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The hybrid mean value of Dedekind sums and two-term exponential sums

  • Chang Leran and Li Xiaoxue EMAIL logo
Published/Copyright: June 27, 2016

Abstract

In this paper, we use the mean value theorem of Dirichlet L-functions, the properties of Gauss sums and Dedekind sums to study the hybrid mean value problem involving Dedekind sums and the two-term exponential sums, and give an interesting identity and asymptotic formula for it.

MSC 2010: 11L03; 11F20

1 Introduction

Let q be a natural number and h an integer prime to q. The classical Dedekind sums

S(h,q)=a=1qaqahq,

where

((x))=0,ifxisaninteger,x[x]12,ifxisnotaninteger;

describes the behaviour of the logarithm of the eta-function (see [1, 2]) under modular transformations. The various arithmetical properties of S(h, q) were investigated by many authors, who obtained a series of results, see [310]. For example, W. P. Zhang and Y. N. Liu [10] studied the hybrid mean value problem of Dedekind sums and Kloosterman sums

K(m,n;q)=a=1qema+na¯q,

where q ≥ 3 is an integer, a=1q denotes the summation over all 1 < a <q with (a, q) = 1, e(y) = e2πiy, and ā denotes the multiplicative inverse of a mod q. They proved the following results:

Theorem A

Let p be an odd prime, then one has the identity

a=1p1b=1p1K(a,1;p)2K(b,1;p)2S2(ab¯,p)=112p2(p1)(p2),ifp1mod4,112p2((p1)(p2)12hp2),ifp3mod4;

where hp denotes the class number of the quadratic field Q(p).

Theorem B

Let p be an odd prime, then one has the asymptotic formula

a=1p1b=1p1K(a,1;p)2K(b,1;p)2S2(ab¯,q)=124p5+0p4exp3InInpInp,

where exp(y) = ey.

On the other hand, W. P. Zhang and D. Han [11] studied the sixth power mean of the two-term exponential sums, and proved that for any prime p > 3 with (3, p — 1) = 1, one has the identity

a=1p1n=0p1en3+anp6=5p48p3p2.

It is natural that one will ask, for the two-term exponential sums

E(r,s)=n=0p1ern3+snp,

whether there exists an identity (or asymptotic formula) similar to Theorem A(or Theorem B. The answer is yes.

The main purpose of this paper is to show this point. That is, we shall use the mean value theorem of Dirichlet L-functions, the properties of Gauss sums and Dedekind sums to prove the following similar conclusions:

Theorem 1.1

Let p > 3 be an odd prime with (3, p — 1) = 1, then we have the identity

a=1p1b=1p1E(a,1)2E(b,1)2S(ab¯,p)=112(p2)(p1)p2p2hp2,ifp3mod4,112(p2)(p1)p2ifp1mod4;

where hp denotes the class number of the quadratic field Q(p).

Theorem 1.2

Let p > 3 be a prime with (3, p — 1) = 1, then we have the asymptotic formula

a=1p1b=1p1E(a,1)2E(b,1)2S2(ab¯,p)=124p5+0p4exp3InInpInp.

It is very interesting that the results in our paper are exactly the same as in reference [10]. This means that there is close relationship between Kloosterman sums and two-term exponential sums. In fact, some close relationships can be found in W. Duke and H. Iwaniec [12].

2 Several lemmas

To complete the proof of our theorems, we need to prove several lemmas. Hereinafter, we shall use some properties of characters mod q and Dirichlet L-functions, all of these can be found in reference [13], so they will not be repeated here.

Lemma 2.1

Let p be an odd prime, a be any integer with (a, p) = 1. For any non-principal character χ mod p, we have the identity

n=0p1Xan2+bn+c=X¯(4)τ(X2)τ(X¯X2)τ(X¯)X4cb2a¯4acb2p,

where X2=p denotes the Legendre symbol mod p.

Proof

From the definition and properties of Gauss sums we have

n=0p1X(an2+bn+c)=1τ(X¯)r=1p1X¯(r)n=0p1er(an2+bn+c)p=X¯(4)τ(X¯)r=1p1X¯(r)e4rcrb2a¯)pn=0p1era(2n+ba¯)2)p=X¯(4)τ(X¯)r=1p1X¯(r)er(4crb2a¯)pn=0p1eran2)p.(1)

For any integer a with (a, p) = 1, from Theorem 7.5.4 of [14] we know that

n=0p1ean2p=apτ(X2).(2)

Combining (1) and (2) we have the identity

n=0p1Xan2+bn+c=X¯(4)τ(X2)τ(X¯)apr=1p1X¯(r)rper(4cb2a¯)p=X¯(4)τ(X2)τ(X¯X2)τ(X¯)X(4cb2a¯)4acb2p.

This proves Lemma 2.1. □

Lemma 2.2

Let p be an odd prime. Then for any non-principal character χ mod p with χ3 ≠ χ0 (the principal character mod p), we have the identity

a=1p1X(a)n=0p1ean3+np2=X(4)X2(3)τ(X2)τ(XX2)τ(X¯3),
Proof

From Lemma 2.1 the definition and properties of Gauss sums we have

a=1p1X(a)n=0p1ean3+np2=m=0p1n=0p1a=1p1X(a)ea(m3n3)+mnp=τ(X)m=0p1n=0p1X¯m3n3emnp=τ(X)m=0p1X¯3(m)emp+τ(X)m=0p1n=1p1X¯m3n3emnp=τ(X)m=0p1X¯3(m)emp+τ(x)m=0p1X¯m31n=1p1X¯3m31n=1p1X¯3(n)en(m1)p=τXτX¯3+τXτX¯3m=0p1Xm13X¯m31=2τXτX¯3+τXτX¯3m=2p1Xm13X¯m31=2τXτX¯3+τXτX¯3m=1p2Xm3X¯m3+3m2+3m=2τXτX¯3+τXτX¯3m=1p2X¯1+3m¯2+3m¯=2τXτX¯3+τXτX¯3m=1p2X¯3m2+3m+1=τXτX¯3m=0p1X¯3m2+3m+1=X(4)X2(3)τ(X2)τ(XX2)τ(X¯3),

This proves Lemma 2.2 □

Lemma 2.3

Let q > 2 be an integer. Then for any integer a with (a, q) = 1, we have the identity

S(a,q)=1π2qd/qd2ϕ(d)XmoddX(1)=1X(a)L(1,X)2,

where L(1,χ) denotes the Dirichlet L-function corresponding to character χ mod d

Proof

See Lemma 2 of [9]. □

Lemma 2.4

For any odd prime p, we have the asymptotic formula

XmoddX(1)=1L(1,X)4=5144π4p+oexp3InInpInp.
Proof

See Lemma 6 of [15]. □

3 Proof of the theorems

In this section, we shall complete the proof of our theorems. First we prove Theorem 1.1. From Lemma 2.3. with q = p (an odd prime) we have

S(a,p)=1π2pp1XmoddX(1)=1X(a)L(1,X)2(3)

and

XmoddX(1)=1L(1,X)2=π2(p1)pa=1p1ap122=π212(p1)2(p2)p2.(4)

It is clear that if (3, p — 1) = 1, then for any odd character χ mod p, we have χ3 ≠ χ0, the principal character mod p. Note that τ(X)=p, if χ ≠χ0. So from (3) and Lemma 2.2. we have

a=1p1b=1p1E(a,1)2E(b,1)2S(ab¯,p)=1π2pp1XmoddX(1)=1X(4)X2(3)τ(X2)τ(XX2)τ(X¯3)2L(1,X)2=1π2p3p1XmoddX(1)=1τ(XX2)2L(1,X)2.(5)

If p ≡ 1 mod 4, then χχ2 ≠ χ0 for all odd character χ mod p. This time, from (4) and (5) we have

a=1p1b=1p1E(a,1)2E(b,1)2S(ab¯,p)=1π2p4p1XmoddX(1)=1L(1,X)2=(p2)(p1)p212.(6)

If p ≡ 1 mod 4, thenχ2(–1) = — 1. This time we have |τ(χ2χ2)| = 1 and τ(XX2)=p,XX2. Combining (5) and (6) we obtain

a=1p1b=1p1E(a,1)2E(b,1)2S(ab¯,p)=1π2p4p1XmoddX(1)=1L(1,X)2+1π2p3p1L(1,X)21π2p4p1L(1,X)2=(p2)(p1)p212p3π2L(1,X)2=(p2)(p1)p212p2hp2,(7)

where we have used the identity L(1,X2)=πhp/p.

Combining (6) and (7) we may immediately deduce Theorem 1.1.

Now we prove Theorem 1.2. From (3) we have

a=1p1b=1p1E(a,1)2E(b,1)2S(ab¯,p)=1π4p4(p1)2XmoddX(1)=1ηmodpη(1)=1a=1p1X(a)η(a)E(a,1)2×b=1p1X¯(b)η¯(b)E(b,1)2L(1,X)2L(1,η)2=1π4p2(p1)2XmodpX(1)=1ηmodpη(1)=1XηXoa=1p1X(a)η(a)E(a,1)22L(1,X)2L(1,η)2+1π4p2(p1)2a=1p1E(a,1)22XmodpX(1)=1L(1,X)4R1+R2.(8)

Now we estimate R1 and R2 in (8) respectively. Note that the identity

a=1p1E(a,1)2=p2,

from Lemma 2.4 we have the asymptotic formula

R2=1π4p6(p1)25π4144p+Oexp3InInpInp=5144p5+Op4exp3InInpInp.(9)

If p ≡ 1 mod 4, then there exist two odd characters χ and ŋ such that χŋχ2 =χ0. This time, we have the estimate

XmodpX(1)=1ηmodpη(1)=1XηX2=XoL(1,X)2L(1,η)2=XmodpX(1)=1L(1,X)2L(1,XX2)2=O(p).

So for any prime p > 3, from (4), Lemma 2.2 and Lemma 2.4 we also have the asymptotic formula

R1=1π4p5(p1)2XmodpX(1)=1ηmodpη(1)=1XηXoL(1,X)2L(1,η)2+1π4p4(p1)2XmodpX(1)=1ηmodpη(1)=1Xη=X2L(1,X)2(L(1,η)21π5p5(p1)2XmodpX(1)=1ηmodpη(1)=1Xη=X2L(1,X)2L(1,η)2=1π4p5(p1)2XmodpX(1)=1L(1,X)22+O(p4)=1π4p5(p1)2π4144(p1)4(p2)2p4+O(p4)1144p5+O(p4).(10)

Combining (8), (9) and (10) we have the asymptotic formula

a=1p1b=1p1E(a,1)2E(b,1)2S2(ab¯,p)=1144p5+5144p5+Op4exp3InInpInp=124p5+Op4exp3InInpInp.

This completes the proofs of our results.

Competing interests

The authors declare that they have no competing interests.

Acknowledgement

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 N. S. F. (11371291) of P. R. China and G.I.C.F. (YZZ15009) of Northwest University.

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Received: 2016-5-12
Accepted: 2016-6-13
Published Online: 2016-6-27
Published in Print: 2016-1-1

© 2016 Leran and Xiaoxue, published by De Gruyter Open.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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