A type of block withholding delay attack and the countermeasure based on type-2 fuzzy inference

Liang Liu; Wen Chen; Lei Zhang; JiaYong Liu; Jian Qin; Liang Liu; Wen Chen; Lei Zhang; JiaYong Liu; Jian Qin

doi:10.3934/mbe.2020017

Mathematical Biosciences and Engineering

2020, Volume 17, Issue 1: 309-327. doi: 10.3934/mbe.2020017

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Research article Special Issues

A type of block withholding delay attack and the countermeasure based on type-2 fuzzy inference

1.
College of Cybersecurity, Sichuan University, Chengdu, China
2.
College of Electronics and Information Engineering, Sichuan University, Chengdu, China
3.
Chengdu No.7 high school, Chengdu, China

Received: 19 August 2019 Accepted: 25 September 2019 Published: 09 October 2019

We proposed a new type of bitcoin withholding attack named block withholding delay (BWD). It is different from the traditional withholding attacks which always drop valid blocks. BWD attackers never discard blocks but they delay the submissions of blocks to the pool managers, resulting the pool failed in the mining competitions and loss of rewards. We analyzed the optimum strategy of a BWD attacker who split its computing power into two parts, one was utilized to launch BWD attacks on the victim pools, while the other part was used for solo mining. We present detailed quantitative analysis of the maximum incentive that an attacker can earn by carefully splitting its computing power, and demonstrated that the attacker can obtain higher incentives than its contribution to the network in different conditions. Furthermore, we proposed a countermeasure against BWD based on the interval type-2 Takagi-Sugeno-Kang fuzzy inference system (IT2-TSK-FIS). The principle is to modify the private payoff scheme of pools to increase the risk of losing revenues of the rogue miners who deliberately delay block submissions. The scheme dealing the uncertain cause of block delay using fuzzy inference, and it is so designed that it does not require modifications of public mining protocols or data structures of the bitcoin network, which makes it applicable in practical pools.

Keywords:

Citation: Liang Liu, Wen Chen, Lei Zhang, JiaYong Liu, Jian Qin. A type of block withholding delay attack and the countermeasure based on type-2 fuzzy inference[J]. Mathematical Biosciences and Engineering, 2020, 17(1): 309-327. doi: 10.3934/mbe.2020017

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Abstract

1. Introduction

Let $\mathbb{F}_{q}$ be the finite field of $q$ elements with characteristic $p$ , where $q = p^r$ , $p$ is a prime number. Let $\mathbb{F}_{q}^* = \mathbb{F}_{q}\backslash \{0\}$ and $\mathbb{Z}^+$ denote the set of positive integers. Let $s\in \mathbb{Z}^+$ and $b\in \mathbb{F}_{q}$ . Let $f(x_1, \ldots, x_s)$ be a diagonal polynomial over $\mathbb{F}_{q}$ of the following form

$\begin{equation*} f(x_1,\ldots,x_s) = a_{1}x_{1}^{m_{1}}+a_{2}x_{2}^{m_{2}}+\dots+a_{s}x_{s}^{m_{s}}, \end{equation*}$

where $a_{i}\in \mathbb{F}_{q}^*,$ $m_{i}\in \mathbb{Z}^{+},$ $i = 1, \ldots, s.$ Denote by $N_{q}(f = b)$ the number of $\mathbb{F}_{q}$ -rational points on the affine hypersurface $f = b$ , namely,

$\begin{equation*} N_{q}(f = b) = \#\{\left(x_{1}, \ldots, x_{s}\right) \in \mathbb{A}^{s}(\mathbb{F}_{q}) \mid f\left(x_{1}, \ldots, x_{s}\right) = b\}. \end{equation*}$

In 1949, Hua and Vandiver ^[1] and Weil ^[2] independently obtained the formula of $N_{q}(f = b)$ in terms of character sum as follows

$\begin{equation} N_q( f = b ) = q^{s-1}+\sum{\psi}_1\left( a_{1}^{-1} \right) \cdots \psi _s\left( a_{s}^{-s}\right) J_q^0\left( \psi _1,\dots ,\psi _s \right), \end{equation}$

(1.1)

where the sum is taken over all $s$ multiplicative characters of $\mathbb{F}_{q}$ that satisfy $\psi_{i}^{m_{i}} = \varepsilon,$ $\psi_{i}\ne \varepsilon,$ $i = 1, \dots, s$ and $\psi_{1}\cdots \psi_{s} = \varepsilon$ . Here $\varepsilon$ is the trivial multiplicative character of $\mathbb{F}_{q}$ , and $J_q^{0}(\psi _1, \dots, \psi _s)$ is the Jacobi sum over $\mathbb{F}_{q}$ defined by

$\begin{equation*} J_q^0\left( \psi _1,\dots ,\psi _s \right) = \sum\limits_{c_1+\cdots +c_s = 0,c_i\in \mathbb{F}_{q}}{\psi _1\left( c_1 \right) \cdots \psi _s\left( c_s \right)}. \end{equation*}$

Though the explicit formula for $N_q(f = b)$ are difficult to obtain in general, it has been studied extensively because of their theoretical importance as well as their applications in cryptology and coding theory; see^{[3,4,5,6,7,8,9]}. In this paper, we use the Jacobi sums, Gauss sums and the results of quadratic form to deduce the formula of the number of $\mathbb{F}_{q^2}$ -rational points on a class of hypersurfaces over $\mathbb{F}_{q^2}$ under certain conditions. The main result of this paper can be stated as

Theorem 1.1. Let $q = 2^r$ with $r\in \mathbb{Z}^+$ and $\mathbb{F}_{q^2}$ be the finite field of $q^2$ elements. Let $f(X) = a_{1}x_{1}^{m_{1}}+a_{2}x_{2}^{m_{2}}+\dots+a_{s}x_{s}^{m_{s}},$ $g(Y) = y_{1}y_{2}+y_{3}y_{4}+\dots+y_{n-1}y_{n}+y_{n-2t-1}^{2}+\dots$ $+y_{n-3}^2+y_{n-1}^{2}+b_{t}y_{n-2t}^{2}+\dots +b_{1}y_{n-2}^{2}+b_{0}y_{n}^{2}$ , and $l(X, Y) = f(X)+g(Y)$ , where $a_{i}, b_j\in \mathbb{F}_{q^{2}}^{*},$ $m_i\ne 1,$ $(m_{i}, m_{k}) = 1,$ $i\ne k,$ $m_{i}|(q+1),$ $m_{i}\in \mathbb{Z}^{+},$ $2|n$ , $n > 2$ , $0\leq t\leq \frac{n}{2}-2$ , ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(b_{j}) = 1$ for $i, k = 1, \dots, s$ and $j = 0, 1, \dots, t$ . For $h\in \mathbb{F}_{q^2}$ , we have

(1) If $h = 0$ , then

$\begin{equation*} \begin{aligned} N_{q^2}(l(X,Y) = 0) = q^{2(s+n-1)}+ \\\sum\limits_{\gamma\in \mathbb{F}_{q^2}^*}{\left( \prod\limits_{i = 1}^s{\left( \left( \frac{\gamma}{a_i} \right) _{m_i}m_i-1 \right) \left( q^{s+2n-3}+(-1)^{t}q^{s+n-3} \right)} \right)}. \end{aligned} \end{equation*}$

(2) If $h\in \mathbb{F}_{q^2}^{*}$ , then

$\begin{equation*} \begin{aligned} &N_{q^2}(l(X,Y) = h)\\ = &q^{2\left( s+n-1 \right)}+\left( q^{s+2n-3}+\left( -1 \right) ^{t+1}\left( q^2-1 \right) q^{s+n-3} \right) \prod\limits_{i = 1}^s{\left( \left( \frac{h}{a_i} \right)_{m_i} m_i-1 \right) }\\ &+\sum\limits_{\substack{ \gamma \in \mathbb{F}_{q^2}^{*}\backslash \{h\}}}{\left[ \prod\limits_{i = 1}^s{\left( \left( \frac{\gamma}{a_i} \right)_{m_i} m_i-1 \right)}\left( q^{2n+s-3}+\left( -1 \right) ^{t}q^{n+s-3} \right) \right]}. \end{aligned} \end{equation*}$

Here,

$\left(\frac{\gamma}{a_{i}}\right)_{m_{i}} = \left\{\begin{array}{ll} 1, & if\; \frac{\gamma}{a_{i}}\; is \;a\; residue\; of\; order\; m_{i}, \\ 0, & otherwise. \end{array}\right.$

2. Prerequisites

To prove Theorem 1.1, we need the lemmas and theorems below which are related to the Jacobi sums and Gauss sums.

Definition 2.1. Let $\chi$ be an additive character and $\psi$ a multiplicative character of $\mathbb{F}_{q}$ . The Gauss sum $G_q(\psi, \chi)$ in $\mathbb{F}_{q}$ is defined by

$G_q(\psi, \chi) = \sum\limits_{x \in \mathbb{F}_{q}^{*}} \psi(x) \chi(x).$

In particular, if $\chi$ is the canonical additive character, i.e., $\chi(x) = e^{2 \pi i {\mathrm{Tr}}_{\mathbb{F}_{q}/ \mathbb{F}_{p}}(x)/p}$ where ${\mathrm{Tr}}_{\mathbb{F}_{q}/ \mathbb{F}_{p}}(y) = y + y^{p} + \dots + y^{p^{r - 1}}$ is the absolute trace of $y$ from $\mathbb{F}_{q}$ to $\mathbb{F}_{p}$ , we simply write $G_q(\psi): = G_q(\psi, \chi)$ .

Let $\psi$ be a multiplicative character of $\mathbb{F}_{q}$ which is defined for all nonzero elements of $\mathbb{F}_{q}$ . We extend the definition of $\psi$ by setting $\psi(0) = 0$ if $\psi\neq \varepsilon$ and $\varepsilon(0) = 1$ .

Definition 2.2. Let $\psi_1, \ldots, \psi_s$ be $s$ multiplicative characters of $\mathbb{F}_{q}$ . Then, $J_q(\psi _1, \dots, \psi _s)$ is the Jacobi sum over $\mathbb{F}_{q}$ defined by

$\begin{equation*} J_q\left( \psi _1,\dots ,\psi _s \right) = \sum\limits_{c_1+\cdots +c_s = 1,c_i\in \mathbb{F}_{q}}{\psi _1\left( c_1 \right) \cdots \psi _s\left( c_s \right)}. \end{equation*}$

The Jacobi sums $J_q(\psi_1, \ldots, \psi_s)$ as well as the sums $J_q^0(\psi_1, \ldots, \psi_s)$ can be evaluated easily in case some of the multiplicative characters $\psi_i$ are trivial.

Lemma 2.3. ([,Theorem $5.19$ ,p. 206]) If the multiplicative characters $\psi_1, \ldots, \psi_s$ of $\mathbb{F}_{q}$ are trivial, then

$\begin{equation*} J_q(\psi_1,\ldots,\psi_s) = J_q^0(\psi_1,\ldots,\psi_s) = q^{s-1}. \end{equation*}$

If some, but not all, of the $\psi_i$ are trivial, then

$\begin{equation*} J_q(\psi_1,\ldots,\psi_s) = J_q^0(\psi_1,\ldots,\psi_s) = 0. \end{equation*}$

Lemma 2.4. ([,Theorem $5.20$ ,p. 206]) If $\psi_1, \ldots, \psi_s$ are multiplicative characters of $\mathbb{F}_{q}$ with $\psi_s$ nontrivial, then

$\begin{equation*} J_q^0(\psi_1,\ldots,\psi_s) = 0 \end{equation*}$

if $\psi_1\cdots\psi_s$ is nontrivial and

$\begin{equation*} J_q^0(\psi_1,\ldots,\psi_s) = \psi_s(-1)(q-1)J_q(\psi_1,\ldots,\psi_{s-1}) \end{equation*}$

if $\psi_1\cdots\psi_s$ is trivial.

If all $\psi_i$ are nontrivial, there exists an important connection between Jacobi sums and Gauss sums.

Lemma 2.5. ([,Theorem $5.21$ ,p. 207]) If $\psi_1, \ldots, \psi_s$ are nontrivial multiplicative characters of $\mathbb{F}_{q}$ and $\chi$ is a nontrivial additive character of $\mathbb{F}_{q}$ , then

$\begin{equation*} J_q(\psi_1,\ldots,\psi_s) = \frac{G_q(\psi_1,\chi)\cdots G_q(\psi_s,\chi)}{G_q(\psi_1\cdots \psi_s,\chi)} \end{equation*}$

if $\psi_1\cdots \psi_s$ is nontrivial and

$\begin{equation*} \begin{aligned} J_q(\psi_1,\ldots,\psi_s)& = -\psi_s(-1)J_q(\psi_1,\ldots,\psi_{s-1})\\ & = -\frac{1}{q}G_q(\psi_1,\chi)\cdots G_q(\psi_s,\chi) \end{aligned} \end{equation*}$

if $\psi_1\cdots \psi_s$ is trivial.

We turn to another special formula for Gauss sums which applies to a wider range of multiplicative characters but needs a restriction on the underlying field.

Lemma 2.6. ([,Theorem $5.16$ ,p. 202]) Let q be a prime power, let $\psi$ be a nontrivial multiplicative character of $\mathbb{F}_{q^2}$ of order m dividing $q+1$ . Then

$\begin{equation*} G_{q^2}\left(\psi\right) = \left\{\begin{array}{ll} q, & if\; m\; odd \;or\; \frac{q+1}{m}\; even, \\ -q, & if\; m\; even \;and\; \frac{q+1}{m}\; odd. \end{array}\right. \end{equation*}$

For $h\in \mathbb{F}_{q^2}$ , define $v(h) = -1$ if $h\in \mathbb{F}_{q^2}^{*}$ and $v(0) = q^2-1$ . The property of the function $v(h)$ will be used in the later proofs.

Lemma 2.7. ([,Lemma $6.23$ ,p. 281]) For any finite field $\mathbb{F}_{q}$ , we have

$\sum\limits_{c\in \mathbb{F}_{q}}v(c) = 0,$

for any $b\in \mathbb{F}_{q}$ ,

$\begin{equation*} \sum\limits_{c_1+\cdots +c_m = b}{v( c_1 )} \cdots v( c_k ) = \begin{cases} 0,& 1\leqslant k < m,\\ v\left( b \right) q^{m-1},& k = m,\\ \end{cases} \end{equation*}$

where the sum is over all $c_1, \ldots, c_m\in \mathbb{F}_{q}$ with $c_1+\dots+c_m = b.$

The quadratic forms have been studied intensively. A quadratic form $f$ in $n$ indeterminates is called nondegenerate if $f$ is not equivalent to a quadratic form in fewer than $n$ indeterminates. For any finite field $\mathbb{F}_q$ , two quadratic forms $f$ and $g$ over $\mathbb{F}_q$ are called equivalent if $f$ can be transformed into $g$ by means of a nonsingular linear substitution of indeterminates.

Lemma 2.8. ([,Theorem $6.30$ ,p. 287]) Let $f\in \mathbb{F}_{q}[x_1, \ldots, x_n]$ , $q$ even, be a nondegenerate quadratic form. If $n$ is even, then $f$ is either equivalent to

$x_1x_2+x_3x_4+\dots+x_{n-1}x_n$

or to a quadratic form of the type

$x_1x_2+x_3x_4+\dots+x_{n-1}x_n+x_{n-1}^{2}+ax_{n}^2,$

where $a\in \mathbb{F}_{q}$ satisfies ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a) = 1.$

Lemma 2.9. ([,Corollary $3.79$ ,p. 127]) Let $a\in \mathbb{F}_{q}$ and let $p$ be the characteristic of $\mathbb{F}_{q}$ , the trinomial $x^p-x-a$ is irreducible in $\mathbb{F}_{q}$ if and only if ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a)\ne 0.$

Lemma 2.10. ([,Lemma $6.31$ ,p. 288]) For even $q$ , let $a\in \mathbb{F}_{q}$ with ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a) = 1$ and $b\in \mathbb{F}_{q}.$ Then

$\begin{equation*} N_q(x_{1}^{2}+x_{1}x_{2}+ax_{2}^{2} = b) = q-v(b). \end{equation*}$

Lemma 2.11. ([,Theorem $6.32$ ,p. 288]) Let $\mathbb{F}_{q}$ be a finite field with $q$ even and let $b\in \mathbb{F}_{q}$ . Then for even $n$ , the number of solutions of the equation

$x_1x_2+x_3x_4+\cdots+x_{n-1}x_n = b$

in $\mathbb{F}_{q}^n$ is $q^{n-1}+v(b)q^{(n-2)/2}$ . For even $n$ and $a\in \mathbb{F}_{q}$ with ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a) = 1$ , the number of solutions of the equation

$x_1x_2+x_3x_4+\cdots+x_{n-1}x_n+x_{n-1}^2+ax_n^2 = b$

in $\mathbb{F}_{q}^n$ is $q^{n-1}-v(b)q^{(n-2)/2}$ .

Lemma 2.12. Let $q = 2^r$ and $h\in \mathbb{F}_{q^2}$ . Let $g(Y)\in \mathbb{F}_{q^2}[y_{1}, y_2, \dots, y_n]$ be a polynomial of the form

$g(Y) = y_{1}y_{2}+y_{3}y_{4}+\dots+y_{n-1}y_{n}+y_{n-2t-1}^{2}+\dots +y_{n-3}^2+y_{n-1}^{2}+\\b_{t}y_{n-2t}^{2}+\dots +b_{1}y_{n-2}^{2}+b_{0}y_{n}^{2},$

where $b_j\in \mathbb{F}_{q^{2}}^{*},$ $2|n$ , $n > 2$ , $0\leq t\leq \frac{n}{2}-2$ , ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(b_{j}) = 1,$ $j = 0, 1, \dots, t.$ Then

$\begin{equation} N_{q^2}(g(Y) = h) = q^{2(n-1)}+(-1)^{t+1}q^{n-2}v(h). \end{equation}$

(2.1)

Proof. We provide two proofs here. The first proof is as follows. Let $q_{1} = q^2$ . Then by Lemmas 2.7 and 2.10, the number of solutions of $g(Y) = h$ in $\mathbb{F}_{q^2}$ can be deduced as

$\begin{array}{l} N_{q^2}(g(Y) = h)\\ = \sum\limits_{c_{1}+c_{2}+\dots+c_{t+2} = h} N_{q^2}\left(y_{1} y_{2}+y_{3} y_{4}+\cdots +y_{n-2t-3} y_{n-2t-2} = c_{1}\right)\\ \cdot N_{q^2}\left(y_{n-2t-1} y_{n-2t}+y_{n-2t-1}^{2}+b_{t}y_{n-2t}^{2} = c_{2}\right)\cdots N_{q^2}\left(y_{n-1} y_{n}+y_{n-1}^{2}+b_{0} y_{n}^{2} = c_{t+2}\right) \\ = \sum\limits_{c_{1}+c_{2}+\dots+c_{t+2} = h}\left(q_{1}^{n-2t-3}+v\left(c_{1}\right) q_{1}^{\left(n-2t-4\right) / 2}\right)\left(q_{1}-v\left(c_{2}\right)\right)\cdots\left(q_{1}-v\left(c_{t+2}\right)\right) \\ = \sum\limits_{c_{1}+c_{2}+\dots+c_{t+2} = h}\left(q_{1}^{n-2t-2}+v\left(c_{1}\right) q_{1}^{\left(n-2t-2\right) / 2}-v\left(c_{2}\right) q_{1}^{n-2t-3}-v\left(c_{1}\right) v\left(c_{2}\right) q_{1}^{\left(n-2t-4\right) / 2}\right)\\ \cdot\left(q_{1}-v\left(c_{3}\right)\right) \cdots\left(q_{1}-v\left(c_{t+2}\right)\right) \\ = \sum\limits_{c_{1}+c_{2}+\dots+c_{t+2} = h}\\\left(q_{1}^{n-t-2}+v\left(c_{1}\right) q_{1}^{\left(n-2\right) / 2}-v\left(c_{2}\right) q_{1}^{n-t-3}+\cdots+(-1)^{t+1}v\left(c_{1}\right) v\left(c_{2}\right) \cdots v\left(c_{t+2}\right) q_{1}^{\left(n-2 t-4\right) / 2}\right) \\ = q_{1}^{n-1}+q_{1}^{\left(n-2\right) / 2} \sum\limits_{c_{1} \in \mathbb{F}_{q^2}} v\left(c_{1}\right)+\cdots +(-1)^{t+1}\sum\limits_{c_{1}+c_{2}+\dots+c_{t+2} = h}v\left(c_{1}\right) v\left(c_{2}\right) \cdots v\left(c_{t+2}\right) q_{1}^{\left(n-2t-4\right)/2}. \end{array}$

(2.2)

By Lamma 2.7 and (2.2), we have

$N_{q^2}(g(Y) = h) = q_{1}^{n-1}+(-1)^{t+1}v(h) q_{1}^{\left(n-2\right) / 2} = q^{2\left(n-1\right)}+(-1)^{t+1}v(h) q^{n-2}.$

Next we give the second proof. Note that if $f$ and $g$ are equivalent, then for any $b\in \mathbb{F}_{q^2}$ the equation $f(x_1, \ldots, x_n) = b$ and $g(x_1, \ldots, x_n) = b$ have the same number of solutions in $\mathbb{F}_{q^2}.$ So we can get the number of solutions of $g(Y) = h$ for $h\in \mathbb{F}_{q^2}$ by means of a nonsingular linear substitution of indeterminates.

Let $k(X)\in \mathbb{F}_{q^2}[x_1, x_2, x_3, x_4]$ and $k(X) = x_1x_2+x_1^2+Ax_2^2+x_3x_4+x_3^2+Bx_4^2$ , where ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(A) = {\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(B) = 1.$ We first show that $k(x)$ is equivalent to $x_1x_2 + x_3x_4$ .

Let $x_3 = y_1+y_3$ and $x_i = y_i$ for $i\neq 3$ , then $k(X)$ is equivalent to $y_1y_2+y_1y_4+y_3y_4+Ay_2^2+y_3^2+By_4^2$ .

Let $y_2 = z_2+z_4$ and $y_i = z_i$ for $i\ne 2$ , then $k(X)$ is equivalent to $z_1z_2+z_3z_4+Az_2^2+z_3^2+Az_4^2+Bz_4^2.$

Let $z_1 = \alpha_1+A\alpha_2$ and $z_i = \alpha_i$ for $i\ne 1$ , then $k(X)$ is equivalent to $\alpha_1\alpha_2+\alpha_3^2+\alpha_3\alpha_4+(A+B)\alpha_4^2.$

Since ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(A+B) = 0$ , we have $\alpha_3^2+\alpha_3\alpha_4+(A+B)\alpha_4^2$ is reducible by Lemma 2.9. Then $k(X)$ is equivalent to $x_1x_2+x_3x_4$ . It follows that if $t$ is odd, then $g(Y)$ is equivalent to $x_{1}x_{2}+x_{3}x_{4}+\cdots+x_{n-1}x_{n}$ , and if $t$ is even, then $g(Y)$ is equivalent to $x_{1}x_{2}+x_{3}x_{4}+\cdots+x_{n-1}x_{n}+x_{n-1}^{2}+ax_{n}^{2}$ with ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(a) = 1.$ By Lemma 2.11, we get the desired result. □

3. Proof of Theorem 1.1

From (1.1), we know that the formula for the number of solutions of $f(X) = 0$ over $\mathbb{F}_{q^2}$ is

$N_{q^2}(f(X) = 0) = q^{2(s-1)}+\sum\limits_{j_1 = 1}^{d_1-1}{\cdots}\sum\limits_{j_s = 1}^{d_s-1}{\overline{\psi _{1}^{j_1}}}\left( a_1 \right) \cdots \overline{\psi _{s}^{j_s}}\left( a_s \right) J_{q^2}^0\left( \psi _{1}^{j_1},\dots ,\psi _{s}^{j_s} \right),$

where $d_{i} = (m_{i}, q^{2}-1)$ and $\psi_i$ is a multiplicative character of $\mathbb{F}_{q^2}$ of order $d_i$ . Since $m_{i}|q+1,$ we have $d_i = m_i.$ Let $H = \{(j_{1}, \dots, j_{s}) \mid 1 \leq j_{i} < m_{i},$ $1 \leq i \leq s\}$ . It follows that $\psi_{1}^{j_{1}} \cdots \psi_{s}^{j_{s}}$ is nontrivial for any $\left(j_{1}, \dots, j_{s}\right) \in H$ as $(m_{i}, m_{j}) = 1$ . By Lemma 2, we have $J_{q^2}^0\left(\psi _{1}^{j_1}, \dots, \psi _{s}^{j_s} \right) = 0$ and hence $N_{q^2}(f(X) = 0) = q^{2(s-1)}.$

Let $N_{q^2}(f(X) = c)$ denote the number of solutions of the equation $f(X) = c$ over $\mathbb{F}_{q^2}$ with $c\in\mathbb{F}_{q^2}^*$ . Let $V = \{(j_1, \ldots, j_s)|0\leq j_i < m_i, \; 1\leq i\leq s\}$ . Then

$\begin{equation*} \begin{aligned} N_{q^2}(f(X) = c)& = \sum\limits_{\gamma_{1}+\cdots+\gamma_{s} = c} N_{q^2}\left(a_{1} x_{1}^{m_{1}} = \gamma_{1}\right) \cdots N_{q^2}\left(a_{s} x_{s}^{m_{s}} = \gamma_{s}\right) \\ & = \sum\limits_{\gamma_{1}+\cdots+\gamma_{s} = c} \sum\limits_{j_{1} = 0}^{m_{1}-1} \psi_{1}^{j_{1}}\left(\frac{\gamma_{1}}{a_{1}}\right) \cdots \sum\limits_{j_{s} = 0}^{m_{s}-1} \psi_{s}^{j_{s}}\left(\frac{\gamma_{s}}{a_{s}}\right). \end{aligned} \end{equation*}$

Since $\psi_i$ is a multiplicative character of $\mathbb{F}_{q^2}$ of order $m_i$ , we have

$\begin{equation*} \begin{aligned} N_{q^2}(f(X) = c) = &\sum\limits_{\frac{\gamma_{1}}{c}+\cdots+\frac{\gamma_{s}}{c} = 1} \sum\limits_{\left(j_{1}, \dots, j_{s}\right) \in V} \psi_{1}^{j_{1}}\left(\frac{\gamma_{1}}{c}\right) \psi_{1}^{j_{1}}\left(\frac{c}{a_{1}}\right) \cdots \psi_{s}^{j_{s}}\left(\frac{\gamma_{s}}{c}\right) \psi_{s}^{j_{s}}\left(\frac{c}{a_{s}}\right) \\ = & \sum\limits_{\left(j_{1}, \dots, j_{s}\right) \in V} \psi_{1}^{j_{1}}\left(\frac{c}{a_{1}}\right) \cdots \psi_{s}^{j_{s}}\left(\frac{c}{a_{s}}\right) \sum\limits_{\frac{\gamma_{1}}{c}+\cdots+\frac{\gamma_{s}}{c} = 1} \psi_{1}^{j_{1}}\left(\frac{\gamma_{1}}{c}\right) \cdots \psi_{s}^{j_{s}}\left(\frac{\gamma_{s}}{c}\right) \\ = & \sum\limits_{\left(j_{1}, \dots, j_{s}\right) \in V} \psi_{1}^{j_{1}}\left(\frac{c}{a_{1}}\right) \cdots \psi_{s}^{j_{s}}\left(\frac{c}{a_{s}}\right) J_{q^2}\left(\psi_{1}^{j_{1}}, \dots, \psi_{s}^{j_{s}}\right) . \end{aligned} \end{equation*}$

By Lemma 2.3,

$\begin{equation*} N_{q^2}(f(X) = c) = q^{2(s-1)}+\sum\limits_{\left(j_{1}, \dots, j_{s}\right) \in H} \psi_{1}^{j_{1}}\left(\frac{c}{a_{1}}\right) \cdots \psi_{s}^{j_{s}}\left(\frac{c}{a_{s}}\right) J_{q^2}\left(\psi_{1}^{j_{1}}, \dots, \psi_{s}^{j_{s}}\right). \end{equation*}$

By Lemma 2.5,

$\begin{equation*} J_{q^2}\left(\psi_{1}^{j_{1}}, \dots, \psi_{s}^{j_{s}}\right) = \frac{G_{q^2}(\psi_{1}^{j_{1}})\cdots G_{q^2}(\psi_{s}^{j_{s}})}{G_{q^2}(\psi_{1}^{j_{1}}\cdots \psi_{s}^{j_{s}})}. \end{equation*}$

Since $m_i|q+1$ and $2\nmid m_i$ , by Lemma 2.6, we have

$G_{q^2}(\psi_{1}^{j_{1}}) = \cdots = G_{q^2}(\psi_{s}^{j_{s}}) = G_{q^2}(\psi_{1}^{j_{1}}\cdots \psi_{s}^{j_{s}}) = q.$

Then

$\begin{align*} &N_{q^2}(f(X) = c)\nonumber \\ = &q^{2(s-1)}+q^{s-1}\sum\limits_{j_1 = 1}^{m_1-1}\psi_{1}^{j_{1}}\left(\frac{c}{a_1}\right)\dots \sum\limits_{j_s = 1}^{m_s-1}\psi_{s}^{j_{s}}\left(\frac{c}{a_s}\right)\nonumber\\ = &q^{2(s-1)}+q^{s-1}\left(\sum\limits_{j_1 = 0}^{m_1-1}\psi_{1}^{j_{1}}\left(\frac{c}{a_1}\right)-1\right)\cdots \left(\sum\limits_{j_s = 0}^{m_s-1}\psi_{s}^{j_{s}}\left(\frac{c}{a_s}\right)-1\right). \end{align*}$

It follows that

$\begin{equation} N_{q^2}(f(X) = c) = q^{2(s-1)}+q^{s-1} \prod\limits_{i = 1}^{s}\left(\left(\frac{c}{a_{i}}\right)_{m_{i}} m_{i}-1\right), \end{equation}$

(3.1)

where

$\left(\frac{c}{a_{i}}\right)_{m_{i}} = \left\{\begin{array}{ll} 1, & {\text{if}}\; \frac{c}{a_{i}} {\text{ is a residue of order}}\; m_{i}, \\ 0, & {\text{otherwise.}} \end{array}\right.$

For a given $h\in \mathbb{F}_{q^2}$ . We discuss the two cases according to whether $h$ is zero or not.

Case 1: $h = 0$ . If $f(X) = 0$ , then $g(Y) = 0$ ; if $f(X)\ne 0$ , then $g(Y)\ne 0$ . Then

$\begin{align} &N_{q^2}(l(X,Y) = 0) \nonumber \\ = &\sum\limits_{c_{1}+c_{2} = 0} N_{q^2}\left(f(X) = c_{1}\right) N_{q^2}\left(g(Y) = c_{2}\right) \\ = &q^{2(s-1)} \left(q^{2\left(n-1\right)}+(-1)^{t+1}(q^{2}-1) q^{n-2}\right)+\sum\limits_{\substack{c_{1}+c_{2} = 0 \\ c_{1}, c_{2} \in \mathbb{F}_{q^2}^{*}}} N_{q^2}\left(f(X) = c_{1}\right) N_{q^2}\left(g(Y) = c_{2}\right). \end{align}$

(3.2)

By Lemma 2.12, (3.1) and (3.2), we have

$\begin{align} &N_{q^2}(l(X,Y) = 0) \\ = &q^{2\left(s+n-2\right)}+(-1)^{t+1}q^{2(s-1)+h_{n}}-(-1)^{t+1}q^{2(s-2)+n}+\sum\limits_{\substack{c_{1} \in \mathbb{F}_{q^2}^{*}}}\left[q^{2\left(s+n-2\right)}-(-1)^{t+1}q^{2(s-2)+n} \right. \\ &\left.+\prod\limits_{\substack{i = 1}}^{s}\left(\left(\frac{c_{1}}{a_{i}}\right)_{m_{i}} m_{i}-1\right)\left(q^{2 n+s-3}-(-1)^{t+1}q^{n+s-3}\right)\right] \\ = &q^{2\left(s+n-2\right)}+(-1)^{t+1}q^{2(s-1)+n}-(-1)^{t+1}q^{2(s-2)+n}+q^{2(s+n-1)}-(-1)^{t+1}q^{2(s-1)+n}-q^{2(s+n-2)} \\ &+(-1)^{t+1}q^{2(s-2)+n}+ \sum\limits_{\substack{c_{1} \in \mathbb{F}_{q^2}^{*}}}\left[\prod\limits_{i = 1}^{s}\left(\left(\frac{c_{1}}{a_{i}}\right)_{m_{i}} m_{i}-1\right)\left(q^{2 n+s-3}-(-1)^{t+1}q^{n+s-3}\right)\right] \\ = &q^{2\left(s+n-1\right)}+\sum\limits_{c_{1} \in \mathbb{F}_{q^2}^{*}}\left[\prod\limits_{i = 1}^{s}\left(\left(\frac{c_{1}}{a_{i}}\right)_{m_{i}}m_{i}-1\right)\left(q^{2 n+s-3}-(-1)^{t+1}q^{n+s-3}\right)\right]. \end{align}$

(3.3)

Case 2: $h\in \mathbb{F}_{q^2}^{*}$ . If $f(X) = h$ , then $g(Y) = 0$ ; if $f(X) = 0$ , then $g(Y) = h$ ; if $f(X)\notin \{0, h\}$ , then $g(Y)\notin \{0, h\}$ . So we have

$\begin{align} &N_{q^2}( l(X,Y)) = h) \\ = &\sum\limits_{c_1+c_2 = h}{N_{q^2}\left( f\left( X \right) = c_1 \right) N_{q^2}\left( g\left( Y \right) = c_2 \right)} \\ = &N_{q^2}(f(X) = 0)N_{q^2}(g(Y) = h)+N_{q^2}(f(X) = h)N_{q^2}(g(Y) = 0) \\ &+\sum\limits_{\substack{c_{1}+c_{2} = h \\ c_{1}, c_{2} \in \mathbb{F}_{q^2}^{*}\backslash \{h\}}}{N_{q^2}\left( f\left( X \right) = c_1 \right) N_{q^2}\left( g\left( Y \right) = c_2 \right)}. \end{align}$

(3.4)

By Lemma 2.12, (3.1) and (3.4),

$\begin{array}{l} N_{q^2}(l(X,Y) = h)\nonumber\\ = 2q^{2\left( s+n-2 \right)}+\left( -1 \right) ^{t+1}q^{2s+n-2}-\left( -1 \right) ^{t+1}2q^{2s+n-4}+ \\\left( q^{s+2n-3}+\left( -1 \right) ^{t+1}\left( q^2-1 \right) q^{s+n-3} \right) \prod\limits_{i = 1}^{s}\left(\left(\frac{h}{a_{i}}\right)_{m_{i}}m_{i}-1\right) \nonumber\\ +\sum\limits_{\substack{c_{1} \in \mathbb{F}_{q^2}^{*}\backslash \{h\}}}\\\left[ q^{2\left( s+n-2 \right)}-\left( -1 \right) ^{t+1}q^{2s+n-4}+\prod\limits_{i = 1}^{s}\left(\left(\frac{c_{1}}{a_{i}}\right)_{m_{i}}m_{i}-1\right)\left( q^{2n+s-3}-\left( -1 \right) ^{t+1}q^{n+s-3} \right) \right]. \end{array}$

It follows that

$\begin{aligned} &N_{q^2}(l(X,Y) = h)\\ = &2q^{2\left( s+n-2 \right)}+\left( -1 \right) ^{t+1}q^{2s+n-2}-\left( -1 \right) ^{t+1}2q^{2s+n-4}+\left( q^{s+2n-3}+\left( -1 \right) ^{t+1}\left( q^2-1 \right) q^{s+n-3} \right) \prod\limits_{i = 1}^{s}\left(\left(\frac{h}{a_{i}}\right)_{m_{i}}m_{i}-1\right) \\ &+\sum\limits_{\substack{ c_{1} \in \mathbb{F}_{q^2}^{*}\backslash \{h\}}}\left[ q^{2\left( s+n-2 \right)}-\left( -1 \right) ^{t+1}q^{2s+n-4}+\prod\limits_{i = 1}^{s}\left(\left(\frac{c_{1}}{a_{i}}\right)_{m_{i}}m_{i}-1\right)( q^{2n+s-3}-( -1 ) ^{t+1}q^{n+s-3} ) \right]\\ = &q^{2\left( s+n-1 \right)}+\left( q^{s+2n-3}+\left( -1 \right) ^{t+1}\left( q^2-1 \right) q^{s+n-3} \right) \prod\limits_{i = 1}^{s}\left(\left(\frac{h}{a_{i}}\right)_{m_{i}}m_{i}-1\right) +\sum\limits_{\substack{ c_{1} \in \mathbb{F}_{q^2}^{*}\backslash \{h\}}}\left[ \prod\limits_{i = 1}^{s}\left(\left(\frac{c_{1}}{a_{i}}\right)_{m_{i}}m_{i}-1\right)\right.\\&\cdot\left.( q^{2n+s-3}+( -1 ) ^{t}q^{n+s-3} ) \right]. \end{aligned}$

(3.5)

By (3.3) and (3.5), we get the desired result. The proof of Theorem 1.1 is complete.□

4. Corollary and examples

There is a direct corollary of Theorem 1.1 and we omit its proof.

Corollary 4.1. Under the conditions of Theorem 1.1, if $a_1 = \cdots = a_s = h\in \mathbb{F}_{q^2}^*$ , then we have

$\begin{equation*} \begin{aligned} &N_{q^2}(l(X,Y) = h)\\ = &q^{2\left( s+n-1 \right)}+\left( q^{s+2n-3}+\left( -1 \right) ^{t+1}\left( q^2-1 \right) q^{s+n-3} \right) \prod\limits_{i = 1}^s{\left( m_i-1 \right) }\\ &+\sum\limits_{\substack{ \gamma \in \mathbb{F}_{q^2}^{*}\backslash \{h\}}}{\left[ \prod\limits_{i = 1}^s{\left( \left( \frac{\gamma}{h} \right)_{m_i} m_i-1 \right)}\left( q^{2n+s-3}+\left( -1 \right) ^{t}q^{n+s-3} \right) \right]}, \end{aligned} \end{equation*}$

where

$\left(\frac{\gamma}{h}\right)_{m_{i}} = \left\{\begin{array}{ll} 1, & if\; \frac{\gamma}{h}\; is\; a \;residue\; of\; order\; m_{i}, \\ 0, & otherwise. \end{array}\right.$

Finally, we give two examples to conclude the paper.

Example 4.2. Let $\mathbb{F}_{2^{10}} = \left < \alpha \right > = \mathbb{F}_2[x]/{(x^{10}+x^3+1)}$ where $\alpha$ is a root of $x^{10}+x^3+1.$ Suppose $l(X, Y) = \alpha^{33}x_{1}^{3}+x_{2}^{11}+y_{3}^{2}+\alpha^{10} y_{4}^{2}+y_{1}y_{2}+y_{3}y_{4}$ . Clearly, ${\mathrm{Tr}}_{\mathbb{F}_{2^{10}}/\mathbb{F}_{2}}(\alpha^{10}) = 1$ , $m_{1} = 3,$ $m_2 = 11$ , $s = 2,$ $n = 4$ , $t = 0,$ $a_{2} = 1$ . By Theorem 1.1, we have

$N_{2^{10}}(l(X,Y) = 0) = 1024^{5}+(32^7+32^3)\times20 = 1126587102265344.$

Example 4.3. Let $\mathbb{F}_{2^{12}} = \left < \beta \right > = \mathbb{F}_2[x]/{(x^{12}+x^6+x^4+x+1)}$ where $\beta$ is a root of $x^{12}+x^6+x^4+x+1.$ Suppose $l(X, Y) = x_{1}^{5}+x_{2}^{13}+y_{3}^{2}+\beta^{10} y_{4}^{2}+y_{1}y_{2}+y_{3}y_{4}$ . Clearly, ${\mathrm{Tr}}_{\mathbb{F}_{2^{12}}/\mathbb{F}_{2}}(\beta^{10}) = 1$ , $m_{1} = 5,$ $m_2 = 13$ , $s = 2,$ $n = 4$ , $t = 0,$ $a_{1} = a_{2} = 1$ . By Corollary 1, we have

$N_{2^{12}}(l(X,Y) = 1) = 2^{5\times12}+(64^7-64^3\times4095)\times48 = 1153132559312355328.$

Acknowledgments

This work was jointly supported by the Natural Science Foundation of Fujian Province, China under Grant No. 2022J02046, Fujian Key Laboratory of Granular Computing and Applications (Minnan Normal University), Institute of Meteorological Big Data-Digital Fujian and Fujian Key Laboratory of Data Science and Statistics.

Conflict of interest

The authors declare there is no conflicts of interest.

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