1.
Introduction
Let p be a prime, k a positive integer, q=pk, and let Fq be the finite field of q elements. Let f be a polynomial over Fq with n variables, and denote by
the number of solutions of f(x1,⋯,xn)=0 over Fq.
Studying the value of N(f,q) is one of the main topics in the theory of finite fields. Generally speaking, it is difficult to give the explicit formula for N(f,q).
The degree d of the polynomial f plays an important role in the estimate of N(f,q). An upper bound for N(f,q) [14] is given by
For any positive integer m, we use ordpm to denote the p-adic valuation of m. Suppose that N(f,q)>0, the classical Chevalley-Warning theorem shows that ordpN(f,q)>0 if n>d. Furthermore, let ⌈x⌉ denote the least integer ≥x and let q=pk, Ax [1] showed that
Finding the explicit formula for N(f,q) under certain conditions has attracted researchers for many years. From [13,14] we know that there exists an explicit formula for N(f,q) satisfying deg(f)≤2 over Fq. Some other works were done by Baoulina [2,3,4,5], Cao et al. [7,8], Hua and Vandiver [12], Hu et al. [10,11], Weil [17], and Zhang and Wan [20,21].
In 1977, Chowla, Cowles and Cowles [9] got a formula for the number of solutions of the hypersurface
over Fp. In 1979, Myerson [15] extended the result in [9] to the field Fq and studied the number of solutions of the equation
over Fq. For q=p2t with pr≡−1(mod d) for a divisor r of t and d∣(q−1), Wolfmann [18] gave an explicit formula of the number of solutions of the equation
over Fq in 1992, where a1,a2,…,an∈F∗q and c∈Fq. In 2018, Zhang and Hu [19] determined an explicit formula of the number of solutions of the equation
over Fp, with p≡1(mod 3) and c∈F∗p. In 2020, J. Zhao et al. [22,23] investigated the number of solutions of the forms
and
over Fq, with c∈F∗q.
In this paper, we consider the problem of finding the number of solutions of the diagonal sextic equation
over Fq, where q=pk and c∈F∗q.
It is well-known that (see [13], p. 105)
over Fq. Let q=pk with p a prime. If p=2, then gcd(6,q−1)=1 or 3 depends on k is odd or even. From Corollary 4 in [18], we can obtain the following result.
Theorem 1.1. Let p=2, k an integer, q=pk and c∈F∗q. Then
if k is odd, and
if k is even and n≥2, with ξ=(−1)k2+1.
If p=3 and k is an integer, or p≡5(mod 6) and k is an odd integer, then gcd(6,q−1)=2. It follows from Theorems 6.26 and 6.27 in [14] that the following result is given.
Theorem 1.2. Let p=3 and k an integer, or p≡5(mod 6) be a prime and k an odd integer, q=pk and c∈F∗q. Then
if n is even, and
if n is odd, where η is the quadratic multiplicative character of Fq.
If p≡5(mod 6) and k is an even integer, Hua and Vandiver [12] studied the number of solutions of some trinomial equations over Fq and Wolfmann [18] also got the number of solutions of certain diagonal equations over Fq. The following result can be deduced from Corollary 4 of [18].
Theorem 1.3. Let p≡5(mod 6) be a prime, k an even integer, q=pk, n≥2 and c∈F∗q. Then
where ξ=(−1)k2+1.
However, the formula for N(x61+x62+⋯+x6n=c) is still unknown when p≡1(mod 6). In this paper, we solve this problem by using Jacobi sums and an analog of Hasse-Davenport theorem. We give an explicit formula for the case with 2 variables. The case with arbitrary n≥3 variables can be deduced from the reduction formula for Jacobi sums.
Let g∈F∗q be a fixed primitive element of Fq. For any β∈F∗q, there exists exactly one integer r∈[1,q−1] such that β=gr. Such r is called the index of β with the primitive element g, and denoted by indgβ:=r.
For any element α∈E=Fpk and F=Fp, the norm of α relative to Fp are defined by (see, for example, [6,14])
For the simplicity, we write N(α) for NE/F(α).
The main result of this paper is stated as follows.
Theorem 1.4. Let t and k be positive integers, p=6t+1 be a prime, q=pk, and let Fq be the finite field with q elements. Let c∈F∗q, g be a primitive element of Fq and Z=indN(g)2. Then
when t or k is even, and
when both t and k are odd, where a+ib√3=(a′+ib′√3)k, u+iv√3=(u′+iv′√3)k, r+is√3=(r′+is′√3)k with a′ and b′ being integers such that
and the integers r′, s′, u′ and v′ are given by
This paper is organized as follows. In Section 2, we recall some useful known results which will be needed later. In Section 3, we prove Theorem 1.4 and then present an example to illustrate the validity of our result.
2.
Preliminary lemmas
In this section, we present some useful lemmas that are needed in the proof of Theorem 1.4. For any multiplicative character λ of Fq, it is now convenient to extend the definition of λ by setting λ(0)=1 if λ is the trivial character and λ(0)=0 otherwise.
Let λ1,⋯,λs be s multiplicative characters of Fq, the Jacobi sum J(λ1,⋯,λs) is defined by
where the summation is taken over all s-tuples (γ1,⋯,γs) of elements of Fq with γ1+⋯+γs=1.
It is clear that if σ is a permutation of {1,⋯,s}, then
Also for any α∈F∗q, we have that
The readers are referred to [6] and [14] for basic facts on Jacobi sums.
The following theorem is an analog of Hasse-Davenport theorem for Jacobi sums which establishes an important relationship between the Jacobi sums in Fq and the Jacobi sums in Fp.
Lemma 2.1. [14] Let χ1,⋯,χs be s multiplicative characters of Fp, not all of which are trivial. Suppose χ1,⋯,χs are lifted to characters λ1,⋯,λs, respectively, of the finite extension field E of Fp with [E:Fp]=k. Then
Let χ be a multiplicative character of F and λ be a multiplicative character of E. Recall that χ can be lifted to E by setting λ(α)=χ(N(α)). The characters of Fp can be lifted to the characters of Fq, but not all the characters of Fq can be obtained by lifting a character of Fp. The following lemma tells us when p≡1(mod 6), then the multiplicative character λ of order 6 of Fq can be lifted by a multiplicative character of order 6 of Fp.
Lemma 2.2. [14] Let Fp be a finite field and Fq be an extension of Fp. A multiplicative character λ of Fq can be lifted by a multiplicative character χ of Fp if and only if λp−1 is trivial.
Let g be a generator of F∗q=F∗pk. Since
and
one knows that N(g) is a generator of F∗p. Then we can state the following lemma.
Lemma 2.3. [6] Let p≡1(mod 6) be a prime, q a power of p, g be a generator of F∗q, and let χ be a multiplicative character of order 6 over Fp. Then
where the integers a′ and b′ are uniquely determined by
Lemma 2.4. [6] Let p=6t+1 be a prime number. Let g′ be the generator of F∗p and χ be a multiplicative character of order 6 over Fp such that χ(g′)=1+i√32. Let the integers a′ and b′ be defined as in Lemma 2.3 and the integers u′, v′, r′, s′ are given as in Theorem 1.4. The values of the 36 Jacobi sums J(χm,χn) (m,n=0,1,2,3,4,5) are given in the following Table 1.
The following lemma gives an explicit formula for the number of solutions of the diagonal equation in terms of Jacobi sums.
Lemma 2.5. [6,12,17] Let k1,⋯,ks be positive integers, a1,⋯,as,c∈F∗q. Set di=gcd(ki,q−1), and let λi be a multiplicative character on Fq of order di, i=1,⋯,s. Then the number N of solutions of the equation a1xk11+⋯+asxkss=c is given by
3.
Proof of Theorem 1.4
In this section, we give the proof of Theorem 1.4.
Proof. Let g be a primitive element of Fq and λ be a multiplicative character of Fq of order 6 with λ(g)=1+i√32. Since q≡1(mod 6), then gcd(6,q−1)=6. Using Lemma 2.5, we have
Since p≡1(mod 6), it follows that λp−1 is trivial. By Lemma 2.2, the multiplicative character λ can be lifted by a multiplicative character χ of order 6 of Fp. By Lemma 2.1, we obtain
Consider the case when t or k is even. Noting that λ(1)=1, from (3.1), one has
From Table 1 of Lemma 2.4, we derive that
and
Thus if indgc≡0(mod 6), then it follows from (3.2)–(3.4) that
From (3.1), we have
Noting that λ(g)=1+i√32 and λ6(g)=1, then from Table 1 of Lemma 2.4, we deduce that
and
Thus if indgc≡1(mod 6), from (3.5)–(3.7), one can deduce that
In the similar way, one can deduce that
The case when both t and k are odd can also be proved by the same argument.
Remark. (Reduction formula for Jacobi sums) [6] Let χ1,⋯,χk be k nontrivial multiplicative characters of Fq. If k≥2, then
One can use the reduction formula for Jacobi sums to give an explicit formula for the number of solutions of the diagonal sextic equation
of n≥3 variables over Fpk, with c∈F∗pk and p≡1(mod 6). But we omit the tedious details here.
By concluding this section, we present an example to demonstrate the validity of Theorem 1.4.
Example. Let q=193. We consider the numbers of solutions of the sextic equation
over Fq, where c∈F∗q.
Since 2 is a primitive root modulo 19, we have Z=ind22=1. Let g be a generator of F∗193 such that
That means
The integers a′ and b′ are determined by
We can get that a′=−4, b′=−1. Since Z≡1(mod 3), we obtain that r′=7, s′=−3, u′=1 and v′=5. Thus we have a=−28, b=−45, r=−224, s=−360, u=−224 and v=−360. By Theorem 1.4, one can get that
4.
Conlusions
Studying the number of solutions of the polynomial equation f(x1,x2,⋯,xn)=0 over Fq is one of the main topics in the theory of finite fields. Generally speaking, it is difficult to give an explicit formula for the number of solutions of the equation f(x1,x2,⋯,xn)=0. There are many researchers who concentrated on finding the formula for the number of solutions of f(x1,x2,⋯,xn)=0 under certain conditions. Exponential sums are important tools for solving problems involving the number of solutions of the equation f(x1,x2,⋯,xn)=0 over Fq. In this paper, by using the Jacobi sums and the Hasse-Davenport theorem for Jacobi sums, we give an explicit formula for the number of solutions of the two-variable diagonal sextic equations x61+x62=c over Fq, with c∈F∗q and p≡1(mod 6), where p is the characteristic of Fq. Furthermore, by using the reduction formula for Jacobi sums, the number of solutions of the diagonal sextic equations x61+x62+⋯+x6n=c of n≥3 variables with c∈F∗q and p≡1(mod 6), can also be deduced.
Acknowledgments
The Authors express their gratitude to the anonymous referee for carefully examining this paper and providing a number of important comments and suggestions. This research was supported by the National Science Foundation of China (No. 12026223 and No. 12026224) and by the National Key Research and Development Program of China (No. 2018YFA0704703).
Conflict of interest
We declare that we have no conflict of interest.