Research article

On the number of solutions of two-variable diagonal sextic equations over finite fields

  • Received: 30 November 2021 Revised: 23 February 2022 Accepted: 02 March 2022 Published: 28 March 2022
  • MSC : 11T06, 11T24

  • Let p be a prime, k a positive integer, q=pk, and Fq be the finite field with q elements. In this paper, by using the 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 cFq and p1(mod 6). Furthermore, by using the reduction formula for Jacobi sums, the number of solutions of the diagonal sextic equations x61+x62++x6n=c of n3 variables with cFq and p1(mod 6), can also be deduced.

    Citation: Shuangnian Hu, Rongquan Feng. On the number of solutions of two-variable diagonal sextic equations over finite fields[J]. AIMS Mathematics, 2022, 7(6): 10554-10563. doi: 10.3934/math.2022588

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  • Let p be a prime, k a positive integer, q=pk, and Fq be the finite field with q elements. In this paper, by using the 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 cFq and p1(mod 6). Furthermore, by using the reduction formula for Jacobi sums, the number of solutions of the diagonal sextic equations x61+x62++x6n=c of n3 variables with cFq and p1(mod 6), can also be deduced.



    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

    N(f,q)=N(f(x1,,xn)=0)=#{(x1,,xn)Fnqf(x1,,xn)=0}

    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

    N(f,q)dqn1.

    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

    ordpN(f,q)kndd.

    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

    x31+x32++x3n=0

    over Fp. In 1979, Myerson [15] extended the result in [9] to the field Fq and studied the number of solutions of the equation

    x41+x42++x4n=0

    over Fq. For q=p2t with pr1(mod d) for a divisor r of t and d(q1), Wolfmann [18] gave an explicit formula of the number of solutions of the equation

    a1xd1+a2xd2++anxdn=c

    over Fq in 1992, where a1,a2,,anFq and cFq. In 2018, Zhang and Hu [19] determined an explicit formula of the number of solutions of the equation

    x31+x32+x33+x34=c

    over Fp, with p1(mod 3) and cFp. In 2020, J. Zhao et al. [22,23] investigated the number of solutions of the forms

    x41+x42=c,
    x41+x42+x43=c,

    and

    x41+x42+x43+x44=c

    over Fq, with cFq.

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

    f(x1,x2,,xn)=x61+x62++x6nc=0

    over Fq, where q=pk and cFq.

    It is well-known that (see [13], p. 105)

    N(x61+x62++x6n=c)=N(xgcd(6,q1)1+xgcd(6,q1)2++xgcd(6,q1)n=c)

    over Fq. Let q=pk with p a prime. If p=2, then gcd(6,q1)=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 cFq. Then

    N(x61+x62++x6n=c)=qn1

    if k is odd, and

    N(x61+x62++x6n=c)={qn1+ξn+1qn21[2nq12(q12+ξ)2n+2(1)n3],if cq13=1,qn1+ξn+1qn21[(1)nq12(q12+ξ)2n+2(1)n3],otherwise

    if k is even and n2, with ξ=(1)k2+1.

    If p=3 and k is an integer, or p5(mod 6) and k is an odd integer, then gcd(6,q1)=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 p5(mod 6) be a prime and k an odd integer, q=pk and cFq. Then

    N(x61+x62++x6n=c)=qn1qn22η((1)n2)

    if n is even, and

    N(x61+x62++x6n=c)=qn1+qn12η((1)n12c)

    if n is odd, where η is the quadratic multiplicative character of Fq.

    If p5(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 p5(mod 6) be a prime, k an even integer, q=pk, n2 and cFq. Then

    N(x61+x62++x6n=c)={qn1+ξn+1qn21[5nq12(q12+ξ)5n+5(1)n6],if cq16=1,qn1+ξn+1qn21[(1)nq12(q12+ξ)5n+5(1)n6],otherwise,

    where ξ=(1)k2+1.

    However, the formula for N(x61+x62++x6n=c) is still unknown when p1(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 n3 variables can be deduced from the reduction formula for Jacobi sums.

    Let gFq be a fixed primitive element of Fq. For any βFq, there exists exactly one integer r[1,q1] 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 αE=Fpk and F=Fp, the norm of α relative to Fp are defined by (see, for example, [6,14])

    NE/F(α):=ααpαpk1=αq1p1.

    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 cFq, g be a primitive element of Fq and Z=indN(g)2. Then

    N(x61+x62=c)={q5+(1)k1(3u+r2k1+12a),if indgc0(mod 6),q5+(1)k(ru3s3v2k+4a12b),if indgc1(mod 6),q5+(12)k(3u+3s+r9v),if indgc2(mod 6),q5+(1)k(ur2k1+4a),if indgc3(mod 6),q5+(12)k(9v+3u+r3s),if indgc4(mod 6),q5+(1)k(3v+r+3su2k+4a+12b),if indgc5(mod 6)

    when t or k is even, and

    N(x61+x62=c)={q+1+r+u2k1+4a,if indgc0(mod 6),q+1+3s+3u+9vr2k,if indgc1(mod 6),q+1+3vr3su2k+4a+12b,if indgc2(mod 6),q+1+r3u2k112a,if indgc3(mod 6),q+1+3su3vr2k12b+4a,if indgc4(mod 6),q+1+3u9vr3s2k,if indgc5(mod 6)

    when both t and k are odd, where a+ib3=(a+ib3)k, u+iv3=(u+iv3)k, r+is3=(r+is3)k with a and b being integers such that

    a2+3b2=p,a1(mod 3),and3ba(2g(q1)/3+1)(mod p),

    and the integers r, s, u and v are given by

    {u=r=2a,v=s=2b,if Z0(mod 3),u=3ba,r=a3b,v=ab,s=ab,if Z1(mod 3),u=a3b,r=3ba,v=ab,s=ab,if Z2(mod 3).

    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.

    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

    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).

    Also for any αFq, we have that

    γ1++γs=αλ1(γ1)λs(γs)=(λ1λs)(α)J(λ1,,λs).

    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

    J(λ1,,λs)=(1)(s1)(k1)J(χ1,,χs)k.

    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 p1(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 λp1 is trivial.

    Let g be a generator of Fq=Fpk. Since

    N(g)p1=(gq1p1)p1=gq1=1

    and

    Nl(g)=(gq1p1)l1 for 1lp1,

    one knows that N(g) is a generator of Fp. Then we can state the following lemma.

    Lemma 2.3. [6] Let p1(mod 6) be a prime, q a power of p, g be a generator of Fq, and let χ be a multiplicative character of order 6 over Fp. Then

    J(χ,χ2)=a+ib3,

    where the integers a and b are uniquely determined by

    a2+3b2=p,a1(mod 3),and3ba(2g(q1)/3+1)(mod p).

    Lemma 2.4. [6] Let p=6t+1 be a prime number. Let g be the generator of Fp and χ be a multiplicative character of order 6 over Fp such that χ(g)=1+i32. 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.

    Table 1.  The values of the Jacobi sums J(χm,χn).
    mn 0 1 2 3 4 5
    0 p 0 0 0 0 0
    1 0 (1)t12(u+iv3) a+ib3 (1)t(a+ib3) 12(u+iv3) (1)t
    2 0 a+ib3 12(r+is3) a+ib3 1 12(uiv3)
    3 0 (1)t(a+ib3) a+ib3 (1)t aib3 (1)t(aib3)
    4 0 12(u+iv3) 1 aib3 12(ris3) aib3
    5 0 (1)t 12(uiv3) (1)t(aib3) aib3 (1)t12(uiv3)

     | Show Table
    DownLoad: CSV

    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,cFq. Set di=gcd(ki,q1), 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

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

    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+i32. Since q1(mod 6), then gcd(6,q1)=6. Using Lemma 2.5, we have

    N(x61+x62=c)=q+5j1=15j2=1λ(cj1+j2)J(λj1,λj2)=q+1i5λ2i(c)J(λi,λi)+21i<j5λi+j(c)J(λi,λj).

    Since p1(mod 6), it follows that λp1 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

    N(x61+x62=c)=q+(1)k1(1i5λ2i(c)J(χi,χi)k+21i<j5λi+j(c)J(χi,χj)k).    (3.1)

    Consider the case when t or k is even. Noting that λ(1)=1, from (3.1), one has

    N(x61+x62=1)=q+(1)k1(1i5J(χi,χi)k+21i<j5J(χi,χj)k). (3.2)

    From Table 1 of Lemma 2.4, we derive that

    1i5J(χi,χi)k=(u+iv32)k+(r+is32)k+(1)k+(ris32)k+(uiv32)k=u+iv32k+r+is32k+(1)k+ris32k+uiv32k=12k1(u+r)+(1)k, (3.3)

    and

    21i<j5J(χi,χj)k=6(a+ib3)k+(u+iv3)k2k1+4(1)k+(uiv3)k2k1+6(aib3)k=6(a+ib3)+u+iv32k1+4(1)k+uiv32k1+6(aib3)=4(3a+u2k+(1)k). (3.4)

    Thus if indgc0(mod 6), then it follows from (3.2)–(3.4) that

    N(x61+x62=c)=q5+(1)k1(3u+r2k1+12a).

    From (3.1), we have

    N(x61+x62=g)=q+(1)k1(1i5λ2i(g)J(χi,χi)k+21i<j5λi+j(g)J(χi,χj)k). (3.5)

    Noting that λ(g)=1+i32 and λ6(g)=1, then from Table 1 of Lemma 2.4, we deduce that

    1i5λ2i(g)J(χi,χi)=1+i32((u+iv32)k+(ris32)k)1+i32((r+is32)k+(uiv32)k)+(1)k=1+i32k+1(u+iv3+ris3)1+i32k+1(r+is3+uiv3)+(1)k=3s3vru2k+(1)k, (3.6)

    and

    21i<j5χi+j(g)J(χi,χj)=4a+12b+u+3v2k1+4(1)k. (3.7)

    Thus if indgc1(mod 6), from (3.5)–(3.7), one can deduce that

    N(x61+x62=c)=q5+(1)k(ru3s3v2k+4a12b).

    In the similar way, one can deduce that

    N(x61+x62=c)={q5+(12)k(3u+3s+r9v),if indgc2(mod 6),q5+(1)k(ur2k1+4a),if indgc3(mod 6),q5+(12)k(9v+3u+r3s),if indgc4(mod 6),q5+(1)k(3vu+r+3s2k+4a+12b),if indgc5(mod 6).

    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 k2, then

    J(χ1,,χk)={qJ(χ1,,χk1),if χ1χk1 is trivial,J(χ1χk1,χk)J(χ1,,χk1),if χ1χk1 is nontrivial.

    One can use the reduction formula for Jacobi sums to give an explicit formula for the number of solutions of the diagonal sextic equation

    x61+x62++x6n=c

    of n3 variables over Fpk, with cFpk and p1(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

    x61+x62=c

    over Fq, where cFq.

    Since 2 is a primitive root modulo 19, we have Z=ind22=1. Let g be a generator of F193 such that

    N(g)=g1931191=2.

    That means

    g19313=(g1931191)1913=26.

    The integers a and b are determined by

    a2+3b2=19,a1(mod 3)and3b15a(mod 19).

    We can get that a=4, b=1. Since Z1(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

    N(x61+x62=c)={6636,if indgc0(mod 6),6264,if indgc1(mod 6),6264,if indgc2(mod 6),7308,if indgc3(mod 6),7344,if indgc4(mod 6),7344,if indgc5(mod 6).

    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 cFq and p1(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 n3 variables with cFq and p1(mod 6), 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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