# One Kind Special Gauss Sums and their Mean Square Values

Authors: Jianhong Zhao; Jiejie Gao
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we introduce one kind special Gauss sums; then, using the elementary and analytic methods to study the mean value properties of these kind sums, we obtain several exact calculating formulae for them.


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## Body

## 1. Introduction

Letq and k be two positive integers with q≥3. For any integer a≥1 and Dirichlet character χmodq, we write Rka=ak, where x denotes the greatest integer less than or equal to x. Then, we define the summations Gn,χ,k;q and Hn,χ,k;q as(1)Gn,χ,k;q=∑a=1qk−1χRkaenaq,Hn,χ,k;q=∑a=1qk−1χRkaena−Rkkaq,where ey=e2πiy and n is any integer.It is clear thatGn,χ,k;q is a generalization of the classical Gauss sums. In fact, if k=1, then we have R1a=a and(2)Gn,χ,1;q=∑a=1q−1χaenaq=Gn,χ;q.That is,Gn,χ,1;q becomes the classical Gauss sums Gn,χ;q. Gauss sums play a very important role in the study of analytic number theory, and many number theory problems are closely related to it, so some scholars have studied the properties of the classical Gauss sums and obtained many meaningful and interesting results; some of them can be found in [1–14]. For this reason, we think that it is necessary and meaningful to study the properties of Gn,χ,k;q.On the contrary,Hn,χ,2;q is related to Dirichlet L-function L1,χ. In fact, if p is an odd prime, χ−1=−1 and n=0, then we have(3)H0,χ,2;p=2⋅∑m=1p−1mχm=2⋅p3/2π⋅L1,χ.Therefore, the study of the sumsGn,χ,n;q and Hn,χ,n;q has extensive theoretical significance and application values.In this study, as an attempt in this direction, we first study the mean value properties ofGn,χ,k;q and Hn,χ,k;q. We shall use the elementary and analytic methods to prove the following several results.Theorem 1.
Letp be an odd prime. Then, for any integer n with n,p=1, we have the identity(4)∑χmodp∑a=1p2−1χR2aenap2=pp−12sin2πn/p−p−1.Theorem 2.
Letp be an odd prime. Then, we have the identity(5)1p−1∑χmodp∑a=1p3−1χR3aenap2=p−p⋅cosπ⋅2¯n/p2sin2πn/p−1,ifp≡1mod12,p+p⋅cosπ⋅2¯n/p2sin2πn/p−1,ifp≡5mod12,p−p⋅sinπ⋅2¯n/p2sin2πn/p−1,ifp≡7mod12,p+p⋅sinπ⋅2¯n/p2sin2πn/p−1,ifp≡11mod12,where a¯ denotes the solution of the congruence equation ax≡1modp.Theorem 3.
Letp be an odd prime. Then, for any nonprincipal character χmodp, we have the identities(6)∑n=1p−1∑a=1p2−1χR2aena−R22ap2=pp2−112,∑n=1p−1∑a=1p2−1χR2aena−R22ap4=p2p2+11p2−1720,∑n=1p−1∑a=1p2−1χR2aena−R22ap6=p32p2−11p2+17p2−160480.
From Theorems1 and 2, we can also deduce the following corollaries:Corollary 1.
Letp be an odd prime. Then, we have the asymptotic formula(7)∑χmodp∑a=1p2−1χR2aeap2=p42⋅π2+Op3.Corollary 2.
Letp be an odd prime. Then, we have the asymptotic formula(8)∑χmodp∑a=1p2−1χR3aeap2=p42⋅π2+Op7/2.
Notes. In fact, for any integer h≥4, using our methods and the results in [15], we can give an exact calculating formula for the general 2hth power mean:(9)∑n=1p−1∑a=1p2−1χR2aena−R22ap2h.
However, whenh is large, the calculation is more complicated, so we do not consider it.


## 2. Several Lemmas

This section, we need to give some simple lemmas, which are necessary in the proofs of our theorems. Of course, the proofs of these lemmas also need some knowledge of elementary and analytic number theory, in particular, the properties of the character sums and the classical Gauss sums modulop. All these can be found in [16, 17], we do not repeat them. First, we have the following.Lemma 1.
Letp be an odd prime and k≥2 be a fixed integer. Then, for any nonprincipal Dirichlet character χmodp, we have the identity(10)∑a=1pk−1χRkaenap=∑m=1p−1χm⋅enm+1k/p−enmk/pen/p−1,∑a=1pk−1χRkaena−Rkkap=∑m=1p−1χm⋅enm+1k−mk/p−1en/p−1,∑a=1p2−1χR2aena−R22ap=χ¯2nen/pτχen/p−1,where τχ=∑a=1p−1χaea/p denotes the classical Gauss sums.Proof.
First, we prove the first formula. Similarly, we can deduce the second one. For any positive integera, it is clear that there are two integers m≥1 and 0≤i≤m+1k−mk−1 such that a=mk+i. This number pair m,i not only exists, but it is unique. In fact, we have m=ak and 0≤i=a−akk≤m+1k−1−mk, where x denotes the greatest integer ≤x. From these results and the properties of geometric series, we have(11)∑a=1pk−1χRkaenap=∑m=1p−1∑i=0m+1k−mk−1χmenmk+ip=∑m=1p−1χmenmkp∑i=0m+1k−mk−1enip=∑m=1p−1χmenmkpenm+1k−mk/p−1en/p−1=∑m=1p−1χmenm+1k/p−enmk/pen/p−1.
Now, we prove the third one. Takingk=2 in the second formula, note that m+12−m2=2m+1 and ∑m=1p−1χm=0, from the properties of the classical Gauss sums, and we have(12)∑a=1p2−1χR2aena−R22ap=∑m=1p−1χm⋅en2m+1/p−1en/p−1=1en/p−1enp∑m=1p−1χme2nmp−∑m=1p−1χm=χ¯2nen/pτχen/p−1.
This proves Lemma1.Lemma 2.
Letp be an odd prime. Then, we have the identity(13)∑m=1p−1em+13/p−em3/pe1/p−12=2p−χ23⋅e4¯/p⋅τχ2−χ2−3⋅e−4¯/p⋅τχ24sin2π/p−1,∑m=1p−1em+12/p−em2/pe1/p−12=p2sin2π/p−1.Proof.
First, for any integern with n,p=1, from the properties of the quadratic Gauss sums, we have(14)∑m=0p−1enm2p=1+∑m=1p−11+χ2menmp=np⋅τχ2,where τ2χ2=χ2−1⋅p and χ2=∗/p denotes Legendre’s symbol modulo p.
From (14) and the properties of the complete residue system modulo p, we have(15)∑m=1p−1e3m2+3m+1p=−e1p+∑m=0p−1e3m2+3m+1p=−e1p+∑m=0p−1e3⋅4¯2m+12+4¯p=−e1p+∑m=0p−1e3⋅4¯⋅m2+4¯p=−e1p+χ23⋅4¯⋅e4¯p⋅τχ2=−e1p+χ23⋅e4¯p⋅τχ2.
Applying formula (15), we may immediately deduce(16)∑m=1p−1em+13/p−em3/pe1/p−12=∑m=1p−1e3m2+3m+1/p−1e1/p−12=14sin2π/p⋅∑m=1p−12−e3m2+3m+1p−e−3m2+3m+1p=14sin2π/p⋅2p−2−∑m=1p−1e3m2+3m+1p−∑m=1p−1e−3m2+3m+1p=2p−2+e1/p+e−1/p−χ23⋅e4¯/p⋅τχ2−χ2−3⋅e−4¯/p⋅τχ24sin2π/p=2p−χ23⋅e4¯/p⋅τχ2−χ2−3⋅e−4¯/p⋅τχ24sin2π/p−1.
This proves the first formula in Lemma2.
Similarly, note that the identity(17)∑m=0p−1enmp=p,ifn,p=p,0,ifn,p=1,and we also have(18)∑m=1p−1em+12/p−em2/pe1/p−12=∑m=1p−1e2m+1/p−1e1/p−12=14sin2π/p⋅∑m=1p−12−e2m+1p−e−2m−1p=14sin2π/p⋅2p−2−∑m=1p−1e2m+1p−∑m=1p−1e−2m−1p=14sin2π/p⋅2p−2+e1p+e−1p=p2sin2π/p−1.
This proves Lemma2.Lemma 3.
Letq≥3 be an integer. Then, we have the identities(19)∑a=1q−11sin2πa/q=q2−13,∑a=1q−11sin4πa/q=q2+11q2−145,∑a=1q−11sin6πa/q=2q2−11q2+17q2−1945.Proof.
See [18] or Corollary 1 in [15].


## 3. Proofs of the Theorems

Applying three simple lemmas in Section2, we can easily complete the proofs of our theorems. First, we prove Theorem 1. From the orthogonality of the characters modulo p and Lemma 1 with k=2, we have(20)1p−1∑χmodp∑a=1p2−1χR2aenap2=14sin2πn/p⋅∑m=1p−1enm+12p−enm2p2=p2sin2πn/p−1.This proves Theorem1.Proof of Theorem2. From the orthogonality of the characters modulo p and Lemma 1 with k=3, we have(21)∑χmodp∑a=1p3−1χR3aenap2=p−14sin2πn/p⋅∑m=1p−1enm+13p−enm3p2=p−1⋅2p−χ23⋅e4¯n/p⋅τχ2−χ2−3⋅e−4¯n/p⋅τχ24sin2πn/p−1.Ifp≡1mod12, then χ23=χ2−1=1 and τχ2=p. From (21), we have(22)∑χmodp∑a=1p3−1χR3aenap2=p−1p−p⋅cosπ⋅2¯n/p2sin2πn/p−p−1.Ifp≡5mod12, then χ23=−1, χ2−1=1, and τχ2=p. From (21), we have(23)∑χmodp∑a=1p3−1χR3aenap2=p−1p+p⋅cosπ⋅2¯n/p2sin2πn/p−p−1.Ifp≡7mod12, then χ23=−1, χ2−1=−1, and τχ2=i⋅p, i2=−1. From (21), we have(24)∑χmodp∑a=1p3−1χR3aenap2=p−1p−p⋅sinπ⋅2¯n/p2sin2πn/p−p−1.Ifp≡11mod12, then χ23=1, χ2−1=−1, and τχ2=i⋅p. From (21), we have(25)∑χmodp∑a=1p3−1χR3aenap2=p−1p+p⋅sinπ⋅2¯n/p2sin2πn/p−p−1.Now, Theorem2 follows from (22), (23), (24), and (25).Now, we prove Theorem3. For any nonprincipal character modulo p, note that τχ=p; from the third formula in Lemma 1, we have(26)∑a=1p2−1χR2aena−R22ap2=p4sin2πn/p.From (26) and Lemma 3, we have(27)∑n=1p−1∑a=1p2−1χR2aena−R22ap2=∑n=1p−1p4sin2πn/p=pp2−112.Similarly, from (26) and Lemma 3, we also have(28)∑n=1p−1∑a=1p2−1χR2aena−R22ap4=∑n=1p−1p216sin4πn/p=p2p2+11p2−1720,(29)∑n=1p−1∑a=1p2−1χR2aena−R22ap6=∑n=1p−1p364⋅sin6πn/p=p32p2−11p2+17p2−160480.Now, Theorem3 follows from (27)– (29).This completes the proofs of all our results.



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