# On a Certain Quadratic Character Sums of Ternary Symmetry Polynomialsmodp

Authors: Yuanyuan Meng
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this article, we are using the elementary methods and the properties of the classical Gauss sums to study the calculating problem of a certain quadratic character sums of a ternary symmetry polynomials modulop and obtain some interesting identities for them.


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## Body

## 1. Introduction

Letp be an odd prime, ∗/p denotes the Legendre symbol modp, i.e., for any integer n, one has(1)np=1,ifnis a quadratic residuemodp,−1,ifnis a quadratic nonresiduemodp,0,ifp|n.Some of the most commonly used properties of the Legendre symbol are as follows (see [1, 2]):(2)−1p=−1p−1/2,2p=−1p2−1/8,qp·pq=−1p−1q−1/4,where p and q are two different odd primes.The introduction of the Legendre symbol has not only enriched the content of number theory but also greatly promoted the development of elementary and analytic number theory, especially the research on the properties of primes. For example, ifp is a prime with p≡1mod4, then for any integers r and s with rs/p=−1, one has the identity (see Theorems 4–11 in [2])(3)p=α2p+β2p=∑a=1p−1/2a+ra¯p2+∑b=1p−1/2b+sb¯p2.From (3), we naturally wonder, for other forms of primes p, can they also be expressed in terms of Legendre’s symbol?In particular, ifp is an odd prime with p≡1mod3, then there are two integers d and b such that the identity (see [3])(4)4p=d2+27b2,where d is uniquely determined by d≡1mod3.In addition, ifp is an odd with p≡3mod4, then there are two integers x and y such that(5)p=x2−2⋅2p⋅y2.Although we have not found the representations ofd and b or x and y in terms of the Legendre symbol modulo p, we found that a certain quadratic character sum of the ternary symmetry polynomials are closely related to the numbers d and b.In this paper, we shall use elementary methods and the properties of the classical Gauss sums to study the calculating problem of a certain quadratic character sum of binary symmetry polynomials modulop and obtain several interesting identities for them. That is, we shall prove the following results.Theorem 1.
Letp be an odd prime with 3,p−1=1. Then, we have(6)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=−−1p⋅p.Theorem 2.
Letp be an odd prime with p≡1mod6. If 2 is a cubic residue modulo p, then we have the identity(7)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=−1p⋅9pd−5p−d2,where d is the same as defined in (4).Theorem 3.
Letp be an odd prime with p≡1mod6. If 2 is not a cubic residue modulo p, then we have the identity(8)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=−12⋅−1p⋅9pd+9pb+10p+2d2,∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=−12⋅−1p⋅9pd−9pb+10p+2d2.
From these theorems, we may immediately deduce the following three corollaries.Corollary 1.
Letp be an odd prime with 3,p−1=1. Then, we have(9)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=p.Corollary 2.
Letp be a prime with p≡1mod6. If 2 is a cubic residue modulo p, then we have(10)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=9pd−5p−d2.Corollary 3.
Ifp is an odd prime with p≡1mod6, if 2 is not a cubic residue modulo p, then we have(11)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=12⋅9pd+9pb+10p+2d2or12⋅9pd−9pb+10p+2d2.
Notes: it is clear that our method is applicable to multivariate symmetry polynomialsfx1,x2,…,xk. But, when k is larger, the calculation is more complicated, so we do not give it.Theorem3 is flawed. In other words, it gives us two possibilities. We do not know for sure which one is the exact value. How to determine its exact value is an open problem. Interested readers are encouraged to join us in the research.


## 2. Several Lemmas

In this section, we first give several simple lemmas. Of course, the proofs of these lemmas need some knowledge of elementary and analytic number theory. They can be found in many number theory books, such as [1, 2]. Other papers related to Gauss sums and character sums can also be found in [4–11]; here, we do not need to list.Lemma 1.
Letp be a prime with p≡1mod3. Then, for any third-order character λmodp, one has the identity(12)τ3λ+τ3λ¯=dp,where d is the same as defined in (4).Proof.
See the work of Zhang and Hu [12].Lemma 2.
Letp be a prime with p≡1mod6. Then, for any sixth-order character ψmodp, one has the identity(13)τ3ψ+τ3ψ¯=p1/2⋅d2−2p,ifp≡1mod12;−i⋅p1/2⋅d2−2p,ifp≡7mod12,where i2=−1 and d is the same as defined in (4).Proof.
This result is Lemma 3 in the work of Chen [4], so we omit the proof process.Lemma 3.
Letp be a prime with p≡1mod6. Then for any third-order character λmodp, we have the identity(14)∑a=1p−1∑b=1p−1χ2ab∑c=1p−11+λc+λ¯cχ2a+b+c+1=−1p⋅−p+λ¯2⋅τ3λ+λ2τ3λ¯,where ∗/p=χ2 denotes the Legendre symbol modulo p.Proof.
Note thatχ22=χ0 and τ2χ2=χ2−1⋅p; from the properties of Gauss sums and the Legendre symbol modulo p, we have(15)∑a=1p−1∑b=1p−1χ2ab∑c=1p−1χ2a+b+c+1=−∑a=1p−1∑b=1p−1χ2abχ2a+b+1=−1τχ2⋅∑d=1p−1χ2d∑a=1p−1∑b=1p−1χ2abeda+b+1p,=−τχ2⋅∑d=1p−1χ2dedp=−τ2χ2.
It is clear thatλ−1=1 and(16)∑a=0p−1eba2p=1+∑a=1p−11+apebap=χ2bτχ2,and from the properties of the Gauss sums, we can get(17)∑a=1p−1∑b=1p−1χ2ab∑c=1p−1λcχ2a+b+c+1=1τχ2·∑d=1p−1χ2d∑a=1p−1∑b=1p−1χ2ab∑c=1p−1λceda+b++c+1p,=τ2χ2·τλτχ2∑d=1p−1χ2dλ¯dedp=τχ2⋅τλ⋅τχ2λ¯,(18)∑a=0p−1χ2λa2−1=1τχ2λ¯∑b=1p−1χ2λ¯b∑a=0p−1eba2−1p,=τχ2τχ2λ¯∑b=1p−1χ2λ¯bχ2be−bp=τχ2⋅τλ¯τχ2λ¯.
On the other hand, note thatλ¯2=λ; we also have(19)∑a=0p−1χ2λa2−1=∑a=0p−1χ2λa+12−1=∑a=1p−1χ2λa2+2a,=1τχ2λ¯∑b=1p−1χ2λ¯b∑a=1p−1χ2λaeba+2p,=τχ2λτχ2λ¯∑b=1p−1χ2λ¯bχ2bλ¯be2bp=λ¯2⋅τλ⋅τχ2λτχ2λ¯.
Note thatτλ⋅τλ¯=p; from (17)–(19), we have the identity(20)∑a=1p−1∑b=1p−1χ2ab∑c=1p−1λcχ2a+b+c+1=−1p⋅λ¯2⋅τ3λ.
Similarly, we also get(21)∑a=1p−1∑b=1p−1χ2ab∑c=1p−1λ¯cχ2a+b+c+1=−1p⋅λ2⋅τ3λ¯.
Now, combining (15), (20), and (21), we have the identity(22)∑a=1p−1∑b=1p−1χ2ab∑c=1p−11+λc+λ¯cχ2a+b+c+1=−−1p⋅p+−1p⋅λ¯2⋅τ3λ+λ2τ3λ¯.
This proves Lemma3.Lemma 4.
Letp be a prime with p≡1mod6. Then, for any third-order character λmodp, we have the identity(23)∑a=1p−1∑b=1p−1χ2abλa∑c=1p−11+λc+λ¯cχ2a+b+c+1=−−1p⋅p+−1p⋅λ2+λ¯2⋅τ3λ¯,∑a=1p−1∑b=1p−1χ2abλ¯a∑c=1p−11+λc+λ¯cχ2a+b+c+1=−−1p⋅p+−1p⋅λ2+λ¯2⋅τ3λ¯.Proof.
Note thatτχ2λ⋅τχ2λ¯=χ2−1⋅p; from the methods of proving Lemma 3, we have(24)∑a=1p−1∑b=1p−1χ2abλa∑c=1p−1χ2a+b+c+1=−∑a=1p−1∑b=1p−1χ2abλaχ2a+b+1,=−1τχ2∑c=1p−1χ2c∑a=1p−1∑b=1p−1χ2abλaeca+b+1p=−τχ2λ⋅τχ2λ¯=−−1p⋅p.
Applying (18) and (19), we have(25)∑a=1p−1∑b=1p−1χ2abλa∑c=1p−1λcχ2a+b+c+1=1τχ2∑d=1p−1χ2d∑a=1p−1∑b=1p−1χ2abλa∑c=1p−1λceda+b++c+1p,=τλ·τχ2λ⋅∑d=1p−1χ2dλdedp=τλ⋅τ2χ2λ,=−1p⋅λ¯2⋅τ3λ¯.
Similarly, we also have(26)∑a=1p−1∑b=1p−1χ2abλa∑c=1p−1λ¯cχ2a+b+c+1=−1p·λ2·τ3λ¯.
Combining (24)–(26), we have(27)∑a=1p−1∑b=1p−1χ2abλa∑c=1p−11+λc+λ¯cχ2a+b+c+1=−−1p⋅p+−1p⋅λ2+λ¯2⋅τ3λ¯.
Taking the conjugate in (27), we have(28)∑a=1p−1∑b=1p−1χ2abλ¯a∑c=1p−11+λc+λ¯cχ2a+b+c+1=−−1p⋅p+−1p⋅λ2+λ¯2⋅τ3λ.
Now, Lemma4 follows from (27) and (28).Lemma 5.
Letp be a prime with p≡1mod6. Then, for any third-order character λmodp, we have the identities(29)∑a=1p−1∑b=1p−1χ2abλab∑c=1p−11+λc+λ¯cχ2a+b+c+1=−τ3χ2λτχ2+−1p⋅λ¯2⋅τ3λ¯+λ2⋅τ3λ,∑a=1p−1∑b=1p−1χ2abλ¯ab∑c=1p−11+λc+λ¯cχ2a+b+c+1=−τ3χ2λτχ2+−1p⋅λ¯2⋅τ3λ¯+λ2⋅τ3λ.Proof.
Note thatλ¯2=λ; from (18), (19), and the methods of proving Lemma 3, we have(30)∑a=1p−1∑b=1p−1χ2abλab∑c=1p−1χ2a+b+c+1=−∑a=1p−1∑b=1p−1χ2abλabχ2a+b+1,=−1τχ2∑c=1p−1χ2c∑a=1p−1∑b=1p−1χ2abλabeca+b+1p=−τ3χ2λτχ2,(31)∑a=1p−1∑b=1p−1χ2abλab∑c=1p−1λcχ2a+b+c+1=1τχ2∑d=1p−1χ2d∑a=1p−1∑b=1p−1χ2abλab∑c=1p−1λceda+b++c+1p,=τλ⋅τ2χ2λ=−1p⋅λ¯2⋅τ3λ¯,(32)∑a=1p−1∑b=1p−1χ2abλab∑c=1p−1λ¯cχ2a+b+c+1=1τχ2∑d=1p−1χ2d∑a=1p−1∑b=1p−1χ2abλab∑c=1p−1λ¯ceda+b++c+1p,=1τχ2⋅τλ¯⋅τ2χ2λ⋅τχ2λ¯=−1p⋅λ2⋅τ3λ.
Combining (30)–(32), we have(33)∑a=1p−1∑b=1p−1χ2abλab∑c=1p−11+λc+λ¯cχ2a+b+c+1=−τ3χ2λτχ2+−1p⋅λ¯2⋅τ3λ¯+λ2⋅τ3λ.
Taking the conjugate given above, we can deduce the other identity. This proves Lemma5.Lemma 6.
Letp be a prime with p≡1mod6. Then, for any third-order character λmodp, we have the identity(34)∑a=1p−1∑b=1p−1χ2abλaλ¯b∑c=1p−11+λc+λ¯cχ2a+b+c+1=−−1p⋅p+−1p⋅λ2⋅τ3λ¯+λ¯2⋅τ3λ.Proof.
From (18), (19), and the methods of proving Lemma 3, we have(35)∑a=1p−1∑b=1p−1χ2abλaλ¯b∑c=1p−1χ2a+b+c+1=−∑a=1p−1∑b=1p−1χ2abλaλ¯bχ2a+b+1,=−1τχ2∑c=1p−1χ2c∑a=1p−1∑b=1p−1χ2abλaλ¯beca+b+1p,=−τχ2λτχ2λ¯=−−1p⋅p,(36)∑a=1p−1∑b=1p−1χ2abλaλ¯b∑c=1p−1λcχ2a+b+c+1=1τχ2∑d=1p−1χ2d∑a=1p−1∑b=1p−1χ2abλaλ¯b∑c=1p−1λceda+b++c+1p,=τλ⋅τ2χ2λ¯⋅τχ2λτχ2=−1p⋅λ¯2⋅τ3λ,(37)∑a=1p−1∑b=1p−1χ2abλaλ¯b∑c=1p−1λ¯cχ2a+b+c+1=1τχ2∑d=1p−1χ2d∑a=1p−1∑b=1p−1χ2abλaλ¯b∑c=1p−1λ¯ceda+b++c+1p,=τλ⋅τ2χ2λ¯⋅τχ2λτχ2=−1p⋅λ2⋅τ3λ.
Combining (35)–(37), we have(38)∑a=1p−1∑b=1p−1χ2abλaλ¯b∑c=1p−11+λc+λ¯cχ2a+b+c+1=−−1p⋅p+−1p⋅λ2⋅τ3λ¯+λ¯2⋅τ3λ.
This proves Lemma6.


## 3. Proofs of the Theorems

Now, we prove our theorems. From the properties of the reduced residue system modulop, we have(39)∑a=1p−1∑b=1p−1∑c=1p−1χ2a4bc+b4ac+c4ab+abc=∑a=1p−1∑b=1p−1∑c=1p−1χ2a4bc6+b4ac6+c6ab+abc3,=∑a=1p−1∑b=1p−1χ2ab∑c=1p−1χ2a3+b3+1+c¯3,=∑a=1p−1∑b=1p−1χ2a3b3∑c=1p−1χ2a3+b3+c3+1.If3,p−1=1, when a passes through a reduced residue system modulo p, then a3 also passes through a reduced residue system modulo p. So, from (39) and (15), we have the identity(40)∑a=1p−1∑b=1p−1∑c=1p−1χ2a4bc+b4ac+c4ab+abc=∑a=1p−1∑b=1p−1χ2ab∑c=1p−1χ2a+b+c+1=−τ2χ2=−−1p⋅p.This proves Theorem1.Ifp≡1mod6, let λ be a third-order character modulo p; then, for any integer a with a,p=1, we have the identity(41)1+λa+λ¯a=3,ifais a cubic residue modulop;0,ifais not a cubic residue modulop.So, from the symmetry ofa and b, (39), and Lemmas 3–6, we have(42)∑a=1p−1∑b=1p−1∑c=1p−1χ2a4bc+b4ac+c4ab+abc=∑a=1p−1∑b=1p−1χ2aχ2b1+λa+λ¯a·1+λb+λ¯b×∑c=1p−11+λc+λ¯cχ2a+b+c+1,=∑a=1p−1∑b=1p−1χ2aχ2b1+2λa+2λ¯a+λab+λ¯ab+2λaλ¯b×∑c=1p−11+λc+λ¯cχ2a+b+c+1,=−1p⋅−p+λ¯2⋅τ3λ+λ2τ3λ¯−2−1p⋅p+2−1pλ¯2+λ2⋅τ3λ¯−2−1p⋅p+2−1pλ¯2+λ2⋅τ3λ−τ3χ2λτχ2+−1p⋅λ¯2τ3λ¯+λ2⋅τ3λ−τ3χ2λ¯τχ2+−1p⋅λ¯2⋅τ3λ¯+λ2⋅τ3λ−2−1p⋅p+2−1p⋅λ2⋅τ3λ¯+λ¯2⋅τ3λ,=−7−1p⋅p+−1p⋅4λ2+5λ¯2⋅τ3λ+5λ2+4λ¯2τ3λ¯−τ3χ2λ+τ3χ2λ¯τχ2.If 2 is a cubic residue modulop, then λ2=λ¯2=1. Note that χ2λ=ψ is a sixth-order character modulo p, χ2−1=1, and τχ2=p, if p≡1mod12; χ2−1=−1 and τχ2=i⋅p, if p≡7mod12. From (42) and Lemmas 1 and 2, we have(43)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=−1p⋅9pd−5p−d2.This proves Theorem2.Now, we prove Theorem3. If 2 is not a cubic residue modulo p, then we have 1+λ2+λ¯2=0 or λ2+λ¯2=−1. From (42) and Lemmas 1 and 2, we have(44)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=−−1p⋅4pd+5p+d2+−1p⋅λ¯2⋅τ3λ+λ2τ3λ¯.Since1+λ2+λ¯2=0 or λ22+λ2+1=0, we have(45)λ2=−12±32⋅i,(46)λ¯2=−12∓32⋅i,wherei2=−1,τ3λ−τ3λ¯2=τ3λ+τ3λ¯2−4p3,=dp2−4p3=p2d2−4p=−27p2b2,or(47)τ3λ−τ3λ¯=±3⋅3⋅p⋅b⋅i.So, from (45)–(47) and Lemma 1, we have(48)λ¯2⋅τ3λ+λ2τ3λ¯=−dp2±9pb2.Combining (44) and (48), we have the identity(49)∑a=1p−1∑b=1p−1∑c=1p−1a4bc+b4ac+c4ab+abcp=−−1p⋅4pd+5p+d2−−1p⋅p2⋅d±9b.This completes the proof of Theorem3.


## 4. Conclusions

The main results of this paper are three identities involving a certain quadratic character sums of ternary symmetry polynomials modulop. Theorem 1 proved an exact identity for the sum, if 3,p−1=1. Theorem 2 discussed the case p≡1mod3 and obtained an exact identity, provided 2 is a cubic residue modulo p. Theorem 3 also discussed the case p≡1mod3, where 2 is not a cubic nonresidue modulo p. In this case, two possibilities are given. These results not only give the exact values of a certain quadratic character sums of ternary polynomials modulo p but also some new contribution to research in related fields.



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