You are viewing the new article page. Let us know what you think. Return to old version

Research | Open | Published:

# 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 $q≥3$ 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  and . For example, from Weil’s classical work  one can deduce the estimate

$|C(m,0,2,χ;p)|≤2⋅ p 1 2$

for $(m,p)=1$.

Recently, Wang  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 $p≠3a+1$. Then, for any integer m with $(m,p)=1$, we have the identity

Wang  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  and , 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

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 , 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 $m ¯ m≡1modp$.

Proof See Lemma 1 in . □

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 $χ 1 (p−1+1)=0$, 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 . 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 ).

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

$∑ m = 0 p − 1 | ∑ a = 1 p − 1 e ( m a 2 + a p ) | 4 = p 3 −3 p 2 +2 ( − 1 p ) p 2 −p−8 ( − 1 p ) p.$
(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

$A= ∑ χ mod p | ∑ a = 1 p − 1 χ 1 (a+1) χ ¯ (1+2 a ¯ ) | 2 =(p−1)(p−3),$
(10)
(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)
$D=| ∑ a = 1 p − 2 χ 1 (a+1) χ 0 (1+2 a ¯ ) | 2 =p−3,$
(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

This completes another proof of our theorem.

The corollary follows from Theorem and Lemma 3.

## References

1. 1.

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

2. 2.

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

3. 3.

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

4. 4.

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

5. 5.

Hua LK: Introduction to Number Theory. Science Press, Beijing; 1979.

6. 6.

Zhang W: Moments of generalized quadratic Gauss sums weighted by L -functions. J. Number Theory 2002, 92: 304–314. 10.1006/jnth.2001.2715

7. 7.

Weil A: On some exponential sums. Proc. Natl. Acad. Sci. USA 1948, 34: 204–207. 10.1073/pnas.34.5.204

8. 8.

Wang T: On the fourth power mean of generalized two-term exponential sums. Bull. Korean Math. Soc. 2013, 50: 233–240. 10.4134/BKMS.2013.50.1.233

9. 9.

Wang J: Hybrid mean value of the generalized Kloosterman sums and Dirichlet character of polynomials. Bull. Korean Math. Soc. 2013, 50: 451–458. 10.4134/BKMS.2013.50.2.451

10. 10.

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

## 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.

## Author information

Correspondence to Xiaoxue Li.

### 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.

## Rights and permissions

Reprints and Permissions 