# A Hybrid Mean Value Involving  the Two-Term Exponential Sums and Two-Term Character Sums

Authors: Liu Miaohua; Li Xiaoxue
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

The main purpose of this paper is using the properties of Gauss sums and the estimate for character sums to study the hybrid mean value problem involving the two-term exponential sums and two-term character sums and give an interesting asymptotic formula for it.


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## Body

## 1. Introduction

Letq≥3 be an integer and χ denotes a Dirichlet character mod⁡q. For any integers m and n with (mn,q)=1, we define the two-term exponential sum C(m,n,k;q) and two-term character sum N(m,n,χ;q) as follows:
(1)C(m,n,k;q)=∑a=1q‍e(mak+naq),N(m,n,k,χ;q)=∑a=1q‍χ(mak+na),
where e(x)=e2πix, χ denotes a nonprincipal Dirichlet character mod⁡q, and k is a fixed positive integer.These sums play a very important role in the study of analytic number theory, so they caused many number theorists’ interest and favor. Some works related toC(m,n,k;q) can be found in [1–5]. For example, Cochrane and Zheng [1] show that
(2)|C(m,n,k;q)|≤kω(q)q1/2,
where ω(q) denotes the number of all distinct prime divisors of q.On the other hand, the sumsN(m,n,k,χ;q) are a special case of the general character sums of the polynomials
(3)∑a=N+1N+M‍χ(f(a)),
where M and N are any positive integers and f(x) is a polynomial. If q=p is an odd prime, then Weil (see [6]) obtained the following important conclusion.Letχ be a qth-order character mod⁡p; if f(x) is not a perfect qth power mod⁡p, then we have the estimate
(4)∑x=N+1N+M‍χ(f(x))≪p1/2ln⁡p,
where “≪” constant depends only on the degree of f(x). Some related results can also be found in [7–10].Now we are concerned about whether there exists an asymptotic formula for the hybrid mean value(5)∑m=1q-1‍|∑a=1q-1‍χ(mak+a)|2·|∑b=1q-1‍e(mbk+bq)|2.In this paper, we will use the analytic method and the properties of character sums to study this problem and give a sharp asymptotic formula for (5) with q=p, an odd prime. That is, we will prove the following.Theorem 1.
Letp be an odd prime, let χ be any nonprincipal even character mod⁡p, and let χ3≠χ0 be the principal character mod⁡p. Then we have the asymptotic formula
(6)∑m=1p-1‍|∑a=1p-1‍χ(ma3+a)|2·|∑b=1p-1‍e(mb3+bp)|2=2p3+E(p),
where E(p) satisfies the inequalities -12p2-2p≤E(p)≤4p2-2p.From this theorem we may immediately deduce the following.Corollary 2.
For any odd primep and any nonprincipal even character χmod⁡p with χ3≠χ0, one has
(7)∑m=1p-1‍|∑a=1p-1‍χ(ma3+a)|2·|∑b=1p-1‍e(mb3+bp)|2=2p3+O(p2).
In the theorem, we only consider the polynomialf(x)=mx3+x. For general polynomial f(x)=mxk+xh with k≥4 and 1≤h<k, whether there exists an asymptotic formula is complex problem for (5), it needs us to further study.
For general positive integerq≥4, whether there exists an asymptotic formula for (5) is also an interesting open problem.


## 2. Several Lemmas

To complete the proof of our theorem, we need the following several lemmas.Lemma 1.
Letp be an odd prime and let χ be any nonprincipal even character mod⁡p. Then for any integer m with (m,p)=1, the identity
(8)∑a=1p-1‍χ(ma3+a)=τ(χ1)τ(χ1¯3)χ1¯(m)τ(χ¯)×(1+(mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)),
where (*/p)=χ2 denotes the Legendre symbol and χ=χ12.Proof.
Sinceχ(-1)=1, there exists one and only one character χ1mod⁡p such that χ=χ12. Thus, from the properties of Gauss sums we have
(9)∑a=1p-1‍χ(ma3+a)=1τ(χ¯)∑a=1p-1‍∑b=1p-1‍χ¯(b)e(b(ma3+a)p)=1τ(χ¯)∑a=1p-1‍∑b=1p-1‍χ¯(ba¯)e(ba¯(ma3+a)p)=1τ(χ¯)∑b=1p-1‍χ¯(b)e(bp)∑a=1p-1‍χ(a)e(bma2p)=1τ(χ¯)∑b=1p-1‍χ¯(b)e(bp)∑a=1p-1‍χ12(a)e(bma2p)=1τ(χ¯)∑b=1p-1‍χ¯(b)e(bp)×∑a=1p-1‍χ1(a)(1+(ap))e(bmap)=1τ(χ¯)∑b=1p-1‍χ¯(b)e(bp)iiiiiiiiiiiiiiii×(χ1¯(bm)τ(χ1)iiiiiiiiiiiiiiiiiiiiii+χ1¯(bm)χ2(bm)τ(χ1χ2))=χ1¯(m)τ(χ¯)((mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)iiiiiiiiiiiiiiii+(mp)τ(χ1χ2)τ(χ1¯3χ2))=τ(χ1)τ(χ1¯3)χ1¯(m)τ(χ¯)×(1+(mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)).
This proves Lemma1.Lemma 2.
Letp be an odd prime, let χ be any nonprincipal even character mod⁡p, χ=χ12, and χ3≠χ0, the principal character mod⁡p. Then for any integer m and any quadratic nonresidue rmod⁡p with (m,p)=1, we have the identity
(10)|∑a=1p-1‍χ(ma3+a)|2=2p+(mp)τ2(χ2)2p∑a=1p-1‍(χ(a)+χ¯(a))×∑b=1p-1‍(1-a2b3p)(1-bp)+(mp)τ2(χ2)2p×∑a=1p-1‍(χ1(r)χ(a)+χ1¯(r)χ¯(a))×∑b=1p-1‍(1-ra2b3p)(1-bp).Proof.
From the properties of Gauss sums we have(11)τ(χ1)¯τ(χ1χ2)=∑a=1p-1‍χ¯1(a)∑b=1p-1‍χ1(b)χ2(b)e(b-ap)=∑a=1p-1‍χ¯1(a)∑b=1p-1‍χ2(b)e(b(1-a)p)=τ(χ2)∑a=1p-1‍χ¯1(a)(1-ap).
So from (11) we have
(12)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)=1p2τ(χ1)τ(χ1¯3)¯τ(χ1χ2)τ(χ1¯3χ2)=τ2(χ2)p2∑a=1p-1‍χ¯1(a)(1-ap)∑b=1p-1‍χ13(b)(1-bp)=τ2(χ2)p2∑a=1p-1‍χ¯1(a)∑b=1p-1‍(1-ab3p)(1-bp)=τ2(χ2)2p2∑a=1p-1‍χ¯(a)∑b=1p-1‍(1-a2b3p)(1-bp)+χ¯1(r)τ2(χ2)2p2∑a=1p-1‍χ¯(a)∑b=1p-1‍(1-ra2b3p)(1-bp).
Note that|τ(χ)|=|τ(χ1)|=|τ(χ13)|=p and τ2(χ2)=±p; from (12) and Lemma 1 we may immediately deduce the identity
(13)|∑a=1p-1‍χ(ma3+a)|2=p·|1+(mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)|2=2p+(mp)τ2(χ2)2p∑a=1p-1‍(χ(a)+χ¯(a))×∑b=1p-1‍(1-a2b3p)(1-bp)+(mp)τ2(χ2)2p×∑a=1p-1‍(χ1(r)χ(a)+χ1¯(r)χ¯(a))×∑b=1p-1‍(1-ra2b3p)(1-bp).
This proves Lemma2.Lemma 3.
Letp be an odd prime, let χ be any nonprincipal even character mod⁡p, χ=χ12, and χ3≠χ0, the principal character mod⁡p. Then for any integer m and any quadratic nonresidue rmod⁡p with (m,p)=1, one has the estimate
(14)|∑a=1p-1‍(χ1(r)χ(a)+χ1¯(r)χ¯(a))∑b=1p-1‍(1-ra2b3p)(1-bp)iiiiiiii+∑a=1p-1‍(χ(a)+χ¯(a))∑b=1p-1‍(1-a2b3p)(1-bp)|≤4p.Proof.
Letn be any integer such that (mn/p)=-1 or (m/p)+(n/p)=0. Then from Lemma 2 we have
(15)|∑a=1p-1‍χ(ma3+a)|2+|∑a=1p-1‍χ(na3+a)|2=4p.
Note that|(m/p)(τ2(χ2)/p)|=1; applying (15) and Lemma 2 we have the estimate
(16)|∑a=1p-1‍(χ1(r)χ(a)+χ1¯(r)χ¯(a))∑b=1p-1‍(1-ra2b3p)(1-bp)iiiiiiiiii+∑a=1p-1‍(χ(a)+χ¯(a))∑b=1p-1‍(1-a2b3p)(1-bp)|=||∑a=1p-1‍χ(ma3+a)|2-|∑a=1p-1‍χ(na3+a)|2|≤|∑a=1p-1‍χ(ma3+a)|2+|∑a=1p-1‍χ(na3+a)|2≤4p.
This proves Lemma3.Lemma 4.
Letp>3 be a prime. Then we have the identity
(17)∑m=1p-1‍(mp)|∑a=1p-1‍e(ma3+ap)|2=-τ2(χ2)(2+(3p)),
where (*/p)=χ2 denotes the Legendre symbol.Proof.
For any odd primep and integer n with (n,p)=1, from Hua's book [11] (Section 7.8, Theorem 8.2) we know that
(18)∑a=1p‍(a2+np)=-1.
From this identity and the definition and properties of Gauss sums we have(19)∑m=1p-1‍(mp)|∑a=1p-1‍e(ma3+ap)|2=∑a=1p-1‍∑b=1p-1‍∑m=1p-1‍(mp)e(m(a3-b3)+a-bp)=∑a=1p-1‍∑b=1p-1‍∑m=1p-1‍(mp)e(mb3(a3-1)+b(a-1)p)=τ(χ2)∑a=1p-1‍∑b=1p-1‍(b3(a3-1)p)e(b(a-1)p)=τ(χ2)∑a=1p-1‍(a3-1p)∑b=1p-1‍(bp)e(b(a-1)p)=τ2(χ2)∑a=1p-1‍((a3-1)(a-1)p)=τ2(χ2)(∑a=1p‍(4a2+4a+4p)-1-(3p))=τ2(χ2)(∑a=1p‍((2a+1)2+3p)-1-(3p))=τ2(χ2)(∑a=1p‍(a2+3p)-1-(3p))=-τ2(χ2)(2+(3p)).
This proves Lemma4.


## 3. Proof of the Theorem

In this section, we will complete the proof of our theorem. Note that the identities|τ(χ2)|2=p and
(20)∑m=1p-1|∑a=1p-1e(ma3+ap)‍|2=∑m=1p|∑a=1p-1‍e(ma3+ap)|2-1=∑a=1p-1∑b=1p-1∑m=1pe(m(a3-b3)+a-bp)-1=∑a=1p-1∑b=1p-1∑m=1pe(m(a3-1)+b(a-1)p)-1={p2-p-1,if3†p-1,p2-3p-1,if3∣p-1.So from (20), Lemmas 2, 3, and 4, and noting that |τ(χ2)|2=p we have
(21)∑m=1p-1‍|∑a=1p-1‍χ(ma3+a)|2·|∑b=1p-1‍e(mb3+bp)|2=2p·∑m=1p-1‍|∑c=1p-1‍e(mc3+cp)|2+τ2(χ2)2p∑a=1p-1‍(χ(a)+χ¯(a))×∑b=1p-1‍(1-a2b3p)(1-bp)×∑m=1p-1‍(mp)|∑c=1p-1‍e(mc3+cp)|2+τ2(χ2)2p×∑a=1p-1‍(χ1(r)χ(a)+χ¯1(r)χ¯(a))×∑b=1p-1‍(1-a2b3p)(1-bp)×∑m=1p-1‍(mp)|∑c=1p-1‍e(mc3+cp)|2=2p3+E(p),
where E(p) satisfies the inequalities -12p2-2p≤E(p)≤4p2-2p.This completes the proof of our theorem.



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*Source: 845845-2014-02-27.xml*