# One-Kind Hybrid Power Means of the Two-Term Exponential Sums and Quartic Gauss Sums

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

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## Abstract

The main purpose of this article is using the analytic methods and the properties of the classical Gauss sums to study the calculating problem of the hybrid power mean of the two-term exponential sums and quartic Gauss sums and then prove two interesting linear recurrence formulas. As applications, some asymptotic formulas are obtained.


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## Body

## 1. Introduction

Letq≥3 be a fixed integer. For any integers k≥2 and m with m,q=1, the two-term exponential sums Gm,k;q and quartic Gauss sums Gm,q are defined by(1)Gm,k;q=∑a=0q−1emak+aq,Gm,q=∑a=0q−1ema4q,where, as usual, ey=e2πiy and i2=−1.These sums play a significant role in the research of analytic number theory, and many number theory problems are closely related to them. Therefore, for the sake of promoting the development of research work in related fields, it is necessary to study the various properties ofGm,k;q and Gm,q. Some research results in these fields can be found in references [1–12]. We will not list all of them. For example, Zhang and Zhang [1] proved the identity(2)∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2,if3∤p−1,2p3−7p2,if3|p−1,where p be an odd prime and n denotes any integer with n,p=1.Zhang and Han [2] obtained the identity:(3)∑a=1p−1∑n=0p−1en3+anp6=5p4−8p3−p2,where p denotes an odd prime with 3∤p−1.Zhang and Zhang [3] derived that, for any prime p, one has the identity:(4)∑m=1p−1χ2m∑a=0p−1ema3+ap4=p2δ−3,ifp≡1mod6,p2δ+3,ifp≡−1mod6,where, as usual, ∗/p=χ2 denotes the Legendre’s symbol modulo p, d⋅d¯≡1modp, and δ=∑d=1p−1d−1+d¯/p is an integer which satisfies the estimate δ≤2p.The author [4] studied the following hybrid power mean:(5)Mkp=∑m=1p−1∑a=0p−1ema4pk⋅∑a=0p−1ema4+ap2,and obtained two interesting fourth-order linear recurrence formula as follows:(1)
Mkp=−2pMk−2p+8pαMk−3p−p9p−4α2Mk−4p, where the first four item in sequence Mkp is M0p=pp−3; M1p=2pα; M2p=−pp2−3p−4α2; and M3p=2p2α3p−14(2)
Mkp=6pMk−2p+8pαMk−3p−pp−4α2Mk−4pThe first four items in the sequenceMkp is M0p=pp−3; M1p=−6pα; M2p=p3p2−17p−4α2; and M3p=6p2αp−8.Inspired by references [1–4], in this paper, we will consider the following 2k-th hybrid power mean:(6)Akp=∑m=1p−1∑a=0p−1ema4p2k⋅∑b=0p−1emb3+bp4,and k-th hybrid power mean(7)Bkp=∑m=1p−1∑a=0p−1ema4pk⋅∑b=0p−1emb3+bp4.Of course, the work in this paper looks a little imaginative with that in [4], but they have different essence, and the main difference lies in the power of the two-term exponential sums. In fact, it is a lot easier that we are dealing with the quadratic power mean of the two-term exponential sum in [4]. In this paper, we are dealing with the fourth power of the two-term exponential sums, and it is very difficult.In this paper, we will give a second-order linear recurrence formula forAkp and a fourth-order linear recurrence formula for Bkp by using the properties of Legendre’s symbol and the classical Gauss sums. That is, we will prove the following results.Theorem 1.
Ifp>3 is an odd prime with p≡1mod4, for any four-order character χ4modp, we have(8)∑m=1p−1χ4m∑a=0p−1ema3+ap4=τ2χ4+χ44⋅p⋅τ2χ4⋅∑u=1p−1χ¯4u2+u+u¯+u¯2+2χ4−36⋅p5/2,where τχ=∑a=1p−1χaea/p denotes the classical Gauss sums.Theorem 2.
Ifp>3 is an odd prime with p≡5mod8, then we have the second-order linear recurrence formula:(9)Akp=6p⋅Ak−1p−9p2−4pα2⋅Ak−2p,k≥2,with the initial values(10)A0p=∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2,if3∤p−1,2p3−7p2,if3|p−1,A1p=3p⋅A0p−2p⋅α⋅∑m=1p−1χ2m∑a=0p−1ema3+nap4,∑m=1p−1χ2m∑a=0p−1ema3+ap4=p2δ−3,ifp≡1mod6,p2δ+3,ifp≡−1mod6,where ∗/p=χ2 denotes Legendre’s symbol modulo p. α=αp=∑a=1p−1/2a+a¯/p and δ=δp=∑d=1p−1d−1+d¯/p are two integers, and they satisfy the estimates α≤p and δ≤2p, d⋅d¯≡1modp.Theorem 3.
Ifp>3 is an odd prime with p≡1mod8, then we have the fourth-order linear recurrence formula:(11)Bkp=6p⋅Bk−2p+8pα⋅Bk−3p+4pα2−p2⋅Bk−4p,with the initial values(12)B0p=∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2,if3∤p−1,2p3−7p2,if3|p−1,B1p=p⋅∑m=1p−1χ2m∑a=0p−1ema3+ap4+τχ¯4⋅∑m=1p−1χ4m∑a=0p−1ema3+ap4+τχ4⋅∑m=1p−1χ¯4m∑a=0p−1ema3+ap4;B2p=3p⋅B0p+2pα⋅∑m=1p−1χ2m∑a=0p−1ema3+ap4+2p⋅τχ¯4⋅∑m=1p−1χ¯4m∑a=0p−1ema3+ap4+2p⋅τχ4⋅∑m=1p−1χ4m∑a=0p−1ema3+ap4,B3p=α⋅B2p+5p⋅Bp+3pα⋅B0p+2p⋅p−α2⋅∑m=1p−1χ2m∑a=0p−1ema3+ap4.From (4), Theorem 1, and Weil’s works [13,14], we have the estimates:(13)∑m=1p−1χ2m∑a=0p−1ema3+ap4=Op5/2,∑m=1p−1χ4m∑a=0p−1ema3+ap4=Op5/2,∑m=1p−1∑a=0p−1ema4p2k−1⋅∑a=0p−1ema3+ap4=Okp2k+3/2.Applying (2), Theorems 2 and 3, the properties of the linear recursive sequences, and these three estimates, we can deduce the following three corollaries.Corollary 1.
Ifp is an odd prime with p≡5mod8, for any positive integer k, we have the asymptotic formula:(14)∑m=1p−1∑a=0p−1ema4p2k⋅∑b=0p−1emb3+bp4=p3⋅3p+2pαk+3p−2pαk+Okp2k+5/2,where Ok denotes the big -O constant depending only on the positive integer k.Especially fork=2, we have the asymptotic formula:(15)∑m=1p−1∑a=0p−1ema4p4⋅∑b=0p−1emb3+bp4=2p4⋅9p+4α2+Op9/2.Corollary 2.
Ifp is an odd prime with p≡1mod8, for any positive integer k, we have the asymptotic formula:(16)∑m=1p−1∑a=0p−1ema4p2k⋅∑b=0p−1emb3+bp4=p3⋅3p+22p2+pα2k+3p−22p2+pα2k+Okp2k+5/2,∑m=1p−1∑a=0p−1ema4p4⋅∑b=0p−1emb3+bp4=2p4⋅17p+4α2+Op9/2.Corollary 3.
Ifp is an odd prime with p≡1mod4, then we have the asymptotic formula:(17)∑m=1p−1∑a=0p−1ema4p2⋅∑b=0p−1emb3+bp4=6p4+Op7/2.


## 2. Several Lemmas

In this section, we will give four basic lemmas that they are all necessary in the proofs of the theorems. Certainly, the proofs of these lemmas need some theoretical knowledge of elementary and analytic number theory. They can be found in references [15–17]. Firstly, we have the following:Lemma 1.
Ifp>3 is an odd prime with p≡1mod4, for any four-order character χ4modp, we have(18)∑m=1p−1χ4m∑a=0p−1ema3+ap2=χ4−36⋅p3/2.Proof.
Letχ2 denotes the Legendre’s symbol modulo p. Then, for any integer b, from the properties of the Legendre’s symbol modulo p, we have(19)∑a=0p−1eba2p=χ2b⋅τχ2,ifp∤b,p,ifp|b.
Note thatχ43=χ¯4 and τχ2=p, and using the definition and properties of Gauss sums, reduced residue system modulo p, and formula (19), we have(20)∑m=1p−1χ4m∑a=0p−1ema3+ap2=∑a=0p−1∑b=0p−1∑m=1p−1χ4mema3+b3+a+bp=τχ4∑a=0p−1∑b=0p−1χ¯4a3+b3ea+bp=τχ4∑a=0p−1χ¯4a3eap+τχ4∑a=0p−1∑b=1p−1χ¯4a3+b3ea+bp=τ2χ4+τ2χ4∑a=0p−1χ¯4a3+1χ¯4a+1=τ2χ4+τ2χ4∑b=0p−1χ¯4b3−3b2+3bχ¯4b=τ2χ4+τ2χ4∑b=1p−1χ¯41−3b¯+3b¯2=τ2χ4∑b=0p−1χ¯43b2−3b+1=χ22τ2χ4∑b=0p−1χ¯432b−12+1=χ¯412τ2χ4∑b=0p−1χ¯4b2+3¯.
Note thatτχ4τχ¯4=χ4−1⋅p, and from the properties of Gauss sums, we have(21)∑a=0p−1χ4a2+r=1τχ¯4∑b=1p−1χ¯4b∑a=0p−1eba2+rp=χ¯4−r⋅τ2χ4p,ifp∤r,0,ifp|r.
From (20) and (21), we have(22)∑m=1p−1χ4m∑a=0p−1ema3+ap2=χ4−36⋅p3/2.
This proves Lemma1.Lemma 2.
Ifp>3 is an odd prime with p≡1mod4, for any four-order character χ4modp, we have(23)∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp=p+1⋅τ2χ4+χ4−36⋅p3/2⋅∑a=0p−1χ¯4a+2χ4a3+2.Proof.
From Lemma1, we have(24)∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp=∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=1p−1e−mb3−bp+∑m=1p−1χ4m∑a=0p−1ema3+ap2=τ2χ4∑a=0p−1∑b=0p−1χ¯4a3+b3−1χ¯4a+b−1+χ4−36⋅p3/2.
Letc=b−1, then from the properties of the complete residue system modulo p, we have(25)∑a=0p−1∑b=0p−1χ¯4a3+b3−1χ¯4a+b−1=∑a=0p−1∑c=0p−1χ¯4a3+c+13−1χ¯4a+c=∑a=0p−1∑c=0p−1χ¯4a3+c3+3c2+3cχ¯4a+c=p+∑a=0p−1∑c=1p−1χ¯4a3c3+c3+3c2+3cχ¯4ac+c=p+∑a=0p−1∑c=1p−1χ¯4a3+1+3c+3c2χ¯4a+1=p+χ44∑a=0p−1∑c=1p−1χ¯44a3+1+32c+12χ¯4a+1=p+χ44∑a=0p−1∑c=0p−1χ¯44a3+1+3c2χ¯4a+1−∑a=0p−1χ¯4a3+1χ¯4a+1.
From (19) and the methods of proving (20), we have(26)∑a=0p−1χ¯4a3+1χ¯4a+1=∑b=1p−1χ¯43b2−3b+1=χ44∑c=0p−1χ¯43c2+1−1=χ44τχ4∑b=1p−1χ4b∑c=0p−1eb3c2+1p−1=χ4−36⋅τ2χ¯4p−1.
From (19), we have(27)∑a=0p−1∑c=0p−1χ¯44a3+1+3c2χ¯4a+1=1τχ4∑a=0p−1χ¯4a+1∑b=1p−1χ4b∑c=0p−1eb4a3+1+3c2p=χ23⋅p⋅τχ¯4τχ4∑a=0p−1χ¯4a+1χ44a3+1=χ4−9⋅τ2χ¯4p⋅∑a=0p−1χ¯4a+2χ4a3+2.
Combining (24)–(27), we have the identity(28)∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp=p+1⋅τ2χ4+χ4−36⋅p3/2⋅∑a=0p−1χ¯4a+2χ4a3+2.
This proves Lemma2.Lemma 3.
Ifp>3 is an odd prime with p≡1mod4, then we have(29)∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp∑c=1p−1e−mc3−cp=χ44⋅p⋅τ2χ4⋅∑u=1p−1χ¯4u2+u+u¯+u¯2+2χ4−36⋅p5/2−p⋅τ2χ4−χ4−36⋅p3/2⋅∑a=0p−1χ¯4a+2χ4a3+2.Proof.
From the properties of the classical Gauss sums and the methods of proving Lemmas1 and 2, we have(30)∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp∑c=1p−1e−mc3−cp=τ2χ4∑a=0p−1∑b=0p−1∑c=0p−1χ¯4a3+b3−c3−1χ¯4a+b−c−1.
Letu=a−1 and v=b−c, then from the properties of the complete residue system modulo p, we have(31)∑a=0p−1∑b=0p−1∑c=0p−1χ¯4a3+b3−c3−1χ¯4a+b−c−1=∑u=0p−1∑v=0p−1∑c=0p−1χ¯4u+13+v+c3−c3−1χ¯4u+v=∑u=0p−1∑v=0p−1∑c=0p−1χ¯4u3+3u2+3u+v3+3v2c+3vc2χ¯4u+v=∑u=0p−1∑v=1p−1∑c=0p−1χ¯4u3+3u2v+3uv2+1+3c+3c2χ¯4u+1+p⋅∑u=1p−1χ¯43u2+3u+1=χ44∑u=0p−1∑v=1p−1∑c=0p−1χ¯4u3+3u2v+u2+1+32c+12χ¯4u+1+χ44⋅p⋅∑u=0p−1χ¯432u+12+1−p=χ44∑u=0p−1∑v=0p−1∑c=0p−1χ¯4u3+3uv2+1+3c2χ¯4u+1−χ44∑u=0p−1∑c=0p−1χ¯44u3+1+3c2χ¯4u+1+χ44⋅p⋅∑u=0p−1χ¯43u2+1−p.
From (19), we have(32)∑u=0p−1χ¯43u2+1=1τχ4⋅∑b=1p−1χ4b∑u=0p−1eb3u2+1p=χ23⋅pτχ4⋅∑b=1p−1χ¯4bebp=χ4−9⋅τ2χ4¯p.
From (19), we also have(33)∑u=0p−1∑v=0p−1∑c=0p−1χ¯4u3+3uv2+1+3c2χ¯4u+1=1τχ4∑u=0p−1χ¯4u+1∑b=1p−1χ4b∑v=0p−1∑c=0p−1ebu3+3uv2+1+3c2p=pτχ4∑u=1p−1χ¯4u3+u2∑b=1p−1χ4bebu3+1p+χ4−9⋅p⋅τ2χ¯4=p⋅∑u=1p−1χ¯4u3+u2χ¯4u3+1+χ4−9⋅p⋅τ2χ¯4=p⋅∑u=1p−1χ¯4u2+u+u¯+u¯2+χ4−9⋅p⋅τ2χ¯4.
Now, combining (27)–(33), we have the identity:(34)∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp∑c=1p−1e−mc3−cp=χ44⋅p⋅τ2χ4⋅∑u=1p−1χ¯4u2+u+u¯+u¯2+2χ4−36⋅p5/2−p⋅τ2χ4−χ4−36⋅p3/2⋅∑a=0p−1χ¯4a+2χ4a3+2.
This proves Lemma3.Lemma 4.
Letp>3 is an odd prime with p≡1mod4, then for any four-order character χ4modp, we have the identity:(35)τ2χ4+τ2χ¯4=2p⋅α,where α=αp=∑a=1p−1/2a+a¯/p is an integer.Proof.
See Lemma 2.2 in Han [5].


## 3. Proofs of the Theorems

In this section, we will complete the proofs of our theorems. In fact, for any primep with p≡1mod4, from Lemmas 2 and 3, we have(36)∑m=1p−1χ4m∑a=0p−1ema3+ap4=∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp+∑m=1p−1χ4m∑a=0p−1ema3+ap2∑b=0p−1e−mb3−bp∑c=1p−1e−mc3−cp=p+1⋅τ2χ4+χ4−36⋅p3/2⋅∑a=0p−1χ¯4a+2χ4a3+2+χ44⋅p⋅τ2χ4⋅∑u=1p−1χ¯4u2+u+u¯+u¯2+2χ4−36⋅p5/2−p⋅τ2χ4−χ4−36⋅p3/2⋅∑a=0p−1χ¯4a+2χ4a3+2=τ2χ4+χ44⋅p⋅τ2χ4⋅∑u=1p−1χ¯4u2+u+u¯+u¯2+2χ4−36⋅p5/2.This proves Theorem1.Now, we prove Theorem2. If p≡5mod8, then for any four-order character χ4modp, we have χ4−1=−1. From the properties of the Gauss sums and the four-order character modulo p, we have(37)Gm,p=∑a=0p−1ema4p=1+∑a=1p−11+χ4a+χ2a+χ¯4a⋅emap=χ¯4m⋅τχ4+χ2m⋅τχ2+χ4m⋅τχ¯4=χ¯4m⋅τχ4+χ2m⋅p+χ4m⋅τχ¯4,(38)Gm,p¯=−χ4m⋅τχ¯4−χ¯4m⋅τχ4+χ2m⋅p.So, from (37), (38), and Lemma 4, we have(39)Gm,p2=p+χ4m⋅τχ¯4+χ¯4m⋅τχ42=3p−χ2mτ2χ4+τ2χ¯4=3p−2χ2mpα,(40)Gm,p4=6pGm,p2+4pα2−9p2.So, for any integerk≥2, from (2), (4), (39), and (40), we have(41)Akp=∑m=1p−1Gm,p2k−4⋅Gm,p4⋅∑b=0p−1emb3+bp4=∑m=1p−1Gm,p2k−4⋅6pGm,p2+4pα2−9p2⋅∑b=0p−1emb3+bp4=6p⋅Ak−1p−9p2−4pα2⋅Ak−2p,with the initial values(42)A0p=∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2,if3∤p−1,2p3−7p2,if3|p−1,A1p=3p⋅A0p−2p⋅α⋅∑m=1p−1χ2m∑a=0p−1ema3+nap4,∑m=1p−1χ2m∑a=0p−1ema3+ap4=p2δ−3,ifp≡1mod6,p2δ+3,ifp≡−1mod6,where δ is the same as the definition in (4). This proves Theorem 2.Now, we prove Theorem3. If p≡1mod8, then note that χ4−1=−1, so from (37) and Lemma 4, we have(43)G2m,p=3p+χ2m⋅2pα+2pχ¯4mτχ¯4+χ4m⋅τχ4,(44)G3m,p=α⋅G2m,p+5p⋅Gm,p+χ2m⋅2p⋅p−α2+3pα,(45)G4m,p=6p⋅G2m,p+8pα⋅Gm,p+4pα2−p2.So, for any integerk≥4, from (43) and (44), we have(46)Bkp=6p⋅Bk−2p+8pα⋅Bk−3p+4pα2−p2⋅Bk−4p,with the initial values(47)B0p=∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2,if3∤p−1,2p3−7p2,if3|p−1,B1p=p⋅∑m=1p−1χ2m∑a=0p−1ema3+ap4+τχ¯4⋅∑m=1p−1χ4m∑a=0p−1ema3+ap4+τχ4⋅∑m=1p−1χ¯4m∑a=0p−1ema3+ap4,B2p=3p⋅B0p+2pα⋅∑m=1p−1χ2m∑a=0p−1ema3+ap4+2p⋅τχ¯4⋅∑m=1p−1χ¯4m∑a=0p−1ema3+ap4+2p⋅τχ4⋅∑m=1p−1χ4m∑a=0p−1ema3+ap4,B3p=α⋅B2p+5p⋅B1p+3pα⋅B0p+2p⋅p−α2⋅∑m=1p−1χ2m∑a=0p−1ema3+ap4.This proves Theorem3.Now, we will give a simple proof for Corollary2. Note the estimate(48)∑m=1p−1∑a=0p−1ema4p2k−1⋅∑a=0p−1ema3+ap4=Okp2k+3/2,and the identity(49)B2kp=6p⋅B2k−2p+8pα⋅B2k−3p+4pα2−p2⋅B2k−4p.From (48) and (49), we have(50)B2kp=6p⋅B2k−1p+4pα2−p2⋅B2k−2p+Okpk+2.Note that the linear recurrence sequenceUn=6p⋅Un−1+4pα2−p2⋅Un−2 has the general terms:(51)Un=C1⋅3p+22p2+pα2n+C2⋅3p−22p2+pα2n,n≥0.IfU0=2p3 and U1=6p4, then we have C1=C2=p3 and(52)Un=p3⋅3p+22p2+pα2n+3p−22p2+pα2n,n≥0.Combining (50) and (52), we have the asymptotic formula:(53)B2kp=p3⋅3p+22p2+pα2k+3p−22p2+pα2k+Okp2k+5/2.This completes the proofs of our all results.


## 4. Conclusion

The main results of this paper are two linear recurrence formulas for the hybrid power means of the two-term exponential sumsGm,k;q and the quartic Gauss sums Gm,q. As some applications of these results, we obtained a sharp asymptotic formula for the hybrid power means:(54)∑m=1p−1∑a=0p−1ema4p2k⋅∑b=0p−1emb3+bp4,for all positive integer k and odd prime p with p≡1mod4. All these results are new contributions to the related fields.



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