# A New Gauss Sum and Its Recursion Properties

Authors: Li Chen
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we introduce a new Gauss sum, and then we use the elementary and analytic methods to study its various properties and prove several interesting three-order linear recursion formulae for it.


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## Body

## 1. Introduction

Letq>1 be an integer. For any Dirichlet character χ modulo q, the classical Gauss sums Gm,χ;q is defined as follows:(1)Gm,χ;q=∑a=1qχaemaq,where m is any integer, ey=e2πiy, and i2=−1.For convenience, we writeτχ=G1,χ;q. The Gauss sum plays a very important role in the study of elementary number theory and analytic number theory, and many number theory problems are closely related to it. Because of this, many scholars have studied its various properties and obtained a series of important results. For example, if m,q=1, then we have the identity (see [1, 2])(2)Gm,χ;q=χ¯mG1,χ;q=χ¯mτχ.Ifχ is any primitive character modulo q, then one has also Gm,χ;q=χ¯mτχ and the identity τχ=q.In addition, Zhang and Hu [3] (or Berndt and Evans [4]) studied the properties of some special Gauss sums and obtained the following interesting results: let p be a prime with p≡1mod3. Then for any third-order character λ modulo p, one has the identity(3)τ3λ+τ3λ¯=dp,where d is uniquely determined by 4p=d2+27b2 and d≡1mod3.Chen and Zhang [5] studied the case of the fourth-order character modulo p and obtained the following conclusion: let p be a prime with p≡1mod4. Then for any four-order character χ4 modulo p, we have the identity(4)τ2χ4+τ2χ¯4=2p⋅α,α=12∑a=1p−1a+a¯p,where ∗/p=χ2 denotes Legendre’s symbol modulo p.The constantα=αp in (4) has a special meaning. In fact, we have the identity (For this, see Theorems 4–11 in [6])(5)p=α2+β2≡12∑a=1p−1a+a¯p2+12∑a=1p−1a+ra¯p2,where r is any quadratic nonresidue modulo p. That is, χ2r=−1.Some other results related to various Gauss sums and their recursion properties can also be found in references [7–10], and we will not list them all here.In this paper, we introduce a new Gauss sumAm=Am,p, which is defined as follows: let p be an odd prime. For any integer m with m,p=1, we define(6)Am=∑a=1p−1χ2aema3p,Gnp=∑m=1p−1A2nm,where χ2=∗/p denotes the Legendre’s symbol modulo p.It is clear that ifp−1,3=1, then note that χ23=χ2; from the properties of the reduced residue system modulo p, we have(7)Am=∑a=1p−1χ2a3ema3p=∑a=1p−1χ2aemap=χ2m⋅τχ2.So this time,Am=Gm,χ2;p=χ2mτχ2 becomes the classical Gauss sum.Ifp≡1mod3, then we only knew that Am is a real number, if p≡1mod12; and Am is a pure imaginary number, if p≡7mod12. In fact if p≡1mod12, then note that χ2−1=1, and this time we have(8)Am¯=∑a=1p−1χ2ae−ma3p=∑a=1p−1χ2−aem−a3p=Am.Ifp≡7mod12, then note that χ2−1=−1, and this time we have(9)Am¯=∑a=1p−1χ2ae−ma3p=−∑a=1p−1χ2−aem−a3p=−Am.But beyond these relatively simple properties, we do not know anything else. In this paper, we shall focus on the calculating problems ofGnp. We shall use the analytic methods to give an interesting three-order linear recursion formula for Gnp. That is, we shall prove the following two results.Theorem 1.
Letp be an odd prime with p≡7mod12. Then for any integer n≥3, we have the recursion formula(10)Gnp=−9pGn−1p−6p4p−d2Gn−2p−p4p−d22Gn−3p,where d is uniquely determined by 4p=d2+27b2 and d≡1mod3, and the three initial values G0p=p−1, G1p=−3p−1p, and G2p=p−1p11p+4d2.Theorem 2.
Letp be an odd prime with p≡1mod12. Then for any integer n≥3, we have the recursion formula(11)Gnp=9pGn−1p−6p4p−d2Gn−2p+p4p−d22Gn−3p,where d is the same as in Theorem 1, and the three initial values G0p=p−1, G1p=3p−1p, and G2p=p−1p11p+4d2.Of course, our theorems are also true for all integersn<0. In particular, we have the following conclusions:Theorem 3.
For any primep with p≡1mod3, we have the identities(12)∑m=1p−1Am4=p−1⋅p⋅11p+4d2,∑m=1p−11Am4=2243⋅p−1b4,where b is the same as defined in (3), i.e., 4p=d2+27b2.


## 2. Several Lemmas

In this section, we first give several simple lemmas. Of course, the proofs of these lemmas and theorems need some knowledge of character sums and analytic number theory. They can be found in many number theory books, such as [1, 2, 6], here we do not need to list.Lemma 1.
Letp be a prime with p≡1mod6. Then for any six-order character ψmodp, we have the identity(13)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, refer the study of Chen [11].Lemma 2.
Letp be a prime with p≡7mod12, χ2 denote Legendre’s symbol modulo p, and λ denote any three-order Dirichlet character modulo p. Then for any integer m with m,p=1, we have the identities(14)Am=χ2mτχ2+χ2m⋅λ¯mτχ2λ+λmτχ2λ¯,A2m=−p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯,A3m=3χ2mτχ2A2m+χ2mτχ24p−d2.Proof.
It is clear that for any integerr with r,p=1, from the properties of the three-order character modulo p, we have(15)1+λr+λ¯r=3,ifris a3rd residue modulop,0,otherwise.
It is clear thatχ2=χ¯2 is a real character modulo p; from (15) and the properties of the classical Gauss sums, we have(16)Am=∑a=1p−1χ2a3ema3p=∑a=1p−1χ2a1+λa+λ¯aemap=∑a=1p−1χ2aemap+∑a=1p−1χ2aλaemap+∑a=1p−1χ2aλ¯aemap=χ2mτχ2+χ2mλ¯mτχ2λ+χ2mλmτχ2λ¯.
Note thatp≡3mod4, χ2−1=−1, λ−1=1, τχ2=i⋅p, λ2=λ¯, and the identity τχ2λτχ2λ¯=χ2−1τχ2λτχ2λ¯=−p; from (16), we also have(17)A2m=χ2mτχ2+χ2mλ¯mτχ2λ+χ2mλmτχ2λ¯2=−p+2τχ2λ¯mτχ2λ+λmτχ2λ¯+λ¯mτχ2λ+λmτχ2λ¯2=−3p+2τχ2χ2mAm−τχ2+λmτ2χ2λ+λ¯mτ2χ2λ¯=−p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯.
From (16) and (17) and Lemma 1, we have(18)A3m=χ2mτχ2+χ2mλ¯mτχ2λ+χ2mλmτχ2λ¯×−p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯=−pAm+2χ2mτχ2A2m+χ2mτχ2λmτ2χ2λ+λ¯mτ2χ2λ¯+χ2mτ3χ2λ+τ3χ2λ¯−pAm−χ2mτχ2=2χ2mτχ2A2m+χ2mτχ2A2m+p−2χ2mτχ2Am−2pAm+χ2mτχ2p−χ2mτχ2d2−2p=3χ2mτχ2A2m+χ2mτχ24p−d2.
Now Lemma2 follows from (16)–(18).Lemma 3.
Letp be an odd prime with p≡1mod12. Then for any m with m,p=1, we have the identities(19)Am=χ2mτχ2+χ2m⋅λ¯mτχ2λ+λmτχ2λ¯,A2m=p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯,A3m=3χ2mpA2m−χ2mp4p−d2.Proof.
Note thatp≡1 modulo 4, χ2−1=1, τχ2=p, λ2=λ¯, and τχ2λτχ2λ¯=p; from (16) and the methods of proving Lemma 2, we also have(20)Am=χ2mτχ2+χ2mλ¯mτχ2λ+χ2mλmτχ2λ¯.(21)A2m=χ2mτχ2+χ2mλ¯mτχ2λ+χ2mλmτχ2λ¯2=p+2τχ2λ¯mτχ2λ+λmτχ2λ¯+λ¯mτχ2λ+λmτχ2λ¯2=3p+2τχ2χ2mAm−τχ2+λmτ2χ2λ+λ¯mτ2χ2λ¯=p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯.(22)A3m=χ2mτχ2+χ2mλ¯mτχ2λ+χ2mλmτχ2λ¯×p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯=pAm+2χ2mτχ2A2m+χ2mτχ2λmτ2χ2λ+λ¯mτ2χ2λ¯+χ2mτ3χ2λ+τ3χ2λ¯+pAm−χ2mτχ2=2χ2mτχ2A2m+χ2mτχ2A2m−p−2χ2mτχ2Am+2pAm+χ2mτχ2p+χ2mτχ2d2−2p=3χ2mpA2m−χ2mp4p−d2.
It is clear that Lemma3 follows from (20)–(22).


## 3. Proofs of the Theorems

Now we shall complete the proofs of our all results. First we prove Theorem1. Let p be an odd prime with p≡7mod12, and then note that τ2χ2=−p; from Lemma 2 and the properties of the character sums modulo p, we have(23)∑m=1p−1A2m=∑m=1p−1−p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯=−pp−1+2∑m=1p−1τχ2τχ2+λ¯mτχ2λ+λmτχ2λ¯=−p−1p+2∑m=1p−1τ2χ2=−3p−1p.From (23) and Lemmas 1 and 2, we have(24)∑m=1p−1A4m=∑m=1p−1−p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯2=p−1p2−4p∑m=1p−1χ2mτχ2Am−4p∑m=1p−1A2m+2p−1p2+4∑m=1p−1χ2mτχ2λmτ2χ2λ+λ¯mτ2χ2λ¯Am=19p−1p2+4p−1τχ2τ3χ2λ+τ3χ2λ¯=19p−1p2+4p−1pd2−2p=p−111p+4d2p.Ifn≥3, then 2n≥6, from Lemma 2, we have(25)A6m=−9pA4m−6p4p−d2A2m−p4p−d22,(26)A2nm=−9pA2n−2m−6p4p−d2A2n−4m−p4p−d22A2n−6m.From (23)–(26) and the definition of Gnp, we may immediately deduce the three-order linear recursion formula(27)Gnp=−9pGn−1p−6p4p−d2Gn−2p−p4p−d22Gn−3p,with the three initial values G0p=p−1, G1p=−3p−1p, and(28)G2p=p−111p+4d2p.This proves Theorem1.Now we prove Theorem2. If p be an odd prime with p≡1mod12, then note that χ2−1=1 and τχ2=p; from Lemma 3, we have(29)A6m=9pA4m−6p4p−d2A2m+p4p−d22,(30)A2nm=9pA2n−2m−6p4p−d2A2n−4m+p4p−d22A2n−6m.It is clear that from Lemmas1 and 3, we have(31)G1p=∑m=1p−1A2m=p⋅∑m=1p−11+2∑m=1p−1χ2mτχ2Am=3p−1p.(32)∑m=1p−1A4m=∑m=1p−1p+2χ2mτχ2Am+λmτ2χ2λ+λ¯mτ2χ2λ¯2=p−1p2+4p−1p2+4p∑m=1p−1A2m+2p−1p2+4p∑m=1p−1χ2mAmλmτ2χ2λ+λ¯mτ2χ2λ¯=19p−1p2+4p−1pτ3χ2λ+τ3χ2λ¯=19p−1p2+4p−1pd2−2p=p−1p4d2+11p.From (30)–(32) and the definition of Gnp, we have the three-order linear recursion formula(33)Gnp=9pGn−1p−6p4p−d2Gn−2p+p4p−d22Gn−3p,n≥3,where the three initial values G0p=p−1, G1p=3p−1p, and G2p=p−1p11p+4d2.This proves Theorem2.Now we prove Theorem3. First for any integer m with m,p=1, we have Am≠0. In fact if Am=0; then from (25), we have(34)A6m=−9pA4m−6p4p−d2A2m−p4p−d22=−p⋅27b22=0.This is impossible. So we haveAm≠0.On the contrary, ifp≡7mod12, then from (25), we have(35)∑m=1p−1A4m=−9p∑m=1p−1A2m−6p4p−d2p−1−p4p−d22∑m=1p−11A2m.Combining (23), (24), and (35), we have(36)p−111p+4d2p=27p2p−1−6p4p−d2p−1−p4p−d22∑m=1p−11A2m,or(37)∑m=1p−11A2m=−2p−14p−d2.From (23), (24), and (37), we also have(38)∑m=1p−1A2m=−9pp−1−6p4p−d2∑m=1p−11A2m−p4p−d22∑m=1p−11A4m,or(39)∑m=1p−11A4m=6p−14p−d22.Similarly, ifp≡1mod12, then from (31), (35), Lemma 3, and the methods of proving (39), we can also deduce that(40)∑m=1p−11A4m=6p−14p−d22.Ifp≡1mod12, then Am is a real number, so this time we have(41)A4m=Am4.Ifp≡7mod12, then Am is a pure imaginary number; this time we also have(42)Am4=A2m⋅Am¯2=A2m⋅−Am2=A4m.It is clear that from (24), (32), (41), and (42), we can deduce the identity(43)∑m=1p−1Am4=p−1⋅p⋅11p+4d2,p≡1mod3.From (39)–(42) and noting that 4p=d2+27b2, we can also deduce(44)∑m=1p−11Am4=2243⋅p−1b4,p≡1mod3.This completes the proofs of our all results.


## 4. Conclusion

The main result of this paper is to prove a three-order linear recursion formula for one kind new Gauss sums. As an application of this result, we obtained following conclusion: for any primep with p≡1mod3, we have the identities(45)∑m=1p−1Am4=p−1⋅p⋅11p+4d2,∑m=1p−11Am4=2243⋅p−1b4.These results not only gave the exact values for the fourth power mean and its inverse fourth power mean of a new Gauss sums, they are also some new contribution to research in related fields.



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