# One Kind New Hybrid Power Mean and Its Computational Formulae

Authors: Li Wang; Xuexia Wang
Journal: Journal of Mathematics
 (2022)

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## Abstract

The main purpose of this study is to use the elementary and analytic methods and the properties of the classical Gauss sums to study the calculation problems of one kind of hybrid power mean involving the quadratic character sums and the two-term exponential sums and give an exact computational formula for it.


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## Body

## 1. Introduction

Letq, r, and s be three positive integers with q>2 and r>s≥1. For any integers m and n, the two-term exponential sums Gm,n,r,s;q is defined as follows:(1)Gm,n,r,s;q=∑a=0q−1emar+nasq,where ey=e2πiy and i2=−1.The study of the properties of the two-term exponential sum plays an important role in analytic number theory, and many famous number theory problems such as Waring’s problem and Goldbach’s conjecture are closely related to it. Because of this, many scholars have studied various properties of the two-term exponential sums and obtained a series of valuable research results [1–5].In addition, for any odd primep, the quadratic character modulo p is called Legendre’s symbol, and its definition is(2)ap=1,ifais a quadratic residue modulop,−1,ifais a quadratic nonresidue modulop,0,ifp|a.It is the introduction of this symbol that greatly facilitates the study of quadratic residue properties and promotes the development of analytic number theory. Some related important works can be seen in [6–15]. For example, let p and q be two distinct odd primes; then, one has the quadratic reciprocal formula (see [16]: Theorem 9.8 or [17]: Theorems 4–6).(3)pq⋅qp=−1p−1q−1/4.For any odd primep with p≡1mod4, there must be two nonzero integers αp and βp such that (see [17]: Theorems 4–11)(4)p=α2p+β2p=12∑a=1p−1a+a¯p2+12∑a=1p−1a+ra¯p2,where r is any quadratic nonresidue modulo p and a¯ is the inverse of amodp. That is, a¯ satisfies the equation x⋅a≡1modp.Now, for any odd primep with p≡1mod6 and integer n with n,p=1, we define quadratic character sums An as follows:(5)An=∑a=0p−1a3+n¯2p.Then, for any integerk, we consider the k-th hybrid power mean(6)∑n=1p−1Akn⋅∑a=0p−1ena4+ap2.In this paper, we tried to give an exact computational formula for (6). That is, we will prove the following several results.Theorem 1.
Letp be a prime with p≡1mod6. If 2 is a cubic residue modulo p, then we have the identity(7)∑n=1p−1An⋅∑a=0p−1ena4+ap2=p2p−d,ifp≡1mod12,−p2p+d,ifp≡7mod12.If 2 is not a cubic residue modulop, then we have the identity(8)∑n=1p−1An⋅∑a=0p−1ena4+ap2=−pp+difp≡1mod12,pp−difp≡7mod12,where d is uniquely determined by 4p=d2+27b2 and d≡1mod3.Theorem 2.
Letp be a prime with p≡1mod6. If 2 is a cubic residue modulo p, then we have the identity(9)∑n=1p−1A2n⋅∑a=0p−1ena4+ap2=p2p2−4p+d3−3pd−d2,ifp≡1mod12,p2p2+3dp−d3−d2,ifp≡7mod12.Theorem 3.
Letp be a prime with p≡1mod6. If 2 is a cubic residue modulo p, then we have the identity(10)∑n=1p−1A3n⋅∑a=0p−1ena4+ap2=pd3p−3dp2+6p2−3d3+6dp,ifp≡1mod12,pd3p−3dp2−6p2−d3,ifp≡7mod12.If 2 is not a cubic residue modulop, then we have the identity(11)∑n=1p−1A3n⋅∑a=0p−1ena4+ap2=pd3p−3dp2−3p2−3d3+6dp,ifp≡1mod12,pd3p−3dp2+3p2−d3,ifp≡7mod12.Theorem 4.
Letp be a prime with p≡1mod6. Then, for any integer k, we have the third-order linear recursive formula(12)∑n=1p−1Akn⋅∑a=0p−1ena4+ap2=3p⋅∑n=1p−1Ak−2n⋅∑a=0p−1ena4+ap2+d3−3dp⋅∑n=1p−1Ak−3n⋅∑a=0p−1ena4+ap2.Note that 2 is a cubic residue modulop if and only if d is an even number (see Corollary 1 in [18]). So from Theorems 1, 2, and 4, we may immediately deduce the following two corollaries.Corollary 1.
Letp be a prime with 4p=d2+27b2 and d≡1mod3. If d is an even number, then we have the identity(13)∑n=1p−1∑a=0p−1ena4+a/p2∑a=0p−1a3+n¯2/p=pp2+3dp−d3+d2−5pd3p−d2,ifp≡1mod12,pp2−3dp+d3+d2−3pd3p−d2,ifp≡7mod12.Corollary 2.
Letp be a prime with 4p=d2+27b2 and d≡1mod3. If d is an even number, then we have the identity(14)∑n=1p−1∑a=0p−1ena4+a/p∑a=0p−1a3+n¯2/p2=p⋅3p3+3dp2−d3p+6d2p−15p2−d4d2⋅3p−d22,ifp≡1mod12,p⋅3p3−3dp2+d3p+6d2p−9p2−d4d2⋅3p−d22,ifp≡7mod12.
Notes: in Theorems1–3, why do we want to distinguish whether 2 is a cubic residue modulo p? The reason is that we need to determine the exact value of λ2+λ¯2, where λ is a third-order character modulo p. If 2 is a cubic residue modulo p, then λ2+λ¯2=2; if 2 is not a cubic residue modulo p, then λ2+λ¯2=−1.
In Theorem2, we only discussed the case p≡1mod6 and 2 is a cubic residue modulo p. In fact, if 2 is not a cubic residue modulo p, then we can also give an exact computational formula for it, but in this case, the result looks very cumbersome and not beautiful. Corollaries 1 and 2 are the same things. Hence, all these results are not listed.
In addition, in this paper, we did not consider primep with 3∤p−1, because in this case, the result is trivial. In fact in this case, if a passes through a complete residue system modulo p, then a3 also passes through a complete residue system modulo p. Hence, we have the identity(15)An=∑a=0p−1a3+n¯2p=∑a=0p−1a+n¯2p=0.


## 2. Some Lemmas

In this section, we will give several simple lemmas and they are all necessary in the proofs of our theorems. In addition, we also need some properties of the classical Gauss sums and character sums; all these can be found in [16, 17, 19], and we will not repeat them here. First we have the following.Lemma 1.
Letp be a prime with p≡1mod3. Then, for any third-order character λ modulo p, we have the identity(16)τ3λ+τ3λ¯=dp,where τχ=∑a=1p−1χaea/p denotes the classical Gauss sums, ey=e2πiy, i2=−1, and d is the same as defined the theorem.Proof.
See [20, 21].Lemma 2.
Letp be an odd prime. Then, for any nonprincipal character χ modulo p, we have the identity(17)τχ2=χ22τχ2⋅τχ⋅τχχ2,where χ2=∗/p denotes Legendre’s symbol modulo p.Proof.
In fact, this is the product formula of Gauss sums. We can see the general results in [20, 22, 23]. Related works can also be found in [24, 25].Lemma 3.
Letp be an odd prime. Then, for any integer n with n,p=1 and any third-order character λ modulo p, we have the identity(18)∑a=0p−1a3+n¯2p=λ4np⋅τ3λ+λ¯4np⋅τ3λ¯.Proof.
χ2=∗/p denotes Legendre’s symbol modulo p. Then, from the properties of the third-order character modulo p and classical Gauss sums, we have(19)∑a=0p−1χ2a3+n=χ2n+∑a=1p−11+λa+λ¯aχ2a+n=∑a=0p−1χ2a+n+λnχ2n∑a=1p−1λaχ2a+1+λ¯nχ2n∑a=1p−1λ¯aχ2a+1=λnχ2n∑a=1p−1λaχ2a+1+λ¯nχ2n∑a=1p−1λ¯aχ2a+1.
Note that the identity(20)∑a=1p−1λaχ2a+1=1τχ2∑b=1p−1χ2b∑a=1p−1λaeba+1p=τλτχ2∑b=1p−1χ2bλ¯bebp=τλ⋅τχ2λ¯τχ2.
Takingχ=λ¯ in Lemma 2, we have(21)τλ=λ2τχ2⋅τλ¯⋅τλ¯χ2.
Note thatτλ⋅τλ¯=p; from (20) and (21), we have(22)∑a=1p−1λaχ2a+1=λ¯2p⋅τ3λ.
Similarly, we also have(23)∑a=1p−1λ¯aχ2a+1=λ2p⋅τ3λ¯.
Note thatλ¯2=λ4; combining (19), (22), and (23), we have the identity(24)∑a=0p−1a3+np=λ4nχ24np⋅τ3λ+λ¯4nχ24np⋅τ3λ¯,∑a=0p−1a3+n¯2p=λ4np⋅τ3λ+λ¯4np⋅τ3λ¯.
This proves Lemma3.Lemma 4.
Letp be any odd prime with p≡1mod6; then, for a three-order character λ modulo p, we have the identity(25)∑m=1p−1λm⋅∑a=0p−1ema4+ap2=p⋅−λ2+λ4⋅τ2χ2⋅τ3λ¯p2.Proof.
For any integerb with b,p=1, note that the identity(26)∑a=0p−1eba2p=1+∑a=1p−11+χ2aebap=χ2b⋅τχ2,and from the properties of the Gauss sums and (21), we have(27)∑m=1p−1λm⋅∑a=0p−1ema4+ap2=τλ∑a=0p−1∑b=0p−1λ¯a4−b4ea−bp=τλ∑a=1p−1λ¯a4eap+∑a=0p−1∑b=1p−1λ¯a4−1λ¯b4eba−1p=τλτλ¯+τλ¯∑a=0p−1λ¯a4−1λa−1=p⋅1+∑a=0p−1λ¯a2+1a+1−λ¯4=p⋅1+∑a=1p−1λ¯a2−2a+2a−λ¯4=p⋅1+∑a=1p−1λ¯2a¯2−2a¯+1−λ¯4=p⋅∑a=0p−1λ¯2a2−2a+1−λ¯4=p⋅∑a=0p−1λ¯2a−2¯2+2¯−λ¯4=p⋅λ2∑a=0p−1λ¯a2+1−λ2=p⋅−λ2+λ2τλ∑b=1p−1λb∑a=0p−1eba2+1p=λ2p⋅−1+τχ2τλ∑b=1p−1λbχ2bebp=λ2p⋅−1+τχ2τχ2λτλ=p⋅−λ2+λ4⋅τ2χ2⋅τ3λ¯p2.
This proves Lemma4.Lemma 5.
Letp be any odd prime with p≡1mod6; then, for any integers k≥3 and n with n,p=1, we have the third-order linear recursive formula(28)Akn=3p⋅Ak−2n+d3−3dp⋅Ak−3n,where d is the same as defined in Theorem 1.Proof.
Note thatλ3=λ¯3=χ0; as the principal character modulo p, from Lemmas 1 and 3, we have(29)A3n=1p3⋅λ4n⋅τ3λ+λ¯4n⋅τ3λ¯3=1p3⋅τ9λ+τ9λ¯+3p3⋅λ4n⋅τ3λ+λ¯4n⋅τ3λ¯=1p3⋅τ3λ+τ3λ¯3−3p3τ3λ+τ3λ¯+3p4⋅An=d3−3dp+3p⋅An.
So, for any integerk≥3, from (29), we have the third-order recursive formula as follows:(30)Akn=Ak−3n⋅A3n=3p⋅Ak−2n+d3−3dp⋅Ak−3n.
This proves Lemma5.Lemma 6.
Letp be any odd prime with p≡1mod6; then, we have(31)∑n=1p−1∑a=0p−1ena4+ap2=pp−3,ifp≡1mod12,pp−1,ifp≡7mod12.Proof.
From the properties of the trigonometrical sums, we have(32)∑n=1p−1∑a=0p−1ena4+ap2=∑n=1p−11+∑a=1p−1ena4+ap+∑a=1p−1e−na4−ap+∑a=1p−1ena4+ap2=p−1+∑a=1p−1∑n=1p−1ena4+ap+∑a=1p−1∑n=1p−1e−na4−ap+∑a=1p−1∑b=1p−1∑n=1p−1enb4a4−1+ba−1p=p−1+2+∑a=1p−1∑n=1p−1ena4−1p∑b=1p−1eba−1p=p+1+p−12−∑a=2p−1∑n=1p−1ena4−1p.
Ifp≡1mod12, then note that the congruence equation a4≡1modp has four solutions 1≤a≤p−1; from (32), we have the identity(33)∑n=1p−1∑a=0p−1ena4+ap2=p+1+p−12−3p−1+p−5=pp−3.
Ifp≡7mod12, note that the congruence equation a4≡1modp has two solutions 1≤a≤p−1; so from (32), we have(34)∑n=1p−1∑a=0p−1ena4+ap2=p+1+p−12−p−1+p−3=pp−1.
Now, Lemma6 follows from formulae (33) and (34).


## 3. Proofs of the Theorems

In this section, we will complete the proofs of our theorems. First we prove Theorem1. Note that τλ⋅τλ¯=p; from Lemmas 1, 3, and 4, we have(35)∑n=1p−1An⋅∑a=0p−1ena4+ap2=∑n=1p−1λ4np⋅τ3λ+λ¯4np⋅τ3λ¯⋅∑a=0p−1ena4+ap2=λ4τ3λp∑n=1p−1λn∑a=0p−1ena4+ap2+λ¯4τ3λ¯p∑n=1p−1λ¯n∑a=0p−1ena4+ap2=−τ3λ−τ3λ¯+λ2⋅p⋅τ2χ2+λ¯2⋅p⋅τ2χ2=−dp+λ2+λ¯2⋅p⋅τ2χ2.If 2 is a cubic residue modulop, then from (35), we have(36)∑n=1p−1An⋅∑a=0p−1ena4+ap2=p2p−d,ifp≡1mod12,−p2p+d,ifp≡7mod12.If 2 is not a cubic residue modulop, then note that the identities λ2+λ¯2=−1 and τ2χ2=χ2−1⋅p, from (35), we have(37)∑n=1p−1An⋅∑a=0p−1ena4+ap2=−pp+d,ifp≡1mod12,pp−d,ifp≡7mod12.Now, Theorem1 follows from identities (36) and (37).Now, we prove Theorem2. From Lemma 3, we have(38)A2n=λ4np⋅τ3λ+λ¯4np⋅τ3λ¯2=λ¯4np2⋅τ6λ+λ4np2⋅τ6λ¯+2p.From (38) and Lemma 4, we have(39)∑n=1p−1A2n⋅∑a=0p−1ena4+ap2=∑n=1p−1λ¯4np2⋅τ6λ+λ4np2⋅τ6λ¯+2p⋅∑a=0p−1ena4+ap2=2p⋅∑n=1p−1∑a=0p−1ena4+ap2−τ6λp+λ¯2⋅τ9λ⋅τ2χ2p3−τ6λ¯p+λ2⋅τ9λ¯⋅τ2χ2p3.If 2 is a cubic residue modulop, then from (39) and Lemmas 1 and 6, we have(40)∑n=1p−1A2n⋅∑a=0p−1ena4+ap2=2p⋅∑n=1p−1∑a=0p−1ena4+ap2−1pd2p2−2p3+τ2χ2p3⋅τ3λ+τ3λ¯3−3p3τ3λ+τ3λ¯=p2p2−4p+d3−3pd−d2,ifp≡1mod12,p2p2+3dp−d3−d2,ifp≡7mod12.This proves Theorem2.Now, we prove Theorem3. If 2 is a cubic residue modulo p, then from Theorem 1 and Lemma 6, we have(41)∑n=1p−1A3n⋅∑a=0p−1ena4+ap2=∑n=1p−1dd2−3p+3p⋅An⋅∑a=0p−1ena4+ap2=dd2−3p∑n=1p−1∑a=0p−1ena4+ap2+3p∑n=1p−1An⋅∑a=0p−1ena4+ap2=pd3p−3dp2+6p2−3d3+6dp,ifp≡1mod12,pd3p−3dp2−6p2−d3,ifp≡7mod12.Similarly, if 2 is not a cubic residue modulop, then from Theorem 1 and Lemma 6, we also have(42)∑n=1p−1A3n⋅∑a=0p−1ena4+ap2=pd3p−3dp2−3p2−3d3+6dp,ifp≡1mod12,pd3p−3dp2+3p2−d3,ifp≡7mod12.Now, Theorem3 follows from (41) and (42).Theorem4 follows from Lemma 5.From Theorems1, 2, and 4 and Lemmas 5 and 6, we may immediately deduce Corollaries 1 and 2.This completes the proofs of our all results.


## 4. Conclusion

The main results of this study give four theorems for the hybrid power mean involving the quadratic character sums and the two-term exponential sums. In addition, we also obtained a third-order linear recursive formula for (6). That is, for any integer k, we have the recursive formula(43)∑n=1p−1Akn⋅∑a=0p−1ena4+ap2=3p⋅∑n=1p−1Ak−2n⋅∑a=0p−1ena4+ap2+d3−3dp⋅∑n=1p−1Ak−3n⋅∑a=0p−1ena4+ap2,where d is uniquely determined by 4p=d2+27b2 and d≡1mod3.So, for any integerk, we have solved the calculation problem of (6) completely.



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