m∖n | 0 | 1 | 2 |
0 | p | 0 | 0 |
1 | 0 | 12(u+iv√3) | −1 |
2 | 0 | −1 | 12(u−iv√3) |
Let p be a prime, k be a positive integer, q=pk, and Fq be the finite field with q elements. Let F∗q be the multiplicative group of Fq, that is F∗q=Fq∖{0}. In this paper, explicit formulae for the numbers of solutions of cubic diagonal equations a1x31+a2x32=c and b1x31+b2x32+b3x33=c over Fq are given, with ai,bj∈F∗q (1≤i≤2,1≤j≤3), c∈Fq and p≡1(mod 3). Furthermore, by using the reduction formula for Jacobi sums, the number of solutions of the cubic diagonal equations a1x31+a2x32+⋯+asx3s=c of s≥4 variables with ai∈F∗q (1≤i≤s), c∈Fq and p≡1(mod 3), can also be deduced.
Citation: Shuangnian Hu, Rongquan Feng. The number of solutions of cubic diagonal equations over finite fields[J]. AIMS Mathematics, 2023, 8(3): 6375-6388. doi: 10.3934/math.2023322
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Let p be a prime, k be a positive integer, q=pk, and Fq be the finite field with q elements. Let F∗q be the multiplicative group of Fq, that is F∗q=Fq∖{0}. In this paper, explicit formulae for the numbers of solutions of cubic diagonal equations a1x31+a2x32=c and b1x31+b2x32+b3x33=c over Fq are given, with ai,bj∈F∗q (1≤i≤2,1≤j≤3), c∈Fq and p≡1(mod 3). Furthermore, by using the reduction formula for Jacobi sums, the number of solutions of the cubic diagonal equations a1x31+a2x32+⋯+asx3s=c of s≥4 variables with ai∈F∗q (1≤i≤s), c∈Fq and p≡1(mod 3), can also be deduced.
Let p be a prime, k be a positive integer, q=pk. Let Fq be the finite field with q elements and F∗q represent the nonzero elements of Fq. Let f(x1,⋯,xs)∈Fq[x1,⋯,xs] be a polynomial over Fq with s variables. A solution (or a rational point) of f(x1,⋯,xs) over Fq is an s-tuple (c1,⋯,cs)∈Fsq such that f(c1,⋯,cs)=0. Denote by
N(f,q)=N(f(x1,⋯,xs)=0)=#{(c1,⋯,cs)∈Fsq∣f(c1,⋯,cs)=0} |
the number of solutions of f(x1,⋯,xs)=0 over Fq.
It is one of the central problems to study the number N(f,q) of rational points over finite fields. From [14,15] we know that there exists an explicit formula for N(f,q) with degree deg(f)≤2. But generally speaking, it is much difficult to give an explicit formula for N(f,q). Finding the explicit formula for N(f,q) under certain conditions has attracted many researchers for many years (See, for instance, [1,3,4,5,6,7,8,9,10,11,12,13,16,17,18,20,21,22,23,24,25,26,27]).
Let k1,…,ks be positive integers. A diagonal equation is an equation of the form
a1xk11+⋯+asxkss=c |
with coefficients a1,…,as∈F∗q and c∈Fq. The special case where all the ki's are equal has extensively been studied (see, for example, [7,8,9,10,11,12,17,18,21,22,23,26,27]).
In 1977, S. Chowla et al. ([7]) investigated a problem about the number of solutions of an equation
x31+x32+⋯+x3s=0 |
over field Fp, where p is a prime with p≡1(mod 3). In 1979, Myerson [17] extended the result in [7] to the field Fq and also studied the number of solutions of the equation
x41+x42+⋯+x4s=0 |
over Fq. When q=p2t with pr≡−1(mod d) for a divisor r of t and d∣(q−1), Wolfmann [22] gave an explicit formula of the number of solutions of the equation
a1xd1+a2xd2+⋯+asxds=c |
over Fq in 1992, where a1,a2,…,as∈F∗q and c∈Fq. In 2018, Zhang and Hu [23] determined the number of solutions of the equation
x31+x32+x33+x34=c |
over Fp, with c∈F∗p and p≡1(mod 3).
In 2020, J. Zhao et al. [26,27] investigated the number of solutions of the equations
x41+x42=c, x41+x42+x43=c and x41+x42+x43+x44=c. |
over Fq, with c∈F∗q.
For any c∈Fq, let An(c) and Bn(c) denote the number of solutions of the equations x31+x32+⋯+x3n=c and x31+x32+⋯+x3n+cx3n+1=0 over Fq respectively. In 2021, by using the generator of F∗q, Hong and Zhu [10] gave the generating functions ∞∑n=1An(c)xn and ∞∑n=1Bn(c)xn. In 2022, W. Ge et al. [9] studied these two generating functions in a different way. Moreover, formulas of the number of solutions of equation a1x31+a2x32=c and a1x31+a2x32+a3x33=0 were also presented in [9].
In this paper, we consider the problem of finding the number of solutions of the diagonal cubic equation
f(x1,x2,⋯,xs)=a1x31+a2x32+⋯+asx3s−c=0 |
over Fq, where q=pk and a1,a2,…,as∈F∗q and c∈Fq.
If p=3 and k is an integer, or p≡2(mod 3) and k is an odd integer, then gcd(3,q−1)=1. It follows that (see [14] pp.105)
N(a1x31+a2x32+⋯+asx3=c)=N(a1x1+a2x2+⋯+asxs=c)=qs−1 |
with a1,a2,…,as∈F∗q, and c∈Fq.
If p≡2(mod 3) and k is an even integer, Hua and Vandiver [13] studied the number of solutions of some trinomial equations over Fq and Wolfmann [22] also got the number of solutions of certain diagonal equations over Fq. The following result can be deduced from Theorem 1 of [22].
Theorem 1.1. Let p≡2(mod 3) be a prime, k an even integer, q=pk, n=q−13, s≥2 and c∈Fq. Let α be a primitive element of Fq. Denote by N the number of solutions of the equation
a1x31+a2x32+⋯+asx3s=c |
over Fq. Then
N=qs−1+13(−1)ks/2qs/2−1(q−1)2∑j=0(−2)v(j) |
if c=0, and
N=qs−1−(−1)ks/2+1qs/2−1[(−2)θ(c)q1/2−13(q1/2−(−1)k/2)2∑j=0(−2)τ(j)] |
if c≠0, where v(j) is the number of i, 1≤i≤s, such that (αj)nani=(−1)k(p+1)/6; θ(c) is the number of i, 1≤i≤s, such that ani=(−c)n and τ(j) is the number of i, 1≤i≤s, such that ani=(αj)n.
However, the explicit formula for N(a1x31+a2x32+⋯+asx3s=c) is still unknown when p≡1(mod 3). 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 number of solutions of diagonal cubic equations
f1(x1,x2)=a1x31+a2x32−c=0 |
and
f2(x1,x2,x3)=b1x31+b2x32+b3x33−c=0 |
over Fq, with a1,a2,b1,b2,b3∈F∗q, c∈Fq and the characteristic p≡1(mod 3). Note that our approach, which applies Jacobi sums, is not the same as that of Ge et al. [9] and Hong and Zhu [10] which mainly applies Gauss sums. The case with arbitrary s≥4 variables can be deduced from the reduction formula for Jacobi sums. But we omit the tedious details here.
Let α∈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=αr. Such an integer r is called the index of β with the primitive element α, and denoted by indαβ:=r. For any nonzero integer m and prime number p, we define νp(m) as the greatest integer t such that pt divides m. Then νp(m) is a nonnegative integer, and νp(m)≥ 1 if and only if p divides m.
The results of this paper are stated as follows.
Theorem 1.2. Let k be a positive integer and q=pk with the prime p≡1(mod 3). Let α be a primitive element of Fq. Denote by N1 the number of solutions of the equation a1x31+a2x32=c over Fq. Then
N1=q−(q−1)δ(a1,a2) |
if c=0, and
N1={q+(−1)k−12k−1E(u,v,k)+δ(a1,a2), if indαca1a2≡0(mod 3),q+(−1)k2k(E(u,v,k)−O(u,v,k))+δ(a1,a2), if indαca1a2≡1(mod 3),q+(−1)k2k(E(u,v,k)+O(u,v,k))+δ(a1,a2), if indαca1a2≡2(mod 3) |
if c≠0, where
δ(a1,a2):={−2, if indαa1a22≡0(mod 3),1, if indαa1a22≢0(mod 3), |
E(u,v,k):=uk−k∑t=1ν2(t)=1(kt)uk−tvt3t2+k∑t=1ν2(t)≥2(kt)uk−tvt3t2, |
O(u,v,k):=k∑t=1ν2(t+1)=1(kt)uk−tvt3t+12−k∑t=1ν2(t+1)≥2(kt)uk−tvt3t+12 |
and the integers u and v are uniquely determined such that
u2+3v2=4p, u≡1(mod 3), v≡0(mod 3) and 3v≡u(2α(q−1)/3+1)(mod p). |
Theorem 1.3. Let k be a positive integer and q=pk with the prime p≡1(mod 3). Let α be a primitive element of Fq. Denote by N2 the number of solutions of the equation b1x31+b2x32+b3x33=c over Fq. Then
N2={q2−(q−1)(−1)k2k−1E(u,v,k), if indαb1b2b3≡0(mod 3),q2−(q−1)(−1)k+12k(E(u,v,k)−O(u,v,k)), if indαb1b2b3≡1(mod 3),q2−(q−1)(−1)k+12k(E(u,v,k)+O(u,v,k)), if indαb1b2b3≡2(mod 3) |
if c=0, and
N2={q2+(−1)k2k−1E(u,v,k)+S(c,b1,b2,b3), if indαb1b2b3≡0(mod 3),q2+(−1)k+12k(E(u,v,k)−O(u,v,k))+S(c,b1,b2,b3), if indαb1b2b3≡1(mod 3),q2+(−1)k+12k(E(u,v,k)+O(u,v,k))+S(c,b1,b2,b3), if indαb1b2b3≡2(mod 3) |
if c≠0, where E(u,v,k), O(u,v,k), u, v are defined as in Theorem 1.2 and
S(c,b1,b2,b3):=δ(c,b1,b2,b3)+q(ω(c,b1,b2,b3)+ω′(c,b1,b2,b3)), |
with
δ(c,b1,b2,b3):={2q, if indαcb21b22b3≡0(mod 3),−q, if indαcb21b22b3≢0(mod 3), |
ω(c,b1,b2,b3):={2, if indαcb21b2b23≡0(mod 3),−1, if indαcb21b2b23≢0(mod 3), |
and
ω′(c,b1,b2,b3):={2, if indαc2b21b2b3≡0(mod 3),−1, if indαc2b21b2b3≢0(mod 3). |
Corollary 1.4. Let k be a positive integer and q=pk with the prime p≡1(mod 3). Let α be a primitive element of Fq. Denote by N3 (resp. N4) the number of solutions of the equation x31+x32=0 (resp. x31+x32+x33=0) over Fq. Then
N3=3q−2 |
and
N4=q2+u(q−1), |
where the integer u is uniquely determined such that
u2+3v2=4p, u≡1(mod 3), v≡0(mod 3) and 3v≡u(2α(q−1)/3+1)(mod p). |
Corollary 1.4 is a special case of [17]. If k=1, then Corollary 1.4 is a special case of [7].
This paper is organized as follows. In Section 2, we present several basic concepts including the Jacobi sums, and give some preliminary lemmas. In Section 3, we prove Theorems 1.2 and 1.3 and finally, in Section 4, we supply some examples to illustrate the validity of our results.
In this section, we present some auxiliary lemmas that are needed in the proof of Theorems 1.2 and 1.3.
If λ is a multiplicative character of Fq, then λ is defined for all nonzero elements 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.
For any element α∈Fq=Fpk, the norm of α relative to Fp is defined by (see, for example, [14,15])
NFq/Fp(α):=ααp⋯αpk−1=αpk−1p−1. |
For the simplicity, we write N(α) for NFq/Fp(α). For any α∈Fq, it is clear that N(α)∈Fp. Furthermore, if α is a primitive element of Fq, then N(α) is a primitive element of Fp.
Let χ be a multiplicative character of Fp. Then χ can be lifted to a multiplicative character λ of Fq by setting λ(α)=χ(N(α)). The characters of Fp can be lifted to the characters of Fq, but not all characters of Fq can be obtained by lifting a character of Fp. The following lemma tells us when p≡1(mod 3), then any multiplicative character λ of order 3 of Fq can be lifted by a multiplicative character of order 3 of Fp.
Lemma 2.1. [15] 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 λ1,…,λs be s multiplicative characters of Fq. The Jacobi sum J(λ1,⋯,λs) is defined by
J(λ1,⋯,λs):=∑γ1+⋯+γs=1λ1(γ1)⋯λs(γs), |
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
J(λσ(1),⋯,λσ(s))=J(λ1,⋯,λs). |
The readers are referred to [2] and [15] 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 2.2. [15] 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 Fpk of Fp. Then
J(λ1,⋯,λs)=(−1)(s−1)(k−1)J(χ1,⋯,χs)k. |
We give the reduction formula for Jacobi sums as follows.
Lemma 2.3. [2] Let λ1,⋯,λs−1,λs be s nontrivial multiplicative characters of Fq. If s≥2, then
J(λ1,⋯,λs−1,λs)={−qJ(λ1,⋯,λs−1), if λ1⋯λs−1 is trivial,J(λ1⋯λs−1,λs)J(λ1,⋯,λs−1), if λ1⋯λs−1 is nontrivial. |
The value of some needed Jacobi sums are listed in the following two lemmas.
Lemma 2.4. [2] Let p≡1(mod 3) be a prime, q=pk, α be a primitive element of Fq, and let χ be a multiplicative character of order 3 over Fp. Then
2J(χ,χ)=u+iv√3, |
where the integers u and v are uniquely determined such that
u2+3v2=4p, u≡1(mod 3), v≡0(mod 3) and 3v≡u(2α(q−1)/3+1)(mod p). |
Lemma 2.5. [2] Let p≡1(mod 3), g be a primitive element of Fp and let χ be a multiplicative character of order 3 over Fp such that χ(g)=−1+i√32. Let the integers u and v be defined as in Lemma 2.4. Then the values of the nine Jacobi sums J(χm,χn) (m,n=0,1,2) are given in the following Table 1.
m∖n | 0 | 1 | 2 |
0 | p | 0 | 0 |
1 | 0 | 12(u+iv√3) | −1 |
2 | 0 | −1 | 12(u−iv√3) |
The following lemma gives the number of solutions of the diagonal equation in terms of Jacobi sums.
Lemma 2.6. [2] Let k1,…,ks be positive integers, a1,…,as∈F∗q and let c∈Fq. Set di=gcd(ki,q−1), and let λi be a multiplicative character of order di of Fq, i=1,…,s. Then the number N of solutions of the equation a1xk11+⋯+asxkss=c is given by
N=qs−1−(q−1)d1−1∑j1=1λj11⋯λjnntrivial⋯ds−1∑js=1λj11(a−11)⋯λjss(a−1s)J(λj11,…,λjss) |
if c=0, and by
N=qs−1+d1−1∑j1=1⋯ds−1∑js=1λj11(ca−11)⋯λjss(ca−1s)J(λj11,⋯,λjss) |
if c≠0.
In this section, we give the proof of Theorems 1.2 and 1.3. First, we begin with a lemma.
Lemma 3.1. Let p≡1(mod 3) be a prime, q=pk, α be a primitive element of Fq and let λ be the multiplicative character of order 3 of Fq such that λ(α)=−1+i√32. Then for any positive integers a, b and β∈F∗q, we have
λ(β)+λ(β2)={2, if indαβ≡0(mod 3),−1, if indαβ≢0(mod 3) |
and
λ(β)(a+ib√3)k+λ(β2)(a−ib√3)k={2E(a,b,k), if indαβ≡0(mod 3),−(E(a,b,k)+O(a,b,k)), if indαβ≡1(mod 3),−(E(a,b,k)−O(a,b,k)), if indαβ≡2(mod 3), |
where E(a,b,k) and O(a,b,k) are defined as in Theorem 1.2.
Proof. The first part of the lemma is obvious. Now we focus on the proof of the second part of the lemma. One can divide this into the following three cases.
If indαβ≡0(mod 3), then λ(β)=λ(β2)=1. One has
(a+ib√3)k+(a−ib√3)k=k∑t=0(kt)ak−t[(ib√3)t+(−ib√3)t]=2ak+k∑t=1(kt)ak−t[(ib√3)t+(−ib√3)t]=2ak−2k∑t=1ν2(t)=1(kt)ak−tbt3t2+2k∑t=1ν2(t)≥2(kt)ak−tbt3t2=2E(a,b,k). | (3.1) |
If indαβ≡1(mod 3), then λ(β)=−1+i√32 and λ(β2)=−1−i√32. One has
12((−1+i√3)(a+ib√3)k−(1+i√3)(a−ib√3)k)=12(−(a+ib√3)k−(a−ib√3)k+i√3(a+ib√3)k−i√3(a−ib√3)k)=−12(2k∑t=0t=even(kt)ak−titbt3t2−2i√3k∑t=1t=odd(kt)ak−titbt3t2)=−ak+k∑t=1ν2(t)=1(kt)ak−tbt3t2−k∑t=1ν2(t)≥2(kt)ak−tbt3t2−k∑t=1ν2(t+1)=1(kt)ak−tbt3t+12 +k∑t=1ν2(t+1)≥2(kt)ak−tbt3t+12=−(E(a,b,k)+O(a,b,k)). | (3.2) |
If indαβ≡2(mod 3), then λ(β)=−1−i√32 and λ(β2)=−1+i√32. One has
12((−1−i√3)(a+ib√3)k+(−1+i√3)(a−ib√3)k)=−k∑t=0t=even(kt)ak−titbt3t2−i√3k∑t=1t=odd(kt)ak−titbt3t2=−ak+k∑t=1ν2(t)=1(kt)ak−tbt3t2−k∑t=1ν2(t)≥2(kt)ak−tbt3t2+k∑t=1ν2(t+1)=1(kt)ak−tbt3t+12 −k∑t=1ν2(t+1)≥2(kt)ak−tbt3t+12=−(E(a,b,k)−O(a,b,k)). | (3.3) |
The result follows immediately from (3.1)–(3.3).
Now we can turn our attention to prove Theorems 1.2 and 1.3.
Proof of Theorem 1.2. Since gcd(3,q−1)=3, by Lemma 2.6, let α be a primitive element of Fq and λ be the multiplicative character of Fq of order 3 with λ(α)=−1+i√32. Then we deduce that
N1=q−(q−1)2∑j1=12∑j2=1λj1λj2 trivialλ(a−j11a−j22)J(λj1,λj2), | (3.4) |
if c=0, and
N1=q+2∑j1=12∑j2=1λ(cj1+j2a−j11a−j22)J(λj1,λj2), | (3.5) |
if c≠0.
Since p≡1(mod 3), it follows that λp−1 is trivial. By Lemma 2.1, the cubic multiplicative character λ can be lifted by a cubic multiplicative character χ of Fp. Combining with the Lemma 2.2, Table 1 of Lemma 2.5 and Lemma 3.1, we obtain
2∑j1=12∑j2=1λj1λj2trivialλ(a−j11a−j22)J(λj1,λj2)=(−1)k−1J(χ,χ2)k(λ(a21a2)+λ(a1a22))=−(λ(a21a2)+λ(a1a22))=δ(a1,a2), | (3.6) |
and
2∑j1=12∑j2=1λ(cj1+j2a−j11a−j22)J(λj1,λj2)=(−1)k−1(λ(c2a21a22)J(χ,χ)k+λ(a21a2)J(χ,χ2)k+λ(a1a22)J(χ2,χ)k +λ(ca1a2)J(χ2,χ2)k)=−(λ(a21a2)+λ(a1a22))+(−1)k−1(λ(c2a21a22)J(χ,χ)k+λ(ca1a2)J(χ2,χ2)k)=δ(a1,a2)+(−1)k−12k(λ(c2a21a22)(u+iv√3)k+λ(ca1a2)(u−iv√3)k)={(−1)k−12k−1E(u,v,k)+δ(a1,a2), if indαca1a2≡0(mod 3),(−1)k2k(E(u,v,k)−O(u,v,k))+δ(a1,a2), if indαca1a2≡1(mod 3),(−1)k2k(E(u,v,k)+O(u,v,k))+δ(a1,a2), if indαca1a2≡2(mod 3). | (3.7) |
Then from (3.4)–(3.7), one can easily deduce the result of Theorem 1.2.
Proof of Theorem 1.3. By the same argument as in the proof of theorem 1.2, let α be a primitive element of Fq and λ be the multiplicative character of Fq of order 3 with λ(α)=−1+i√32. We deduce that
N2=q2−(q−1)2∑j1=12∑j2=1λj1λj2λj3 trivial2∑j3=1λ(b−j11b−j22b−j33)J(λj1,λj2,λj3) | (3.8) |
if c=0, and
N2=q2+2∑j1=12∑j2=12∑j3=1λ(cj1+j2+j3b−j11b−j22b−j33)J(λj1,λj2,λj3) | (3.9) |
if c≠0.
Similarly, the cubic multiplicative character λ can be lifted by a cubic multiplicative character χ of Fp. By using the Lemmas 2.2, 2.3, Table 1 of Lemma 2.5 and Lemma 3.1, we get
2∑j1=12∑j2=1λj1λj2λj3trivial2∑j3=1λ(b−j11b−j22b−j33)J(λj1,λj2,λj3)=λ(b21b22b23)J(χ,χ)kJ(χ2,χ)k+λ(b1b2b3)J(χ,χ2)kJ(χ2,χ2)k=(−12)k(λ(b21b22b23)(u+iv√3)k+λ(b1b2b3)(u−iv√3)k)={(−1)k2k−1E(u,v,k), if indαb1b2b3≡0(mod 3),(−1)k+12k(E(u,v,k)−O(u,v,k)), if indαb1b2b3≡1(mod 3),(−1)k+12k(E(u,v,k)+O(u,v,k)), if indαb1b2b3≡2(mod 3) | (3.10) |
and
2∑j1=12∑j2=12∑j3=1λ(cj1+j2+j3b−j11b−j22b−j33)J(λj1,λj2,λj3)=λ(b1b2b3)J(χ,χ2)kJ(χ2,χ2)k+λ(b21b22b23)J(χ2,χ)kJ(χ,χ)k +λ(cb21b22b3)J(χ2,χ2)kJ(χ,χ)k+λ(c2b1b2b23)J(χ,χ)kJ(χ2,χ2)k +(−1)kq(λ(cb21b2b23)J(χ,χ2)k+λ(c2b1b22b3)J(χ2,χ)k +λ(c2b21b2b3)J(χ,χ2)k+λ(cb1b22b23)J(χ2,χ)k)=(−12)k(λ(b1b2b3)(u−iv√3)k+λ(b21b22b23)(u+iv√3)k) +122k(u2+3v2)k(λ(cb21b22b3)+λ(c2b1b2b23)) +q(λ(cb21b2b23)+λ(c2b1b22b3)+λ(c2b21b2b3)+λ(cb1b22b23)). | (3.11) |
By second part of Lemma 3.1, we derive that
(−12)k(λ(b1b2b3)(u−iv√3)k+λ(b21b22b23)(u+iv√3)k)={(−1)k2k−1E(u,v,k), if indαb1b2b3≡0(mod 3),(−1)k+12k(E(u,v,k)−O(u,v,k)), if indαb1b2b3≡1(mod 3),(−1)k+12k(E(u,v,k)+O(u,v,k)), if indαb1b2b3≡2(mod 3). | (3.12) |
Note that Lemma 2.4 tells us u2+3v2=4p. Then from the first part of Lemma 3.1, one has
122k(u2+3v2)k(λ(cb21b22b3)+λ(c2b1b2b23))={2q, if indαcb21b22b3≡0(mod 3),−q, if indαcb21b22b3≢0(mod 3) | (3.13) |
and
λ(cb21b2b23)+λ(c2b1b22b3)+λ(c2b21b2b3)+λ(cb1b22b23))=ω(c,b1,b2,b3)+ω′(c,b1,b2,b3). | (3.14) |
Thus from (3.8)–(3.14), the desired result of Theorem 1.3 follows immediately.
In this section, we present some examples to demonstrate the validity of our results.
Example 4.1. Let q=134. One can check that 2 is a primitive element of F13. Let ω be a primitive element of Fq such that N(ω)=ω134−113−1=2. We consider the numbers of solutions of the cubic equations
x31+ω2x32=0 |
and
x31+ω2x32=ω |
over Fq.
Since ω134−13=(ω134−113−1)13−13=24, the integers u and v in Lemma 2.4 are determined by
u2+3v2=52, u≡1(mod 3), v≡0(mod 3) and 3v≡u(2×24+1)(mod 13). |
We can get that u=−5 and v=−3. Therefore, by Theorem 1.2, we have
N(x31+ω2x32=0)=1 |
and
N(x31+ω2x32=ω)=28899. |
Example 4.2. Let q=312. One can check that 3 is a primitive element of F31. Let ω be a primitive element of Fq such that N(ω)=ω312−131−1=3. We consider the numbers of solutions of the cubic equations
ω4x31+x32+ωx33=0 |
and
ω4x31+x32+ωx33=ω |
over Fq.
Since ω312−13=(ω312−131−1)31−13=310, the integers u and v are determined by
u2+3v2=124, u≡1(mod 3), v≡0(mod 3) and 3v≡u(2×310+1)(mod 31). |
We get u=4 and v=6. Thus by Theorem 1.3, we deduce that
N(ω4x31+x32+ωx33=0)=936001 |
and
N(ω4x31+x32+ωx33=ω)=920625. |
Example 4.3. Let q=73. It is clear that 3 is a primitive element of F7. Let ω be a primitive element of Fq such that N(ω)=ω73−17−1=3. We consider the the numbers of solutions of the cubic equations
x31+ω2x32+ω3x33=0 |
and
x31+ω2x32+ω3x33=ω |
over Fq.
Similarly, since ω73−13=(ω73−17−1)7−13=32, the integers u and v are determined by
u2+3v2=28, u≡1(mod 3), v≡0(mod 3) and 3v≡u(2×32+1)(mod 7). |
We deduce that u=1 and v=−3. Thus by Theorem 1.3, we have
N(x31+ω2x32+ω3x33=0)=111835 |
and
N(x31+ω2x32+ω3x33=ω)=117666. |
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 formulae for the numbers of solutions of cubic diagonal equations a1x31+a2x32=c and b1x31+b2x32+b3x33=c over Fq are given, with ai,bj∈F∗q (1≤i≤2,1≤j≤3), c∈Fq and p≡1(mod 3). Furthermore, by using the reduction formula for Jacobi sums, the number of solutions of the cubic diagonal equations a1x31+a2x32+⋯+asx3s=c of s≥4 variables with ai∈F∗q (1≤i≤s), c∈Fq and p≡1(mod 3), can also be deduced.
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).
We declare that we have no conflict of interest.
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1. | Shuangnian HU, Shihan WANG, Yanyan LI, Yujun NIU, Note on the Number of Solutions of Cubic Diagonal Equations over Finite Fields, 2023, 28, 1007-1202, 369, 10.1051/wujns/2023285369 |
m∖n | 0 | 1 | 2 |
0 | p | 0 | 0 |
1 | 0 | 12(u+iv√3) | −1 |
2 | 0 | −1 | 12(u−iv√3) |