# A New Sum Analogous to Gauss Sums and Its Fourth Power Mean

Authors: Shaofeng Ru; Wenpeng Zhang
Journal: The Scientific World Journal
 (2014)

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## Abstract

The main purpose of this paper is to use the analytic methods and the properties of Gauss sums to study the computational problem of one kind of new sum analogous to Gauss sums and give an interesting fourth power mean and a sharp upper bound estimate for it.


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## Body

## 1. Introduction

Letq≥3 be an integer, and let χ be a Dirichlet character mod⁡q. Then for any integer n, famous Gauss sum G(χ,n) is defined as follows:
(1)G(χ,n)=∑a=1q‍χ(a)·e(naq),
where e(y)=e2πiy.This sum plays a very important role in the study of analytic number theory; many famous number theoretic problems are closely related to it. For example, the distribution of primes, Goldbach problem, the estimate of character sums, and the properties of DirichletL-functions are some good examples.It is clear that ifχ is a primitive Dirichlet character mod⁡q, then we have G(χ,n)=χ¯(n)·G(χ,1)≡χ¯(n)·τ(χ) and |τ(χ)|=q. Many properties of G(χ,n) and τ(χ) can be found in [1–3].In this paper, we introduce a new sum analogous to Gauss sums as follows:(2)G(χ,b,c,m;q)=∑a=0q-1‍χ(a2+ba+c)·e(maq),
where χ is a character mod⁡q, and b, c, and  m are any integers with (m,q)=1.The main purpose of this paper is to study the properties ofG(χ,b,c,m;q), such as the following two problems:(A)
giving an upper bound estimate ofG(χ,b,c,m;q);(B)
the problem of whether there exists a computational formula for the2kth power mean
(3)∑c=0q-1‍|∑a=0q-1‍χ(a2+ba+c)·e(maq)|2k,k≥2.It seems that no one has studied these problems yet; at least we have not seen any related results in the existing literature. The problems are interesting, because there exists a close relationship between the sumG(χ,b,c,m;q) and the generalized Kloosterman sums; they can also help us to further understand the properties of hybrid mean value of an exponential sum and character of a polynomial.For general integerq>3, these two problems seem to be hard to make progress. But if q=p>2 is a prime and k=2, then we can prove the following two conclusions.Theorem 1.
Letp be an odd prime and χ any nonprincipal character modp. Then for any integers b, c, and m with (m,p)=1, one has the estimate
(4)|∑a=0p-1‍χ(a2+ba+c)·e(map)|≤2p.Theorem 2.
Letp be an odd prime and χ any nonprincipal character modp. Then for any integers b and m with (m,p)=1, one has the identity
(5)∑c=0p-1‍|∑a=0p-1‍χ(a2+ba+c)·e(map)|4={2p3-3p2-3p,ifχistheLegendresymbolmodp;2p3-6p2,ifχisanonrealcharactermodp.Some Notes. If χ=χ0 is the principal character modp, then note that χ0(a2+ba+c)=0 or 1 and |G(χ0,b,c,m;p)|≤3. So, in this case, the result is trivial; we do not need to discuss problems (A) and (B).For any integerk≥3, whether there exists an exact computational formula for the 2kth power mean
(6)∑c=0p-1‍|∑a=0p-1‍χ(a2+ba+c)·e(map)|2k,
is an open problem.Furthermore, for general integerq>3, whether there exists a nontrivial upper bound estimate for G(χ,b,c,m;q) is also an interesting problem.


## 2. Several Lemmas

In this section, we will give several lemmas, which are necessary in the proof of our theorems. Hereinafter, we will use many properties of character sums and Gauss sums; all of these can be found in [1, 4]. So they will not be repeated here. First we have the following.Lemma 3.
Letp be an odd prime; then, for any integer c with (c,p)=1, one has the identity
(7)∑a=0p-1‍e(ca2p)=(cp)·τ(χ2),
where χ2=(*/p) denotes the Legendre symbol.Proof.
From the properties of Gauss sums and quadratic residuemodp we have
(8)∑a=0p-1‍e(ca2p)=1+∑a=1p-1‍e(ca2p)=1+∑a=1p-1‍(1+(ap))·e(cap)=∑a=0p-1‍e(cap)+∑a=1p-1‍(ap)·e(cap)=(cp)∑a=1p-1‍(ap)·e(ap)=(cp)·τ(χ2).
This proves Lemma 3.Lemma 4.
Letp be an odd prime and χ any nonprincipal Dirichlet character modp. Then for any integers b and c, one has the identity
(9)|∑a=0p-1‍χ(a2+ba+c)·e(ap)|=|∑r=1p-1‍χ¯χ2(r)·e((4c-b2)r-16¯·r¯p)|,
where r¯ denotes the solution of the congruence equation r·x≡1 modp.Proof.
Sinceχ is a nonprincipal Dirichlet character modp, from Lemma 3 and the properties of Gauss sums we have
(10)∑a=0p-1‍χ(a2+ba+c)·e(ap)=1τ(χ¯)·∑a=0p-1‍∑r=1p-1‍χ¯(r)e(r(a2+ba+c)p)·e(ap)=1τ(χ¯)·∑r=1p-1‍χ¯(r)×∑a=0p-1‍e(ra2+(br+1)a+crp)=1τ(χ¯)·∑r=1p-1‍χ¯(r)×∑a=0p-1‍e(4¯r(2a+r¯(br+1))2+cr-4¯r¯(br+1)2p)=1τ(χ¯)·∑r=1p-1‍χ¯(r)×∑u=0p-1‍e(4¯ru2+4¯·(4c-b2)r-2¯·b-4¯·r¯p)=1τ(χ¯)·∑r=1p-1‍χ¯(r)(4¯·rp)·τ(χ2)·e(4¯·(4c-b2)r-2¯·b-4¯·r¯p)=τ(χ2)τ(χ¯)·χ¯(4)·e(-2¯·bp)·∑r=1p-1‍χ¯(r)(rp)·e((4c-b2)r-16¯·r¯p).
For any nonprincipal character χ modp, we have |τ(χ)|=p. So, from (10) and noting that |χ¯(4)|=|e(-2¯·b/p)|=1, we have
(11)|∑a=0p-1‍χ(a2+ba+c)·e(ap)|=|∑r=1p-1‍χ¯χ2(r)·e((4c-b2)r-16¯·r¯p)|.
This proves Lemma 4.Lemma 5.
Letp be an odd prime and χ any Dirichlet character modp. Then for any integers m and n, one has the estimate
(12)∑a=1p-1‍χ(a)·e(ma+na¯p)≤2p·(m,n,p)1/2,
where (m,n,p) denotes the greatest common divisor of m, n, and p.Proof.
This estimate is by Weil [5], Chowla [6], Malyshev [7], and Estermann [8] with some minor modifications.Lemma 6.
Letp be an odd prime; then, for any integer n with (n,p)=1, one has the calculating formula
(13)∑m=1p-1‍|∑a=1p-1‍χ(a)·e(ma+na¯p)|4={2p3-3p2-3p-1,ifχistheprincipalcharactermodp;3p3-8p2,ifχistheLegendre’ssymbolmodp;p2(2p-7),ifχisanonrealcharactermodp.Proof.
See [9] or Corollary 2 of [10].


## 3. Proof of the Theorems

In this section, we will complete the proof of our theorems. First note that ifa passes through a reduced residue system modp, then ma also passes through a reduced residue system modp, if (m,p)=1. From these properties we have
(14)|∑a=0p-1‍χ(a2+ba+c)·e(map)|=|∑a=0p-1‍χ(m¯2a2+m¯ba+c)·e(ap)|=|χ2(m¯)∑a=0p-1‍χ(a2+mba+cm2)·e(ap)|=|∑a=0p-1‍χ(a2+mba+cm2)·e(ap)|.
So without loss of generality we can assume that m=1. Now we prove Theorem 1. From Lemmas 4 and 5 we have
(15)|∑a=0p-1‍χ(a2+ba+c)·e(ap)|=|∑r=1p-1‍χ¯χ2(r)·e((4c-b2)r-16¯·r¯p)|≤2p·(4c-b2,16¯,p)1/2=2p.
This proves Theorem 1.Now we prove Theorem2. From the properties of a complete residue system modp we know that if c passes through a complete residue system modp, then 4c-b2 also passes through a complete residue system modp. So from Lemma 4 we have
(16)∑c=0p-1‍|∑a=0p-1‍χ(a2+ba+c)·e(ap)|4=∑c=0p-1‍|∑r=1p-1‍χ¯χ2(r)·e((4c-b2)r-16¯·r¯p)|4=∑m=0p-1‍|∑r=1p-1‍χ¯χ2(r)·e(mr-16¯·r¯p)|4=∑m=0p-1‍|∑r=1p-1‍χ¯χ2(r)·e(mr+r¯p)|4.
If χ=χ2 is the Legendre symbol, then χχ2=χ0 is the principal character modp, so from (16) and Lemma 6 we have
(17)∑c=0p-1‍|∑a=0p-1‍χ(a2+ba+c)·e(ap)|4=∑m=0p-1‍|∑r=1p-1‍χ¯χ2(r)·e(mr+r¯p)|4=1+∑m=1p-1‍|∑r=1p-1‍e(mr+r¯p)|4=1+2p3-3p2-3p-1=2p3-3p2-3p.
If χ is not a real character modp, then from (16) and Lemma 6 we have
(18)∑c=0p-1‍|∑a=0p-1‍χ(a2+ba+c)·e(ap)|4=∑m=0p-1‍|∑r=1p-1‍χ¯χ2(r)·e(mr+r¯p)|4=|∑r=1p-1‍χ¯χ2(r)·e(r¯p)|4+∑m=1p-1‍|∑r=1p-1‍χ¯χ2(r)·e(mr+r¯p)|4=p2+p2(2p-7)=2p2(p-3).
Now combining (14), (17), and (18) we may immediately deduce the identity
(19)∑c=0p-1‍|∑a=0p-1‍χ(a2+ba+c)·e(map)|4={2p3-3p2-3p,ifχis the Legendre symbol modp;2p3-6p2,ifχis a nonreal character modp.
This completes the proof of Theorem 2.



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*Source: 139725-2014-05-21.xml*