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Research article

The number of solutions of cubic diagonal equations over finite fields

  • Received: 17 October 2022 Revised: 16 December 2022 Accepted: 19 December 2022 Published: 03 January 2023
  • MSC : 11T06, 11T24

  • Let p be a prime, k be a positive integer, q=pk, and Fq be the finite field with q elements. Let Fq be the multiplicative group of Fq, that is Fq=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,bjFq (1i2,1j3), cFq and p1(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 s4 variables with aiFq (1is), cFq and p1(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 Fq be the multiplicative group of Fq, that is Fq=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,bjFq (1i2,1j3), cFq and p1(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 s4 variables with aiFq (1is), cFq and p1(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 Fq 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)Fsqf(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,,asFq and cFq. 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 p1(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 pr1(mod d) for a divisor r of t and d(q1), 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,,asFq and cFq. In 2018, Zhang and Hu [23] determined the number of solutions of the equation

    x31+x32+x33+x34=c

    over Fp, with cFp and p1(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 cFq.

    For any cFq, 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 Fq, 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++asx3sc=0

    over Fq, where q=pk and a1,a2,,asFq and cFq.

    If p=3 and k is an integer, or p2(mod 3) and k is an odd integer, then gcd(3,q1)=1. It follows that (see [14] pp.105)

    N(a1x31+a2x32++asx3=c)=N(a1x1+a2x2++asxs=c)=qs1

    with a1,a2,,asFq, and cFq.

    If p2(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 p2(mod 3) be a prime, k an even integer, q=pk, n=q13, s2 and cFq. 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=qs1+13(1)ks/2qs/21(q1)2j=0(2)v(j)

    if c=0, and

    N=qs1(1)ks/2+1qs/21[(2)θ(c)q1/213(q1/2(1)k/2)2j=0(2)τ(j)]

    if c0, where v(j) is the number of i, 1is, such that (αj)nani=(1)k(p+1)/6; θ(c) is the number of i, 1is, such that ani=(c)n and τ(j) is the number of i, 1is, such that ani=(αj)n.

    However, the explicit formula for N(a1x31+a2x32++asx3s=c) is still unknown when p1(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+a2x32c=0

    and

    f2(x1,x2,x3)=b1x31+b2x32+b3x33c=0

    over Fq, with a1,a2,b1,b2,b3Fq, cFq and the characteristic p1(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 s4 variables can be deduced from the reduction formula for Jacobi sums. But we omit the tedious details here.

    Let αFq be a fixed primitive element of Fq. For any βFq, there exists exactly one integer r[1,q1] 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 p1(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(q1)δ(a1,a2)                                                                                               

    if c=0, and

    N1={q+(1)k12k1E(u,v,k)+δ(a1,a2),   if indαca1a20(mod 3),q+(1)k2k(E(u,v,k)O(u,v,k))+δ(a1,a2),  if indαca1a21(mod 3),q+(1)k2k(E(u,v,k)+O(u,v,k))+δ(a1,a2),  if indαca1a22(mod 3)

    if c0, where

    δ(a1,a2):={2, if indαa1a220(mod 3),1, if indαa1a220(mod 3),                       
    E(u,v,k):=ukkt=1ν2(t)=1(kt)uktvt3t2+kt=1ν2(t)2(kt)uktvt3t2,            
    O(u,v,k):=kt=1ν2(t+1)=1(kt)uktvt3t+12kt=1ν2(t+1)2(kt)uktvt3t+12            

    and the integers u and v are uniquely determined such that

    u2+3v2=4p, u1(mod 3), v0(mod 3) and 3vu(2α(q1)/3+1)(mod p).

    Theorem 1.3. Let k be a positive integer and q=pk with the prime p1(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(q1)(1)k2k1E(u,v,k),  if indαb1b2b30(mod 3),q2(q1)(1)k+12k(E(u,v,k)O(u,v,k)),  if indαb1b2b31(mod 3),q2(q1)(1)k+12k(E(u,v,k)+O(u,v,k)),  if indαb1b2b32(mod 3)

    if c=0, and

    N2={q2+(1)k2k1E(u,v,k)+S(c,b1,b2,b3), if indαb1b2b30(mod 3),q2+(1)k+12k(E(u,v,k)O(u,v,k))+S(c,b1,b2,b3),  if indαb1b2b31(mod 3),q2+(1)k+12k(E(u,v,k)+O(u,v,k))+S(c,b1,b2,b3),  if indαb1b2b32(mod 3)

    if c0, 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 indαcb21b22b30(mod 3),q,  if indαcb21b22b30(mod 3),
    ω(c,b1,b2,b3):={2,    if indαcb21b2b230(mod 3),1,   if indαcb21b2b230(mod 3),

    and

    ω(c,b1,b2,b3):={2,    if indαc2b21b2b30(mod 3),1,   if indαc2b21b2b30(mod 3).

    Corollary 1.4. Let k be a positive integer and q=pk with the prime p1(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=3q2

    and

    N4=q2+u(q1),

    where the integer u is uniquely determined such that

    u2+3v2=4p, u1(mod 3), v0(mod 3) and 3vu(2α(q1)/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 αFq=Fpk, the norm of α relative to Fp is defined by (see, for example, [14,15])

    NFq/Fp(α):=ααpαpk1=αpk1p1.

    For the simplicity, we write N(α) for NFq/Fp(α). For any αFq, 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 p1(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 λp1 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)(s1)(k1)J(χ1,,χs)k.

    We give the reduction formula for Jacobi sums as follows.

    Lemma 2.3. [2] Let λ1,,λs1,λs be s nontrivial multiplicative characters of Fq. If s2, then

    J(λ1,,λs1,λs)={qJ(λ1,,λs1),                      if λ1λs1 is trivial,J(λ1λs1,λs)J(λ1,,λs1),  if λ1λs1 is nontrivial.

    The value of some needed Jacobi sums are listed in the following two lemmas.

    Lemma 2.4. [2] Let p1(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+iv3,

    where the integers u and v are uniquely determined such that

    u2+3v2=4p, u1(mod 3), v0(mod 3) and 3vu(2α(q1)/3+1)(mod p).

    Lemma 2.5. [2] Let p1(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+i32. 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.

    Table 1.  the values of the Jacobi sums J(χm,χn).
    mn 0 1 2
    0 p 0 0
    1 0 12(u+iv3) 1
    2 0 1 12(uiv3)

     | Show Table
    DownLoad: CSV

    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,,asFq and let cFq. Set di=gcd(ki,q1), 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=qs1(q1)d11j1=1λj11λjnntrivialds1js=1λj11(a11)λjss(a1s)J(λj11,,λjss)

    if c=0, and by

    N=qs1+d11j1=1ds1js=1λj11(ca11)λjss(ca1s)J(λj11,,λjss)

    if c0.

    In this section, we give the proof of Theorems 1.2 and 1.3. First, we begin with a lemma.

    Lemma 3.1. Let p1(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+i32. Then for any positive integers a, b and βFq, we have

    λ(β)+λ(β2)={2,    if indαβ0(mod 3),1,   if indαβ0(mod 3)

    and

    λ(β)(a+ib3)k+λ(β2)(aib3)k={2E(a,b,k),                  if indαβ0(mod 3),(E(a,b,k)+O(a,b,k)),   if indαβ1(mod 3),(E(a,b,k)O(a,b,k)),   if indαβ2(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 indαβ0(mod 3), then λ(β)=λ(β2)=1. One has

    (a+ib3)k+(aib3)k=kt=0(kt)akt[(ib3)t+(ib3)t]=2ak+kt=1(kt)akt[(ib3)t+(ib3)t]=2ak2kt=1ν2(t)=1(kt)aktbt3t2+2kt=1ν2(t)2(kt)aktbt3t2=2E(a,b,k). (3.1)

    If indαβ1(mod 3), then λ(β)=1+i32 and λ(β2)=1i32. One has

    12((1+i3)(a+ib3)k(1+i3)(aib3)k)=12((a+ib3)k(aib3)k+i3(a+ib3)ki3(aib3)k)=12(2kt=0t=even(kt)aktitbt3t22i3kt=1t=odd(kt)aktitbt3t2)=ak+kt=1ν2(t)=1(kt)aktbt3t2kt=1ν2(t)2(kt)aktbt3t2kt=1ν2(t+1)=1(kt)aktbt3t+12    +kt=1ν2(t+1)2(kt)aktbt3t+12=(E(a,b,k)+O(a,b,k)). (3.2)

    If indαβ2(mod 3), then λ(β)=1i32 and λ(β2)=1+i32. One has

    12((1i3)(a+ib3)k+(1+i3)(aib3)k)=kt=0t=even(kt)aktitbt3t2i3kt=1t=odd(kt)aktitbt3t2=ak+kt=1ν2(t)=1(kt)aktbt3t2kt=1ν2(t)2(kt)aktbt3t2+kt=1ν2(t+1)=1(kt)aktbt3t+12    kt=1ν2(t+1)2(kt)aktbt3t+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,q1)=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+i32. Then we deduce that

    N1=q(q1)2j1=12j2=1λj1λj2 trivialλ(aj11aj22)J(λj1,λj2), (3.4)

    if c=0, and

    N1=q+2j1=12j2=1λ(cj1+j2aj11aj22)J(λj1,λj2), (3.5)

    if c0.

    Since p1(mod 3), it follows that λp1 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

    2j1=12j2=1λj1λj2trivialλ(aj11aj22)J(λj1,λj2)=(1)k1J(χ,χ2)k(λ(a21a2)+λ(a1a22))=(λ(a21a2)+λ(a1a22))=δ(a1,a2), (3.6)

    and

    2j1=12j2=1λ(cj1+j2aj11aj22)J(λj1,λj2)=(1)k1(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)k1(λ(c2a21a22)J(χ,χ)k+λ(ca1a2)J(χ2,χ2)k)=δ(a1,a2)+(1)k12k(λ(c2a21a22)(u+iv3)k+λ(ca1a2)(uiv3)k)={(1)k12k1E(u,v,k)+δ(a1,a2),                 if indαca1a20(mod 3),(1)k2k(E(u,v,k)O(u,v,k))+δ(a1,a2),  if indαca1a21(mod 3),(1)k2k(E(u,v,k)+O(u,v,k))+δ(a1,a2),  if indαca1a22(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+i32. We deduce that

    N2=q2(q1)2j1=12j2=1λj1λj2λj3 trivial2j3=1λ(bj11bj22bj33)J(λj1,λj2,λj3) (3.8)

    if c=0, and

    N2=q2+2j1=12j2=12j3=1λ(cj1+j2+j3bj11bj22bj33)J(λj1,λj2,λj3) (3.9)

    if c0.

    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

    2j1=12j2=1λj1λj2λj3trivial2j3=1λ(bj11bj22bj33)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+iv3)k+λ(b1b2b3)(uiv3)k)={(1)k2k1E(u,v,k),                if indαb1b2b30(mod 3),(1)k+12k(E(u,v,k)O(u,v,k)),   if indαb1b2b31(mod 3),(1)k+12k(E(u,v,k)+O(u,v,k)),   if indαb1b2b32(mod 3) (3.10)

    and

    2j1=12j2=12j3=1λ(cj1+j2+j3bj11bj22bj33)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)(uiv3)k+λ(b21b22b23)(u+iv3)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)(uiv3)k+λ(b21b22b23)(u+iv3)k)={(1)k2k1E(u,v,k),                    if indαb1b2b30(mod 3),(1)k+12k(E(u,v,k)O(u,v,k)),  if indαb1b2b31(mod 3),(1)k+12k(E(u,v,k)+O(u,v,k)),  if indαb1b2b32(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 indαcb21b22b30(mod 3),q,  if indαcb21b22b30(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)=ω1341131=2. We consider the numbers of solutions of the cubic equations

    x31+ω2x32=0

    and

    x31+ω2x32=ω

    over Fq.

    Since ω13413=(ω1341131)1313=24, the integers u and v in Lemma 2.4 are determined by

    u2+3v2=52, u1(mod 3), v0(mod 3) and 3vu(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)=ω3121311=3. We consider the numbers of solutions of the cubic equations

    ω4x31+x32+ωx33=0

    and

    ω4x31+x32+ωx33=ω

    over Fq.

    Since ω31213=(ω3121311)3113=310, the integers u and v are determined by

    u2+3v2=124, u1(mod 3), v0(mod 3) and 3vu(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)=ω73171=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 ω7313=(ω73171)713=32, the integers u and v are determined by

    u2+3v2=28, u1(mod 3), v0(mod 3) and 3vu(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,bjFq (1i2,1j3), cFq and p1(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 s4 variables with aiFq (1is), cFq and p1(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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