m∖n | 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 | u−vi |
3 | 0 | (−1)t+1 | u−vi | (−1)t(u−vi) |
Let Fq be the finite field of order q where q=pk, k is a positive integer and p is an odd prime. Let F∗q 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 c∈F∗q. 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,bj∈F∗q(1≤i≤2,1≤j≤3) and c∈Fq. Furthermore, by using the reduction formula for Jacobi sums, the number of rational points of the quartic diagonal hypersurface a1x41+a2x42+⋯+anx4n=c of n≥4 variables with ai∈F∗q (1≤i≤n), c∈Fq and p≡1(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 F∗q 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 c∈F∗q. 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,bj∈F∗q(1≤i≤2,1≤j≤3) and c∈Fq. Furthermore, by using the reduction formula for Jacobi sums, the number of rational points of the quartic diagonal hypersurface a1x41+a2x42+⋯+anx4n=c of n≥4 variables with ai∈F∗q (1≤i≤n), c∈Fq and p≡1(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 F∗q 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,⋯,an∈F∗q and c∈Fq. It is clear that
N(a1x1+a2x2+⋯+anxn=c)=qn−1. |
For e=2, there is an explicit formula for N(a1x21+⋯+anx2n=c) in [13]. When q=p2t with pr≡−1(mod e) for a divisor r of t and e∣(q−1), 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 p≡1(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 p≡1(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 c∈F∗q. For any c∈Fq, 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+⋯+anx4n−c=0 |
over Fq, where q=pk, a1,a2,⋯,an∈F∗q and c∈Fq.
If p≡3(mod 4) and k is an odd integer, then gcd(4,q−1)=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 p≡3(mod 4) and k be an odd integer. Let ψ(c)=−1 for c∈F∗q and ψ(0)=q−1. Then the number of rational points of the hypersurface
f(x1,x2,⋯,xn)=a1x41+a2x42+⋯+anx4n−c |
over Fq is
qn−1+ψ(c)qn−22η((−1)n2a1a2⋯an) |
if n is even, and is
qn−1+qn−12η((−1)n−12ca1a2⋯an) |
if n is odd.
If p≡3(mod 4) and k is an even integer, the following result can be derived from [17, Theorem 1].
Theorem 1.2. Let p≡3(mod 4) be a prime, k an even integer, q=pk, s=q−14, n≥2 and c∈Fq. 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=qn−1+14(−1)kn/2qn/2−1(q−1)3∑j=0(−3)v(j) |
if c=0, and
N=qn−1−(−1)k(n+1)/2qn/2−1[(−3)θ(c)q1/2−14(q1/2−(−1)k/2)3∑j=0(−3)τ(j)] |
if c≠0, where v(j) is the number of i, 1≤i≤n, such that
(αj)sasi=(−1)k(p+1)/8; |
θ(c) is the number of i, 1≤i≤n, such that asi=(−c)s and τ(j) is the number of i, 1≤i≤n, such that asi=(αj)s.
However, the explicit formula for N(a1x41+a2x42+⋯+anx4n=c) is still unknown when p≡1(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+a2x42−c=0 | (1.1) |
and
f2(x1,x2,x3)=b1x41+b2x42+b3x43−c=0 | (1.2) |
over Fq, with a1,a2,b1,b2,b3∈F∗q, c∈Fq and the characteristic p≡1(mod 4). The case with arbitrary n≥4 variables can be deduced from the reduction formula for Jacobi sums, but we omit the tedious details here.
For a generator α of F∗q, we define the index of β∈F∗q with respect to α, denoted by indαβ, to be the unique integer r∈[1,q−1] 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 F∗q 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 indαβ≡0(mod 4),0, if indαβ≡1(mod 4) or indαβ≡3(mod 4),−2, if indαβ≡2(mod 4). |
Definition 1.2. Let r, s and k be positive integers, and we define
E(r,s,k):=rk−k∑m=1ν2(m)=1(km)rk−msm+k∑m=1ν2(m)≥2(km)rk−msm |
and
O(r,s,k):=k∑m=1ν2(m+1)=1(km)rk−msm−k∑m=1ν2(m+1)≥2(km)rk−msm. |
Moreover, let α be a primitive element of Fq and β∈F∗q. Then associated to r, s and k, we define
W(r,s,k)(β):={2E(r,s,k), if indαβ≡0(mod 4),−2O(r,s,k), if indαβ≡1(mod 4),−2E(r,s,k), if indαβ≡2(mod 4),2O(r,s,k), if indαβ≡3(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+(q−1)((−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)+3∑i=2W(u,−v,k)(αi)) |
if c≠0, 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 v≡uα(q−1)/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 F∗q, 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−(q−1)(−1)(t+1)k3∑i=1W(u,−v,k)(βi) |
if c=0, and
N2=q2+(−1)(t+1)k3∑i=1W(u,−v,k)(βi)+(−1)ktW(u2−v2,−2uv,k)(cb1b2b3)+q((−1)kt6∑j=1Sλ(γi)+9∑j=7Sλ(γi)+η(cb1b2b3)) |
if c≠0, 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 β∈E=Fq and F=Fp, the norm of β relative to Fp is defined by (see, for example, [1,13])
NE/F(β):=ββp⋯βpk−1=βq−1p−1. |
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 α∈Fq, 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 λp−1 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)(n−1)(k−1)J(χ1,⋯,χn)k. |
Lemma 2.3. (Reduction formula for the Jacobi sums) [1] 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. |
Lemma 2.4. [1] Let p≡1(mod 4) be a prime, q a power of p, α be a generator of F∗q, 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 v≡uα(q−1)/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.
m∖n | 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 | u−vi |
3 | 0 | (−1)t+1 | u−vi | (−1)t(u−vi) |
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,⋯,an∈F∗q and c∈Fq. Set
di=gcd(ki,q−1), |
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=qn−1−(q−1)d1−1∑j1=1⋯dn−1∑jn=1λj11⋯λjnntrivialλj11(a−11)⋯λjnn(a−1n)J(λj11,…,λjnn) |
if c=0, and by
N=qn−1+d1−1∑j1=1⋯dn−1∑jn=1λj11(ca−11)⋯λjnn(ca−1n)J(λj11,⋯,λjnn) |
if c≠0.
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 β∈F∗q, we have
λ(β)(r+si)k+λ(β3)(r−si)k=W(r,s,k)(β), |
where the function W(r,s,k)(β) is defined as in Definition 1.2.
Proof. If indαβ≡0(mod 4), then
λ(β)=λ(β3)=1. |
Thus, one has
λ(β)(r+si)k+λ(β3)(r−si)k=(r+si)k+(r−si)k=k∑m=0(km)rk−m[(si)m+(−si)m]=2rk+k∑m=1m is even(km)rk−m[(si)m+(−si)m]=2rk−2k∑m=1ν2(m)=1(km)rk−msm+2k∑m=1ν2(m)≥2(km)rk−msm=2E(r,s,k). |
If indαβ≡1(mod 4), then
λ(β)=iandλ(β3)=−i. |
Thus, one has
λ(β)(r+si)k+λ(β3)(r−si)k=(r+si)ki−(r−si)ki=k∑m=0(km)rk−m[smim+1+sm(−i)m+1]=2k∑m=1m is odd(km)rk−msmim+1=−2k∑m=1ν2(m+1)=1(km)rk−msm+2k∑m=1ν2(m+1)≥2(km)rk−msm=−2O(r,s,k). |
If indαβ≡2(mod 4), then
λ(β)=λ(β3)=−1, |
and if indαβ≡3(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 q≡1(mod 4), then
gcd(4,q−1)=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−(q−1)3∑j1=13∑j2=1λj1λj2trivialλ(a−j11a−j22)J(λj1,λj2) | (3.1) |
if c=0, and by
N1=q+3∑j1=13∑j2=1λ(cj1+j2a−j11a−j22)J(λj1,λj2) | (3.2) |
if c≠0.
Since p≡1(mod 4), it follows that λp−1 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 c≠0.
If c=0, we derive that
3∑j1=1λj1λj2trivial3∑j2=1λ(a−j11a−j22)J(λj1,λj2)=λ(a31a2)J(λ,λ3)+λ(a21a22)J(λ2,λ2)+λ(a1a32)J(λ3,λ)=(−1)k−1(λ(a31a2)J(χ,χ3)k+λ(a1a32)J(χ3,χ)k+λ(a21a22)J(χ2,χ2)k)=(−1)tk−1(λ(a31a2)+λ(a1a32))−λ(a21a22)=(−1)tk−1Sλ(a31a2)−η(a1a2). | (3.3) |
Thus, from (3.1) and (3.3), the first part of Theorem 1.3 follows immediately.
If c≠0, we obtain
3∑j1=13∑j2=1λ(cj1+j2a−j11a−j22)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)tk−1Sλ(a31a2)−η(a1a2)+(−1)k−1(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)tk−1Sλ(a31a2)−η(a1a2)+(−1)k−1((−1)tk(λ(c2a31a32)(u+vi)k +λ(c2a1a2)(u−vi)k)+λ(c3a31a22)(u+vi)k+λ(ca1a22)(u−vi)k +λ(c3a21a32)(u+vi)k+λ(ca21a2)(u−vi)k)=(−1)tk−1Sλ(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−(q−1)3∑j1=13∑j2=1λj1λj2λj3 trivial3∑j3=1λ(b−j11b−j22b−j33)J(λj1,λj2,λj3) | (3.5) |
if c=0, and
N2=q2+3∑j1=13∑j2=13∑j3=1λ(cj1+j2+j3b−j11b−j22b−j33)J(λj1,λj2,λj3) | (3.6) |
if c≠0.
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
3∑j1=13∑j2=1λj1λj2λj3trivial3∑j3=1λ(b−j11b−j22b−j33)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)(u−vi)k+λ(b31b22b33)(u+vi)k +λ(b1b22b3)(u−vi)k+λ(b21b32b33)(u+vi)k+λ(b21b2b3)(u−vi)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,
3∑j1=13∑j2=13∑j3=1λ(cj1+j2+j3b−j11b−j22b−j33)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)(u2−v2+2uvi)k+λ(cb1b2b3)(u2−v2−2uvi)k]=(−1)ktW(u2−v2,−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)=−1andω55−14=N(ω)=2. |
Thus, the integers u and v are determined by
u2+v2=5, u≡1(mod 4) and v≡2u(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)=−1andω132−14=N(ω)3=23. |
Thus, the integers u and v are determined by
u2+v2=13, u≡1(mod 4) and v≡23u(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,bj∈F∗q(1≤i≤2,1≤j≤3) |
and c∈Fq. Furthermore, by using the reduction formula for Jacobi sums, the number of rational points of the quartic diagonal hypersurface
a1x41+a2x42+⋯+anx4n=c |
of n≥4 variables with
ai∈F∗q(1≤i≤n), c∈Fq |
and p≡1(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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m∖n | 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 | u−vi |
3 | 0 | (−1)t+1 | u−vi | (−1)t(u−vi) |