# On the Hybrid Power Mean of Two-Term Exponential Sums and Cubic Gauss Sums

Authors: Shaofan Cao; Tingting Wang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, an interesting third-order linear recurrence formula is presented by using elementary and analytic methods. This formula is concerned with the calculating problem of the hybrid power mean of a certain two-term exponential sums and the cubic Gauss sums. As an application of this result, some exact computational formulas for one kind hybrid power mean of trigonometric sums are obtained.


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## Body

## 1. Introduction

As usual, letp be an odd prime. For any integer m, we define the cubic Gauss sums Am,p=Am as follows:(1)Am=∑a=0p−1ema3p,where ey=e2πiy and i2=−1.In recent years, several scholars have studied the hybrid power mean problems of various trigonometric sums and proved many interesting results. For example, Chen and Hu [1] studied the computational problem of(2)Skp=∑m=1p−1∑a=0p−1ema3pk⋅∑c=1p−1emc+c¯p2,where c¯ denotes the multiplicative inverse of cmodp. That is, c⋅c¯≡1modp. For p≡1mod3, they obtained a third-order linear recurrence formula for Skp.Li and Hu [2] studied another hybrid power mean(3)∑b=1p−1∑a=0p−1eba4p2⋅∑c=1p−1ebc+c¯p2,and gave an exact computational formula for (3).Some other related papers can also be found in references [3–11]. We would not have to repeat them here.Very recently, Chen and Chen [12] studied the recursive properties of the hybrid power mean(4)Hkc,p=∑m=1p−1∑a=0p−1ecma3pk⋅∑b=0p−1emb3+bp3,and obtained a third-order linear recurrence formula for it with c=1 and p≡1mod3. That is, they proved the following result:Letp be an odd prime with p≡1mod3. If 3 is a cubic residue modp, then for any integer k≥3, one has the third-order linear recurrence formula:(5)Hk1,p=3pHk−21,p+dpHk−31,p,where the first three terms are H01,p=2p2−pd, H11,p=p2d−6, and H21,p=p26p−5d.Note that Zhang and Zhang [13] proved the identity(6)∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2if3∤p−1,2p3−7p2if3|p−1.Perhaps this is the best result for which there is no conditional requirement on the primep. The interesting results of the above work motivate us to ask such a problem of whether there exists a similar recursive formula for the hybrid power mean(7)Ukp=∑m=1p−1∑a=0p−1ecma3pk⋅∑b=0p−1emb3+bp4,where p is an odd prime with p≡1mod3.Obviously, the problem in (7) is much harder than the problem in [12] because we are dealing with the fourth power mean of the two-term exponential sums in (7). Our main contribution is to obtain an identity for the fourth power mean of the two-term exponential sums weighted by third-order character modulo p, i.e., the following Lemma 3. Then, we use this lemma to derive several interesting recursion formulas for Ukp. In this way, the continuities and the value distribution properties of this kind of trigonometric sums can be described from different views. Of course, the reason why we focus on the calculation of (7) is that the problem is closely related to the number of the solutions of some congruence equation. These contents play a very important role in study of some famous analytic number theory problems, such as Waring problem and Goldbach conjecture.Through the study, it is found that the problem we studied is closely related to integer 3. If 3 is a cubic residue modulop, then there exists a beautiful third-order linear recurrence formula for Ukp, and the first three terms U0p, U1p, and U2p are integers. If 3 is not a cubic residue modp, then we can get the exact value of U2p. For any other positive integer k, we can only give a more complex mathematical representation for Ukp. That is, we have the following three results:Theorem 1.
Letp be a prime with p≡1mod3. If 3 is a cubic residue modulo p, then for any integer k≥3, we have the third-order linear recurrence formula(8)Ukp=3pUk−2p+dpUk−3p,where the first three terms are U0p=2p3−7p2, U1p=−dp6p−1, and U2p=p24p2−22p−d2+2 and d is uniquely determined by 4p=d2+27b2 and d≡1mod3.Theorem 2.
Letp be a prime with p≡1mod3. If 3 is a cubic residue modulo p, then for any integer k≥3, we also have the third-order linear recurrence formula(9)U−kp=−3d⋅U−k−1p+1dp⋅U−k−3p,where the first three terms are U0p=2p3−7p2, U−1p=−p⋅2p2+p+d2−2/d, and U−2p=1−3p+3p2p2+p−2/d2.Theorem 3.
Letp be an odd prime with p≡1mod3. If 3 is not a cubic residue modulo p, then we have the identity(10)∑m=1p−1∑a=0p−1ema3p2⋅∑a=0p−1ema3+ap4=p24p2−19p−d2−1.From our theorems, we may immediately deduce the following three corollaries:Corollary 1.
Letp be an odd prime with p≡1mod3, then we have(11)∑m=1p−1∑a=0p−1ema3p2⋅∑a=0p−1ema3+ap4=p24p2−22p−d2+2if3is a cubic residue modulop,p24p2−19p−d2−1if3is not a cubic residue modulop.Corollary 2.
Letp be a prime with p≡1mod3. If 3 is a cubic residue modulo p, then we have(12)∑m=1p−1∑a=0p−1ema3p4⋅∑a=0p−1ema3+ap4=p212p3−66p2−9d2p+6p+d2.Corollary 3.
Letp be a prime with p≡1mod3. If 3 is a cubic residue modulo p, then we have(13)∑m=1p−1∑a=0p−1ema3+a/p∑a=0p−1ema3/p4=27d4⋅2p3+p2−2p−1d2⋅8p2+7p−11−1.Some notes: first in Theorem2, if 3,p−1=1, then the question we are discussing is trivial. Because in this case, we have(14)∑a=0p−1ema3p=∑a=0p−1emap=0.Second, ifp≡1mod3 and 3 is not a cubic residue modulo p, then we can only get the exact value of U2p.Third, the advantage of our work is that we completely solve the calculation problem ofUkp with p≡1mod3.Fourth, the mean value estimation of the exponential sums is closely related to the upper and lower bounds of the individual exponential sums. So, by studying the mean value of the positive exponential sums, we can obtain a better upper bound estimation of the exponential sums. If we want to get its lower bound estimation of the exponential sums, we should study the negative power of the exponential sums. Our Theorems1 and 2 address both types of problems.Finally, for any fixed positive integerh≥5, whether there is a third-order linear recurrence formula for the hybrid power mean(15)Wkh,p=∑m=1p−1∑a=0p−1ema3pk⋅∑b=0p−1emb3+bph,is an open problem, which is the limitation of our work. The other drawback, of course, is that we cannot compute all Ukp when 3 is not a cubic residue modulo p. In fact, our ultimate goal is to obtain a precise calculation formula for Wkh,p for all positive integers h≥5. In the future, we will continue to improve the research in this aspect. It also requires us to continue to study.


## 2. Several Lemmas

To complete the proofs of our theorems, several simple lemmas are necessary. Hereafter, we will use many properties of the classical Gauss sums and the third-order character modulop, all of which can be found in books concerning about Elementary Number Theory or Analytic Number Theory, such as references [14–16], so the related contents will not be repeated here. First we have the following:Lemma 1.
Ifp is a prime with p≡1mod3, then for any third-order character ψmodp, we have the identity(16)∑m=1p−1ψm∑a=0p−1ema3+ap3=ψ¯3τ2ψ¯−3pτψ.Proof.
First applying trigonometric identity(17)∑m=1qenmq=qifq|n,0ifq∤n,and noting that ψ3=χ0, the principal character modulo p, we have(18)∑m=1p−1ψm∑a=0p−1ema3+ap3=∑m=1p−1ψm∑a=0p−1ema3+ap2+∑m=1p−1ψm∑a=0p−1ema3+ap2∑a=1p−1ema3+ap=2∑m=1p−1ψm∑a=1p−1ema3+ap+∑m=1p−1ψm∑a=1p−1ema3+ap2+τψ∑a=0p−1∑b=0p−1∑c=1p−1ψ¯a3+b3+c3ea+b+cp=−2τψ+τψ∑a=1p−1ψ¯a3+1∑b=1p−1eba+1p+τψ∑a=0p−1∑b=0p−1ψ¯a3+b3+1∑c=1p−1eca+b+1p=−2τψ−τψ∑a=1p−1ψ¯a3+1+pτψ∑a=0p−1∑b=0p−1a+b+1≡0modpψ¯a3+b3+1−τψ∑a=0p−1∑b=0p−1ψ¯a3+b3+1=−2τψ−τψ∑a=1p−1ψ¯a3+1+pτψ∑a=0p−1ψ¯a3−a+13+1−τψ∑a=0p−1∑b=0p−1ψ¯a3+b3+1.
Note thatψ2=ψ¯ and τψτψ¯=p, and from the properties of Gauss sums and the characteristic function of the third-order character modulo p(19)1+ψa+ψ¯a=3,ifais a cubic residue modulop,0,ifais not a cubic residue modulop,we have(20)∑a=1p−1ψ¯a3+1=∑a=1p−1ψ¯a+11+ψa+ψ¯a=∑a=1p−1ψ¯a+1+∑a=1p−1ψ¯1+a¯+∑a=1p−1ψ¯a2+a=−2+1τψ∑b=1p−1ψb∑a=1p−1ψ¯aeba+1p=−2+τ2ψ¯τψ=−2+τ3ψ¯p,(21)∑a=0p−1ψ¯a3−a+13+1=∑a=0p−1ψ¯−3aa+1=ψ¯3∑a=1p−1ψ¯aa+1=ψ¯3τ3ψ¯p.
Sinceψ is a third-order character modulo p, for any integer c with c,p=1, from the properties of the classical Gauss sums, we have(22)∑a=0p−1eca3p=1+∑a=1p−11+ψa+ψ¯aecap=ψ¯cτψ+ψcτψ¯.
From (22) and the properties of Gauss sums, then we can get(23)∑a=0p−1∑b=0p−1ψ¯a3+b3+1=1τψ∑c=1p−1ψc∑a=0p−1∑b=0p−1eca3+cb3+cp=1τψ∑c=1p−1ψcecp∑a=0p−1eca3p2=1τψ∑c=1p−1ψcecpψcτ2ψ+2p+ψ¯cτ2ψ¯=τψτψ¯+2p−τ3ψ¯p=3p−τ3ψ¯p.
Combining (18), (20), (21), and (23), we have the identity(24)∑m=1p−1ψm∑a=0p−1ema3+ap3=ψ¯3τ2ψ¯−3pτψ.
This proves Lemma1.Lemma 2.
If p is a prime withp≡1mod3 and ψ is any third-order character modulo p, then we have the identity(25)τ3ψ+τ3ψ¯=dp,where τψ denotes the classical Gauss sums, d is uniquely determined by 4p=d2+27b2, and d≡1mod3.Proof.
See [3] or [11].Lemma 3.
Ifp is a prime with p≡1mod3, then for any third-order character ψmodp, we have the identity(26)∑m=1p−1ψm∑a=0p−1ema3+ap4=ψ¯3−3p−ψ¯3pτ2ψ¯−dpτψ.Proof.
Note that the two-term exponential sums satisfies(27)∑a=0p−1ema3+ap=∑a=0p−1e−ma3−ap.
So, from the properties of Gauss sums and Lemma1, we have(28)∑m=1p−1ψm∑a=0p−1ema3+ap4=∑m=1p−1ψm∑a=0p−1ema3+ap3∑d=1p−1emd3+dp+∑m=1p−1ψm∑a=0p−1ema3+ap3=τψ∑a=0p−1∑b=0p−1∑c=0p−1ψ¯a3+b3−c3−1∑d=1p−1eda+b−c−1p+ψ¯3τ2ψ¯−3pτψ=pτψ∑a=0p−1∑b=0p−1∑c=0p−1a+b≡c+1modpψ¯a3+b3−c3−1+ψ¯3τ2ψ¯−3pτψ−τψ∑a=0p−1∑b=0p−1∑c=0p−1ψ¯a3+b3−c3−1.
From (22) and Lemma 2, we have(29)A3m=∑a=0p−1ema3p3=ψ¯cτψ+ψcτψ¯3=τ3ψ+τ3ψ¯+3pAm=dp+3pAm.
It is clear thatAm is a real number, so from the properties of Gauss sums and (29), we have(30)τψ∑a=0p−1∑b=0p−1∑c=0p−1ψ¯a3+b3−c3−1=∑m=1p−1ψm∑a=0p−1∑b=0p−1∑c=0p−1ema3+b3−c3−1p=∑m=1p−1ψme−mpA3m=∑m=1p−1ψme−mpdp+3pAm=dpτψ+3p∑m=1p−1ψmψ¯mτψ+ψmτψ¯e−mp=dpτψ−3pτψ+3pτ2ψ¯=pd−3τψ+3pτ2ψ¯.
Note that the congruencea+b≡c+1modp implies the congruence(31)a3+b3−c3−1≡−3a+ba−1b−1modp.
So, we have(32)∑a=0p−1∑b=0p−1∑c=0p−1a+b≡c+1modpψ¯a3+b3−c3−1=∑a=0p−1∑b=0p−1ψ¯−3a+ba−1b−1=ψ¯3∑a=1p−1∑b=1p−1ψ¯a+b+2ab=ψ¯3∑a=1p−1ψ¯a∑b=1p−1ψ¯ba+b+2.
It is clear that(33)∑b=1p−1ψ¯ba+b+2=1τψ∑c=1p−1ψc∑b=1p−1ψ¯becb+a+2p=τψ¯τψ∑c=1p−1ψ2ceca+2p=τ2ψ¯τψ⋅ψa+2.
From (32) and (33), we have(34)∑a=1p−1∑b=1p−1ψ¯a+b+2ab=τ2ψ¯τψ∑a=1p−1ψ¯aψa+2=τ2ψ¯τψ∑a=1p−1ψ1+2a¯=τ2ψ¯τψ∑a=1p−1ψ1+2a=−τ2ψ¯τψ.
Combining (28), (30), (32), and (34), we may immediately deduce that(35)∑m=1p−1ψm∑a=0p−1ema3+ap4=ψ¯3−3p−ψ¯3pτ2ψ¯−dpτψ.
This proves Lemma3.Lemma 4.
Ifp is a prime with p≡1mod3, then we have the identity(36)∑m=1p−1∑a=0p−1ema3+ap4=2p3−7p2.Proof.
See Zhang and Zhang [13].


## 3. Proofs of the Theorems

Now, we prove our main results. First, we prove Theorem1. If 3 is a cubic residue modp, then ψ3=1. From (22), Lemmas 2 and 3, we have(37)U1p=∑m=1p−1Am∑a=0p−1ema3+ap4=∑m=1p−1ψ¯mτψ+ψmτψ¯∑a=0p−1ema3+ap4=τψ¯1−4pτ2ψ¯−dpτψ+τψ1−4pτ2ψ−dpτψ¯=1−4pτ3ψ+τ3ψ¯−2dp2=−dp6p−1.Applying (22), Lemmas 2–4, we have(38)U2p=∑m=1p−1∑a=0p−1ema3p2∑a=0p−1ema3+ap4=∑m=1p−1ψmτ2ψ+2p+ψ¯mτ2ψ¯∑a=0p−1ema3+ap4=2p2p3−7p2+τ2ψ1−4pτ2ψ¯−dpτψ+τ2ψ¯1−4pτ2ψ−dpτψ¯=p24p2−22p−d2+2.Applying (29), (37), Lemmas 3 and 4, we also have(39)U3p=∑m=1p−1∑a=0p−1ema3p3∑a=0p−1ema3+ap4=∑m=1p−1dp+3pAm∑a=0p−1ema3+ap4=dp2p3−7p2−3dp26p−1=dp22p2−25p+3.Ifk≥3, then from (29) and the definition of Ukp, we have(40)Ukp=∑m=1p−1Akm∑a=0p−1ema3+ap4=∑m=1p−1Ak−3mdp+3pAm∑a=0p−1ema3+ap4=3pUk−2p+dpUk−3p.Combining Lemma4, (37)–(40), we complete the proof of Theorem 1.Note that identityA3m=dp+3pAm, so we have A−1m=A2m−3p/dp, A−2m=Am−3pA−1m/dp, and A−3m=1/dp−3/d⋅A−2m.From Lemma4 and Theorem 1, we have the third-order linear recurrence formula:(41)U0p=2p3−7p2,U−1p=1dpU2p−3pU0p=−p⋅2p2+p+d2−2d,U−2p=1dpU1p−3pU−1p=1−3pd2+3p2p2+p−2d2.Ifk≥3, then we have the third-order linear recurrence formula(42)U−kp=−3d⋅U−k−1p+1dp⋅U−k−3p.This proves Theorem2.Now we prove Theorem3. If p≡1mod3 and 3 is not a cubic residue modulo p, then we have ψ3+ψ¯3+1=0 or ψ3+ψ¯3=−1. From (22), Lemmas 2–4, we have(43)∑m=1p−1∑a=0p−1ema3p2⋅∑a=0p−1ema3+ap4=∑m=1p−1ψmτ2ψ+2p+ψ¯mτ2ψ¯∑a=0p−1ema3+ap4=τ2ψψ¯3−3p−ψ¯3pτ2ψ¯−dpτψ+2p2p3−7p2+τ2ψ¯ψ3−3p−ψ3pτ2ψ−dpτψ¯=ψ3+ψ¯3−6p−ψ3+ψ¯3pp2−dpτ3ψ+τ3ψ¯+2p32p−7=−5p+1p2−d2p2+2p32p−7=p24p2−19p−d2−1.This completes the proofs of all our results.


## 4. Conclusion

This paper mainly proposed three theorems, which are all closely related to the hybrid power mean of the two-term exponential sums and the classical cubic Gauss sums. Theorems1 and 2 obtained two interesting third-order linear recurrence formulas of Ukp for k≥0 and k≤0, respectively, providing 3 is a cubic residue modulo p. However, when 3 is not a cubic residue modulo p, we gave an exact calculating formula for U2p in Theorem 3. In general, this work not only generalized the results in reference [12] but also provided some new ideas and methods for the further study of such problems.



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