On the Sixth Residues and Some New Properties of Their Distribution

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’ modulo 
 
 p
 
 and give two exact calculating formulas for them.


Introduction
Let p be an odd prime and k be a fixed positive integer. For any integer a with (a, p) � 1, if the congruence equation x k ≡ a mod p 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: if a is a quadraticre sidue modulo p, −1, if a is a quadratic nonresidue modulo p, 0, if p|a.
Sometimes, we write Legendre's symbol ( * /p) as χ 2 for the sake of writing. is 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. is is especially true in the study of primes and related problems. For example, if p is a prime with p ≡ 1 mod 4, then one has (see eorem 4-11 in [1]) where a denotes the inverse of a. at is, a · a ≡ 1 mod p, and (rs/p) � −1.
In this paper, we are concerned with the problem of whether the special integers a + a and a − a both are kth residues' modulo p. Let N k (p) denote the number of all integers 1 < a < p − 1 such that a + a and a − a both are kth residues' modulo p.
en, how are the values of N k (p) distributed?
Very recently, some authors had studied the calculating problem of N k (p) and obtained a series of interesting results. For example, Wang and Lv [10] obtained the identity Hu and Chen [11] proved the following result: let p be an odd prime with p ≡ 7 mod 12. If 2 is a cubic residue mod p, then one has the identity If 2 is not a cubic residue mod p, then one has the asymptotic formula where d is defined in (7) and E(p) satisfies the estimates Su and Zhang [12] considered the case p ≡ 5 mod 8 and proved the identity As an extension of [10][11][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.
at is, any quadratic residue a modulo p is a sixth residue modulo p. So, we just consider the nontrivial case p ≡ 1 mod 6. In this case, we know that there are two integers d and b such that the identity where d is uniquely determined by d ≡ 1 mod 3 (see [13]). And, it is clear from (7) that the value of N 6 (p) 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 of N 6 (p) and give an exact calculating formula for it. at is, we will prove the following two results. Theorem 1. Let p be an odd prime with p ≡ 7 mod 12. If 2 is a cubic residue modulo p, then we have the identity where d is the same as defined in (7).

Theorem 2.
Let p be an odd prime with p ≡ 7 mod 12. If 2 is not a cubic residue modulo p, then we have From our theorems, we may immediately deduce the following two corollaries.
First, in eorems 1 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, if p is a prime with p ≡ 1 mod 12, 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 N 6 (p). at is, ird, if p is an odd prime with p ≡ 7 mod 12 and 2 is not a cubic residue modulo p, then our eorem 2 also obtained an exact calculating formula for N 6 (p), which is obviously better than the corresponding result in [11].
Of course, our eorem 2 is flawed, and it presents two possibilities. How to determine its correct value is an interesting open problem.

Conjecture.
Let p be an odd prime with p ≡ 7 mod 12. If 2 is not a cubic residue modulo p, then we have the identity

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 2 Journal of Mathematics only three necessary references [1,16,17] are provided here. First, we have the following.

Lemma 1.
Let p be an odd prime with p ≡ 1 mod 3. en, for any third-order character λ modulo p (i.e., λ ≠ χ 0 and λ 3 � χ 0 , the principal character modulo p), we have the identity where Proof. For the proof of this lemma, see Zhang and Hu [18] or Berndt and Evans [19]. □ Lemma 2. Let p be a prime with p ≡ 1 mod 6. en, for any third-order character λ mod p and sixth-order character ψ � χ 2 λ (i.e., ψ i ≠ χ 0 , 1 ≤ i ≤ 5, and ψ 6 � χ 0 ), we have the identity Proof. From the properties of the Gauss sums and the reduced residue system modulo p, note that the identity and we have On the contrary, we also have Note that identity τ(λ) · τ(λ) � p, and from (17) and (18), we deduce the identity is proves Lemma 2.
Similarly, we can also deduce that It is clear that Lemma 3 follows from (21) and (22). □ Lemma 4. Let p be an odd prime with p ≡ 7 mod 12. en, we have the identity Proof. From the methods of proving Lemma 2 and the properties of the Gauss sums, we have So, from (24), we have is proves Lemma 4. □ Lemma 5. Let p be an odd prime with p ≡ 7 mod 12. en, we have Proof. It is the same as the proof of Lemma 4, so it is omitted. Proof. Note that ψ(−1) � ψ(−1) � −1, and from the properties of the reduced residue system modulo p, we have which implies that is proves Lemma 6. □ Lemma 7. Let p be an odd prime with p ≡ 7 mod 12. For any sixth-order character ψ mod p, we have Proof. Since ψ � χ 2 λ, χ 2 (−1) � −1, so, from the reduced residue system modulo p, we have which implies that Similarly, we can also deduce the identity is proves Lemma 7.
□ Lemma 8. Let p be an odd prime with p ≡ 7 mod 12. en, for any third-order character λ mod p and ψ � χ 2 λ, we have Proof. Note that λ(−1) � 1 and ψ(−1) � −1, we have So, we have the identity Similarly, we can also deduce the identity is proves Lemma 8.
□ Lemma 9. Let p be an odd prime with p ≡ 7 mod 12. en, for any third-order character λ mod p, we have Proof. From the properties of Legendre's symbol mod p, we have Similarly, we also have Combining (39)

Proofs of the Theorems
In this section, we shall complete the proofs of our main results. First, we prove eorem 1. For any prime p with p ≡ 7 mod 12, let λ denote a third-order character modulo p; then, ψ � χ 2 λ is a sixth-order character modulo p. So, for any integer a with (a, p) � 1, from the characteristic function Now, if 2 is a cubic residue modulo p, then 1 + λ(2) + λ(2) � 3. Combining (43)-(46), Lemma 3, and Lemma 6-9, we have