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Research article

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

  • Received: 16 November 2019 Accepted: 16 March 2020 Published: 20 March 2020
  • MSC : 11T23, 11T24

  • Let p be a odd prime number and let Fq be the finite field of characteristic p with q elements. In this paper, by using the Gauss sum and Jacobi sum, we give an explicit formula for the number N(x41+x42=c) of solutions of the following two-variable diagonal quartic equations over Fq: x41+x42=c with cFq. From this result, one can deduce that N(x41+x42=c)=q+O(q12).

    Citation: Junyong Zhao, Yang Zhao, Yujun Niu. On the number of solutions of two-variable diagonal quartic equations over finite fields[J]. AIMS Mathematics, 2020, 5(4): 2979-2991. doi: 10.3934/math.2020192

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  • Let p be a odd prime number and let Fq be the finite field of characteristic p with q elements. In this paper, by using the Gauss sum and Jacobi sum, we give an explicit formula for the number N(x41+x42=c) of solutions of the following two-variable diagonal quartic equations over Fq: x41+x42=c with cFq. From this result, one can deduce that N(x41+x42=c)=q+O(q12).


    Let p be an odd prime number with q=ps, sZ+. Let Fq be the finite field of q elements. For any polynomial f(x1,,xn) over Fq with n variables, we let N(f=0) stand for the number of Fq-rational points on the affine hypersurface f(x1,,xn)=0 over Fnq. That is, we have

    N(f=0)={(x1,,xn)Fnq|f(x1,,xn)=0}.

    Calculating the value of N(f=0) is a main topic in finite fields. Weil [15] proposed his famous conjecture on the number of rational points of the nonsingular projective hypersurface over Fnq. However, it is difficult to give an exact formula for N(f=0). Studying the explicit formula for N(f=0) under certain conditions has attracted a lot of authors for many years. Some works were done by Ax [3], Adolphson and Sperber [1,2], Carlitz [5], Hong [7,8,9], Hu, Hong and Zhao [11], Zhao, Hong and Zhu [17]. It is noticed that the p-adic method is used by Hong et al. in [10] to establish the universal Kummer congruences.

    On the other hand, in 1977, Chowla, Cowles and Cowles [6] determined the number of solutions of the equation

    x31+x32++x3n=0

    in Fp. In 1981, Myerson [13] extend the result in [6] to the field Fq and first studied the number of solution of the equation

    x41+x42++x4n=0

    over Fq. In 2018, Zhang and Hu [16] determined an explicit formula of equation

    x31+x32+x33+x34=c,cFp

    with p1(mod3) as follows: Let N(c) be the number of solutions of x31+x32+x33+x34=c,cFp with p1(mod3), and Fp=g. Then they proved the following formula:

    N(c)={p36p12p(5d27b), if cg3m+1(modp),p36p12p(5d±27b), if cg3m+2(modp),p36p+5dp, if cg3m(modp).

    In this paper, we investigate the question of counting the number of solutions of the following equation:

    x41+x42=c

    with cFq. Actually, we obtain the following result.

    Theorem 1.1. Let F=Fq be the finite field with q=ps where p is an odd prime and sZ+. Let cFq and g be a primitive element of Fq.

    (i). If p1(mod8) or p5(mod8) and s is even, then

    N(x41+x42=c)={q+6a(1)s13,if  indg(c)0(mod4),q+2(1)s(a2b)3,if  indg(c)1(mod4),q+2a(1)s3,if  indg(c)2(mod4),q+2(1)s(a+2b)3,if  indg(c)3(mod4).

    If p5(mod8) and s is odd, then

    N(x41+x42=c)={q2a(1)s1+1,if  indg(c)0(mod4),q+2(1)s(a+2b)+1,if  indg(c)1(mod4),q+6a(1)s1+1,if  indg(c)2(mod4),q+2(1)s(a2b)+1,if  indg(c)3(mod4),

    where a+bi=(a+bi)s with a and b being integers such that

    a2+b2=p, a1(mod4), bagq14(modp).

    (ii). If p3(mod4) and q1(mod4), then

    N(x41+x42=c)=(q12φ(1)+6rη(c)(1)s1)+2q9r(1)s1((±¯φ(c))+(φ(c)))i,

    where r is uniquely determined by

    q=r2+4t2,r1(mod4),and,if p1(mod4), then (r,p)=1.

    (iii). If p3(mod4) and q3(mod4), then

    N(x41+x42=c)=q+1.

    We notice that Theorem 1.1 (ⅰ) for the special case q=p has been mentioned in the book of Jacobsthal's book. Furthermore, Theorem 1.1 (ⅱ) is a special case of Wolfmann [14], but we here get it by a different method. From Theorem 1.1, we can easily deduce the following statement.

    Corollary 1.1. Let F=Fq be the finite field with q=ps, where p1(mod4) is an odd prime and sZ+. Let cFq. Then each of the following is true.

    (i). If p1(mod4), then |N(x41+x42=c)q|7q for each q9.

    (ii). If p3(mod4) and q1(mod4), then |N(x41+x42=c)q|(7+42)q for each q81.

    This paper is organized as follows. First of all, in Section 2, we present several basic concepts including the Gauss sums, and give some preliminary lemmas. Then in Section 3, we give the proof of our main result Theorem 1.1 and Corollary 1.2. Finally, in Section 4, we supply two examples.

    In this section, we present several definitions and auxiliary lemmas that are needed in the proof of Theorem 1.1. We begin with three definitions.

    Definition 2.1. Let p be a prime number and q=ps with s being a positive integer. Let α be an element of Fq. Then the trace and norm of α relative to Fp are defined by

    TrFq/Fp(α):=α+αp++αps1

    and

    NFq/Fp(α):=ααpαps1=αq1p1,

    respectively. For the simplicity, we write Tr(α) and N(α) for TrFq/Fp(α) and NFq/Fp(α), respectively.

    Definition 2.2. Let χ be a multiplicative character of Fq and ψ an additive character of Fq. Then we define the Gauss sum G(χ,ψ) by

    G(χ,ψ):=xFqχ(x)ψ(x).

    Definition 2.3. Let χ1 and χ2 be multiplicative characters of Fq. Then the sum

    J(χ1,χ2):=xFqχ1(x)χ2(1x)

    is called a Jacobi sum in Fq.

    The character ψ0 represents the trivial additive character such that ψ0(x)=1 for all xFq and χ0 represents the trivial multiplicative character such that χ0(x)=1 for all xFq. For any xFq, let

    ψ1(x):=exp(2πiTr(x)p).

    Then we call ψ1 the canonical additive character of Fq. Let aFq. Then we define

    ψa(x):=exp(2πiTr(ax)p)

    for all xFq. For each character ψ of Fq there is associated the conjugate character ¯ψ defined by ¯ψ(x)=¯ψ(x) for all xFq. Let η be the quadratic character of Fq.

    We give several basic identities about Gauss sums as follows.

    Lemma 2.1. [12] Each of the following is true:

    (i). G(χ,ψab)=¯χ(a)G(χ,ψb) for aFq, bFq.

    (ii). G(¯χ,ψ)=χ(1)¯G(χ,ψ).

    (iii). |G(χ,ψ)|=q1/2 for χχ0 and ψψ0.

    Lemma 2.2. [12] Let Fq be a finite field with q=ps, where p is an odd prime and sN+. Then

    G(η,ψ1)={(1)s1q12,if  p1(mod4),(1)s1isq12,if  p3(mod4).

    If χ1 and χ2 are nontrivial, there exists an important connection between Jacobi sums and Gauss sums that will allow us to determine the value of Jacobi sums.

    Lemma 2.3. [12] If χ1 and χ2 are multiplicative characters of Fq and ψ is a nontrivial additive character of Fq, then

    J(χ1,χ2)=G(χ1,ψ)G(χ2,ψ)G(χ1χ2,ψ)

    if χ1χ2 is nontrivial.

    For a multiplicative character χ of Fq, we obviously have χ(1)=±1. The value χ(1) is of interest. The following result is regarding the sign of χ(1).

    Lemma 2.4. [12] Let χ be a multiplicative character of Fq of order n. Then χ(1)=1 if and only if n is even and q1n is odd.

    Clearly, we have the following consequence.

    Corollary 2.1. Let p1(mod4) be an odd prime and q=ps with s being a positive integer. Let φ be a multiplicative character of Fq of order 4. Then

    φ(1)={1,if  p1(mod8) or p5(mod8) and s is even,1,if  p5(mod8) and s is odd.

    Proof. This corollary follows immediately from Lemma 2.4.

    Let ^Fq be the dual group consisting of all multiplicative characters of Fq with the generator φ. Then ord(φ)=q1 and the multiplicative character λ with order d with d|(q1) has the expression λ=φq1dt, where 0t<d and gcd(t,d)=1. Furthermore, the number of multiplicative character λ with order d is ϕ(d), where ϕ is Euler's totient function. We also need the following result.

    Lemma 2.5. Let λ be a multiplicative character of Fq with order gcd(4,q1). Then

    N(x4=b)=gcd(4,q1)1j=0λj(b).

    Proof. We divide this into the following three cases. Let λ be any multiplicative character of Fq with order d:=gcd(4,q1).

    CASE 1. b=0. Then x4=0 has only zero solution x=0 in Fq. That is, one has N(x4=0)=1. Since λ0(0)=1 and λj(0)=0 for 1jd1, it follows that

    d1j=0λj(0)=1=N(x4=0)

    as desired. So part (ⅰ) is proved in this case.

    CASE 2. b0 and x4=b has a solution in Fq. Let b=gk and x=gy. Then x4=b is equivalent to the congruence

    4yk(modq1). (2.1)

    Then the congruence (2.1) has exactly d=gcd(4,q1) solutions y. Hence x4=b has exactly d solutions in Fq. Namely, N(x4=b)=d.

    Let x0 be an element of Fq with x40=b. For any integer j with 0jd1, since d|4 implying that λ4=χ0, the trivial multiplicative character, we have

    λj(b)=λj(x40)=(λ4(x0))j=1.

    Therefore one derives that

    d1j=0λj(b)=d1j=01=d=N(x4=b)

    as desired. Hence part (ⅰ) holds in this case.

    CASE 3. b0 and x4=b has no solution in Fq. Then N(x4=b)=0 and (2.1) has no solution in Fq. Let b=gk. Then d and \lambda(b) = {\lambda}^k(g)\ne 1 since \lambda(g) is a d -th primitive root of unity. Then

    \lambda(b)\sum\limits_{j = 0}^{d-1}{\lambda}^j(b) = \sum\limits_{j = 0}^{d-1}{\lambda}^{j+1}(b) = \sum\limits_{j = 0}^{d-1}{\lambda}^j(b),

    which implies that

    (\lambda(b)-1)\sum\limits_{j = 0}^{d-1}{\lambda}^j(b) = 0.

    Since \lambda(b)\ne 1 , we have

    \sum\limits_{j = 0}^{d-1}{\lambda}^j(b) = 0 = N(x^4 = b)

    as required. Part (ⅰ) is proved in this case.

    This finishes the proof of Lemma 2.5.

    The following relation between the Gauss sum G(\chi ^\prime, \psi^\prime) of \mathbb{F}_p and the Gauss sum G(\chi, \psi) of \mathbb{F}_q is due to Hasse and Davenport.

    Lemma 2.6. [12] Let \psi^\prime be an additive and \chi ^\prime a multiplicative character of \mathbb{F}_p , not both of them trivial. Suppose that \psi^\prime and \chi ^\prime are lifting to characters \psi and \chi , respectively, of the finite extension field \mathbb{F}_q of \mathbb{F}_p with [\mathbb{F}_q:\mathbb{F}_p] = s . Then

    G(\chi , \psi) = (-1)^{s-1}G^s(\chi ^\prime, \psi^\prime).

    For a certain special multiplicative character of \mathbb{F}_p, the following result gives an explicit formula about the associated Jacobi sums.

    Lemma 2.7. [4] Let p\equiv1\pmod4 be an odd prime number and let \varphi' be a multiplicative character of \mathbb{F}_p with ord(\varphi') = 4. If \theta is a generator of \mathbb{F}_p^* with \varphi'(\theta) = {\rm i} , then

    J(\varphi', \varphi') = a'+b'{\rm i},

    where a' and b' are integers such that a'^2+b'^2 = p , a'\equiv-1\pmod4 and b'\equiv a'{\theta}^{\frac{p-1}{4}} \pmod p .

    The characters of \mathbb{F}_p can be lifted to the characters of \mathbb{F}_q , but not all the characters of \mathbb{F}_q can be obtained by lifting a character of \mathbb{F}_p . The following result characterizes all the characters of \mathbb{F}_q that can be obtained by lifting a character of \mathbb{F}_p .

    Lemma 2.8. [12] Let \chi be a multiplicative character of \mathbb{F}_q with q = p^s . Then \chi can be lifted from a multiplicative character \chi' of \mathbb{F}_p if and only if \chi ^{p-1} is trivial.

    In this section, we present the proof of Theorem 1.1 as follows.

    Proof of Theorem 1.1. {\rm(i)}. Let p\equiv1\pmod4 . For x\in\mathbb{F}_q, from the trigonometric identity

    \begin{equation*} \sum\limits_{y\in\mathbb{F}_q}\exp\Big(\frac{2\pi {\rm i}\text{Tr}(xy)}{p}\Big) = \begin{cases} q, & \mbox{if }\ x = 0, \\ 0, & \mbox{if }\ x\neq 0, \end{cases} \end{equation*}

    we can deduce that

    \begin{align} N(x_1^4+x_2^4 = c)& = {\frac{1}{q}}\sum\limits_{x\in\mathbb{F}_q}\sum\limits_{(x_1, x_2)\in\mathbb{F}_q^2} \exp\Big({\frac{2\pi{\rm i}\ \text{Tr}(x(x_1^4+x_2^4-c))}{p}}\Big) \\ & = {\frac{1}{q}}\sum\limits_{x\in\mathbb{F}_q}\bigg(\sum\limits_{y\in\mathbb{F}_q} \exp\Big({\frac{2\pi{\rm i}\ \text{Tr}(xy^4)}{p}}\Big)\bigg)^2 \exp\Big({\frac{2\pi{\rm i}\ \text{Tr}(-xc)}{p}}\Big) \\ & = q+{\frac{1}{q}}\sum\limits_{x\in\mathbb{F}_q^*}\bigg(\sum\limits_{y\in\mathbb{F}_q} \exp\Big({\frac{2\pi{\rm i}\ \text{Tr}(xy^4)}{p}}\Big)\bigg)^2\psi_1(-xc). \end{align} (3.1)

    Denote

    \begin{equation*} R_x: = \sum\limits_{y\in\mathbb{F}_q}\exp\Big({\frac{2\pi{\rm i}\ \text{Tr}(xy^4)}{p}}\Big). \end{equation*}

    Then by Lemma 2.5, we get

    \begin{align*} R_x& = 1+\sum\limits_{z\in\mathbb{F}_q^*}N(y^4 = z)\exp\Big({\frac{2\pi{\rm i}\ \text{Tr}(xz)}{p}}\Big)\\ & = 1+\sum\limits_{z\in\mathbb{F}_q^*}\Big(1+\varphi(z)+\varphi^2(z)+\varphi^3(z)\Big)\psi_1(xz), \end{align*}

    where \varphi is a multiplicative character of \mathbb{F}_q with {\rm ord}(\varphi) = 4. Then \varphi(g) = \pm{\rm i} . WLOG, in what follows, we set \varphi(g) = {\rm i} .

    Note that \varphi^2 = \eta and \varphi^3 = \overline{\varphi}, we know that

    \begin{equation*} R_x = \sum\limits_{z\in\mathbb{F}_q}\psi_1(xz)+\sum\limits_{z\in\mathbb{F}_q^*}\varphi(z)\psi_1(xz) +\sum\limits_{z\in\mathbb{F}_q^*}\eta(z)\psi_1(xz)+\sum\limits_{z\in\mathbb{F}_q^*}\overline{\varphi}(z)\psi_1(xz). \end{equation*}

    Since

    \begin{equation*} \sum\limits_{z\in\mathbb{F}_q}\psi_1(xz) = 0, \end{equation*}

    it follows from the definition of Gauss sum and Lemma 2.1 that

    \begin{align*} R_x& = \sum\limits_{z\in\mathbb{F}_q^*}\varphi(z)\psi_1(xz)+\sum\limits_{z\in\mathbb{F}_q^*}\eta(z)\psi_1(xz) +\sum\limits_{z\in\mathbb{F}_q^*}\overline{\varphi}(z)\psi_1(xz)\\ & = G(\varphi, \psi_x)+G(\eta, \psi_x)+G(\overline{\varphi}, \psi_x) \\ & = \overline{\varphi}(x)G(\varphi, \psi_1)+{\overline{\eta}}(x)G(\eta, \psi_1)+\varphi(x)G(\overline{\varphi}, \psi_1). \end{align*}

    Noticing that the value of \eta is real and \eta(x) = \Big(\frac{\mathbb{N}(x)}{p}\Big) , from the Lemmas 2.1 and 2.2 we deduce that

    \begin{equation} R_x = \overline{\varphi}(x)G(\varphi, \psi_1)+\Big(\frac{\mathbb{N}(x)}{p}\Big)(-1)^{s-1} \sqrt{q}+\varphi(-x)\overline{G(\varphi, \psi_1)}. \end{equation} (3.2)

    From (3.1)) and ((3.2), we derive that

    \begin{align} &N(x_1^4+x_2^4 = c)\\ & = q+{\frac{1}{q}}\sum\limits_{x\in\mathbb{F}_q^*}\bigg(\overline{\varphi}(x)G(\varphi, \psi_1) +\Big(\frac{\mathbb{N}(x)}{p}\Big)(-1)^{s-1}\sqrt{q} +\varphi(-x)\overline{G(\varphi, \psi_1)}\bigg)^2\psi_1(-xc)\\ &: = q+{\frac{1}{q}}\sum\limits_{x\in\mathbb{F}_q^*}T_x. \end{align} (3.3)

    Since \varphi^3 = \overline{\varphi} , \overline{\varphi^2} = \varphi^2 = \eta , \varphi^2(x) = \Big(\frac{\mathbb{N}(x)}{p}\Big) and Lemma 2.1 implying that

    G(\varphi, \psi_1)\overline{G(\varphi, \psi_1)} = q , it follows that

    \begin{align*} T_x = &\Big(\eta(x)\big(G^2(\varphi, \psi_1)+\overline{G^2(\varphi, \psi_1)}\big) +2(-1)^{s-1}\sqrt{q}\big(\varphi(x)G(\varphi, \psi_1)+\varphi(-1)\overline{\varphi}(x)\overline{G(\varphi, \psi_1)}\big)\\ &+\big(1+2\varphi(-1)\big)q\Big)\psi_1(-xc). \end{align*}

    By using the following simple facts

    \begin{equation*} \eta(x) = \frac{\eta(-xc)}{\eta(-c)}, \quad \varphi(x) = {\frac{\varphi(-1)}{\varphi(c)}}\varphi(-xc), \quad \overline{\varphi}(x) = {\frac{\varphi(-1)}{\overline{\varphi}(c)}}\overline{\varphi}(-xc), \end{equation*}

    we deduce that

    \begin{align} T_x = &\Big(\frac{\eta(-xc)}{\eta(-c)}\big(G^2(\varphi, \psi_1)+\overline{G^2(\varphi, \psi_1)}\big)+\big(1+2\varphi(-1)\big)q\\ &+2(-1)^{s-1}\sqrt{q}\big({\frac{\varphi(-1)}{\varphi(c)}}\varphi(-xc)G(\varphi, \psi_1) +{\frac{1}{\overline{\varphi}(c)}}\overline{\varphi}(-xc)\overline{G(\varphi, \psi_1)}\big)\Big)\psi_1(-xc). \end{align} (3.4)

    Since -xc runs over \mathbb{F}_q^* as x runs through \mathbb{F}_q^* , it follows from (3.3), (3.4) and the fact of \sum\limits_{x\in\mathbb{F}_q^*}\psi_1(-xc) = -1 that

    \begin{align*} &N(x_1^4+x_2^4 = c)\\ & = q+{\frac{1}{q}}\Big(\frac{G^2(\varphi, \psi_1)+\overline{G^2(\varphi, \psi_1)}}{\eta(-c)}G(\eta, \psi_1)-(1+2\varphi(-1))q\\ &+2(-1)^{s-1}\sqrt{q}\frac{\varphi(-1)}{\varphi(c)}G^2(\varphi, \psi_1) +(-1)^{s-1}\sqrt{q}\frac{2}{\overline{\varphi}(c)}\overline{G(\varphi, \psi_1)}G(\overline{\varphi}, \psi_1)\Big). \end{align*}

    Note that

    \begin{equation*} \frac{1}{\varphi(c)} = \overline{\varphi}(c), \quad \frac{1}{\eta(-c)} = \eta(-c) = \eta(-1)\eta(c) = \eta(c). \end{equation*}

    From Lemmas 2.1 and 2.2, we have

    \begin{align} &N(x_1^4+x_2^4 = c) = q+\frac{1}{q}\Big((-1)^{s-1}\sqrt{q}\eta(c)\big(G^2(\varphi, \psi_1) +\overline{G^2(\varphi, \psi_1)}\big)-(1+2\varphi(-1))q\\ &+2(-1)^{s-1}\sqrt{q}\varphi(-1)\big(\overline{\varphi}(c)G^2(\varphi, \psi_1) +\varphi(c)\overline{G^2(\varphi, \psi_1)}\big)\Big). \end{align} (3.5)

    Noting that \varphi(1) = \overline{\varphi}(1) = 1 = \eta(1) , it follows from (3.5) that

    \begin{align*} &N(x_1^4+x_2^4 = 1) = q+\frac{1}{q}\big(1+2\varphi(-1)\big)\Big((-1)^{s-1}\sqrt{q} \big(G^2(\varphi, \psi_1)+\overline{G^2(\varphi, \psi_1)}\big)-q\Big). \end{align*}

    From Lemmas 2.2, 2.3 and 2.6-2.8, we can deduce that

    \begin{equation*} G^2(\varphi, \psi_1) = (a'+b'{\rm i})^s q^{\frac{1}{2}}, \end{equation*}

    where a' and b' are integers such that

    \begin{equation*} a'^2+b'^2 = p, \quad a'\equiv-1\pmod4, \quad b'\equiv a'g^{\frac{q-1}{4}} \pmod p. \end{equation*}

    Letting a+b{\rm i} = (a'+b'{\rm i})^s gives us that

    \begin{equation} G^2(\varphi, \psi_1)+\overline{G^2(\varphi, \psi_1)} = 2a\sqrt{q}, \quad G^2(\varphi, \psi_1)-\overline{G^2(\varphi, \psi_1)} = 2b{\rm i}\sqrt{q}. \end{equation} (3.6)

    Thus

    \begin{equation*} N(x_1^4+x_2^4 = 1) = q+\big(1+2\varphi(-1)\big)\big(2a(-1)^{s-1}-1\big). \end{equation*}

    By Corollary 2.1, we get

    \begin{equation} N(x_1^4+x_2^4 = 1) = \begin{cases} q+6a(-1)^{s-1}-3, & \mbox{if either}\ p\equiv1\pmod8, \ \\ &{\rm or}\ p\equiv 5 \pmod 8 \ \text{and $s$ is even}, \\ q-2a(-1)^{s-1}+1, & \mbox{if }\ p\equiv 5 \pmod 8 \ \text{and $s$ is odd}. \end{cases} \end{equation} (3.7)

    From (3.5), (3.6), \varphi(g) = {\rm i} and \eta(g) = -1 we obtain that

    \begin{equation*} N(x_1^4+x_2^4 = g) = q+2(-1)^s\big(a-2b\varphi(-1)\big)-\big(1+2\varphi(-1)\big). \end{equation*}

    By Corollary 2.1, we have

    \begin{equation} N(x_1^4+x_2^4 = g) = \begin{cases} q+2(-1)^{s}(a-2b)-3, & \mbox{if either}\ p\equiv1\pmod8, \\ &{\rm or} \ p\equiv 5 \pmod 8 \ \text{and $s$ is even}, \\ q+2(-1)^s(a+2b)+1, & \mbox{if }\ p\equiv 5 \pmod 8 \; \text{and $s$ is odd}. \end{cases} \end{equation} (3.8)

    From (3.5), (3.6), \varphi(g^2) = -1 and \eta(g^2) = 1 , we have

    \begin{equation*} N(x_1^4+x_2^4 = g^2) = q+2a(-1)^{s-1}\big(1-2\varphi(-1)\big)-\big(1+2\varphi(-1)\big). \end{equation*}

    By Corollary 2.1 it follows that

    \begin{equation} N(x_1^4+x_2^4 = g^2) = \begin{cases} q+2a(-1)^s-3, & \mbox{if either}\ p\equiv1\pmod8, \\ &{\rm or}\ p\equiv 5 \pmod 8 \ \text{and $s$ is even}, \\ q+6a(-1)^{s-1}+1, & \mbox{if }\ p\equiv 5 \pmod 8 \ \text{and $s$ is odd}. \end{cases} \end{equation} (3.9)

    From (3.5), (3.6), \varphi(g^3) = -{\rm i} and \eta(g^3) = -1 , we deduce

    \begin{equation*} N(x_1^4+x_2^4 = g^3) = q+2(-1)^s\big(a+2b\varphi(-1)\big)-\big(1+2\varphi(-1)\big). \end{equation*}

    By Corollary 2.1 we have

    \begin{equation} N(x_1^4+x_2^4 = g^3) = \begin{cases} q+2(-1)^s(a+2b)-3, & \mbox{if either}\ p\equiv1\pmod8, \\ &{\rm or} \ p\equiv 5 \pmod 8 \ \text{and $s$ is even}, \\ q+2(-1)^s(a-2b)+1, & \mbox{if }\ p\equiv 5 \pmod 8 \ \text{and $s$ is odd}. \end{cases} \end{equation} (3.10)

    From (3.7), (3.8), (3.9) and (3.10), we can conclude the proof of part \rm(i) .

    {\rm(ii)}. Let p\equiv3\pmod4 and q\equiv1\pmod4. Since p\equiv3\pmod4 implying that

    \begin{equation} q = p^s\equiv \begin{cases} 1\pmod4, & \mbox{if}\ \text{$s$ is even}, \\ 3\pmod4, & \mbox{if}\ \text{$s$ is odd}, \end{cases} \end{equation} (3.11)

    one must have that s is even. Let N_k be the number of solutions of x_1^4+x_2^4+\cdots +x_k^4 = 0 over \mathbb{F}_q . Myerson [13] gave the value of

    N_2 = 4q-3, \quad N_3 = q^2-6rq+6r,

    where r is uniquely determined by

    q = r^2+4t^2, r\equiv1\pmod4, \ {\rm and \ if} \ p\equiv1\pmod4, \ {\rm then} \ (r, p) = 1.

    Since

    N_3 = \sum\limits_{(x_1, x_2, x_3)\in\mathbb{F}_q^3\atop x_1^4+x_2^4+x_3^4 = 0}1 = \sum\limits_{(x_1, x_2)\in\mathbb{F}_q^2\atop x_1^4+x_2^4 = 0}1 +\sum\limits_{x_3\in\mathbb{F}_q^*\atop x_1^4+x_2^4 = -x_3^4}1
    = N_2+(q-1)N(x_1^4+x_2^4 = -1),

    one has

    \begin{align} N(x_1^4+x_2^4 = -1) = \frac{N_3-N_2}{q-1} = \frac{q^2-(6r+4)q+6r+3}{q-1}. \end{align} (3.12)

    From (3.11) and Corollary 2.1, we have \varphi(-1) = 1 if s is even.

    By (3.5), one has

    \begin{align} &N(x_1^4+x_2^4 = -1)\\ & = q+\frac{1}{q}\Big((-1)^{s-1}\sqrt{q}\big(G^2(\varphi, \psi_1) +\overline{G^2(\varphi, \psi_1)}\big)-\big(1+2\varphi(-1)\big)q\Big)\\ & = q-\frac{1}{q}\Big(\sqrt{q}\big(G^2(\varphi, \psi_1)+\overline{G^2(\varphi, \psi_1)}\big)+3q\Big) \end{align} (3.13)

    From (3.12) and (3.13), we can deduce

    G^2(\varphi, \psi_1)+\overline{G^2(\varphi, \psi_1)} = 6r\sqrt{q}.

    Let G^2(\varphi, \psi_1) = 3r\sqrt{q}+B{\rm i}. By G^2(\varphi, \psi_1)\overline{G^2(\varphi, \psi_1)} = q^2 , one has

    B = \pm\sqrt{q^2-9r^2q}.

    So we can write G^2(\varphi, \psi_1) = 3r\sqrt{q}\pm\sqrt{q^2-9r^2q}\ {\rm i}. From (3.5), one has

    \begin{align*} &N(x_1^4+x_2^4 = c)\\ & = \big(q-1-2\varphi(-1)+6r\eta(c)(-1)^{s-1}\big) +2\sqrt{q-9r}(-1)^{s-1}\Big(\big(\pm\overline{\varphi}(-c)\big) +\big(\mp\varphi(-c)\big)\Big){\rm i}. \end{align*}

    This finishes the proof of \rm(ii) .

    {\rm(iii)}. Let p\equiv3\pmod4 and q\equiv3\pmod4. Since \gcd(4, q-1) = \gcd(2, q-1). By Lemma 2.5, one has N(x^4 = A) = N(x^2 = A). It follows that

    \begin{align*} &N(x_1^4+x_2^4 = c)\\ & = \sum\limits_{(x_1, x_2)\in\mathbb{F}_q^2\atop c_1+c_2 = c}N(x_1^4 = c_1)N(x_2^4 = c_2)\\ & = \sum\limits_{(x_1, x_2)\in\mathbb{F}_q^2\atop c_1+c_2 = c}N(x_1^2 = c_1)N(x_2^2 = c_2)\\ & = \sum\limits_{(x_1, x_2)\in\mathbb{F}_q^2\atop c_1+c_2 = c}\big(1+\eta(c_1)\big)\big(1+\eta(c_2)\big)\\ & = q+\sum\limits_{c_1\in\mathbb{F}_q}\eta(c_1)+\sum\limits_{c_1\in\mathbb{F}_q}\eta(c_2)+\sum\limits_{c_1\in\mathbb{F}_q}\eta(c_1c_2)\\ & = q+\sum\limits_{x\in\mathbb{F}_q}\eta(cx-x^2) = q+\sum\limits_{x\in\mathbb{F}_q^{*}}\eta(cx^{-1}-1).\\ & = q+\sum\limits_{y\in\mathbb{F}_q^{*}}\eta(y-1) = q+1. \end{align*}

    This finishes the proof of part \rm(iii) , and hence that of Theorem 1.1.

    Proof of Corollary 1.2. {\rm(i)}. Let q\equiv1\pmod4 . From Theorem 1.1, one has |a|\le \sqrt{q} and |b|\le \sqrt{q} , therefore one deduces that |a\pm 2b|\le 3\sqrt{q} . Then we derive the desired result |N(x_1^4+x_2^4 = c)-q|\le7\sqrt{q} by triangle inequality.

    {\rm(ii)}. Let p\equiv1\pmod4 and q\equiv3\pmod4 . From Theorem 1.1, one has |r|\le \sqrt{q} , \sqrt{q-9\sqrt{q}}\le\sqrt{q-9r}\le\sqrt{q+9\sqrt{q}} and |\Big(\big(\pm\overline{\varphi}(-c)\big)+\big(\mp\varphi(-c)\big)\Big){\rm i}|\le 2 . From this, we deduce that

    \begin{align*} |N(x_1^4+x_2^4 = c)-q|&\le|3+6r+4\sqrt{q-4r}|\\ &\le3+6r+4\sqrt{q+9r}\\ &\le3+6r+4\sqrt{2q}\\ &\le(7+4\sqrt{2})\sqrt{q} \end{align*}

    as required. The proof of Corollary 1.2 is complete.

    In this final section, we provide two examples to demonstrate the validity of our main result Theorem 1.1.

    Example 4.1. For finite field \mathbb{F}_5 , it is easy to see that 2 is a generator of \mathbb{F}_5^* . Note that s = 1 . Then a' = a = -1 and b' = b = -2 .

    Since {\rm ind}_2(1)\equiv 0\pmod 4 , {\rm ind}_2(2)\equiv 1\pmod 4 , {\rm ind}_2(3)\equiv 3\pmod 4 , {\rm ind}_2(2)\equiv 2\pmod 4 . From Theorem 1.1, one can compute and get that

    N(x_1^4+x_2^4 = 1) = 8, \quad N(x_1^4+x_2^4 = 2) = 16,
    N(x_1^4+x_2^4 = 3) = N(x_1^4+x_2^4 = 4) = 0.

    Example 4.2. Observe that x^2-2 is irreducible over \mathbb{F}_5 . Let \alpha be a root of x^2-2 over its split field. Then \mathbb{F}_5(\alpha) is an extension field of \mathbb{F}_5 with order 25 , and we denote it by \mathbb{F}_{25} , where

    \mathbb{F}_{25} = \{ x+y\alpha\mid x\in\mathbb{F}_5, y\in\mathbb{F}_5\}.

    For any {x_i+y_i}\in \mathbb{F}_{25} with i = 1, 2 , we define

    (x_1+y_1\alpha)+(x_2+y_2\alpha): = ((x_1+x_2)\pmod 5 +\Big((y_1+y_2)\pmod 5\Big)\alpha

    and

    (x_1+y_1\alpha)(x_2+y_2\alpha): = (x_1x_2+2y_1y_2)\pmod 5 +\Big((x_1y_2+x_2y_1)\pmod 5\Big)\alpha.

    By matlab programme, we confirm that the element 2+4 \alpha is a generator of \mathbb{F}_{25} .

    Note that s = 2 , a' = -1 , b' = -2 . Then a = -3 and b = 4 . Since {\rm ind}_{2+4\alpha}(1)\equiv 0\pmod 4 and {\rm ind}_{2+4\alpha}(\alpha)\equiv 3\pmod 4 , it follows from Theorem 1.1 that

    N(x_1^4+x_2^4 = 1) = 40, \quad N(x_1^4+x_2^4 = \alpha) = 32.

    The authors are thankful for the anonymous referees for their careful reading of the manuscript and helpful comments. They also thank Dr. Chaoxi Zhu and Dr. Lingfeng Ao for many helpful suggestions.

    We declare that we have no conflict of interest.



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