# On the Fourth Hybrid Power Mean Involving the Generalized Gauss Sums

Authors: Xiaoge Liu; Yuanyuan Meng
Journal: Journal of Mathematics
 (2023)

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## Abstract

In this paper, we use the elementary and analytic methods to study the fourth hybrid power mean involving the generalized Gauss sums and prove several interesting identities for them.


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## Body

## 1. Introduction

Letq>1 be an integer, the generalized kth Gauss sums Gm,k,χ;q is defined as follows:(1)Gm,k,χ;q=∑a=1qχaemakq,where χ is a Dirichlet character modulo q and m and k are any integers with k≥1, ey=e2πiy, and i2=−1.Gm,k,χ;q is used to solve many problems related to number theory. Thus, many scholars at home and abroad have conducted various research studies on Gm,k,χ;q. When m=k=1, G1,1,χ;q is called as classical Gauss sums and represented by τχ. The most classical and important property of Gm,k,χ;q is that if χ is a primitive character modulo q, then Gm,1,χ;q=χ¯mτχ and τχ=q (see [1,2]). Besides, some mathematicians have obtained identities related to classical Gauss sums. For example, when p is a prime number, p≡1mod3. Berndt and Evans [3] obtained the following identity, i.e.,(2)τ3λ+τ3λ¯=dp,where λ is any third-order character modulo p, d is uniquely determined by 4p=d2+27b2, and d≡1mod3.Whenp is a prime number, p≡1mod4. Chen and Zhang [4] obtained(3)τ2χ4+τ2χ¯4=2p⋅α,where χ4 is any fourth-order character modulo p, α=1/2∑a=1p−1a+a¯/p, and ∗/p=χ2 denotes Legendre’s symbol modulo p.Furthermore,α=αp and p go hand in hand. When p≡1mod4, p can be written as the sum of two square numbers (see Theorem 4–11 in [5]), that is,(4)p=α2+β2≡12∑a=1p−1a+a¯p2+12∑a=1p−1a+ra¯p2,where r is any quadratic nonresidue modulo p, i.e., χ2r=−1.There are a number of studies with regard to the Gauss sums and their recursion properties, which can be found in [6–15].Understanding the relationship between differentkth Gauss sums will help us have a deeper understanding of Gauss sums. Thus, the aim of this paper is to consider the hybrid power mean of Gm,3,χ4;p and Gm,4,λ;p, i.e.,(5)∑m=1p−1∑a=1p−1χ4aema3p2k⋅∑b=1p−1λbemb4p2h,where p is a prime number with p≡1mod12, χ4 and λ denote any fourth-order and third-order character modulo p, respectively, and k and h are two nonnegative integers.So far, we have not seen anyone studying the above problem. Thus, in this paper, using analytic methods and the properties of the Gauss sums, we obtain several interesting computational formulae for (5) with k=h=1 and k=2 and h=0. That is, we give the proof of the following two results.Theorem 1.
Letp be an odd prime number with p≡1mod12. Then, for any third-order character λ and fourth-order character χ4 modulo p, we have the identity as(6)∑m=1p−1∑a=1p−1λaema4p2⋅∑a=1p−1χ4aema3p2=12p2p−1.Theorem 2.
Letp be an odd prime number with p≡1mod12. Then, for any fourth-order character χ4 modulo p, we have the identity as(7)∑m=1p−1∑a=1p−1χ4aema3p4=15p2p−1+4χ43χ4−1pp−12α2−p+4pp−1d⋅Cp,where the definition of d is the same as in (2), Cp=±α or Cp=±β, and the definitions of α and β are the same as defined in (4).Obviously, Theorem2 has its limitations. That is, it is not clear what value Cp is. How to determine the exact value of Cp in Theorem 2 is an interesting open problem. We will start to study above works in the future.


## 2. Several Lemmas

In this section, we are going to give five lemmas, which are very important in the proof of the theorems. In the process of proof, we will use some knowledge of analytic number theory, which can be found in [1, 2, 5].Lemma 1.
Letp be a prime number with p≡1mod6. Then, for any sixth-order character ψmodp, we have the identity as(8)τ3ψ+τ3ψ¯=p1/2⋅d2−2p,ifp≡1mod12;−i⋅p1/2⋅d2−2p,ifp≡7mod12,where i2=−1, d is uniquely determined by 4p=d2+27b2, and d≡1mod3.Proof.
For this, see Chen [16].Lemma 2.
Letp be an odd prime number with p≡1mod3. Then, for any character χ modulo p, we have the identity as(9)τχ3=1p⋅χ33⋅τχ⋅τχλ⋅τχλ¯,where λ is a third-order character modulo p.Proof.
For this, see [17] or [18]. The general result can also be found in [19].Lemma 3.
Letp be a prime number with p≡1mod12, χ4 be any fourth-order character, and λ be any third-order character modulo p. Then, we have the identity as(10)τ3χ4λ¯τ3χ4λ+τ3χ4λτ3χ4λ¯=d2−2pp.Proof.
Note thatλ2=λ¯ and χ42=χ¯42=χ2. From the properties of the classical Gauss sums, we obtain(11)∑a=0p−1χ4λa2−1=∑a=0p−1χ4λa+12−1=∑a=1p−1χ4λaχ4λa+2=1τχ¯4λ¯∑b=1p−1χ¯4λ¯b∑a=1p−1χ4λaeba+2p=τχ4λτχ¯4λ¯∑b=1p−1χ¯4λ¯bχ¯4λ¯be2bp=τχ4λτχ¯4λ¯∑b=1p−1χ2λbe2bp=λ¯2χ22⋅τχ4λ⋅τχ2λτχ¯4λ¯.
On the other hand, we have(12)∑a=0p−1eba2p=1+∑a=1p−11+χ2aebap=∑a=1p−1χ2aebap=χ2b⋅p,where b,p=1. Note that λ−1=1 and χ¯4χ2=χ4, and we also have(13)∑a=0p−1χ4λa2−1=1τχ¯4λ¯∑a=0p−1∑b=1p−1χ¯4λ¯beba2−1p=1τχ¯4λ¯∑b=1p−1χ¯4λ¯be−bp∑a=0p−1eba2p=pτχ¯4λ¯∑b=1p−1χ¯4λ¯bχ2be−bp=χ4−1p⋅τχ4λ¯τχ¯4λ¯.
From (11) and (13) and noting that λ32=χ222=χ42−1=1, we have(14)τχ2λ=λ2χ22χ4−1p⋅τχ4λ¯τχ4λ,or(15)τ3χ2λ=χ22⋅χ4−1⋅p3/2⋅τ3χ4λ¯τ3χ4λ.
Sinceχ22=−1p2−1/8 and χ4−1=−1p−1/4,(16)χ22⋅χ4−1=−1p2−1/8⋅−1p−1/4=−1p−1/2=1.
Combining (15), (16), and Lemma 1, we have the identity as(17)p3/2⋅τ3χ4λ¯τ3χ4λ+τ3χ4λτ3χ4λ¯=τ3χ2λ+τ3χ2λ¯=p1/2⋅d2−2p,orτ3χ4λ¯τ3χ4λ+τ3χ4λτ3χ4λ¯=d2−2pp.
This proves Lemma3.Lemma 4.
Letp be a prime number with p≡1mod12, χ4 be any fourth-order character, and λ be any third-order character modulo p. Then, we have the identity(18)τ3χ4λ¯+τ3χ4λτχ4+τ3χ¯4λ¯+τ3χ¯4λτχ¯42=4d2p2α2or4d2p2β2,where the definitions of α and β are given in (4).Proof.
Note thatτχ4τχ¯4=χ¯4−1⋅p, and taking χ=χ4 in Lemma 2, we have(19)τχ¯4=χ¯43pτχ4τχ4λτχ4λ¯.
So, from Lemma3 and (19), we have(20)τ3χ4λ¯+τ3χ4λ2τ2χ4=d2pτ3χ4λ¯τ3χ4λτ2χ4=d2p⋅χ¯43⋅p3⋅τ3χ¯4τχ4=χ¯43χ4−1⋅d2⋅p⋅τ4χ¯4.
From Lemma3, we have(21)τ3χ4λ¯+τ3χ4λτχ4τ3χ¯4λ¯+τ3χ¯4λτχ¯4=p⋅τ3χ4λ¯+τ3χ4λ2=p4⋅2+τ3χ4λ¯τ3χ4λ+τ3χ4λτ3χ4λ¯=p4⋅d2p=d2⋅p3.
Note thatχ¯43χ4−1=±1. Combining (20), (21), and (3), we have the identity as(22)τ3χ4λ¯+τ3χ4λτχ4+τ3χ¯4λ¯+τ3χ¯4λτχ¯42=χ¯43χ4−1⋅d2⋅p⋅τ4χ4¯+χ¯43χ4−1⋅d2⋅p⋅τ4χ4+2d2p3=χ¯43χ4−1⋅d2⋅p⋅τ2χ4+τ2χ¯42−2p2+2d2p3=2χ¯43χ4−1⋅d2⋅p2⋅2α2−p+2d2p3=4d2p2α2or4d2p2β2.
From (22), we can obtain Lemma 4.Lemma 5.
Letp be an odd prime number with p≡1mod12. Then, for any third-order character λ and fourth-order character χ4 modulo p and any integer m with m,p=1, we have the identity as(23)λχ¯4mAmBm=τχ¯4τλ+χ¯4mτχ¯4τχ4λ+χ2mτχ¯4τχ2λ+χ4mτχ¯4τχ¯4λ+λ¯mτχ¯4λτλ+λ¯χ¯4mτχ¯4λτχ4λ+λ¯χ2mτχ¯4λτχ2λ+λ¯χ4mτχ¯4λτχ¯4λ+λmτχ¯4λ¯τλ+λχ¯4−m⋅p+λχ2mτχ¯4λ¯τχ2λ+λχ4mτχ¯4λ¯τχ¯4λ.Proof.
Using the properties of the Gauss sums and Dirichlet character, we have(24)Am=∑a=1p−1χ4aema3p=∑a=1p−1χ¯4a3ema3p=∑a=1p−1χ¯4a1+λa+λ¯aemap=χ4mτχ¯4+λ¯mτχ¯4λ+λmτχ¯4λ¯,(25)Bm=∑a=1p−1λaema4p=∑a=1p−1λa4ema4p=∑a=1p−1λa1+χ4a+χ2a+χ¯4aemap=λ¯mτλ+χ¯4mτχ4λ+χ2mτχ2λ+χ4mτχ¯4λ.
Note thatχ4−1=χ¯4−1, τχ4λτχ¯4λ¯=χ4−1⋅p, and λ−1=1. From (24) and (25), we have(26)λχ¯4mAmBm=τχ¯4τλ+χ¯4mτχ¯4τχ4λ+χ2mτχ¯4τχ2λ+χ4mτχ¯4τχ¯4λ+λ¯mτχ¯4λτλ+λ¯χ¯4mτχ¯4λτχ4λ+λ¯χ2mτχ¯4λτχ2λ+λ¯χ4mτχ¯4λτχ¯4λ+λmτχ¯4λ¯τλ+λχ¯4−m⋅p+λχ2mτχ¯4λ¯τχ2λ+λχ4mτχ¯4λ¯τχ¯4λ.
This proves Lemma5.


## 3. Proofs of the Theorems

Now, let us begin to prove the theorems. Letp be an odd prime number with p≡1mod12, and then, for any nonprincipal character χ modulo p, we have(27)∑m=1p−1χm=0.From (27) and Lemma 5, we have(28)∑m=1p−1Am2⋅Bm2=∑m=1p−1AmBm⋅AmBm¯=∑m=1p−1|τχ¯4τλ+χ¯4mτχ¯4τχ4λ+χ2mτχ¯4τχ2λ+χ4mτχ¯4τχ¯4λ+λ¯mτχ¯4λτλ+λ¯χ¯4mτχ¯4λτχ4λ+λ¯χ2mτχ¯4λτχ2λ+λ¯χ4mτχ¯4λτχ¯4λ+λmτχ¯4λ¯τλ+λχ¯4−m⋅p+λχ2mτχ¯4λ¯τχ2λ+λχ4mτχ¯4λ¯τχ¯4λ|2=p2p−1+p2p−1+p2p−1+p2p−1+p2p−1+p2p−1+p2p−1+p2p−1+p2p−1+p2p−1+p2p−1+p2p−1=12p2p−1.This proves Theorem1.Now, we prove Theorem2. Note that λ2=λ¯ and χ43χ4−1=±1, and from (24), we have(29)χ2mA2m=τχ¯4+λ¯mτχ¯4λ+λmτχ¯4λ¯2=τ2χ¯4+λmτ2χ¯4λ+λ¯mτ2χ¯4λ¯+2λ¯mτχ¯4τχ¯4λ+2λmτχ¯4τχ¯4λ¯+2τχ¯4λτχ¯4λ¯.Applying Lemma 2, we have(30)τ2χ4=χ43χ¯4−1τχ¯4λτχ¯4λ¯.From (27), (29), (30), and Lemma 4, we have(31)∑m=1p−1Am4=∑m=1p−1τ2χ¯4+λmτ2χ¯4λ+λ¯mτ2χ¯4λ¯+2λ¯mτχ¯4τχ¯4λ+2λmτχ¯4τχ¯4λ¯+2τχ¯4λτχ¯4λ¯2=p−1τ2χ¯4+2τχ¯4λτχ¯4λ¯2+p−1τ2χ¯4λ+2τχ¯4τχ¯4λ¯2+p−1τ2χ¯4λ¯+2τχ¯4τχ¯4λ2=p−15p2+2τ2χ4τχ¯4λτχ¯4λ¯+2τ2χ¯4τχ4λ¯τχ4λ+p−15p2+2τ2χ4λ¯τχ¯4τχ¯4λ¯+2τ2χ¯4λτχ4τχ4λ+p−15p2+2τ2χ4λτχ¯4τχ¯4λ+2τ2χ¯4λ¯τχ4τχ4λ¯=15p2p−1+2p−1χ43χ4−1τ4χ4+τ4χ¯4+2χ¯43p−1pτ3χ4λ¯+τ3χ4λτχ4λτχ¯4λ¯τχ4+2χ43p−1pτ3χ¯4λ¯+τ3χ¯4λτχ¯4λτχ4λ¯τχ¯4=15p2p−1+4χ43χ4−1pp−12α2−p+2χ43χ4−1p−1τ3χ4λ¯+τ3χ4λτχ4+2χ43χ4−1p−1τ3χ¯4λ¯+τ3χ¯4λτχ¯4=15p2p−1+4χ43χ4−1pp−12α2−p±4pp−1dαor15p2p−1+4χ43χ4−1pp−12α2−p±4pp−1dβ.This completes the proof of Theorem2.


## 4. Conclusion

The main results of this article are two theorems. Theorem 1 studied an exact identity for the hybrid power mean involving two generalized Gauss sums. Theorem 2 studied one kind of fourth power mean of the generalized Gauss sums and proved that it has four possibilities. These results, on the one hand, give the exact values of one kind of hybrid power mean of the generalized Gauss sums, and on the other hand, they give new contribution to the research of related fields.



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