JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 845845 10.1155/2014/845845 845845 Research Article A Hybrid Mean Value Involving the Two-Term Exponential Sums and Two-Term Character Sums Miaohua Liu 1 Xiaoxue Li 2 Bahn Olivier 1 Institute of Science Air Force Engineering University Xi’an, Shaanxi 710051 China 2 Department of Mathematics Northwest University Xi’an, Shaanxi 710127 China nwu.edu.cn 2014 2722014 2014 16 10 2013 20 01 2014 27 2 2014 2014 Copyright © 2014 Liu Miaohua and Li Xiaoxue. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction

Let q3 be an integer and χ denotes a Dirichlet character modq. 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 modq, 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 to C(m,n,k;q) can be found in . For example, Cochrane and Zheng  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 sums N(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 ) obtained the following important conclusion.

Let χ be a qth-order character modp; if f(x) is not a perfect qth power modp, then we have the estimate (4)x=N+1N+Mχ(f(x))p1/2lnp, where “” constant depends only on the degree of f(x). Some related results can also be found in .

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.

Let p be an odd prime, let χ be any nonprincipal even character modp, and let χ3χ0 be the principal character modp. 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 -12p2-2pE(p)4p2-2p.

From this theorem we may immediately deduce the following.

Corollary 2.

For any odd prime p and any nonprincipal even character χmodp with χ3χ0, 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 polynomial f(x)=mx3+x. For general polynomial f(x)=mxk+xh with k4 and 1h<k, whether there exists an asymptotic formula is complex problem for (5), it needs us to further study.

For general positive integer q4, 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.

Let p be an odd prime and let χ be any nonprincipal even character modp. 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.

Since χ(-1)=1, there exists one and only one character χ1modp such that χ=χ12. Thus, from the properties of Gauss sums we have (9)a=1p-1χ(ma3+a)=1τ(χ¯)a=1p-1b=1p-1χ¯(b)e(b(ma3+a)p)=1τ(χ¯)a=1p-1b=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 Lemma 1.

Lemma 2.

Let p be an odd prime, let χ be any nonprincipal even character modp, χ=χ12, and χ3χ0, the principal character modp. Then for any integer m and any quadratic nonresidue rmodp with (m,p)=1, we have the identity (10)|a=1p-1χ(ma3+a)|2=2p+(mp)τ2(χ2)2pa=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)p2a=1p-1χ¯1(a)(1-ap)b=1p-1χ13(b)(1-bp)=τ2(χ2)p2a=1p-1χ¯1(a)b=1p-1(1-ab3p)(1-bp)=τ2(χ2)2p2a=1p-1χ¯(a)b=1p-1(1-a2b3p)(1-bp)+χ¯1(r)τ2(χ2)2p2a=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)2pa=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 Lemma 2.

Lemma 3.

Let p be an odd prime, let χ be any nonprincipal even character modp, χ=χ12, and χ3χ0, the principal character modp. Then for any integer m and any quadratic nonresidue rmodp 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.

Let n 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)|24p.

This proves Lemma 3.

Lemma 4.

Let p>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 prime p and integer n with (n,p)=1, from Hua's book  (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-1b=1p-1m=1p-1(mp)e(m(a3-b3)+a-bp)=a=1p-1b=1p-1m=1p-1(mp)e(mb3(a3-1)+b(a-1)p)=τ(χ2)a=1p-1b=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 Lemma 4.

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-1b=1p-1m=1pe(m(a3-b3)+a-bp)-1=a=1p-1b=1p-1m=1pe(m(a3-1)+b(a-1)p)-1={p2-p-1,if  3p-1,p2-3p-1,if3p-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)2pa=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 -12p2-2pE(p)4p2-2p.

This completes the proof of our theorem.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to thank the referee for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the NSF (11371291) of China.

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