Computational methods for recognition of cancer protein markers in saliva

Ying Sun; Wei Du; Lili Yang; Min Dai; Ziying Dou; Yuxiang Wang; Jining Liu; Gang Zheng; Ying Sun; Wei Du; Lili Yang; Min Dai; Ziying Dou; Yuxiang Wang; Jining Liu; Gang Zheng

doi:10.3934/mbe.2020134

Mathematical Biosciences and Engineering

2020, Volume 17, Issue 3: 2453-2469. doi: 10.3934/mbe.2020134

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Research article Special Issues

Computational methods for recognition of cancer protein markers in saliva

1.
Tianjin Key Laboratory of Intelligence Computing and Novel Software Technology, School of Computer Science and Engineering, Tianjin University of Technology, Tianjin 300384, China
2.
Information Technology Research Base of Civil Aviation Administration of China, Civil Aviation University of China, Tianjin 300300, China
3.
Key Laboratory of Symbol Computation and Knowledge Engineering of Ministry of Education, College of Computer Science and Technology, Jilin University, Changchun 130012, China
4.
Department of Obstetrics, The First Hospital of Jilin University, Changchun 130012, China

Received: 28 February 2018 Accepted: 17 February 2019 Published: 25 February 2020

In recent years, many studies have supported that cancer tissues can make disease-specific changes in some salivary proteins through some mediators in the pathogenesis of systemic diseases. These salivary proteins have the potential to become cancer-specific biomarkers in the early diagnosis stage. How to effectively identify these potential markers is one of the challenging issues. In this paper, we propose novel machine learning methods for recognition cancer biomarkers in saliva by two stages. In the first stage, salivary secreted proteins are recognized which are considered as candidate biomarkers of cancers. We picked up 557 salivary secretory proteins from 20379 human proteins by public databases and published literatures. Then, we present a training set construction strategy to solve the imbalance problem in order to make the classification methods get better accuracy. From all human protein set, the proteins belonging to the same families as salivary secretory proteins are removed. After that, we use SVC-KM method to cluster the remaining proteins, and select negative samples from each cluster in proportion. Next, the features of proteins are calculated by tools. We collect 24 protein properties such as sequence, structure and physicochemical properties, a total of 1087 features. An innovative procedure based on the local samples is proposed for selecting the appropriate features, in order to further improve the performance of SVM classifier. Experimental results show that the average sensitivity, specificity and accuracy of salivary secretory protein recognition using selected 32 features in training set are 97.09%, 98.10%, 97.61%, respectively. The use of these methods can improve the accuracy of recognition by solving the problems of unbalanced sample size and uneven distribution in training set. In the second stage, we apply the best model to dig out the salivary secreted proteins from 58 reported cancer markers, and get a total of 42 proteins which are considered to be used for salivary diagnosis. We analyze the gene expression data of three types of cancer, and predict that 33 genes will appear in saliva after they are translated into proteins. This study provides an important computational tool to help biologists and researchers reduce the number of candidate proteins and the cost of research. So as to further accelerate the discovery of cancer biomarkers in saliva and promote the development of saliva diagnosis.

Keywords:

salivary secretory protein,
cancer biomarker,
feature selection based on local samples,
SVC-KM,
computational methods

Citation: Ying Sun, Wei Du, Lili Yang, Min Dai, Ziying Dou, Yuxiang Wang, Jining Liu, Gang Zheng. Computational methods for recognition of cancer protein markers in saliva[J]. Mathematical Biosciences and Engineering, 2020, 17(3): 2453-2469. doi: 10.3934/mbe.2020134

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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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