# On the Sixth Residues and Some New Properties of Their Distribution

Authors: Juanli Su; Jin Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we use the analytic methods, the properties of the sixth-order characters, and the classical Gauss sums to study the computational problems of a certain special sixth residues’ modulop and give two exact calculating formulas for them.


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## Body

## 1. Introduction

Letp be an odd prime and k be a fixed positive integer. For any integer a with a,p=1, if the congruence equation xk≡amodp has a solution x, then we call a is a kth residue modulo p. Otherwise, a is called a kth nonresidue modulo p. In particular, if k=2, 3, and 4, we call a is a quadratic residue, cubic residue, and quartic residue modulo p, respectively. Undoubtedly, the research of quadratic residue is the most concerned topic. Legendre first introduced the characteristic function of the quadratic residues a/p modulo p, which later was called Legendre’s symbol. It is defined as follows:(1)ap=1,ifais a quadraticre sidue modulop,−1,ifais a quadratic nonresidue modulop,0,ifp|a.Sometimes, we write Legendre’s symbol∗/p as χ2 for the sake of writing. This is because the introduction of this function greatly facilitated the study of quadratic residue properties and promoted the development of elementary number theory and analytic number theory. This is especially true in the study of primes and related problems. For example, if p is a prime with p≡1mod4, then one has (see Theorem 4–11 in [1])(2)p=12∑a=1p−1a+ra¯p2+12∑b=1p−1b+sb¯p2,where a¯ denotes the inverse of a. That is, a⋅a¯≡1modp, and rs/p=−1.Of course, there are many papers involving quadratic residues and primes, so we cannot cover all of them. Those who are interested can refer to [2–9].In this paper, we are concerned with the problem of whether the special integersa+a¯ and a−a¯ both are kth residues’ modulo p. Let Nkp denote the number of all integers 1<a<p−1 such that a+a¯ and a−a¯ both are kth residues’ modulo p. Then, how are the values of Nkp distributed?Very recently, some authors had studied the calculating problem ofNkp and obtained a series of interesting results. For example, Wang and Lv [10] obtained the identity(3)N2p=18p−3,ifp≡3mod8,18p−7,ifp≡7mod8.Hu and Chen [11] proved the following result: let p be an odd prime with p≡7mod12. If 2 is a cubic residue modp, then one has the identity(4)N3p=19⋅p+4d−11.If 2 is not a cubic residuemodp, then one has the asymptotic formula(5)N3p=19⋅p−5+Ep,where d is defined in (7) and Ep satisfies the estimates Ep≤2/3⋅p.Su and Zhang [12] considered the case p≡5mod8 and proved the identity(6)N4p=116⋅p−7−2∑a=1p−1/2a+a¯p.As an extension of [10–12], a natural problem is what about sixth residues modulo p? It is clear that if p−1,6=2, then the problem is trivial. That is, any quadratic residue a modulo p is a sixth residue modulo p. So, we just consider the nontrivial case p≡1mod6. In this case, we know that there are two integers d and b such that the identity(7)4p=d2+27⋅b2,where d is uniquely determined by d≡1mod3 (see [13]).And, it is clear from (7) that the value of N6p must be related to d and b.In this paper, we will use the analytic methods, the properties of the classical Gauss sums, and the estimate for some special character sums to study the computational problems ofN6p and give an exact calculating formula for it. That is, we will prove the following two results.Theorem 1.
Letp be an odd prime with p≡7mod12. If 2 is a cubic residue modulo p, then we have the identity(8)N6p=136⋅p+4d−11,where d is the same as defined in (7).Theorem 2.
Letp be an odd prime with p≡7mod12. If 2 is not a cubic residue modulo p, then we have(9)N6p=172⋅2p+5d−10+9bor172⋅2p+5d−10−9b.
From our theorems, we may immediately deduce the following two corollaries.Corollary 1.
Letp be an odd prime with p≡7mod12. If 2 is a cubic residue modulo p, then we have the congruence(10)p+4d≡11mod36.Corollary 2.
Letp be an odd prime with p≡7mod12. If 2 is not a cubic residue modulo p, then we have the congruence(11)2p+5d≡10+9bmod72or2p+5d≡10−9bmod72.First, in Theorems1 and 2, we must distinguish whether 3 is a cubic residue modulo p because of the need to calculate the character sums. In different cases, the values of character sums are different.Second, ifp is a prime with p≡1mod12, then, for some character sums, we can only use Weil’s classical work [14, 15] to get some upper bound estimates and we cannot get their exact values. So, in this case, we can only deduce a sharp asymptotic formula for N6p. That is,(12)N6p=136⋅p+Op1/2.Third, ifp is an odd prime with p≡7mod12 and 2 is not a cubic residue modulo p, then our Theorem 2 also obtained an exact calculating formula for N6p, which is obviously better than the corresponding result in [11].Of course, our Theorem2 is flawed, and it presents two possibilities. How to determine its correct value is an interesting open problem.Finally, ifp is a prime with p≡7mod12, then we know that 2 is a cubic residue modulo p if and only if 2|d. That is, d is an even number. Otherwise, d is an odd number. Especially for primes p=7,19,67,79,103,139,151, after some simple calculations, we have 4×7=12+27⋅12, 4×19=72+27⋅12, 4×67=−52+27⋅32, 4×79=−172+27⋅12, 4×103=132+27⋅32, 4×139=−232+27⋅12, and 4×151=192+27⋅32. Since N6p is an integer, so applying Corollary 2, we can get the congruences: 2⋅7+5⋅1≡10+9⋅1mod72, 2⋅19+5⋅7≡10−9⋅1mod72, 2⋅67+5⋅−5≡10+9⋅3mod72, 2⋅79+5⋅−17≡10−9⋅1mod72, 2⋅103+5⋅13≡10−9⋅3mod72, 2⋅139+5⋅−23≡10+9⋅1mod72, and 2⋅151+5⋅19≡10+9⋅3mod72.Now, we consider Legendre’s symbold+b/p. Note that 1+1/7=1, 7+1/19=−1, −5+3/67=1, −17+1/79=−1, 13+3/103=−1, −23+1/139=1, and 19+3/151=1. From the above congruences and these values, we have a reason to believe the following.Conjecture.
Letp be an odd prime with p≡7mod12. If 2 is not a cubic residue modulo p, then we have the identity(13)N6p=172⋅2p+5d−10+d+bp⋅9b.


## 2. Several Lemmas

In this section, we decompose the proofs of our theorems into the following several lemmas. For the sake of simplicity, the basic knowledge required in this section is not listed, and only three necessary references [1, 16, 17] are provided here. First, we have the following.Lemma 1.
Letp be an odd prime with p≡1mod3. Then, for any third-order character λ modulo p (i.e., λ≠χ0 and λ3=χ0, the principal character modulo p), we have the identity(14)τ3λ+τ3λ¯=dp,where 4p=d2+27⋅b2, d is uniquely determined by d≡1mod3, τλ=∑a=1p−1λaea/p denotes the classical Gauss sums, and ey=e2πiy.Proof.
For the proof of this lemma, see Zhang and Hu [18] or Berndt and Evans [19].Lemma 2.
Letp be a prime with p≡1mod6. Then, for any third-order character λmodp and sixth-order character ψ=χ2λ (i.e., ψi≠χ0, 1≤i≤5, and ψ6=χ0), we have the identity(15)τψ=λ2⋅τχ2⋅τ2λ¯p.Proof.
From the properties of the Gauss sums and the reduced residue system modulop, note that the identity(16)∑a=0p−1eba2p=1+∑a=1p−11+χ2aebap=χ2b⋅τχ2,and we have(17)∑a=0p−1χ2λa2−1=1τχ2λ¯∑b=1p−1χ2λ¯b∑a=0p−1eba2−1p=τχ2τχ2λ¯∑b=1p−1χ2bλ¯bχ2be−bp=τχ2⋅τλ¯τχ2λ¯.
On the contrary, we also have(18)∑a=0p−1χ2λa2−1=∑a=1p−1χ2λa2+2a=1τχ2λ¯∑b=1p−1χ2λ¯b∑a=1p−1χ2aλaeba+2p=τχ2λτχ2λ¯∑b=1p−1χ2bλ¯bχ2bλ¯be2bp=λ¯2·τχ2λ·τλτχ2λ¯.
Note that identityτλ⋅τλ¯=p, and from (17) and (18), we deduce the identity(19)τχ2λ=τψ=λ2⋅τχ2⋅τ2λ¯p.
This proves Lemma2.Lemma 3.
Letp be an odd prime with p≡7mod12. Then, for any third-order character λmodp, we have the identity(20)∑a=1p−1λa2−a¯2+λ¯a2−a¯2=1+λ2p·τ3λ+λ¯2⋅τ3λ¯.Proof.
Note thatλ2=λ¯, λ−1=1, and χ2−1=−1, and from Lemma 2, properties of the Gauss sums, and Legendre’s symbol modp, we have(21)∑a=1p−1λa2−a¯2=∑a=1p−1λ¯2aλa4−1=∑a=1p−11+χ2a⋅λ¯aλa2−1=∑a=1p−1λ¯aλa2−1+∑a=1p−1χ2−a⋅λ¯−aλ−a2−1=∑a=1p−1λ2aλa2−1=∑a=1p−11+χ2aλaλa−1=1τλ¯∑a=1p−1λa∑b=1p−1λ¯beba−1p+1τλ¯∑a=1p−1ψa∑b=1p−1λ¯beba−1p=τ2λτλ¯−τ2ψτλ¯=τ3λp−τ2ψ⋅τλp=τ3λ+λ¯2⋅τ3λ¯p,where we have used the identity τ2χ2=χ2−1⋅p=−p.
Similarly, we can also deduce that(22)∑a=1p−1λ¯a2−a¯2=τ3λ¯+λ2⋅τ3λp.
It is clear that Lemma3 follows from (21) and (22).Lemma 4.
Letp be an odd prime with p≡7mod12. Then, we have the identity(23)∑a=1p−1λa+a¯+λ¯a+a¯=1−λ2⋅τ3λ+1−λ¯2·τ3λ¯p.Proof.
From the methods of proving Lemma2 and the properties of the Gauss sums, we have(24)∑a=1p−1λa+a¯=∑a=1p−1λ¯aλa2+1=∑a=1p−1λa2λa2+1=∑a=1p−11+χ2a⋅λaλa+1=τ3λp+τ2χ2λ⋅τλp=τ3λp−λ¯2⋅τ3λ¯p.
So, from (24), we have(25)∑a=1p−1λa+a¯+λ¯a+a¯=τ3λp−λ¯2⋅τ3λ¯p+τ3λ¯p−λ2⋅τ3λp=1−λ2⋅τ3λ+1−λ¯2⋅τ3λ¯p.
This proves Lemma4.Lemma 5.
Letp be an odd prime with p≡7mod12. Then, we have(26)∑a=1p−1λa−a¯+λ¯a−a¯=1+λ2⋅τ3λ+1+λ¯2⋅τ3λ¯p.Proof.
It is the same as the proof of Lemma4, so it is omitted.Lemma 6.
Letp be an odd prime with p≡7mod12. Then, we have the identities(27)∑a=1p−1ψa+a¯ψ¯a−a¯=∑a=1p−1ψa−a¯ψ¯a+a¯=0.Proof.
Note thatψ−1=ψ¯−1=−1, and from the properties of the reduced residue system modulo p, we have(28)∑a=1p−1ψa+a¯ψ¯a−a¯=∑a=1p−1ψa2+1ψ¯a2−1=∑a=1p−1ψa¯2+1ψ¯a¯2−1=∑a=1p−1ψa2+1ψ¯1−a2=−∑a=1p−1ψa2+1ψ¯a2−1=−∑a=1p−1ψa+a¯ψ¯a−a¯,which implies that(29)∑a=1p−1ψa+a¯ψ¯a−a¯=0.
This proves Lemma6.Lemma 7.
Letp be an odd prime with p≡7mod12. For any sixth-order character ψmodp, we have(30)∑a=1p−1ψa+a¯⋅χ2a−a¯=∑a=1p−1ψ¯a+a¯⋅χ2a−a¯=0,∑a=1p−1ψa−a¯⋅χ2a+a¯=∑a=1p−1ψ¯a−a¯⋅χ2a+a¯=0.Proof.
Sinceψ=χ2λ, χ2−1=−1, so, from the reduced residue system modulo p, we have(31)∑a=1p−1ψa+a¯⋅χ2a−a¯=∑a=1p−1λ¯aψa2+1⋅χ2a2−1=∑a=1p−1λaψa¯2+1⋅χ2a¯2−1=∑a=1p−1λ2aψa2+1⋅χ21−a2=−∑a=1p−1λ¯aψa2+1⋅χ2a2−1=−∑a=1p−1ψa+a¯⋅χ2a−a¯,which implies that(32)∑a=1p−1ψa+a¯⋅χ2a−a¯=0.
Similarly, we can also deduce the identity(33)∑a=1p−1ψa−a¯⋅χ2a+a¯=0.
This proves Lemma7.Lemma 8.
Letp be an odd prime with p≡7mod12. Then, for any third-order character λmodp and ψ=χ2λ, we have(34)∑a=1p−1ψa+a¯⋅λa−a¯=∑a=1p−1ψa+a¯⋅λ¯a−a¯=0,∑a=1p−1ψa−a¯⋅λa+a¯=∑a=1p−1ψa−a¯⋅λ¯a+a¯=0.Proof.
Note thatλ−1=1 and ψ−1=−1, we have(35)∑a=1p−1ψa+a¯⋅λa−a¯=∑a=1p−1ψ−a−a¯⋅λ−a+a¯=−∑a=1p−1ψa+a¯⋅λa−a¯.
So, we have the identity(36)∑a=1p−1ψa+a¯⋅λa−a¯=0.
Similarly, we can also deduce the identity(37)∑a=1p−1ψa−a¯⋅λa+a¯=0.
This proves Lemma8.Lemma 9.
Letp be an odd prime with p≡7mod12. Then, for any third-order character λmodp, we have(38)∑a=1p−1λa+a¯⋅λ¯a−a¯+λ¯a+a¯⋅λa−a¯=−4.Proof.
From the properties of Legendre’s symbolmodp, we have(39)∑a=1p−1λa+a¯⋅λ¯a−a¯=∑a=1p−1λa2+1⋅λ¯a2−1=∑a=1p−11+χ2aλa+1⋅λ¯a−1=−1+∑a=0p−1λa+1⋅λ¯a−1+∑a=1p−1χ2aλa+1⋅λ¯a−1=−1+∑a=1p−1λa+2⋅λ¯a−∑a=1p−1χ2aλa−1⋅λ¯a+1=−1+∑a=1p−1λ1+2⋅a¯−∑a=1p−1χ2aλa−1⋅λ¯a+1=−2+∑a=0p−1λ1+2·a−∑a=1p−1χ2aλa−1⋅λ¯a+1=−2−∑a=1p−1χ2aλa−1⋅λ¯a+1.
Similarly, we also have(40)∑a=1p−1λ¯a+a¯·λa−a¯=∑a=1p−1λ¯a2+1·λa2−1=∑a=1p−11+χ2aλ¯a+1⋅λa−1=−2+∑a=1p−1χ2aλ¯a+1·λa−1.
Combining (39) and (40), we have the identity(41)∑a=1p−1λa+a¯⋅λ¯a−a¯+λ¯a+a¯·λa−a¯=−4.
This proves Lemma9.


## 3. Proofs of the Theorems

In this section, we shall complete the proofs of our main results. First, we prove Theorem1. For any prime p with p≡7mod12, let λ denote a third-order character modulo p; then, ψ=χ2λ is a sixth-order character modulo p. So, for any integer a with a,p=1, from the characteristic function(42)1+ψa+ψ2a+ψ3a+ψ¯2a+ψ¯a=6,ifais a sixth residue modulop,0,ifais not a sixth residue modulop,and note that ψ2=λ¯, ψ3=χ2, χ2−1=−1, λ−1=1, ψ−1=−1, and a2+1,p=1, and we have the identity(43)N6p=136∑a=2p−21+ψa+a¯+λ¯a+a¯+χ2a+a¯+λa+a¯+ψ¯a+a¯×1+ψa−a¯+λ¯a−a¯+χ2a−a¯+λa−a¯+ψ¯a−a¯=136∑a=1p−11+ψa+a¯+λ¯a+a¯+χ2a+a¯+λa+a¯+ψ¯a+a¯×1+ψa−a¯+λ¯a−a¯+χ2a−a¯+λa−a¯+ψ¯a−a¯−118⋅1+λ2+λ¯2=p−136+136∑a=1p−1ψa+a¯+λ¯a+a¯+χ2a+a¯+λa+a¯+ψ¯a+a¯+136∑a=1p−1ψa−a¯+λ¯a−a¯+χ2a−a¯+λa−a¯+ψ¯a−a¯+136∑a=1p−1ψa2−a¯2+λ¯a2−a¯2+χ2a2−a¯2+λa2−a¯2+ψ¯a2−a¯2+136∑a=1p−1ψa+a¯λ¯a−a¯+χ2a−a¯+λa−a¯+ψ¯a−a¯+136∑a=1p−1λ¯a+a¯ψa−a¯+λa−a¯+χ2a−a¯+ψ¯a−a¯+136∑a=1p−1χ2a+a¯ψa−a¯+λ¯a−a¯+λa−a¯+ψ¯a−a¯+136∑a=1p−1λa+a¯ψa−a¯+λ¯a−a¯+χ2a−a¯+ψ¯a−a¯+136∑a=1p−1ψ¯a+a¯ψa−a¯+λ¯a−a¯+χ2a−a¯+λa−a¯−118⋅1+λ2+λ¯2.Note thatp≡7mod12 and χ2−1=−1, so we have(44)∑a=1p−1ψa±a¯=∑a=1p−1ψ¯a±a¯=∑a=1p−1χ2a±a¯=0,(45)∑a=1p−1ψa2−a¯2=∑a=1p−1ψ¯a2−a¯2=∑a=1p−1χ2a2−a¯2=0.From Lemma 1, 4, and 5, we have(46)∑a=1p−1λa+a¯+λ¯a+a¯+λa−a¯+λ¯a−a¯=2p⋅τ3λ+τ3λ¯=2d.Now, if 2 is a cubic residue modulop, then 1+λ2+λ¯2=3. Combining (43)–(46), Lemma 3, and Lemma 6–9, we have(47)N6p=p−136+136⋅2d+2d−4−318=136⋅p+4d−11.This proves Theorem1.Now, we prove Theorem2. If 2 is not a cubic residue modulo p, then 1+λ2+λ¯2=0. That is to say,(48)λ2=−12±32⋅i,λ¯2=−12∓32⋅i,where i2=−1.In this case, from Lemma1, we have(49)τ3λ−τ3λ¯2=τ3λ+τ3λ¯2−4⋅τ3λ⋅τ3λ¯=d2⋅p2−4p3=−p2⋅4p−d2=−27⋅p2⋅b2,or(50)τ3λ−τ3λ¯=±3⋅3⋅p⋅b⋅i.From (48), (50), and Lemma 3, we have(51)∑a=1p−1λa2−a¯2+λ¯a2−a¯2=1+λ2p⋅τ3λ+λ¯2⋅τ3λ¯=−1p·λ¯2⋅τ3λ+λ2·τ3λ¯=12p⋅pd±9⋅p⋅b=12⋅d±9b.Combining (43)–(46), (51), and Lemma 6–9, we have(52)N6p=p−136+136⋅2d−4+172⋅d+9b=172⋅2p+5d−10+9b,or(53)N6p=p−136+136⋅2d−4+172⋅d−9b=172⋅2p+5d−10−9b.This completes the proofs of our all results.


## 4. Conclusion

The main results of this paper are two theorems and two corollaries. Theorem1 gives an exact computing formula for N6p with p≡7mod12 and 2 is a cubic residue modulo p. If p≡7mod12 and 2 is not a cubic residue modulo p, then Theorem 2 established an identity for N6p and there are two possibilities. As some applications of these theorems, we also deduced two interesting congruences. For example, one of them is(54)p+4d≡11mod36,where p≡7mod12 and 2 is a cubic residue modulo p.In addition, ifp≡7mod12 and 2 is not a cubic residue modulo p, then we also have an interesting conjecture. That is,(55)N6p=172⋅2p+5d−10+d+bp⋅9b.



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