 χ(nb+ b ¯ )|. The main purpose of this paper is to show this point. That is, we shall prove the following theorem.

Theorem Let p be an odd prime. Then, for any character χ 1 modp, we have the identity

m = 1 p | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 4 = { p 3 3 p 2 + 2 ( 1 p ) p 2 p 8 ( 1 p ) p if  χ 1 = χ 0 ; 2 p 3 3 p 2 if χ 1 ( 1 ) = 1 ; 2 p 3 4 ( 1 p ) p 2 3 p 2 p | a = 1 p 1 χ 1 ( a + a ¯ ) | 2 if χ 1 χ 0 and χ 1 ( 1 ) = 1 ,

where χ 0 denotes the principal character modp, a a ¯ 1modp, and ( p ) is the Legendre symbol.

From this theorem we may immediately deduce the following corollary.

Corollary Let p be an odd prime. Then, for any non-principal character χ 1 modp, we have the inequalities

2 p 3 11 p 2 m = 1 p | a = 1 p 1 χ 1 (a)e ( m a 2 + a p ) | 4 2 p 3 + p 2 .

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 character sums and Gauss sums, all of these can be found in reference [10], so they will not be repeated here. First we have the following.

Lemma 1 Let p be an odd prime. Then, for any integers m and n with (mn,p)=1, we have the identity

b = 0 p 1 e ( m b 2 + n b p ) = ( m p ) e ( 4 m ¯ n 2 p ) a = 0 p 1 e ( a 2 p ) ,

where ( x p ) denotes the Legendre symbol, and m ¯ m1modp.

Proof See Lemma 1 in [8]. □

Lemma 2 Let p be an odd prime, χ 1 be any fixed character modp. Then, for any non-real character χmodp, we have the identity

| m = 1 p 1 χ ( m ) | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 2 | 2 = p 2 | a = 1 p 1 χ 1 ( a + 1 ) χ ¯ ( 1 + 2 a ¯ ) | 2 .

Proof From the properties of Gauss sums, we have

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

Note that χ 1 (p1+1)=0, χ is a non-real character modp, so χ ¯ 2 is also a non-principal character modp. Therefore, |τ(χ)|=|τ( χ ¯ 2 )|= p , so from (1) we may immediately deduce Lemma 2. □

Lemma 3 Let p be an odd prime, χ be any non-principal character modp with χ(1)=1. Then we have

| a = 1 p 1 χ(a+ a ¯ ) | 2 = ( 1 p ) ( a = 1 p 1 χ ( a ) ( a 2 1 p ) ) 2

and

| a = 1 p 1 χ(a+ a ¯ )|2 p .

Proof From the properties of quadratic residue modp, we have

a = 1 p 1 χ ( a + a ¯ ) = b = 1 p 1 χ ( b ) a = 1 p 1 a + a ¯ b mod p 1 = b = 1 p 1 χ ( b ) a = 0 p 1 a 2 b a + 1 0 mod p 1 = b = 1 p 1 χ ( b ) a = 0 p 1 ( 2 a b ) 2 b 2 4 mod p 1 = χ ( 2 ) b = 1 p 1 χ ( b ) a = 0 p 1 a 2 b 2 1 mod p 1 = χ ( 2 ) b = 1 p 1 χ ( b ) ( 1 + ( b 2 1 p ) ) = χ ( 2 ) b = 1 p 1 χ ( b ) ( b 2 1 p ) .
(2)

Note that

b = 1 p 1 χ ¯ (b) ( b 2 1 p ) = b = 1 p 1 χ(b) ( b ¯ 2 1 p ) = ( 1 p ) b = 1 p 1 χ(b) ( b 2 1 p ) ,

so from (2) we may immediately deduce the identity

| a = 1 p 1 χ(a+ a ¯ ) | 2 = ( 1 p ) ( a = 1 p 1 χ ( a ) ( a 2 1 p ) ) 2 .

The estimate

| a = 1 p 1 χ(a+ a ¯ )|2 p

follows from Lemma 1 of [9]. This proves Lemma 3. □

3 Proof of Theorem

In this section, we shall give two different proofs of our theorem. First, if χ 1 is a non-principal character modp, then from Lemma 1 we have

| a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 2 = a = 1 p 1 b = 1 p 1 χ 1 ( a b ¯ ) e ( m ( a 2 b 2 ) + a b p ) = a = 1 p 1 b = 1 p 1 χ 1 ( a ) e ( m b 2 ( a 2 1 ) + b ( a 1 ) p ) = p 1 + χ 1 ( 1 ) b = 1 p 1 e ( 2 b p ) + a = 2 p 2 χ 1 ( a ) b = 1 p 1 e ( m b 2 ( a 2 1 ) + b ( a 1 ) p ) = p 1 χ 1 ( 1 ) + a = 2 p 2 χ 1 ( a ) b = 0 p 1 e ( m b 2 ( a 2 1 ) + b ( a 1 ) p ) a = 2 p 2 χ 1 ( a ) = p + a = 2 p 2 χ 1 ( a ) b = 0 p 1 e ( m ( a + 1 ) a 1 ¯ b 2 + b p ) a = 1 p 1 χ 1 ( a ) = p + G ( p ) a = 2 p 2 χ 1 ( a ) ( m ( a + 1 ) a 1 ¯ p ) e ( 4 m ¯ a + 1 ¯ ( a 1 ) p ) = p + G ( p ) a = 2 p 2 χ 1 ( a ) ( m ( a 2 1 ) p ) e ( 4 m ¯ a + 1 ¯ ( a 1 ) p ) ,
(3)

where G(p)= a = 0 p 1 e( a 2 p ) and G 2 (p)=( 1 p )p (see Theorem 7.5.4 of [5]).

From (3) and the definition of Gauss sums, we may immediately deduce

m = 1 p 1 | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 4 = p 2 ( p 1 ) + 2 p G ( p ) m = 1 p 1 a = 2 p 2 χ 1 ( a ) ( m ( a 2 1 ) p ) e ( 4 m ¯ a + 1 ¯ ( a 1 ) p ) + G 2 ( p ) a = 2 p 2 b = 2 p 2 χ 1 ( a b ) ( ( a 2 1 ) ( b 2 1 ) p ) × m = 1 p 1 e ( 4 m ¯ ( a + 1 ¯ ( a 1 ) + b + 1 ¯ ( b 1 ) ) p ) = p 2 ( p 1 ) + 2 p G 2 ( p ) a = 2 p 2 χ 1 ( a ) ( ( a 2 1 ) ( a 1 ) a + 1 ¯ p ) + p G 2 ( p ) a = 2 p 2 χ 1 ( 1 ) ( ( a 2 1 ) ( a ¯ 2 1 ) p ) G 2 ( p ) ( a = 2 p 2 χ 1 ( a ) ( a 2 1 p ) ) 2 = p 2 ( p 1 ) + 2 p G 2 ( p ) a = 2 p 2 χ 1 ( a ) + p G 2 ( p ) ( 1 p ) ( p 3 ) G 2 ( p ) ( a = 2 p 2 χ 1 ( a ) ( a 2 1 p ) ) 2 .
(4)

If χ 1 is a non-principal character modp with χ 1 (1)=1, then note that

a = 2 p 2 χ 1 ( a ) ( a 2 1 p ) = a = 1 p 1 χ 1 ( a ) ( a 2 1 p ) = 0 , G 2 ( p ) = ( a = 0 p 1 e ( a 2 p ) ) 2 = ( 1 p ) p

and

| a = 1 p 1 χ 1 (a)e ( 0 a 2 + a p ) | 4 =| a = 1 p 1 χ 1 (a)e ( a p ) | 4 = p 2 .

From (4) we have

m = 0 p 1 | a = 1 p 1 χ 1 (a)e ( m a 2 + a p ) | 4 =2 p 3 3 p 2 .
(5)

If χ 1 is a non-principal character modp with χ 1 (1)=1, then from (4) and Lemma 3 we have

m = 0 p 1 | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 4 = 2 p 3 4 ( 1 p ) p 2 3 p 2 ( 1 p ) p ( a = 1 p 1 χ 1 ( a ) ( a 2 1 p ) ) 2 = 2 p 3 4 ( 1 p ) p 2 3 p 2 p | a = 1 p 1 χ 1 ( a + a ¯ ) | 2 .
(6)

If χ 1 = χ 0 is the principal character modp, then from the method of proving (3) and (4) we have

m = 0 p 1 | a = 1 p 1 e ( m a 2 + a p ) | 4 = p 3 3 p 2 +2 ( 1 p ) p 2 p8 ( 1 p ) p.
(7)

Combining (5), (6) and (7), we may immediately deduce our theorem.

The second proof of Theorem. First, from the orthogonality of characters modp, we have

χ mod p | m = 1 p 1 χ ( m ) | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 2 | 2 = ( p 1 ) m = 1 p 1 | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 4 .
(8)

On the other hand, from Lemma 2 we have

χ mod p | m = 1 p 1 χ ( m ) | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 2 | 2 = p 2 χ mod p | a = 1 p 1 χ 1 ( a + 1 ) χ ¯ ( 1 + 2 a ¯ ) | 2 + ( m = 1 p 1 | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 2 ) 2 + | m = 1 p 1 ( m p ) | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 2 | 2 p 2 | a = 1 p 2 χ 1 ( a + 1 ) χ 0 ( 1 + 2 a ¯ ) | 2 p 2 | a = 1 p 2 χ 1 ( a + 1 ) ( 1 + 2 a ¯ p ) | 2 p 2 A + B + C p 2 D p 2 E .
(9)

Applying the orthogonality of characters modp, we can easily deduce that

A= χ mod p | a = 1 p 1 χ 1 (a+1) χ ¯ (1+2 a ¯ ) | 2 =(p1)(p3),
(10)
B = ( m = 1 p 1 | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 2 ) 2 B = { ( p 2 2 p 1 ) 2 if  χ 1 = χ 0 ; p 2 ( p 2 χ 1 ( 1 ) ) 2 if  χ 1 χ 0 .
(11)

From the definition and properties of Gauss sums, we have

C=| m = 1 p 1 ( m p ) | a = 1 p 1 χ 1 (a)e ( m a 2 + a p ) | 2 | 2 =p| a = 1 p 1 χ 1 (a) ( a 2 1 p ) | 2 ,
(12)
D=| a = 1 p 2 χ 1 (a+1) χ 0 (1+2 a ¯ ) | 2 =p3,
(13)
E=| a = 1 p 2 χ 1 (a+1) ( 1 + 2 a ¯ p ) | 2 =| a = 1 p 1 χ 1 (a) ( a 2 1 p ) | 2 .
(14)

Note that if χ 1 (1)=1, then

a = 1 p 1 χ 1 (a) ( a 2 1 p ) =0.

Combining (7)-(14) and Lemma 3, we may immediately deduce the identity

m = 1 p 1 | a = 1 p 1 χ 1 ( a ) e ( m a 2 + a p ) | 4 = { p 3 3 p 2 + 2 ( 1 p ) p 2 p 8 ( 1 p ) p 1 if  χ 1 = χ 0 ; 2 p 3 4 p 2 if  χ 1 ( 1 ) = 1 ; 2 p 3 4 ( 1 p ) p 2 4 p 2 p | a = 1 p 1 χ 1 ( a + a ¯ ) | 2 if  χ 1 χ 0  and  χ 1 ( 1 ) = 1 .

This completes another proof of our theorem.

The corollary follows from Theorem and Lemma 3.

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Acknowledgements

The authors would like to thank the referee for his/her very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the N.S.F. (11071194) of P.R. China.

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Correspondence to Xiaoxue Li.

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Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

LX carried out the part of Introduction, XZ carried out the proof of some lemmas, LX with XZ carried out the theorem’s proof. All authors read and approved the final manuscript.

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Keywords

  • generalized two-term exponential sums
  • fourth power mean
  • computational formula