# A Four-Order Linear Recurrence Formula Involving the Quartic Gauss Sums and One Kind Two-Term Exponential Sums

Authors: Lan Qi; Xingxing Lv
Journal: Journal of Mathematics
 (2021)

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## Abstract

The main purpose of this paper is using the analytic method and the properties of the classical Gauss sums to study the computational problem of one kind hybrid power mean involving the quartic Gauss sums and two-term exponential sums and give an interesting four-order linear recurrence formula for it. As an application, we can obtain all values of this kind hybrid power mean with mathematica software.


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## Body

## 1. Introduction

In the number theory textbooks, especially the elementary number theory and the analytic number theory textbooks, there are many contents related to primes, such as famous prime number theorem and Dirichlet’s theorem. However, maybe the most important and most profound property related to primes is that any primep with p=4k+1 can be expressed as the square sums of two positive integers. That is, p=α2+β2. And, more precisely (see Theorems 4–11 of [1]),(1)p=α2+β2≡∑a=1p−1/2a+ra¯p2+∑a=1p−1/2a+na¯p2,where ∗/p denotes Legendre’s symbol modulo p, r and n are any two integers such that rn/p=−1, and a¯ satisfies the congruence equation a·a¯≡1modp.This conclusion has been extended by Professor W. P. Zhang in an unpublished paper. That is, he proved the following two conclusions. Letp be an odd prime with p≡1mod4. Then, for any nonprincipal even character χmodp, one has the identity(2)p=∑a=1p−1/2χa+ra¯2+∑a=1p−1/2χa+na¯2,where r and n are any two integers such that rn/p=−1.Letp be an odd prime with p≡7mod12. Then, for any nonreal character χmodp with χ3≠χ0, the principal character modp, one has(3)p=∑a=1p−1/3χ3a2+a¯2+∑a=1p−1/3χ3a2+ta¯2+∑a=1p−1/3χ3a2+t2a¯2,where t is any integer such that p|t3−1 and p∤t−1.Unfortunately, formulas (1) and (2) are only suitable for the primes p with p=4k+1. If p=4k+3, we still have not found any similar representation. Some related works can also be found in [2–12].On the other hand, in [10, 11], X. X Li and S. M. Shen studied the mean value properties of the quartic Gauss sums and two-term exponential sums and proved an interesting recurrence formula for it. That is, they give a recurrence formula for(4)Mkp=∑m=1p−1∑b=0p−1emb4pk·∑a=1p−1ema4+aph,with h=2 and 3, respectively.Inspired by literatures [10, 11], we considered the calculating problem of the hybrid power mean(5)Hkp=∑m=1p−1∑b=0p−1emb4pk·∑a=1p−1ema4+ap2.Of course, the contents in [10] and Hkp look a little similar, but they are very different. The main difference between them is the absolute value of the two-term exponential sums. In fact, for all positive integers k≥1, we have(6)∑a=1p−1ema4+apk≠∑a=1p−1ema4+apk.Our goal is to obtain a sharp asymptotic formula forHkp. About this problem, it seems that none had studied it yet, at least we have not seen related papers before. The problem is interesting because it may reveal the regularity of the distribution of values related to the quartic Gauss sums and two-term exponential sums.In this paper, we will use the analytic methods and the properties of the classical Gauss sums to study this problem and prove an interesting four-order linear recurrence formula for it. That is, we will prove the following.Theorem 1.
Letp be an odd prime with p≡1mod8. Then, for any integer k≥1, we have the four-order linear recurrence formula:(7)Hk+4p=6pHk+2p+8pαHk+1p−p2−4pα2Hkp,where the first four terms of Hkp are H1p=6p·1−α,(8)H2p=3p3−20p2−4pα2+12pα−3p,H3p=6p3α−66p2α−6pα+8pα2+34p2,H4p=17p4+4p3α2−88p2α2−116p3−17p2+120p2α−4pα2.Theorem 2.
Letp be an odd prime with p≡5mod8. Then, for any integer k≥1, we have the recurrence formula(9)Hk+4p=−2pHk+2p+8pαHk+1p−9p2−4pα2Hkp,where the first four terms of Hkp are H1p=2p·α+1,(10)H2p=−p3+4p2+p+4pα2+4pα,H3p=−10p2+2α·3p3−17p2−3p,H4p=4p3α2−7p4−8p2α2+28p3−4pα2+7p2+8p2α.Ifp≡1mod8, then there exists an integer c such that c4≡−1modp. So, from Theorem 1, we may immediately deduce the following two corollaries.Corollary 1.
Letp be an odd prime with p≡1mod8. Then, we have(11)∑m=1p−1∑b=0p−1emb4p2·∑a=1p−1ema4+ap2=3p3−20p2−4pα2+12pα−3p.Corollary 2.
Letp be an odd prime with p≡1mod8. Then, we have(12)∑m=1p−1∑b=0p−1emb4p4·∑a=1p−1ema4+ap2=17p4+4p3α2−88p2α2−116p3−17p2+120p2α−4pα2.Notes: in Theorems1 and 2, we only considered the two-term exponential sums:(13)Sm,k,p=∑a=1p−1emak+ap,with k=4. Maybe we can use a similar method to study the general case k≥3, but it is hard to get an exact value for the hybrid power mean as in Hkp, except k=4. Therefore, we will not discuss the general integer k in this paper.Papers [10, 11] and our results are independent of each other. That is, they cannot be deduced from each other.Even if our corollaries are used, they cannot be deduced with the results in [10].In addition, our methods are not suitable for studying the mean value,(14)∑m=1p−1∑b=0p−1emb4pk·∑a=1p−1ema4+ap3,because we cannot deal with the cubic power mean ∑a=1p−1ema4+a/p3.Maybe we can use our methods to study the fourth power mean:(15)∑m=1p−1∑b=0p−1emb4pk·∑a=1p−1ema4+ap4.This is definitely a challenge, which is the goal of our further study.


## 2. Several Lemmas

In this section, we give some lemmas which are necessary in the proof of our theorems. Hereinafter, we shall use some properties of the classical Gauss sums and quadratic residuemodp, and all of them can be found in [1, 13], so they will not be repeated here. First, we have the following.Lemma 1.
Letp be an odd prime with p≡1mod4. Then, for any four-order character ψ modulo p, we have the identity(16)τ2ψ+τ2ψ¯=2p·α,where τχ=∑a=1p−1χaea/p denotes the classical Gauss sums, α=αp=1/2∑a=1p−1a+a¯/p, and ∗/p=χ2 denotes Legendre’s symbol modulo p.Proof.
See Lemma 2.2 in [9].Lemma 2.
Letp be an odd prime with p≡1mod4, and ∗/p=χ2 denotes the Legendre symbol modp. Then, for any integer b with b,p=1, we have the identities(17)Ab=∑a=0p−1eba4p=bpp+∑a=1p−1apeba2p.
Ifp=8k+5, then we have(18)A2b=−p+2bpp∑a=1p−1apeba2p+bpp∑a=1p−1a+a¯p,A3b=−5bpp3/2+p∑a=1p−1apeba2p+3p∑a=1p−1a+a¯p+bpp∑a=1p−1apeba2p∑a=1p−1a+a¯p,A4b=−2pA2b+4pAb∑a=1p−1a+a¯p−9p2+p∑a=1p−1a+a¯p2.
Ifp=8k+1, then we have(19)A2b=3p+2bpp∑a=1p−1apeba2p+bpp∑a=1p−1a+a¯p,A3b=7bpp3/2+5p∑a=1p−1apeba2p+3p∑a=1p−1a+a¯p+bpp∑a=1p−1apeba2p∑a=1p−1a+a¯p,A4b=6pA2b+4pAb∑a=1p−1a+a¯p−p2+p∑a=1p−1a+a¯p2.Proof.
See Lemmas1 and 3 in [12].Lemma 3.
Letp be an odd prime with p≡1mod4. Then, we have(20)∑m=1p−1∑a=1p−1ema4+ap2=p2−4p−1.Proof.
Sincep≡1mod4, the congruence equation x4≡1modp has four solutions. From the properties of reduced residue system modp and the trigonometric identity,(21)∑m=1p−1enmp=p−1,ifp|n,−1,ifp∤n,we have(22)∑m=1p−1∑a=1p−1ema4+ap2=∑a=1p−1∑b=1p−1∑m=1p−1emb4a4−1+ba−1p=∑a=1p−1∑m=1p−1ema4−1p∑b=1p−1eba−1p=p−12−3p−1+p−5=p2−4p−1.
This proves Lemma3.Lemma 4.
Letp be an odd prime with p≡1mod4. Then, for any four-order character ψmodp if p=8k+1, we have the identity(23)∑m=1p−1ψm∑a=1p−1ema4+ap2=2τψ−2p·τψ¯.
Ifp=8k+5, then we have(24)∑m=1p−1ψm∑a=1p−1ema4+ap2=0.Proof.
For any integerm with m,p=1, note that 1+ψm+ψ2m+ψ¯m=4 if m4≡1modp; 1+ψm+ψ2m+ψ¯m=0, otherwise. From (21) and the properties of Gauss sums, we have(25)∑m=1p−1ψm∑a=1p−1ema4+ap2=∑a=1p−1∑b=1p−1∑m=1p−1ψmemb4a4−1+ba−1p=τψ∑a=1p−1ψ¯a4−1∑b=1p−1eba−1p=−τψ∑a=1p−1ψ¯a4−1=−τψ∑a=1p−11+ψa+ψ2a+ψ¯aψ¯a−1=−τψ∑a=1p−1ψ¯a−1+∑a=1p−1ψ¯1−a¯−τψ∑a=1p−1χ2aψ¯a−1−τψ∑a=1p−1ψ¯aa−1,where we have used the identity χ2=ψ2.
Ifp=8k+1, then ψ−1=1 and(26)∑a=1p−1ψ¯a−1+∑a=1p−1ψ¯1−a¯=2∑a=1p−1ψ¯a−1=−2.
From the properties of Gauss sums,τχ2=p and ψ3=ψ¯, we have(27)∑a=1p−1χ2aψ¯a−1=1τψ∑b=1p−1ψb∑a=1p−1χ2aeba−1p=pτψ∑b=1p−1ψbχ2be−bp=p·τψ¯τψ,(28)∑a=1p−1ψ¯aa−1=1τψ∑b=1p−1ψb∑a=1p−1ψ¯aeba−1p=τψ¯τψ∑b=1p−1ψ2be−bp=p·τψ¯τψ.
Ifp=8k+1, then combining (25)–(28), we have(29)∑m=1p−1ψm∑a=1p−1ema4+ap2=2τψ−2p·τψ¯.
Ifp=8k+5, then ψ−1=−1. Thus,(30)∑a=1p−1ψ¯a−1+∑a=1p−1ψ¯1−a¯=∑a=1p−1ψ¯a−1+∑a=1p−1ψ¯1−a=0,(31)∑a=1p−1χ2aψ¯a−1=pτψ∑b=1p−1ψbχ2be−bp=−p·τψ¯τψ,(32)∑a=1p−1ψ¯aa−1=τψ¯τψ∑b=1p−1ψ2be−bp=p·τψ¯τψ.
Combining (25) and (30)–(32), we know that if p=8k+5, then(33)∑m=1p−1ψm∑a=1p−1ema4+ap2=0.
Now, Lemma4 follows from (26) and (33).Lemma 5.
Letp be an odd prime with p≡1mod4; then, we have(34)∑m=1p−1χ2m∑a=1p−1ema4+ap2=2p−ψ−1τ2ψ+τ2ψ¯.Proof.
Note thatχ2 is a real character modp and τχ2=p, and from the method of proving Lemma 3, we have(35)∑m=1p−1χ2m∑a=1p−1ema4+ap2=∑a=1p−1∑b=1p−1∑m=1p−1χ2memb4a4−1+ba−1p=p·∑a=1p−1χ2a4−1∑b=1p−1eba−1p=−p·∑a=1p−1χ2a4−1=−p·∑a=1p−11+ψa+χ2a+ψ¯aχ2a−1=−p·−2+∑a=1p−1ψaχ2a−1+∑a=1p−1ψ¯aχ2a−1=2p−ψ−1τ2ψ+τ2ψ¯.
This proves Lemma5.


## 3. Proofs of the Theorems

In this section, we shall complete the proofs of our theorems. First, we prove Theorem1. For any integer m with m,p=1, note that ψ2=χ2 and τχ2=p, and we have(36)Am=∑a=0p−1ema4p=1+∑a=1p−11+ψa+χ2a+ψ¯aemap=∑a=0p−1emap+ψ¯mτψ+χ2mp+ψmτψ¯=ψ¯mτψ+χ2mp+ψmτψ¯.Ifp is an odd prime with p≡1mod8, then, applying (36) and Lemmas 1, 4, and 5 and noting that τψτψ¯=p, we have(37)H1p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯∑a=1p−1ema4+ap2=τψ2τψ¯−2pτψ+τψ¯2τψ−2pτψ¯+p2p−τ2ψ−τ2ψ¯=6p−3p·τ2ψ+τ2ψ¯=6p⋅1−α.Similarly, note that the identity(38)τ4ψ+τ4ψ¯=τ2ψ+τ2ψ¯2−2p2=2p⋅2α2−p.From the method of proving (37) and Lemmas 1 and 2, we also have(39)H2p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯2∑a=1p−1ema4+ap2=∑m=1p−13p+χ2mτ2ψ+χ2mτ2ψ¯+2pψmτψ+2pψ¯mτψ¯×∑a=1p−1ema4+ap2=3pp2−4p−1+τ2ψ+τ2ψ¯2p−τ2ψ−τ2ψ¯+2pτψ2τψ−2pτψ¯+2pτψ¯2τψ¯−2pτψ=3p3−22p2−3p+6pτ2ψ+τ2ψ¯−τ4ψ+τ4ψ¯=3p3−20p2−4pα2+12pα−3p,(40)H3p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯3∑a=1p−1ema4+ap2=∑m=1p−1ψmτ3ψ+ψ¯mτ3ψ¯+χ2mpp+3pψ¯mτψ+3pψmτψ¯+3pτ2ψ+3pτ2ψ¯+3pψ¯mτψ+3pψmτψ¯+6p3/2χ2m×∑a=1p−1ema4+ap2=τ3ψ2τψ−2pτψ¯+τ3ψ¯2τψ¯−2pτψ+p3/22p−τ2ψ−τ2ψ¯+3pp2−4p−1τ2ψ+τ2ψ¯+6pτψ2τψ¯−2pτψ+6pτψ¯2τψ−2p·τψ¯+6p3/22p−τ2ψ−τ2ψ¯=3p1/2p2−11p−1τ2ψ+τ2ψ¯+2τ4ψ+τ4ψ¯+38p2=6p3α−66p2α−6pα+8pα2+34p2,(41)H4p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯4∑a=1p−1ema4+ap2=∑m=1p−1τ4ψ+τ4ψ¯+p2+6p2+6pχ2mτ2ψ+6pχ2mτ2ψ¯+12p2+12p3/2ψmτψ+12p3/2ψ¯mτψ¯+4pψ¯mτ3ψ+4pψmτ3ψ¯+4pχ2mτ2ψ+4pχ2mτ2ψ¯+4p3/2ψmτψ+4p3/2ψ¯mτψ¯∑a=1p−1ema4+ap2=τ4ψ+τ4ψ¯+19p2p2−4p−1+10pτ2ψ+τ2ψ¯2p−τ2ψ−τ2ψ¯+16p3/2τψ2τψ−2p·τψ¯+16p3/2τψ¯2τψ¯−2p·τψ+4pτ3ψ¯2τψ−2p·τψ¯+4pτ3ψ2τψ¯−2p·τψ=τ4ψ+τ4ψ¯p2−22p−1+60p3/2τ2ψ+τ2ψ¯+p219p2−160p−19=17p4+4p3α2−88p2α2−116p3−17p2+120p2α−4pα2.Ifk≥1, then, from Lemmas 1 and 2, we have(42)Hk+4p=∑m=1p−1Akm⋅A4m·∑a=1p−1ema4+ap2=∑m=1p−1Akm⋅6pA2m+8pαAm−p2−4pα2⋅∑a=1p−1ema4+ap2=6pHk+2p+8pαHk+1p−p2−4pα2Hkp.Now, Theorem1 follows from (37)–(42).Proof of Theorem 2.
If the primep satisfies p=8k+5, then note that ψ−1=−1, and from (13) and Lemmas 1, 3, 4, and 5, we have(43)H1p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯∑a=1p−1ema4+ap2=p2p+τ2ψ+τ2ψ¯=2p⋅α+1.
Similarly, note thatτψτψ¯=−p; from Lemmas 1–5, we also have(44)H2p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯2∑a=1p−1ema4+ap2=∑m=1p−1−p+χ2mτ2ψ+χ2mτ2ψ¯+2pψmτψ+2pψ¯mτψ¯×∑a=1p−1ema4+ap2=−pp2−4p−1+τ2ψ+τ2ψ¯2p+τ2ψ+τ2ψ¯=−p3+4p2+p+4pα2+4pα,(45)H3p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯3∑a=1p−1ema4+ap2=∑m=1p−1ψmτ3ψ+ψ¯mτ3ψ¯+χ2mpp−3pψ¯mτψ−3pψmτψ¯+3pτ2ψ+3pτ2ψ¯+3pψ¯mτψ+3pψmτψ¯−6p3/2χ2m×∑a=1p−1ema4+ap2=−5p3/22p+G2,ψ+3pG2,ψp2−4p−1=−10p2+2α⋅3p3−17p2−3p,(46)H4p=∑m=1p−1ψ¯mτψ+χ2mp+ψmτψ¯4∑a=1p−1ema4+ap2=∑m=1p−1τ4ψ+τ4ψ¯+p2+6p2+6pχ2mτ2ψ+6pχ2mτ2ψ¯−12p2−12p3/2ψmτψ−12p3/2ψ¯mτψ¯+4pψ¯mτ3ψ+4pψmτ3ψ¯−4pχ2mτ2ψ−4pχ2mτ2ψ¯+4p3/2ψmτψ+4p3/2ψ¯mτψ¯∑a=1p−1ema4+ap2=τ4ψ+τ4ψ¯−5p2p2−4p−1+2pτ2ψ+τ2ψ¯2p+τ2ψ+τ2ψ¯=G4,ψp2−2p−1+4p3/2G2,ψ−p25p2−24p−5=4p3α2−7p4−8p2α2+28p3−4pα2+7p2+8p2α.
Ifk≥1, then, from Lemmas 1 and 2, we have(47)Hk+4p=∑m=1p−1Akm⋅A4m⋅∑a=1p−1ema4+ap2=∑m=1p−1Akm⋅−2pA2m+8pαAm−9p2−4pα2⋅∑a=1p−1ema4+ap2=−2pHk+2p+8pαHk+1p−9p2−4pα2Hkp.Now, Theorem2 follows from (43)–(47).This completes the proofs of our all theorems.


## 4. Conclusion

The main results of this paper are two theorems. Theorem1 establishes a four-order linear recurrence formula for the hybrid power mean involving the quartic Gauss sums and the two-term exponential sums for the case p≡1mod8. Theorem 2 establishes a similar conclusion for the case p≡5mod8. These achievements represent new contributions to research in the relevant fields, and it also has a good reference function to the research of related problems.



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