Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 28, Number 5, October 2023
Page(s) 369 - 372
DOI https://doi.org/10.1051/wujns/2023285369
Published online 10 November 2023

© Wuhan University 2023

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Let be the finite field, , with being a prime and being a positive integer. Let be the multiplicative group of , that is . Counting the number of zeros of the equation is an important and fundamental topic in number theory and finite field. From Refs. [1,2], we know that there exists an explicit formula for with degree But generally speaking, it is much difficult to give an explicit formula for .

Let be positive integers. A diagonal equation is an equation of the form

with coefficients and . Counting the number of solutions of the diagonal equation is a difficult problem. The special case where all the are equal has extensively been studied(see, for instance, Refs. [3-14]). This is the example chosen by Weil[10] to illustrate his renowned conjecture on projective varieties over finite fields.

For any , let denote the number of zeros of the following diagonal equation

over . In 1977, Chowla et al[3] initiated the investigation of over . When , it is easy to see that if . However, When , the situation becomes complicated. Chowla et al[3] got that the generating function is a rational function of . In 1979, Myerson[9] extended the result in Ref. [3] to the field . When with for a divisor of and , Wolfmann[11] gave an explicit formula of the number of solutions of the equation

over in 1992, where and . In 2018, Zhang and Hu[12] determined the number of solutions of the equation

over , with and . In 2021, by using the generator of , Hong and Zhu[6] gave the generating functions . In 2022, Ge et al[5] studied the generating functions in a different way.

In this paper, we consider the problem of finding the number of solutions of the diagonal cubic equation

over , where and .

If and is an integer, or and is an odd integer, then . It follows that (see Ref. [2], p.105)

with .

If and is an even integer, Hu and Feng[7] presented an explicit formula for by using the Theorem 1 of Ref. [11]. However, the explicit formula for is still unknown when and . In this paper, we solve this problem by using Jacobi sums and an analog of Hasse-Davenport theorem.

The main result of this paper can be stated as follows.

Theorem 1   Let be a positive integer and with the prime . Let (resp. ) be a generator of (resp. ). Let (resp. ) be a multiplicative character of order 3 over (resp.) given by (resp. . Let and be the integers uniquely determined by

and

Set

Let denote the number of rational points of over . Then

where

and

This paper is organized as follows. In Section 1, we present several basic concepts and give some preliminary lemmas. In Section 2, we prove Theorem 1. In Section 3, we supply an example to illustrate the validity of our result.

1 Preliminary Lemmas

In this section, we present some useful lemmas that are needed in the proof of Theorem 1. We begin with two definitions.

Definition 1[2,15] Let be a prime number and with being a positive integer. For any element , the norm of relative to is defined by

For the simplicity, we write for

For any , it is clear that . Furthermore, if is a primitive element of , then is a primitive element of .

Definition 2[2,15] Let be multiplicative characters of . The Jacobi sum is defined by

where the summation is taken over all n-tuples of elements of with

Let be a multiplicative character of . Then can be lifted to a multiplicative character of by setting . The characters of can be lifted to the characters of , but not all the characters of can be obtained by lifting a character of . The following lemma tells us when , then any multiplicative character of order 3 of can be lifted by a multiplicative character of order 3 of .

Lemma 1[2]Let be a finite field and be a extension of . A multiplicative character of can be lifted by a multiplicative character of if and only if is trivial.

The following lemma provides an important relationship between the Jacobi sums in and the Jacobi sums in .

Lemma 2[2] Let be multiplicative characters of , not all of which are trivial. Suppose are lifted to characters , respectively, of the finite extension field of with . Then

Lemma 3[15] Let be a prime and let be a generator of . Let be a multiplicative character of order 3 over given by . Let and be nonnegative integers with . Set

Then

where and are defined as in Theorem 1.

The following lemma gives an explicit formula for the number of solutions of the diagonal equation in terms of Jacobi sums.

Lemma 4[15] Let be positive integers,. Set , and let be a multiplicative character on of order , Then the number of solutions of the equation is given by

2 Proof of Theorem 1

In this section, we give the proof of Theorem 1.

Proof of Theorem 1   Let be a multiplicative character on of order 3 with . Since , by using Lemma 3, we deduce that the number of solutions in is given by

For integers

we need to calculate the sum over with of the equal to 1 and of the equal to 2. That is

and

Since is a multiplicative character on of order 3 and , thus is trivial. Then from Lemma 1, we can deduce that the cubic multiplicative character of can be lifted by a cubic multiplicative character of . By using Lemma 2 and Lemma 3, one get

So that

Using Lemma 3 for the value of , considering the three cases separately, we obtain

and

Then the desired result

follows immediately.

3 Example

In this section, we present an example to demonstrate the validity of our result Theorem 1.

Example 1 Let be a generator of . Now we use Theorem 1 to obtain the number of zeros of the cubic equation:

It is easy to see that 3 is a generator of , and then we obtain That means

Since , then the integers and are determined by and that is

Thus

and

Further,

and

Thus by Theorem 1, we get

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