# The Recursive Properties of the Error Term of the Fourth Power Mean of the Generalized Cubic Gauss Sums

Authors: Zhang Jin; Zhang Jiafan
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we use the analytic methods and the properties of the classical Gauss sums to study the properties of the error term of the fourth power mean of the generalized cubic Gauss sums and give two recurrence formulae for it.


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## Body

## 1. Introduction

For any integerq≥2 and any Dirichlet character χmodq, the definition of the classical Gauss sums Gm,χ;q is(1)Gm,χ;q=∑a=1qχaemaq,where m is an integer and ey=e2πiy.This sum and its properties are of great significance to the analytic number theory, and many number theory problems are closely related to them. Therefore, it is necessary to study the various properties ofGm,χ;q and related sums. In this paper, we consider the generalized k-th Gauss sums Gm,k,χ;q, which is defined as follows:(2)Gm,k,χ;q=∑a=1qχaemakq,where k is any positive integer and m is an integer with m,q=1.It is clear that this sum is a generalization of the classical Gauss sumsGm,χ;q. In fact, Gm,1,χ;q=Gm,χ;q. Of course, the value of Gm,k,χ;q is irregular as χ varies. However, some scholars have found that Gm,k,χ;q has good value distribution properties in some problems of weighted mean value, even if we can get their exact calculation formulae for some 2kth power mean. In addition, there are some good upper bound estimates for Gm,k,χ;q. For example, for any integer n with n,q=1, from the general result of Cochrane and Zheng [1], we can deduce(3)Gm,k,χ;q≤2ωqq1/2,where ωq denotes the number of distinct prime divisors of q. The case that q is a prime is due to Weil [2].Fork=2, by the results of Zhang [3], let n be any integer with n,p=1, and there are the following two identities:(4)1p−1∑χmodpGn,2,χ;p4=3p2−6p−1+4npp,ifp≡1mod4,3p2−6p−1,ifp≡3mod4,1p−1∑χmodpGn,2,χ;p6=10p3−25p2−4p−1,ifp≡3mod4,where ∗/p denotes the Legendre symbol modulo p.Zhang and Liu [4] have studied the sum(5)∑χmodpGn,3,χ;q4and obtained the following calculation formula:(6)∑χmodpG1,3,χ;p4=5p3−18p2+20p+1+U5p+5pU−5U3−4U2+4U,where p is a prime with 3|p−1 and U=∑a=1pea3/p is a real constant.However, the value ofU was not given in [4], and the form of U was not concise. Now, for any integer 1≤m≤p−1, we let(7)Em,p=1p−1∑χmodpGm,3,χ;p4−5p2+13p+1.In this paper, we use the analytic methods and the properties of the classical Gauss sums to study the calculating problem of thenth power mean of Em,p and give two recurrence formulae for it. That is, we shall prove the following main results.Theorem 1.
Letp be an odd prime with 3|p−1. Then, for any positive integer n and integer m with m,p=1, we have the third-order linear recurrence formula:(8)Enm,p=Up,d⋅En−3m,p+Vp,d⋅En−2m,p,where Up,d and Vp,d are defined as(9)Up,d=64dp+d5−12d4+60d3−160d2+240d−384p2−5d3−48d2+180d−320p3+5d−24p4,Vp,d=3p⋅p2+8p−4dp+4d2−16d+16,and 4p=d2+27b2, where d is uniquely determined by d≡1mod3 (see [5]).Theorem 2.
Letp be an odd prime with 3|p−1. Then, for any positive integer n≥3, we have the recurrence formula(10)∑m=1p−1Enm,p=Up,d∑m=1p−1En−3m,p+Vp,d∑m=1p−1En−2m,p,where the first three terms of ∑m=1p−1Enm,p are(11)∑m=1p−1Em,p=0,∑m=1p−1E2m,p=2pp−1p−2d+42,∑m=1p−1E3m,p=p−1Up,d.Theorem 3.
Letp be an odd prime with 3|p−1. Then, for any positive integer n≥1, we have the third-order linear recurrence formula(12)∑m=1p−1E−nm,p=1Up,d⋅∑m=1p−1E−n−3m,p−Vp,dUp,d⋅∑m=1p−1E−n−1m,p,where the initial values of ∑m=1p−1E−nm,p are(13)∑m=1p−1E−1m,p=2pp−1p−2d+42−p−1Vp,dUp,d,∑m=1p−1E−2m,p=p−1V2p,d−2pp−1p−2d+42Vp,dU2p,d,∑m=1p−1E−3m,p=p−1Up,d+2pp−1p−2d+42V2p,d−p−1V3p,dU3p,d.Takingn=4 in Theorem 3, we may immediately deduce the following corollary.Corollary 1.
Letp be an odd prime with 3|p−1; then, we have the identity(14)1p−1∑m=1p−11Em,p4=2pp−2d+42−2Vp,dU2p,d+V4p,d−2pp−2d+42V3p,dU4p,d.


## 2. Several Lemmas

In this section, we give three lemmas which are necessary in the proofs of our theorems. In the process of proving our lemmas, we need some knowledge of the analytic number theory; all of which can be found in [6–8], so it is not necessary to repeat them here.Lemma 1.
Letp be an odd prime with p≡1mod3. Then, for any third-order character λmodp, we have(15)τ3λ+τ3λ¯=dp,where d is uniquely determined by 4p=d2+27b2 and d≡1mod3.Proof.
This result can be found in [9] or [10].Lemma 2.
Letp be an odd prime with 3|p−1. Then, for any cubic character λmodp, we have the identities(16)Em,p=λ¯m∑a=1p−1λ¯a3−12⋅τλ+λm∑b=1p−1λb3−12⋅τλ¯=λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−22.Proof.
For any integer1≤a≤p−1, it is easy to show that(17)1+λa+λ2a=3,ifais a cubic residuemodp,0,otherwise.
From the properties of the cubic character modulop, we have(18)λ2=λ¯,λ−1=1,τλ¯=τλ¯.
So, we have the identity(19)∑a=1p−1λa3−1=1τλ¯∑b=1p−1∑a=1p−1λ¯beba3−1p=1τλ¯∑b=1p−1λ¯be−bp∑a=1p−11+λa+λ¯aebap=1τλ¯∑b=1p−1λ¯be−bp−1+λ¯bτλ+λbτλ¯=1τλ¯−λ−1τλ¯+τλ∑b=1p−1λbe−bp+τλ¯∑b=1p−1e−bp=1τλ¯−τλ¯+τ2λ−τλ¯=τ3λp−2.
Therefore,(20)Em,p=λ¯mτλ∑a=1p−1λ¯a3−12+λmτλ¯∑a=1p−1λa3−12=λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−22.
This proves Lemma2.Lemma 3.
Letp be an odd prime with 3|p−1. Then, for any cubic character λ modulo p, we have the identity(21)E3m,p=Up,d+Vp,d⋅Em,p,where(22)Up,d=64dp+d5−12d4+60d3−160d2+240d−384p2−5d3−48d2+180d−320p3+5d−24p4,Vp,d=3p⋅p2+8p−4dp+4d2−16d+16,and Em,p is defined as the same as in Lemma 2.Proof.
From Lemma2, we have(23)E3m,p=λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−223=τλτ3λ¯p−223+τλ¯τ3λp−223+3p⋅τ3λ¯p−22τ3λp−22⋅Em,p.
Note thatτλτλ¯=p, and from (17), we have(24)τ6λ+τ6λ¯=τ3λ+τ3λ¯2−2τ3λτ3λ¯=d2p2−2p3.
Therefore,(25)τ3λ¯p−22τ3λp−22=4−4τ3λ¯p+τ6λ¯p2⋅4−4τ3λp+τ6λp2=16−16pτ3λ+τ3λ¯+4p2τ6λ+τ6λ¯+16p2τ3λτ3λ¯−4p3τ3λτ6λ¯+τ3λ¯τ6λ+1p4τ6λτ6λ¯=16−16p⋅dp+4p2⋅d2p2−2p3+16p2⋅p3−4p3⋅p3⋅dp+1p4⋅p6=p2+8p−4dp+4d2−16d+16.
In addition,(26)τλτ3λ¯p−223=τ3λτ3λ¯p−26=τ3λτ18λ¯p6+36τ12λ¯p4+144τ6λ¯p2+64−12τ15λ¯p5+24τ12λ¯p4−16τ9λ¯p3−144τ9λ¯p3+96τ6λ¯p2−192τ3λ¯p=64τ3λ+240pτ3λ¯−160τ6λ¯+60pτ9λ¯−12p2τ12λ¯+1p3τ15λ¯−192p2.
Using the method similar to (24), we obtain(27)τ9λ+τ9λ¯=τ6λ+τ6λ¯τ3λ+τ3λ¯−p3τ3λ+τ3λ¯=d3p3−3dp4,τ12λ+τ12λ¯=d4p4−4d2p5+2p6,τ15λ+τ15λ¯=d5p5−5d3p6+5dp7.
Combining formulae (23)–(27), we have(28)E3m,p=64τ3λ+τ3λ¯+240pτ3λ¯+τ3λ−160τ6λ¯+τ6λ+60pτ9λ¯+τ9λ−12p2τ12λ¯+τ12λ+1p3τ15λ¯+τ15λ−384p2+3p⋅p2+8p−4dp+4d2−16d+16⋅Em,p=5d−24p4−5d3−48d2+180d−320p3+d5−12d4+60d3−160d2+240d−384p2+64dp+3p⋅p2+8p−4dp+4d2−16d+16⋅Em,p=Up,d+Vp,d⋅Em,p.
This proves Lemma3.


## 3. Proofs of the Theorems

Now, we shall complete the proofs of our main results. Firstly, we prove Theorem1. Let p be an odd prime with 3|p−1, χ be any Dirichlet character modulo p, and λ be a cubic character modulo p. Then, from the properties of the classical Gauss sums and (17), we have(29)∑χmodpGm,3,χ;p4=∑χmodp∑a=1p−1χaema3p4=∑χmodp∑a=1p−1∑b=1p−1∑c=1p−1∑d=1p−1χabcd¯ema3+b3−c3−d3p=p−1∑a=1p−1∑b=1p−1∑c=1p−1∑d=1p−1ab≡cdmodpema3+b3−c3−d3p=p−1∑a=1p−1∑b=1p−1∑c=1p−1∑d=1p−1ab≡cmodpemd3a3+b3−c3−1p=p−1∑a=1p−1∑b=1p−1∑d=1p−1emd3a3−1b3−1p=p−1∑a=1p−1∑b=1p−1∑d=1p−11+λd+λ¯demda3−1b3−1p=p−1∑a=1p−1∑b=1p−1∑d=1p−1emda3−1b3−1p+p−1λ¯mτλ∑a=1p−1∑b=1p−1λ¯a3−1b3−1+p−1λmτλ¯∑a=1p−1∑b=1p−1λa3−1b3−1=p−1∑a=1p−1∑b=1p−1∑d=1p−1emda3−1b3−1p+p−1Em,p,where Em,p is the same as in Lemma 2.For any integern, we have the trigonometric identity(30)∑a=1p−1enap=p−1,ifn,p=p,−1,ifn,p=1.From (30), we have the identity(31)∑a=1p−1∑b=1p−1∑d=1p−1emda3−1b3−1p=2p−1∑a=1p−1∑b=1p−11−p−1a3≡1modp∑a=1p−1∑b=1p−11a3≡1modp−b3≡1modp∑a=1p−1∑b=1p−11a3≡1modpb3≡1modp=6p−12−9p−1−p−42=5p2−13p−1.Combining (29)–(31), we have(32)∑χmodpGm,3,χ;p4=∑χmodp∑a=1p−1χaema3p4=p−15p2−13p−1+p−1Em,p.For any positive integern, from Lemma 3, we have(33)Enm,p=E3m,p⋅En−3m,p=Up,d+Vp,d⋅Em,p⋅En−3m,p=Up,d⋅En−3m,p+Vp,d⋅En−2m,p,where(34)Up,d=64dp+d5−12d4+60d3−160d2+240d−384p2−5d3−48d2+180d−320p3+5d−24p4,Vp,d=3p⋅p2+8p−4dp+4d2−16d+16.This proves Theorem1.Now, we prove Theorem2. From Theorem 1, we have(35)∑m=1p−1Enm,p=Up,d∑m=1p−1En−3m,p+Vp,d∑m=1p−1En−2m,p.From Lemma2, we have(36)∑m=1p−1Em,p=∑m=1p−1λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−22=0,(37)∑m=1p−1E2m,p=∑m=1p−1λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−222=∑m=1p−1λmτ2λτ3λ¯p−24+∑m=1p−1λ¯mτ2λ¯τ3λp−24+∑m=1p−12pτ3λ¯p−22τ3λp−22=2pp−1p−2d+42.From (36) and Lemma 3, we have(38)∑m=1p−1E3m,p=∑m=1p−1λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−223=∑m=1p−1Up,d+Vp,dEm,p=p−1Up,d.Now, Theorem2 follows from (35)–(38).Finally, we prove Theorem3. For any integer n≥1, from Lemma 3, we have(39)∑m=1p−1E−nm,p=1Up,d⋅∑m=1p−1E−n−3m,p−Vp,dUp,d⋅∑m=1p−1E−n−1m,p,(40)E2m,p=Up,d⋅E−1m,p+Vp,d,Em,p=Up,d⋅E−2m,p+Vp,d⋅E−1m,p,E0m,p=Up,d⋅E−3m,p+Vp,d⋅E−2m,p,E−1m,p=Up,d⋅E−4m,p+Vp,d⋅E−3m,p.Therefore, we have(41)∑m=1p−1E−1m,p=∑m=1p−1E2m,p−Vp,dUp,d=2pp−1p−2d+42−p−1Vp,dUp,d,(42)∑m=1p−1E−2m,p=∑m=1p−1Em,p−Vp,d⋅E−1m,pUp,d=p−1V2p,d−2pp−1p−2d+42Vp,dU2p,d,(43)∑m=1p−1E−3m,p=∑m=1p−11−Vp,d⋅E−2m,pUp,d=p−1Up,d+2pp−1p−2d+42V2p,d−p−1V3p,dU3p,d.Combining (39)–(43), we may immediately complete the proof of Theorem 3.


## 4. Conclusion

The main results of this paper give two third-order linear recurrence formulae for the error term of the fourth power mean of the generalized cubic Gauss sums, and these results are the improvement and generalization of [4]. They are some new contributions in the relevant fields.



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