# A Hybrid Power Mean Involving the Dedekind Sums and Cubic Gauss Sums

Authors: Jiayuan Hu; Yu Zhan; Qin Si
Journal: Journal of Mathematics
 (2021)

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## Abstract

The main purpose of this paper is using analytic methods and the properties of the Dedekind sums to study one kind hybrid power mean calculating problem involving the Dedekind sums and cubic Gauss sum and give some interesting calculating formulae for it.


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## Body

## 1. Introduction

For any integerq≥3 and integer m, the classical cubic Gauss sum Am,q is defined as follows:(1)Am,q=∑a=0q−1ema3q,where as usual, ex=e2πix and i2=−1.This sum plays a very important role in the study of the elementary number theory and analytic number theory, so there are many people who had studied the arithmetical properties ofAm,q and related contents (see [1–6]). For example, if q=p is an odd prime, Cao and Wang [5] proved the following interesting conclusion: let p be an odd prime with p≡1mod3. If 3 is not a cubic residue modulo p, then we have the identity(2)∑m=1p−1∑a=0p−1ema3p2⋅∑a=0p−1ema3+ap4=p24p2−19p−d2−1,where 4p=d2+27⋅b2 and d is uniquely determined as follows: d≡1mod3On the other hand, we define the Dedekind sumsSh,q as follows.Letq be a natural number and h be an integer prime to q. The classical Dedekind sum Sh,q is defined as(3)Sh,q=∑a=1qaqahq,where(4)x=x−x−12,ifxis not an integer,0,ifxis an integer.About the various properties ofSh,q, many authors had studied them and obtained a series of interesting results, and related works can be found in [7–14].The main purpose of this paper is using the analytic method and the properties of Dedekind sums to study the computational problem of the hybrid power mean:(5)∑m=1q′Shm,qAkm,q,and give some exact computational formulae for (5) with q=p, an odd prime, where h and k are any two fixed positive integers.About this problem, so far no one seems to consider; at least, we have not seen any related papers before. Of course, this problem is meaningful, and it can describe the mean value distribution properties of the two different sums. It is clear that ifp−1,3=1, then from the properties of the complete residue system modulo p, we have Am=0. This time (5) is meaningless. So, in the following, we only consider the case p≡1mod3. The main purpose of this paper is to prove the following several results.Theorem 1.
Letp be a prime with p=12h+1. Then, for any positive integers k and h with 2,h=1, we have the identity(6)∑m=1p−1Sm2,pAkm,p=∑m=1p−1Shm,pAkm,p=0.Theorem 2.
Letp be a prime with p=12h+7. Then, for any positive integer k, we have the identity(7)∑m=1p−1Sm2,pAkm,p=1π2⋅pp−1⋅p−1Ak1,p−Hk,p⋅L1,χ2λ2+hp2p−1⋅Hk,p,where χ2 is Legendre’s symbol modulo p, λ denotes any third-order character modulo p, hp=p/π⋅L1,χ2 denotes the class number of the quadratic field Fp−p, and(8)Hk,p=∑m=1p−11Akm,p,satisfies the third-order linear recursive formula as follows:(9)Hk,p=−3dHk−1,p+1dpHk−3,p,with the initial values H0,p=p−1, H1,p=−p−1/d, and H2,p=3p−1/d2, and 4p=d2+27⋅b2, where d is uniquely determined by d≡1mod3.Theorem 3.
Letp be a prime with p≡1mod3. Then, for any positive integer k, we have the asymptotic formula as follows:(10)∑m=1p−1S2m,pAkm,p=5p144+pπ4⋅∑n=1∞∑d|nλd2n2⋅Hk,p+Opdk⋅exp4lnplnlnp,where expy=ey and λ is a three-order character modulo p.From Theorem3, we may immediately deduce the following two corollaries.Corollary 1.
Let Letp be a prime with p≡1mod3, then we have(11)∑m=1p−1S2m,pAm,p=−p2d⋅5144+1π4⋅∑n=1∞∑d|nλd2n2+Opd⋅exp4lnplnlnp.Corollary 2.
Letp be a prime with p≡1mod3, then we have(12)∑m=1p−1Sm,pAm,p2=3p2d2⋅5144+1π4∑n=1∞∑d|nλd2n2+Opd2⋅exp4lnplnlnp.


## 2. Several Lemmas

To complete the proofs of our all theorems, we need to prove several simple lemmas. Hereinafter, we shall use some properties of the character sums and Gauss sums, and all of these contents can be found in [15], so they will not be repeated here.Lemma 1.
Letp be an odd prime with p≡1mod3 and λ be any third-order character modulo p, then we have the identity(13)τ3λ+τ3λ¯=dp,where τλ=∑a=1p−1λaea/p denotes the classical Gauss sums, and 4p=d2+27⋅b2, where d is uniquely determined by d≡1mod3.Proof.
The proof of this lemma is shown in the study by Zhang and Hu [2] or Berndt and Evans [16].Lemma 2.
Letp be an odd prime with p≡1mod3. Then, for any integer k≥3 and(14)Hk,p=∑m=1p−11Akm,p,we have the third-order linear recursive formula as follows:(15)Hk,p=−3dHk−1,p+1dpHk−3,p,with the initial values H0,p=p−1, H1,p=−p−1/d, and H2,p=3p−1/d2, where d is defined as in Lemma 1.Proof.
Letλ be any third-order character modulo p; then, for any integer 1≤m≤p−1, from the properties of the third-order characters and the classical Gauss sums modulo p, we have the identity(16)Am,p=∑a=0p−1ema3p=1+∑a=1p−11+λa+λ¯aemap,=∑a=0p−1emap+λ¯mτλ+λmτλ¯=λ¯mτλ+λmτλ¯.
Note thatτλτλ¯=p and λ2=λ¯; from (16), we have(17)A2m,p=λmτ2λ+2p+λ¯mτ2λ¯.
From (16) and Lemma 1, we also have(18)A3m,p=λ¯mτλ+λmτλ¯3,=τ3λ+τ3λ¯+3pAm,p=dp+3pAm,p.
From (17) and the properties of the character sums modulo p, we have(19)∑m=1p−1A2m,p=∑m=1p−1λmτ2λ+2p+λ¯mτ2λ¯=2pp−1.
Combining (18) and (19), we can deduce that(20)∑m=1p−1A2m,p=dp⋅∑m=1p−11Am,p+3p∑m=1p−11,or(21)∑m=1p−11Am,p=1dp2pp−1−3pp−1=−p−1d.
Similarly, from (16), (18), and (21), we have(22)0=∑m=1p−1Am,p=∑m=1p−1dpA2m,p+∑m=1p−13pAm,p=∑m=1p−1dpA2m,p−3pp−1d,or(23)∑m=1p−11A2m,p=3p−1d2.
Ifk≥3, then from (18), we have(24)1Ak−3m,p=dp⋅1Akm,p+3p⋅1Ak−1m,p,or(25)Hk,p=−3dHk−1,p+1dpHk−3,p.
Now, Lemma2 follows from (21)–(25) and H0,p=p−1.Lemma 3.
Letp be a prime with p≡1mod3. For any positive integer k and any character χmodp, if χ3≠χ0, then we have the identity(26)∑m=1p−1χmAkm,p=0,where χ0 denotes the principal character modulo p.Proof.
Letg be a primitive root modulo p. Then, from the definition of Am,p and the properties of the complete residue system modulo p, we have(27)Amg3,p=∑a=0p−1emg3a3p=∑a=0p−1emga3p=∑a=0p−1ema3p=Am,p.
From (27), we have(28)∑m=1p−1χmAkm,p=∑m=1p−1χmg3Akmg3,p=χ3g⋅∑m=1p−1χmAkm,p.or(29)1−χ3g⋅∑m=1p−1χmAkm,p=0.
Ifχ3≠χ0, then χ3g≠1. So, from (29), we have the identity(30)∑m=1p−1χmAkm,p=0.
This proves Lemma3.Lemma 4.
Letp be a prime with p≡1mod3 and λ be any third-order character modulo p. Then, for any positive integer k, we have the identity(31)∑m=1p−1λm+λ¯mAkm,p=p−1Ak1,p−Hk,p.Proof.
From the properties of the third-order character modulop, we have(32)1+λm+λ¯m=3,ifmisa3−the residue modulop,0,otherwise.
From (32), the properties of the 3-th residue modulo p, and the identity Akm3,p=Ak1,p where m,p=1, we have(33)∑m=1p−1λm+λ¯mAkm,p=∑m=1p−11+λm+λ¯mAkm,p−∑m=1p−11Akm,p=∑m=1p−11Akm3,p−∑m=1p−11Akm,p=p−1Ak1,p−Hk,p.
This proves Lemma4.Lemma 5.
Letq>2 be an integer; then, for any integer a with a,q=1, we have the identity(34)Sa,q=1π2q∑d|qd2ϕd∑χmoddχ−1=−1χaL1,χ2,where Ls,χ denotes the Dirichlet L-function corresponding to the character χ and ∑χmodpχ−1=−1 denotes the summation over all odd characters modulo d.Proof.
See Lemma2 of [14].


## 3. Proofs of the Theorems

In this section, we shall complete the proofs of our all theorems. Ifp be a prime, then from Lemma 5, we have(35)Sa,p=1π2⋅pp−1⋅∑χmodpχ−1=−1χaL1,χ2.It is clear that ifh is an integer with 2,h=1, then note that the identities Sh−m,p=−Shm,p and A−m,p=Am,p, and we have(36)∑m=1p−1Shm,pAkm,p=∑m=1p−1Sh−m,pAk−m,p=−∑m=1p−1Shm,pAkm,p=0.Now, for any positive integerk, from (35), we have the identity(37)∑m=1p−1Sm2,pAkm,p=1π2⋅pp−1⋅∑χmodpχ−1=−1∑m=1p−1χ2mAkm,pL1,χ2.Ifp=12h+1, then for any odd character χmodp, χ6=χ0 if and only if χ=χ0,χ2,χ2λ or χ2λ2, where χ2 is Legendre’s symbol modulo p. In this time, χ is not an odd character modulo p. So, from (37) and Lemma 3, we have(38)∑m=1p−1Sm2,pAkm,p=0.Ifp=12h+7, then for any odd character χmodp, χ6=χ0 if and only if χ=χ2, χ2λ or χ=χ2λ¯. So, in this time, note that L1,χ2λ2=L1,χ2λ¯2; applying (37) and Lemma 4, we have(39)∑m=1p−1Sm2,pAkm,p=1π2⋅pp−1⋅∑m=1p−1λmAkm,p+∑m=1p−1λ¯mAkm,p⋅L1,χ2λ2+hp2p−1∑m=1p−11Akm,p,=1π2⋅pp−1⋅p−1Ak1,p−Hk,p⋅L1,χ2λ2+hp2p−1⋅Hk,p,where λ is any third-order character modulo p.It is clear that Theorem1 follows from (36) and (38).Theorem2 follows from (39) and Lemma 2.Now, we prove Theorem3. Note that p≡1mod3 and (see [17] or [18]):(40)∑χmodpχ−1=−1L1,χ4=5π4144⋅p+Oexp4lnplnlnp,(41)∑χmodpχ−1=−1L1,χ2⋅L1,χλ2=p2⋅∑n=1∞∑d|nλd2n2+Oexp4lnplnlnp.From (35), (40), and (41) and Lemma 3, we have the asymptotic formula as follows:(42)∑m=1p−1S2m,pAkm,p=1π4p2p−12∑χmodpχ−1=−1∑χmodpχ−1=−1∑m=1p−1χmηmAkm,pL1,χ2L1,η2,=1π4⋅p2p−12⋅∑m=1p−11Akm,p⋅∑χmodpχ−1=−1L1,χ4+2L1,χ2L1,χλ2,=5p144+pπ4⋅∑n=1∞∑d|nλd2n2⋅Hk,p+OHk,p⋅exp4lnplnlnp.Applying Lemma2 and mathematical induction, we can easily deduce the estimate(43)Hk,p≪pdk.In fact, from Lemma2, we have H0,p=p−1, H1,p=−p−1/d, and H2,p=3p−1/d2. So, (43) is true for k=0,1, and 2. Note that the third-order linear recursive formula is as follows:(44)Hk,p=−3dHk−1,p+1dpHk−3,p.So, fork≥3, from the mathematical induction, we can easily deduce the estimate(45)Hk,p≤3d⋅Hk−1,p+1d⋅p⋅Hk−3,p≪pdk.Combining (42) and (43), we complete the proof of Theorem 3.


## 4. Conclusion

The main results in this paper are three theorems, which are closely related to Dedekind sums and cubic Gauss sums. They describe that whenp is a prime, the hybrid power mean of the Dedekind sums and cubic Gauss sums has good mean distribution properties. In fact, we can give some exact calculating formulae (see Theorems 1 and 2) or asymptotic formula (see Theorem 3) for them.



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