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Research article

Counting rational points of quartic diagonal hypersurfaces over finite fields

  • Received: 06 November 2023 Revised: 09 December 2023 Accepted: 12 December 2023 Published: 21 December 2023
  • MSC : 11T06, 11T24

  • Let Fq be the finite field of order q where q=pk, k is a positive integer and p is an odd prime. Let Fq represent the nonzero elements of Fq. For f(x1,,xn)Fq[x1,,xn], we used N(f(x1,,xn)=0) to denote the number of Fq-rational points of the affine hypersurface f(x1,,xn)=0. In 2020, Zhao et al. obtained the explicit formulae for N(x41+x42=c), N(x41+x42+x43=c) and N(x41+x42+x43+x44=c) over Fq, with cFq. In this paper, by using Jacobi sums and an analog of the Hasse-Davenport theorem, we arrived at explicit formulae for N(a1x41+a2x42=c) and N(b1x41+b2x42+b3x43=c) with ai,bjFq(1i2,1j3) and cFq. Furthermore, by using the reduction formula for Jacobi sums, the number of rational points of the quartic diagonal hypersurface a1x41+a2x42++anx4n=c of n4 variables with aiFq (1in), cFq and p1(mod 4), can also be deduced. These extended and improved earlier results.

    Citation: Shuangnian Hu, Yanyan Li, Rongquan Feng. Counting rational points of quartic diagonal hypersurfaces over finite fields[J]. AIMS Mathematics, 2024, 9(1): 2167-2180. doi: 10.3934/math.2024108

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  • Let Fq be the finite field of order q where q=pk, k is a positive integer and p is an odd prime. Let Fq represent the nonzero elements of Fq. For f(x1,,xn)Fq[x1,,xn], we used N(f(x1,,xn)=0) to denote the number of Fq-rational points of the affine hypersurface f(x1,,xn)=0. In 2020, Zhao et al. obtained the explicit formulae for N(x41+x42=c), N(x41+x42+x43=c) and N(x41+x42+x43+x44=c) over Fq, with cFq. In this paper, by using Jacobi sums and an analog of the Hasse-Davenport theorem, we arrived at explicit formulae for N(a1x41+a2x42=c) and N(b1x41+b2x42+b3x43=c) with ai,bjFq(1i2,1j3) and cFq. Furthermore, by using the reduction formula for Jacobi sums, the number of rational points of the quartic diagonal hypersurface a1x41+a2x42++anx4n=c of n4 variables with aiFq (1in), cFq and p1(mod 4), can also be deduced. These extended and improved earlier results.



    Let Fq be the finite field of order q where q=pk, k is a positive integer and p is an odd prime. Let Fq represent the nonzero elements of Fq. For f(x1,,xn)Fq[x1,,xn], we use Nq(f;n) to denote the number of Fq-rational points of the affine hypersurface f(x1,,xn)=0, namely,

    Nq(f;n)=N(f(x1,,xn)=0)=#{(x1,,xn)Fnq|f(x1,,xn)=0}.

    To evaluate the values of Nq(f;n) is a fundamental problem in algebra, number theory and arithmetic geometry. Generally speaking, it is difficult to give explicit formulae for Nq(f;n). An explicit formula for Nq(f;n) is known when deg(f)2 (see [13]). Finding the explicit formula for Nq(f;n) under certain conditions has attracted many researchers for many years (see, for instance, [2,3,4,5,6,7,8,9,10,11,14,15,16,17,18,19,20,21]).

    A special diagonal hypersurface over Fq is given by an equation of the type

    a1xe1+a2xe2++anxen=c,

    with e being a positive integer, coefficients a1,a2,,anFq and cFq. It is clear that

    N(a1x1+a2x2++anxn=c)=qn1.

    For e=2, there is an explicit formula for N(a1x21++anx2n=c) in [13]. When q=p2t with pr1(mod e) for a divisor r of t and e(q1), Wolfmann [17] gave an explicit formula of the number of rational points of the hypersurface

    a1xe1+a2xe2++anxen=c

    over Fq in 1992.

    For the special case where a1=a2==an=1, we denote by

    M(e)n(c)=N(xe1+xe2++xen=c).

    In 1977, Chowla et al. [5] got the generating function n=1M(3)n(0)xn over Fp with p1(mod 3). In 1979, Myerson [14] extended the result in [5] to the field Fq, and also showed that the generating function n=1M(4)n(0)xn over Fq with p1(mod 4) is a rational function in x.

    In 2020, Zhao et al. [19,20] investigated the number of rational points of the hypersurfaces

    x41+x42=c,  x41+x42+x43=c  and  x41+x42+x43+x44=c

    over Fq, with cFq. For any cFq, in 2022, by using the cyclotomic theory and exponential sums, Zhao et al. [21] showed that the generating function n=1M(4)n(c)xn is a rational function in x.

    In this paper, we consider the problem of finding the explicit formula for the number of rational points of the diagonal quartic hypersurface

    f(x1,x2,,xn)=a1x41+a2x42++anx4nc=0

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

    If p3(mod 4) and k is an odd integer, then gcd(4,q1)=2. It follows that (see [12])

    N(a1x41+a2x42++anx4n=c)=N(a1x21+a2x22++anx2n=c).

    Throughout this paper, we let η be the quadratic multiplicative character of Fq. Then from Theorems 6.26 and 6.27 in [13], the following result is deduced.

    Theorem 1.1. Let q=pk with p3(mod 4) and k be an odd integer. Let ψ(c)=1 for cFq and ψ(0)=q1. Then the number of rational points of the hypersurface

    f(x1,x2,,xn)=a1x41+a2x42++anx4nc

    over Fq is

    qn1+ψ(c)qn22η((1)n2a1a2an)

    if n is even, and is

    qn1+qn12η((1)n12ca1a2an)

    if n is odd.

    If p3(mod 4) and k is an even integer, the following result can be derived from [17, Theorem 1].

    Theorem 1.2. Let p3(mod 4) be a prime, k an even integer, q=pk, s=q14, n2 and cFq. Let α be a primitive element of Fq. Denote by N the number of rational points of the hypersurface

    a1x41+a2x42++anx4n=c

    over Fq, then

    N=qn1+14(1)kn/2qn/21(q1)3j=0(3)v(j)

    if c=0, and

    N=qn1(1)k(n+1)/2qn/21[(3)θ(c)q1/214(q1/2(1)k/2)3j=0(3)τ(j)]

    if c0, where v(j) is the number of i, 1in, such that

    (αj)sasi=(1)k(p+1)/8;

    θ(c) is the number of i, 1in, such that asi=(c)s and τ(j) is the number of i, 1in, such that asi=(αj)s.

    However, the explicit formula for N(a1x41+a2x42++anx4n=c) is still unknown when p1(mod 4). In this paper, we solve this problem by using the Jacobi sums and an analog of the Hasse-Davenport theorem. We give an explicit formula for the number of rational points of diagonal quartic hypersurface

    f1(x1,x2)=a1x41+a2x42c=0 (1.1)

    and

    f2(x1,x2,x3)=b1x41+b2x42+b3x43c=0 (1.2)

    over Fq, with a1,a2,b1,b2,b3Fq, cFq and the characteristic p1(mod 4). The case with arbitrary n4 variables can be deduced from the reduction formula for Jacobi sums, but we omit the tedious details here.

    For a generator α of Fq, we define the index of βFq with respect to α, denoted by indαβ, to be the unique integer r[1,q1] such that β=αr (see, for instance, [13]). For any nonzero integer n and prime number p, we define νp(n) as the greatest integer t such that pt divides n. Then νp(n) is a nonnegative integer, and νp(n) 1 if and only if p divides n.

    To give the main results, we need two concepts as follows.

    Definition 1.1. Let λ be a multiplicative character of Fq. Associated to λ, we define the function Sλ over Fq as follows

    Sλ(β):=λ(β)+λ(β3).

    Clearly, if λ is the multiplicative character of order 4 of Fq with λ(α)=i=1, then we have

    Sλ(β)={2,     if indαβ0(mod 4),0,     if indαβ1(mod 4) or indαβ3(mod 4),2,   if indαβ2(mod 4).

    Definition 1.2. Let r, s and k be positive integers, and we define

    E(r,s,k):=rkkm=1ν2(m)=1(km)rkmsm+km=1ν2(m)2(km)rkmsm

    and

    O(r,s,k):=km=1ν2(m+1)=1(km)rkmsmkm=1ν2(m+1)2(km)rkmsm.

    Moreover, let α be a primitive element of Fq and βFq. Then associated to r, s and k, we define

    W(r,s,k)(β):={2E(r,s,k),      if indαβ0(mod 4),2O(r,s,k),   if indαβ1(mod 4),2E(r,s,k),   if indαβ2(mod 4),2O(r,s,k),     if indαβ3(mod 4).

    Now we can state the main results of this paper as follows.

    Theorem 1.3. Let k be a positive integer and q=pk with p=4t+1. Let α be a primitive element of Fq, η be the quadratic multiplicative character of Fq and λ be a multiplicative character of order 4 of Fq such that λ(α)=i. Let N1 denote the number of rational points of the hypersurface over Fq defined by (1.1). Then

    N1=q+(q1)((1)tkSλ(a1a32)+η(a1a2))

    if c=0, and

    N1=q(1)tkSλ(a1a32)η(a1a2)+(1)k+1((1)ktW(u,v,k)(α1)+3i=2W(u,v,k)(αi))

    if c0, with α1=c2a1a2, α2=ca21a2, α3=ca1a22 and the integers u and v being defined uniquely by

    u2+v2=p, u(2p)(mod 4)  and  vuα(q1)/4(mod p),

    where (2p) is the Legendre symbol.

    Theorem 1.4. Let k be a positive integer and q=pk with p=4t+1. Let α be a primitive element of Fq, η be the quadratic multiplicative character of Fq and λ be a multiplicative character of order 4 of Fq such that λ(α)=i. Let N2 denote the number of rational points of the hypersurface over Fq defined by (1.2). Then

    N2=q2(q1)(1)(t+1)k3i=1W(u,v,k)(βi)

    if c=0, and

    N2=q2+(1)(t+1)k3i=1W(u,v,k)(βi)+(1)ktW(u2v2,2uv,k)(cb1b2b3)+q((1)kt6j=1Sλ(γi)+9j=7Sλ(γi)+η(cb1b2b3))

    if c0, with

    β1=b1b2b23,  β2=b1b22b3,  β3=b21b2b3,  γ1=cb31b32b3,
    γ2=c2b31b22b3,  γ3=cb31b2b33,  γ4=c2b31b2b23,  γ5=c3b31b2b3,  
    γ6=c2b21b32b3,  γ7=cb31b22b23,  γ8=c3b21b2b23,  γ9=cb21b22b33

    and the integers u and v being defined as in Theorem 1.3.

    This paper is organized as follows. In Section 2, we recall some useful known lemmas that will be needed later. Subsequently, in Section 3, we prove Theorems 1.3 and 1.4. Finally, in Section 4, we supply two examples to illustrate the validity of our results.

    In this section, we present some useful lemmas that are needed in the proof of Theorems 1.3 and 1.4.

    Let q=pk, where k is a positive integer and p is a prime. For any element βE=Fq and F=Fp, the norm of β relative to Fp is defined by (see, for example, [1,13])

    NE/F(β):=ββpβpk1=βq1p1.

    For the simplicity, we write N(β) for NE/F(β). It is clear that if α is a primitive element of Fq, then N(α) is a primitive element of Fp. Let χ be a multiplicative character of Fq. For any αFq, if χ(α)=1, then we call the character χ is trivial. Let χ be a multiplicative character of Fp. Then χ can be lifted to a multiplicative character λ of Fq by setting λ(β)=χ(N(β)). Any characters of Fp can be lifted to characters of Fq, but not all the characters of Fq can be obtained by lifting a character of Fp. The following lemma characterizes all the characters of Fq that can be obtained by lifting a character of Fp.

    Lemma 2.1. [1] 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, , λn be n multiplicative characters of Fq. The Jacobi sum J(λ1,,λn) is defined by

    J(λ1,,λn):=γ1++γn=1λ1(γ1)λn(γn),

    where the summation is taken over all n-tuples (γ1,,γn) of elements of Fq with γ1++γn=1. It is clear that if σ is a permutation of {1,,n}, then

    J(λσ(1),,λσ(n))=J(λ1,,λn).

    The readers are referred to [1,13] for the basic facts on the Jacobi sum.

    The following lemma is an analog of the Hasse-Davenport theorem which establishes an important relationship between the Jacobi sums in Fq and the Jacobi sums in Fp.

    Lemma 2.2. [13] Let χ1,,χn be multiplicative characters of Fp, not all of which are trivial. Suppose χ1,,χn are lifted to characters λ1,,λn, respectively, of the extension field Fq of Fp with degree [Fq:Fp]=k. Then

    J(λ1,,λn)=(1)(n1)(k1)J(χ1,,χn)k.

    Lemma 2.3. (Reduction formula for the Jacobi sums) [1] Let λ1,,λs1,λs be s nontrivial multiplicative characters of Fq. If s2, then

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

    Lemma 2.4. [1] Let p1(mod 4) be a prime, q a power of p, α be a generator of Fq, and let χ be a multiplicative character of order 4 of Fp with χ(N(α))=i. Then

    J(χ,χ2)=u+iv,

    where the integers u and v are uniquely determined by

    u2+v2=p, u(2p)(mod 4)  and  vuα(q1)/4(mod p).

    Lemma 2.5. [1] Let p=4t+1 be a prime number. Let g be a primitive element of Fp and χ be a multiplicative character of order 4 over Fp such that χ(g)=i. Let the integers u and v be defined as in Lemma 2.4. Then the values of the 16 Jacobi sums J(χm,χn) (m,n=0,1,2,3) of order 4 are given in Table 1.

    Table 1.  The values of the Jacobi sums J(χm,χn).
    mn 0 1 2 3
    0 p 0 0 0
    1 0 (1)t(u+vi) u+vi (1)t+1
    2 0 u+vi 1 uvi
    3 0 (1)t+1 uvi (1)t(uvi)

     | Show Table
    DownLoad: CSV

    The following lemma gives a formula for the number of rational points of a diagonal hypersurface in terms of the Jacobi sums.

    Lemma 2.6. [1] Let k1,,kn be positive integers. Let a1,,anFq and cFq. Set

    di=gcd(ki,q1),

    and let λi be a multiplicative character of Fq of order di, for i=1,,n. Then the number N of rational points of the equation

    a1xk11++anxknn=c

    over Fq is given by

    N=qn1(q1)d11j1=1dn1jn=1λj11λjnntrivialλj11(a11)λjnn(a1n)J(λj11,,λjnn)

    if c=0, and by

    N=qn1+d11j1=1dn1jn=1λj11(ca11)λjnn(ca1n)J(λj11,,λjnn)

    if c0.

    In this section, we give the proofs of Theorems 1.3 and 1.4. First, we begin with a lemma.

    Lemma 3.1. Let α be a primitive element of Fq and λ be a multiplicative character of order 4 of Fq=Fpk such that λ(α)=i. Then for any positive integers r, s and βFq, we have

    λ(β)(r+si)k+λ(β3)(rsi)k=W(r,s,k)(β),

    where the function W(r,s,k)(β) is defined as in Definition 1.2.

    Proof. If indαβ0(mod 4), then

    λ(β)=λ(β3)=1.

    Thus, one has

    λ(β)(r+si)k+λ(β3)(rsi)k=(r+si)k+(rsi)k=km=0(km)rkm[(si)m+(si)m]=2rk+km=1m is even(km)rkm[(si)m+(si)m]=2rk2km=1ν2(m)=1(km)rkmsm+2km=1ν2(m)2(km)rkmsm=2E(r,s,k).

    If indαβ1(mod 4), then

    λ(β)=iandλ(β3)=i.

    Thus, one has

    λ(β)(r+si)k+λ(β3)(rsi)k=(r+si)ki(rsi)ki=km=0(km)rkm[smim+1+sm(i)m+1]=2km=1m is odd(km)rkmsmim+1=2km=1ν2(m+1)=1(km)rkmsm+2km=1ν2(m+1)2(km)rkmsm=2O(r,s,k).

    If indαβ2(mod 4), then

    λ(β)=λ(β3)=1,

    and if indαβ3(mod 4), then

    λ(β)=iandλ(β3)=i.

    The results in these two cases can be proved similarly.

    We can now give the proof of Theorem 1.3.

    Proof of Theorem 1.3. Let α be a primitive element of Fq and λ be a multiplicative character of Fq of order 4 with λ(α)=i. Since q1(mod 4), then

    gcd(4,q1)=4.

    Using Lemma 2.6, by setting λ1=λ2=λ, one can deduce that the number N1 of rational points

    a1x41+a2x42=c

    in F2q is given by

    N1=q(q1)3j1=13j2=1λj1λj2trivialλ(aj11aj22)J(λj1,λj2) (3.1)

    if c=0, and by

    N1=q+3j1=13j2=1λ(cj1+j2aj11aj22)J(λj1,λj2) (3.2)

    if c0.

    Since p1(mod 4), it follows that λp1 is trivial. Thus, from Lemma 2.1, we know that the quartic multiplicative character λ can be lifted by a quartic multiplicative character χ of Fp.

    Using Lemmas 2.2, 2.5 and 3.1 and Definition 1.1, we have the following two cases, depending on c=0 or c0.

    If c=0, we derive that

    3j1=1λj1λj2trivial3j2=1λ(aj11aj22)J(λj1,λj2)=λ(a31a2)J(λ,λ3)+λ(a21a22)J(λ2,λ2)+λ(a1a32)J(λ3,λ)=(1)k1(λ(a31a2)J(χ,χ3)k+λ(a1a32)J(χ3,χ)k+λ(a21a22)J(χ2,χ2)k)=(1)tk1(λ(a31a2)+λ(a1a32))λ(a21a22)=(1)tk1Sλ(a31a2)η(a1a2). (3.3)

    Thus, from (3.1) and (3.3), the first part of Theorem 1.3 follows immediately.

    If c0, we obtain

    3j1=13j2=1λ(cj1+j2aj11aj22)J(λj1,λj2)=λ(a31a2)J(λ,λ3)+λ(a21a22)J(λ2,λ2)+λ(a1a32)J(λ3,λ) +λ(c2a31a32)J(λ,λ)+λ(c2a1a2)J(λ3,λ3)+λ(c3a31a22)J(λ,λ2) +λ(ca1a22)J(λ3,λ2)+λ(c3a21a32)J(λ,λ2)+λ(ca21a2)J(λ2,λ3)=(1)tk1Sλ(a31a2)η(a1a2)+(1)k1(λ(c2a31a32)J(χ,χ)k +λ(c2a1a2)J(χ3,χ3)k+λ(c3a31a22)J(χ,χ2)k+λ(ca1a22)J(χ3,χ2)k +λ(c3a21a32)J(χ,χ2)k+λ(ca21a2)J(χ2,χ3)k)=(1)tk1Sλ(a31a2)η(a1a2)+(1)k1((1)tk(λ(c2a31a32)(u+vi)k +λ(c2a1a2)(uvi)k)+λ(c3a31a22)(u+vi)k+λ(ca1a22)(uvi)k +λ(c3a21a32)(u+vi)k+λ(ca21a2)(uvi)k)=(1)tk1Sλ(a31a2)η(a1a2)+(1)k+1((1)tkW(u,v,k)(c2a1a2) +W(u,v,k)(ca1a22)+W(u,v,k)(ca21a2)). (3.4)

    Thus, from (3.2) and (3.4), the desired result follows immediately.

    Now, we can turn our attention to prove Theorem 1.4.

    Proof of Theorem 1.4. By the same argument as in the proof of Theorem 1.3, let α be a primitive element of Fq and λ be the multiplicative character of Fq of order 4 with λ(α)=i. One has

    N2=q2(q1)3j1=13j2=1λj1λj2λj3 trivial3j3=1λ(bj11bj22bj33)J(λj1,λj2,λj3) (3.5)

    if c=0, and

    N2=q2+3j1=13j2=13j3=1λ(cj1+j2+j3bj11bj22bj33)J(λj1,λj2,λj3) (3.6)

    if c0.

    Clearly, the quartic multiplicative character λ can be lifted by a quartic multiplicative character χ of Fp. Thus, from Lemmas 2.2, 2.3, 2.5 and 3.1 and Definition 1.2, one gets that

    3j1=13j2=1λj1λj2λj3trivial3j3=1λ(bj11bj22bj33)J(λj1,λj2,λj3)=λ(b31b32b23)J(χ2,χ2)kJ(χ,χ)k+λ(b31b22b33)J(χ3,χ)kJ(χ,χ2)k +λ(b21b32b33)J(χ3,χ)kJ(χ,χ2)k+λ(b21b2b3)J(χ,χ3)kJ(χ2,χ3)k +λ(b1b22b3)J(χ,χ3)kJ(χ2,χ3)k+λ(b1b2b23)J(χ2,χ2)kJ(χ3,χ3)k=(1)(t+1)k[λ(b31b32b23)(u+vi)k+λ(b1b2b23)(uvi)k+λ(b31b22b33)(u+vi)k +λ(b1b22b3)(uvi)k+λ(b21b32b33)(u+vi)k+λ(b21b2b3)(uvi)k]=(1)(t+1)k(W(u,v,k)(b1b2b23)+W(u,v,k)(b1b22b3)+W(u,v,k)(b21b2b3)). (3.7)

    Then, from (3.5) and (3.7), the first part of Theorem 1.4 follows immediately.

    We can now turn our attention to prove the second part of Theorem 1.4. Clearly,

    3j1=13j2=13j3=1λ(cj1+j2+j3bj11bj22bj33)J(λj1,λj2,λj3)=(1)(t+1)k(W(u,v,k)(b1b2b23)+W(u,v,k)(b1b22b3)+W(u,v,k)(b21b2b3)) +λ(c3b31b32b33)J(χ2,χ)kJ(χ,χ)k+λ(cb1b2b3)J(χ2,χ3)kJ(χ3,χ3)k +(1)kqJ(χ,χ3)k[λ(cb31b32b3)+λ(c3b1b2b33)+λ(c2b31b22b3)+λ(c2b1b22b33) +λ(cb31b2b33)+λ(c3b1b32b3)+λ(c2b31b2b23)+λ(c2b1b32b23) +λ(c3b31b2b3)+λ(cb1b32b33)+λ(c2b21b32b3)+λ(c2b21b2b33)] +(1)kqJ(χ2,χ2)k[λ(cb31b22b23)+λ(c3b1b22b23)+λ(c3b21b2b23)+λ(cb21b32b23) +λ(cb21b22b33)+λ(c3b21b22b3)]+(1)kqJ(χ2,χ2)kλ(c2b21b22b23). (3.8)

    By Lemmas 2.5 and 3.1 and Definition 1.2, we derive that

    λ(c3b31b32b33)J(χ2,χ)kJ(χ,χ)k+λ(cb1b2b3)J(χ2,χ3)kJ(χ3,χ3)k=(1)kt[λ(c3b31b32b33)(u2v2+2uvi)k+λ(cb1b2b3)(u2v22uvi)k]=(1)ktW(u2v2,2uv,k)(cb1b2b3). (3.9)

    Using Lemma 2.5, one has

    (1)kqJ(χ,χ3)k=(1)tkq and (1)kqJ(χ2,χ2)k=q. (3.10)

    Thus, from (3.6), (3.8)–(3.10) and Definition 1.1, the desired result of the second part of Theorem 1.4 follows immediately.

    In this section, we present two examples to demonstrate the validity of Theorems 1.3 and 1.4. We have validated these two examples by using the Magma which is a powerful algebraic computation program package.

    Example 4.1. Let p=5 and k=5. It can be checked easily that 2 is a primitive element of F5. Choose a primitive element ω of F55 with N(ω)=2. We consider the numbers of rational points (x1,x2)F255 of the quartic hypersurfaces

    x41+ω2x42=0  and  x41+ω2x42=ω

    over F55.

    Now, the Legendre symbol

    (25)=1andω5514=N(ω)=2.

    Thus, the integers u and v are determined by

    u2+v2=5, u1(mod 4)  and  v2u(mod 5).

    Then, one has u=1, v=2. Thus, by Theorem 1.3, we obtain

    N(x41+ω2x42=0)=12497 and N(x41+ω2x42=ω)=3040.

    Example 4.2. Let p=13 and k=2. We know that 2 is a primitive element of F13. Choose a primitive element ω of F132 with N(ω)=2. We consider the numbers of rational points (x1,x2,x3)F3132 of the quartic hypersurfaces

    x41+ωx42+ω2x43=0  and  x41+ωx42+ω2x43=ω

    over F132.

    Now, the Legendre symbol

    (213)=1andω13214=N(ω)3=23.

    Thus, the integers u and v are determined by

    u2+v2=13, u1(mod 4)  and  v23u(mod 13).

    Therefore,

    u=3, v=2.

    By Theorem 1.4, we have

    N(x41+ωx42+ω2x43=0)=26881 and N(x41+ωx42+ω2x43=ω)=28164.

    Studying the number of rational points of the polynomial equation

    f(x1,x2,,xn)=0

    over Fq is a fundamental problem in algebra, number theory and arithmetic geometry. 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 an analog of the Hasse-Davenport theorem, we arrived at explicit formulae for

    N(a1x41+a2x42=c)

    and

    N(b1x41+b2x42+b3x43=c),

    with

    ai,bjFq(1i2,1j3)

    and cFq. Furthermore, by using the reduction formula for Jacobi sums, the number of rational points of the quartic diagonal hypersurface

    a1x41+a2x42++anx4n=c

    of n4 variables with

    aiFq(1in), cFq

    and p1(mod 4), can also be deduced.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This research was supported by the National Science Foundation of China (Nos. 12026223, 12026224 and 12231015), the National Key Research and Development Program of China (No. 2018YFA0704703), the Natural Science Foundation of Beijing (No. M23017), the Natural Science Foundation of Henan Province (No. 232300420123) and the Research Center of Mathematics and Applied Mathematics, Nanyang Institute of Technology.

    We declare that we have no conflicts of interest.



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