Zero-watermarking Algorithm for Audio and Video Matching Verification

Wenxue Sun; Huiyuan Zhao; Xiao Zhang; Yuchao Sun; Xiaoxin Liu; Xueling Lv; Di Fan; Wenxue Sun; Huiyuan Zhao; Xiao Zhang; Yuchao Sun; Xiaoxin Liu; Xueling Lv; Di Fan

doi:10.3934/math.2022468

AIMS Mathematics

2022, Volume 7, Issue 5: 8390-8407. doi: 10.3934/math.2022468

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Research article Special Issues

Zero-watermarking Algorithm for Audio and Video Matching Verification

1.
Shandong University of Science and Technology, Qingdao, 266590, Shandong, China
2.
Department of Management and Economics, Tianjin University, Tianjin, 300072, China

Received: 04 November 2021 Revised: 26 January 2022 Accepted: 06 February 2022 Published: 28 February 2022
MSC : 00A69

For the needs of tamper-proof detection and copyright identification of audio and video matching, this paper proposes a zero-watermark algorithm that can be used for audio and video matching verification. The algorithm segments audio and video in smaller time units, generates a video frame feature matrix based on NSCT, DCT, and SVD, and generates a sound watermark based on methods such as DWT and K-means. The zero watermark combines video, audio and copyright information. The experimental results show that the zero watermark generated by this algorithm can not only realize highly accurate matching detection and positioning of audio and video, but also well resist common single attack and combination attacks such as noise, scaling, rotation, frame attack and format conversion, which has good robustness.

Keywords:

Citation: Wenxue Sun, Huiyuan Zhao, Xiao Zhang, Yuchao Sun, Xiaoxin Liu, Xueling Lv, Di Fan. Zero-watermarking Algorithm for Audio and Video Matching Verification[J]. AIMS Mathematics, 2022, 7(5): 8390-8407. doi: 10.3934/math.2022468

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Abstract

1. Introduction

Let $p$ be a prime, $f$ be a polynomial with k variable and $\mathbf{F}_p = \mathbf{Z}/(p)$ be the finite field, where $\mathbf{Z}$ is the integer ring, and let

$N(f;p) = \#\{(x_1, x_2, \cdots, x_k)\in\mathbf{F}_p^k|f(x_1, x_2, \cdots, x_k) = 0\}.$

Many scholars studied the exact formula (including upper bound and lower bound) for $N(f; p)$ for many years, it is one of the main topics in the finite field theory, the most elementary upper bounds was given as follows (see ^[14])

$N(f;p)\leq p^{k-1}deg f.$

Let $ord_p$ denote the p-adic additive valuation normalized such that $ord_pp = 1$ . The famous Chevalley-Warning theorem shows that $ord_pN(f; p) > 0$ if $n > deg f$ . Let [x] denote the least integer $\geq x$ and $e$ denote the extension degree of $\mathbf{F}_q /\mathbf{F}_p$ . Ax (see ^[2]) showed that

$ord_pN(f;q)\geq e\left[\frac{n-deg f}{deg f}\right].$

In 1977, S. Chowla et al. (see ^[7]) investigated a problem about the number of solutions of a equation in finite field $\mathbf{F}_p$ as follow,

$x_1^3+x_2^3+\cdot\cdot\cdot+x_k^3\equiv0,$

where $p$ is a prime with $p\equiv1\bmod\ 3$ and $x_i\in \mathbf{F}_p$ , $1\leq i\leq k$ .

Let $M_k$ denotes the number of solutions of the above equation. They proved that

$\begin{eqnarray} M_3 = p^2+d(p-1), M_4 = p^2+6(p^2-p), \\ \sum\limits_{s = 1}^{\infty}M_sx^s = \frac{x}{1-px}+\frac{x^2(p-1)(2+dx)}{1-3px^2-pdx^3}, \\ \end{eqnarray}$

where $d$ is uniquely determined by $4p = d^2+27y^2$ and $d\equiv1\ mod\ 3$ .

Myerson ^[12] extended the result in ^[2] to the field $\mathbf{F}_q$ and first studied the following equation over $\mathbf{F}_q$ ,

$x_1^3+x_2^3+\cdot\cdot\cdot+x_k^3\equiv0.$

Recently J. Zhao et al. (see ^[17]) investigated the following equations over field $\mathbf{F}_p$ ,

$\begin{eqnarray} &&\ \ \ f_1 = x_1^4+x_2^4+x_3^4, \\ &&f_2 = x_1^4+x_2^4+x_3^4+x_4^4. \end{eqnarray}$

And they give exact value of $N(f_1;p)$ and $N(f_2;p)$ . For more general problem about this issue interested reader can see ^[6,9,10,11].

In this paper, let $A(k, p)$ denotes the number of solutions of the following equation in $\mathbf{F}_p$ ,

$x_1^6+x_2^6+\cdot\cdot\cdot+x_k^6\equiv0,$

where $p$ is a prime with $p\equiv1\bmod\ 3$ and $x_i\in \mathbf{F}_p$ , $1\leq i\leq k$ , and for simplicity, in the rest of this paper, we assume there exists an integer $z$ such that $z^3\equiv2\bmod\ p$ , we use analytic methods to give a recurrence formula for the number of solutions of the above equation. And our method is based on the properties of Gauss sum. It is worth noting that we used a novel method to simplify the steps and avoid a lot of complicated calculations. We proved the following:

Theorem 1. For any positive integer $k\geq1$ , we have the recurrence formula

$\begin{eqnarray} A(k+6, p)& = &5pA(k+4, p)+10dpA(k+3, p)\\&&+(46p^2+5d^2p+dp)A(k+2, p)\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)A(k+1, p)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\ &&+(129d^2+11d+6)p^2+d^4p)A(k, p)\\ &&+p^{k+5}-p^{k+4}-(10dp+2d^2)p^{k+3}-64p^{k+5/2}\\&&-(429+121d+5d^2)p^{k+2}\\ &&-2d^2p^{k+3/2}-(3d^3+130d^2+12d+6)p^{k+1}-d^4p^k, \end{eqnarray}$

with the initial condition

$\begin{eqnarray} &&A(1, p) = 1, A(2, p) = 4(p-1)+p, A(3, p) = 10d(p-1)+p^2, \\ &&A(4, p) = 56p(p-1)+10d^2(p-1)+p^3, \\ &&A(5, p) = 188dp(p-1)+5d^3(p-1)+16dC(p)(p-1)+p^4, \\ &&A(6, p) = p^5+1400p^2(p-1)+(388d^2+8d-576)p(p-1)+d^2p-d^2, \end{eqnarray}$

where $d$ is uniquely determined by $4p = d^2+27y^2$ and $d\equiv1\ mod\ 3$ , and $C(p) = \sum\limits_{a = 1}^{p}e_p(a^3)$ .

Remark. Our method is suitable to calculus the number of solutions of the following equation in $\mathbf{F}_p$ ,

$x_1^t+x_2^t+\cdot\cdot\cdot+x_k^t\equiv0,$

where $p$ satisfied a certain congruence conditions, and $t$ is any nature number.

Our Theorem 2 can be deduced from Theorem 1 and the theory of the Difference equations.

Theorem 2. Let $t_i$ ( $1\leq i\leq k$ ) be the real root of the below equation with multiplicity $s_i$ ( $1\leq i\leq k$ ) respectively, and $\rho_je^{\pm iw_j}$ ( $1\leq j\leq h$ ) be the complex root of the below equation with multiplicity $r_j$ ( $1\leq j\leq h$ ) respectively,

$\begin{eqnarray} x^6& = &5px^4+10dpx^3+(46p^2+5d^2p+dp)x^2\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)x\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\ &&+(129d^2+11d+6)p^2+d^4p). \end{eqnarray}$

We have

$\begin{eqnarray*} A(n, p)& = &\sum\limits_{i = 1}^k\sum\limits_{a = 1}^{s_i}C_{ia}n^{s_i-a}t_i^n+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}D_{jb}n^{r_j-b}\rho_j^n\cos nw_j+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}E_{jb}n^{r_j-b}\rho_j^n\sin nw_j, \end{eqnarray*}$

where $C_{ia}, D_{jb}, E_{jb}$ , are determined by

$\begin{eqnarray} A(6, p)& = &\sum\limits_{i = 1}^k\sum\limits_{a = 1}^{s_i}C_{ia}6^{s_i-a}t_i^6+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}D_{jb}6^{r_j-b}\rho_j^6\cos 6w_j+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}E_{jb}6^{r_j-b}\rho_j^6\sin 6w_j, \\ A(5, p)& = &\sum\limits_{i = 1}^k\sum\limits_{a = 1}^{s_i}C_{ia}5^{s_i-a}t_i^5+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}D_{jb}5^{r_j-b}\rho_j^5\cos 5w_j+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}E_{jb}5^{r_j-b}\rho_j^5\sin 5w_j, \\ A(4, p)& = &\sum\limits_{i = 1}^k\sum\limits_{a = 1}^{s_i}C_{ia}4^{s_i-a}t_i^4+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}D_{jb}4^{r_j-b}\rho_j^4\cos 4w_j+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}E_{jb}4^{r_j-b}\rho_j^4\sin 4w_j, \\ A(3, p)& = &\sum\limits_{i = 1}^k\sum\limits_{a = 1}^{s_i}C_{ia}3^{s_i-a}t_i^3+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}D_{jb}3^{r_j-b}\rho_j^3\cos 3w_j+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}E_{jb}3^{r_j-b}\rho_j^3\sin 3w_j, \\ A(2, p)& = &\sum\limits_{i = 1}^k\sum\limits_{a = 1}^{s_i}C_{ia}2^{s_i-a}t_i^2+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}D_{jb}2^{r_j-b}\rho_j^2\cos 2w_j+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}E_{jb}2^{r_j-b}\rho_j^2\sin 2w_j, \\ A(1, p)& = &\sum\limits_{i = 1}^k\sum\limits_{a = 1}^{s_i}C_{ia}t_i+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}D_{jb}\rho_j\cos w_j+\sum\limits_{j = 1}^{h}\sum\limits_{b = 1}^{r_j}E_{jb}\rho_j\sin w_j.\\ \end{eqnarray}$

(1.1)

2. Some Lemmas

Before we prove these lemmas, we give some notations, $\chi_2$ denotes the second-order character of $\mathbf{F}_p$ , $\chi$ denotes the third-order character of $\mathbf{F}_p$ , $\psi$ denotes the sixth order character of $\mathbf{F}_p$ .

$\begin{eqnarray} e_p(x) = e^{\frac{2\pi ix}{p}}, \tau(\chi) = \sum\limits_{m = 1}^{p}\chi(m)e_p(m), G(\chi, m) = \sum\limits_{a = 1}^{p}\chi(a)e_p(am). \end{eqnarray}$

We call $G(\chi, m)$ the Gauss sum, and we have the following:

$\begin{eqnarray} G(\chi, m) = \tau(\chi)\overline{\chi}(m), (m, p) = 1. \end{eqnarray}$

(2.1)

And also we have

$\begin{eqnarray} |\tau(\chi)| = \sqrt{p}, \end{eqnarray}$

(2.2)

where $\chi$ is a primitive character of $\mathbf{F}_p$ . And let $G(m, 6;p) = \sum\limits_{a = 0}^{p-1}e_p(ma^6)$ . For the property of the exponential sum and the general Gauss sum, interested readers can see ^{[1,4,5,8,13,15]}.

Lemma 1. Let $p$ be a prime with $p\equiv1\bmod\ 3$ . Then for any third-order character $\chi$ of $\mathbf{F}_p$ , we have the identity

$\tau^3(\chi)+\tau^3(\overline{\chi}) = dp,$

where $d$ is uniquely determined by $4p = d^2+27y^2$ and $d\equiv1\bmod\ 3$ .

Proof. For the proof of this lemma see ^[3].

Lemma 2. Let $\chi$ be a third-order character of $\mathbf{F}_p$ with $p\equiv1\bmod\ 3$ , and $C(p) = \tau(\chi)+\tau(\overline{\chi})$ , then $C(p) = \sum\limits_{a = 1}^{p}e_p(a^3)$ .

Proof.

$A = \tau(\chi)+\tau(\overline{\chi}) = \sum\limits_{a = 1}^p(1+\chi(a)+\overline{\chi}(a))e(\frac{a}{p}) = \sum\limits_{a = 1}^pe(\frac{a^3}{p}).$

Lemma 3. Let $p\equiv1\bmod\ 6$ , $2\equiv z^3\bmod\ p$ for some $z$ , and let $\chi$ be a third-order character of $\mathbf{F}_p$ , $\psi$ be a sixth-order character of $\mathbf{F}_p$ , then we have the identity

$\tau(\psi) = \frac{\tau^2(\chi)}{\sqrt{p}}.\\$

Proof. This is Lemma 3 in ^[16].

Lemma 4. As the definition above, we have the identity

$G(m, 6;p) = \sqrt{p}\chi_2(m)+\frac{\overline{x^2}}{\sqrt{p}}\psi(m)+\frac{x^2}{\sqrt{p}}\overline{\psi}(m)+\overline{x}\chi(m)+x\overline{\chi}(m),$

where $(m, p) = 1$ and $x = \tau(\chi)$ .

Proof. Firstly we have the identity

$\begin{eqnarray} 1+\chi_2\left(m\right)+\chi\left(m\right)+\overline{\chi}\left(m\right)+\psi\left(m\right)+\overline{\psi}\left(m\right) = \left\{ \begin{array}{ll} 6, & \text{ if }\ m\equiv a^6\ \text{mod}\ p; \\ 0, & \text{otherwise.}\end{array}\right. \end{eqnarray}$

So we have

$\begin{eqnarray} G(m, 6;p)& = &\sum\limits_{a = 0}^{p-1}(1+\chi_2\left(a\right)+\chi\left(a\right)+\overline{\chi}\left(a\right)+\psi\left(a\right)+\overline{\psi}\left(a\right))e_p(ma)\\ & = &G(\chi_2, m)+G(\psi, m)+G(\overline{\psi}, m)+G(\chi, m)+G(\overline{\chi}, m) \end{eqnarray}$

By (2.1) and Lemma 3, we have

$\begin{eqnarray} G(m, 6;p)& = &\tau(\chi_2)\chi_2(m)+\tau(\overline{\psi})\psi(m)+\tau(\psi)\overline{\psi}(m)+\tau(\overline{\chi})\chi(m)+\tau(\chi)\overline{\chi}(m)\\ & = &\sqrt{p}\chi_2(m)+\frac{\overline{x^2}}{\sqrt{p}}\psi(m)+\frac{x^2}{\sqrt{p}}\overline{\psi}(m)+\overline{x}\chi(m)+x\overline{\chi}(m).\\ \end{eqnarray}$

(2.3)

By (2.3), we complete the proof of our lemma.

Next we let,

$\begin{eqnarray} G^n(m, 6;p) = a_n+b_n\chi_2(m)+c_n\psi(m)+d_n\overline{\psi}(m)+e_n\chi(m)+f_n\overline{\chi}(m). \end{eqnarray}$

(2.4)

Then we have following Lemma 5.

Lemma 5. Let $a_n, b_n, c_n, d_n, e_n, f_n$ are defined as above, then we have that $a_n, b_n, c_n, d_n, e_n, f_n$ are uniquely determined by $n$ , where $n\geq1$ .

Proof. By the orthogonality of characters of $\mathbf{F}_p$ , we have

$\begin{eqnarray} \sum\limits_{a = 1}^{p-1}\chi\left(a\right) = \left\{ \begin{array}{ll} p-1, & \text{ if}\ χ = χ_0; \\ 0, & \text{otherwise.}\end{array}\right. \end{eqnarray}$

(2.5)

By (2.4) and (2.5) we have

$\begin{eqnarray} \sum\limits_{m = 1}^{p-1}G^n(m, 6;p)& = &(p-1)a_n+b_n\sum\limits_{m = 1}^{p-1}\chi_2\left(m\right)+c_n\sum\limits_{m = 1}^{p-1}\psi\left(m\right)+d_n\sum\limits_{m = 1}^{p-1}\overline{\psi}\left(m\right)\\ &&+e_n\sum\limits_{m = 1}^{p-1}\chi\left(m\right)+f_n\sum\limits_{m = 1}^{p-1}\overline{\chi}\left(m\right)\\ & = &(p-1)a_n. \end{eqnarray}$

So we have

$\begin{eqnarray} &&a_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}G^n(m, 6;p). \end{eqnarray}$

(2.6)

By the same method, we have

$\begin{eqnarray} &&b_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\chi_2(m)G^n(m, 6;p), \\ &&c_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\overline{\psi}(m)G^n(m, 6;p), \\ &&d_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\psi(m)G^n(m, 6;p), \\ &&e_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\overline{\chi}(m)G^n(m, 6;p), \\ &&f_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\chi(m)G^n(m, 6;p). \end{eqnarray}$

So now it is easy to see the conclusion of the lemma.

Lemma 6. The sequences $\{a_n\}$ , $\{b_n\}$ , $\{c_n\}$ , $\{d_n\}$ , $\{e_n\}$ , $\{f_n\}$ are defined above, then they satisfied the following recurrence formulae ( $n\geq0$ ):

$\begin{eqnarray} &&a_{n+1} = \sqrt{p}b_n+\frac{\overline{x^2}}{\sqrt{p}}d_n+\frac{x^2}{\sqrt{p}}c_n+xe_n+\overline{x}f_n, \end{eqnarray}$

(2.7)

$\begin{eqnarray} &&b_{n+1} = \sqrt{p}a_n+\frac{\overline{x^2}}{\sqrt{p}}e_n+\frac{x^2}{\sqrt{p}}f_n+xd_n+\overline{x}c_n, \end{eqnarray}$

(2.8)

$\begin{eqnarray} &&c_{n+1} = \sqrt{p}f_n+\frac{\overline{x^2}}{\sqrt{p}}a_n+\frac{x^2}{\sqrt{p}}e_n+xb_n+\overline{x}d_n, \end{eqnarray}$

(2.9)

$\begin{eqnarray} &&d_{n+1} = \sqrt{p}e_n+\frac{\overline{x^2}}{\sqrt{p}}f_n+\frac{x^2}{\sqrt{p}}a_n+xc_n+\overline{x}b_n, \end{eqnarray}$

(2.10)

$\begin{eqnarray} &&e_{n+1} = \sqrt{p}d_n+\frac{\overline{x^2}}{\sqrt{p}}c_n+\frac{x^2}{\sqrt{p}}b_n+xf_n+\overline{x}a_n, \end{eqnarray}$

(2.11)

$\begin{eqnarray} &&f_{n+1} = \sqrt{p}c_n+\frac{\overline{x^2}}{\sqrt{p}}b_n+\frac{x^2}{\sqrt{p}}d_n+xa_n+\overline{x}e_n, \end{eqnarray}$

(2.12)

with the initial condition

$a_0 = 1, b_0 = c_0 = d_0 = e_0 = f_0 = 0.$

Proof. We only prove (2.7), the rest can be proved in the same way. By Lemma 5, we know $a_n$ is unique determined by $n$ . We can compare the coefficient of the equation

$G^{n+1}(m, 6;p) = G^{n}(m, 6;p)G(m, 6;p).$

We have

$a_{n+1} = \sqrt{p}b_n+\frac{\overline{x^2}}{\sqrt{p}}d_n+\frac{x^2}{\sqrt{p}}c_n+xe_n+\overline{x}f_n.$

So we complete the proof of the lemma.

Lemma 7. Let $a_n$ is defined as above, then we have

$\begin{eqnarray} &&a_0 = 1, a_1 = 0, a_2 = 5p, a_3 = 10dp, a_4 = 56p^2+10d^2p, \\&&a_5 = 188dp^2+5d^3p+16dpC(p). \end{eqnarray}$

Proof. By Lemma 4 and after some elementary calculations we have

$\begin{eqnarray} G^2(m, 6;p)& = &5p+2dp^{1/2}\chi_2(m)+4p^{1/2}x\psi(m)+4p^{1/2}\overline{x}\overline{\psi}(m)\\ &&+(p^{-1}\overline{x^4}+3x^2)\chi(m)+(p^{-1}x^4+3\overline{x^2})\overline{\chi}(m), \\ G^3(m, 6;p)& = &10dp+(16p^{3/2}+dp^{1/2})\chi_2(m)+(15p\overline{x}+2dx^2+p^{-1}x^5)\chi(m)\\ &&+(15px+2d\overline{x^2}+p^{-1}\overline{x^5})\overline{\chi}(m)\\ &&+(4p^{-1/2}x^4+12p^{1/2}\overline{x^2}+2dp^{1/2}x)\psi(m)\\ &&+(4p^{-1/2}\overline{x^4}+12p^{1/2}x^2+2dp^{1/2}\overline{x})\overline{\psi}(m), \\ G^4(m, 6;p)& = &60p^2+9d^2p+dp+48dp^{3/2}\chi_2(m)\\ &&+(p^{-2}x^8+17\overline{x^4}+46px^2+16dp)\chi(m)\\ &&+(p^{-2}\overline{x^8}+17x^4+46p\overline{x^2}+16dp)\overline{\chi}(m)\\ &&+(56p^{3/2}x+4dp^{-1/2}x^4+12dp^{1/2}\overline{x^2}+8p^{-1/2}\overline{x^5})\psi(m)\\ &&+(56p^{3/2}\overline{x}+4dp^{-1/2}\overline{x^4}+12dp^{1/2}x^2+8p^{-1/2}x^5)\overline{\psi}(m), \\ G^5(m, 6;p)& = &188dp^2+5d^3p+16dpC(p)\\ &&+(52d^2p^{3/2}+208p^{5/2}+16dp^{1/2}(x^2+\overline{x^2}))\chi_2(m)\\ &&+(p^{-2/5}x^{10}+p^{-3/2}\overline{x^8}+4dp^{-1/2}\overline{x^5}+71p^{1/2}x^4\\&&+(46p^{3/2}+16p^{1/2})x^2\\ &&+(129p^{3/2}+10d^2p^{1/2})\overline{x^2}+60dp^{3/2}x+16dp^{3/2})\psi(m)\\ &&+(p^{-2/5}\overline{x^{10}}+p^{-3/2}x^8+4dp^{-1/2}x^5+71p^{1/2}\overline{x^4}\\&&+(46p^{3/2}+16p^{1/2})\overline{x^2}\\ &&+(129p^{3/2}+10d^2p^{1/2})x^2+60dp^{3/2}\overline{x}+16dp^{3/2})\overline{\psi}(m)\\ &&+(8p^{-1}\overline{x^7}+p^{-1}x^7+25x^5+52dpx^2\\&&+(28dp+46p^2)x+16d\overline{x^4}+112p^2\overline{x})\chi(m)\\ &&+(8p^{-1}x^7+p^{-1}\overline{x^7}+25\overline{x^5}+52dp\overline{x^2}\\&&+(28dp+46p^2)\overline{x}+16dx^4+112p^2x)\overline{\chi}(m), \end{eqnarray}$

and comparing the above formulae with (2.6), we have

$a_0 = 1, a_1 = 0, a_2 = 5p, a_3 = 10dp, a_4 = 60p^2+9d^2p+dp, a_5 = 188dp^2+5d^3p+16dpC(p).\\$

Lemma 8. Let $a_n$ , $b_n$ , $c_n$ , $d_n$ , $e_n$ , $f_n$ are defined as above, then we have

$\begin{eqnarray} a_6& = &5pa_4+10dpa_3+(46p^2+5d^2p+dp)a_2+(2p^2+120dp^2+3d^3p+d^2p+dp)a_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)a_0\\ b_6& = &5pb_4+10dpb_3+(46p^2+5d^2p+dp)b_2+(2p^2+120dp^2+3d^3p+d^2p+dp)b_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)b_0\\ c_6& = &5pc_4+10dpc_3+(46p^2+5d^2p+dp)c_2+(2p^2+120dp^2+3d^3p+d^2p+dp)c_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)c_0\\ d_6& = &5pd_4+10dpd_3+(46p^2+5d^2p+dp)d_2+(2p^2+120dp^2+3d^3p+d^2p+dp)d_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)d_0\\ e_6& = &5pe_4+10dpe_3+(46p^2+5d^2p+dp)e_2+(2p^2+120dp^2+3d^3p+d^2p+dp)e_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)e_0\\ f_6& = &5pf_4+10dpf_3+(46p^2+5d^2p+dp)f_2+(2p^2+120dp^2+3d^3p+d^2p+dp)f_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)f_0 \end{eqnarray}$

Proof. We only proof the first formula, the rest can be proof in the same way. By Lemma 6, we have

$\begin{eqnarray} a_6& = &\sqrt{p}b_5+\frac{\overline{x^2}}{\sqrt{p}}d_5+\frac{x^2}{\sqrt{p}}c_5+xe_5+\overline{x}f_5\\ & = &5pa_4+2dp^{1/2}b_4+4p^{1/2}\overline{x}c_4+4p^{1/2}xd_4\\&&+(3\overline{x^2}+p^{-1}x^4)e_4+(3x^2+p^{-1}\overline{x^4})f_4\\ & = &5pa_4+10dpa_3+(d^2p^{1/2}+12p^{3/2})b_3+(2dp^{1/2}\overline{x}\\&&+8p^{1/2}x^2+p^{-1/2}\overline{x^4})c_3\\ &&+(2dp^{1/2}x+8p^{1/2}\overline{x^2}+p^{-1/2}x^4)d_3+(11px+\overline{x^2}+p^{-1}\overline{x^5})e_3\\ &&+(11p\overline{x}+x^2+p^{-1}x^5)f_3\\ & = &5pa_4+10dpa_3+(46p^2+5d^2p+dp)a_2+(25dp^{3/2}+2p^{3/2})b_2\\ &&+(p^{-3/2}\overline{x^7}+2p^{-1/2}x^5+p^{-1/2}\overline{x^4}+42p^{3/2}\overline{x}+2dp^{1/2}x^2\\&&+(d^2+1)p^{1/2}\overline{x})c_2\\ &&+(p^{-3/2}x^7+2p^{-1/2}\overline{x^5}+p^{-1/2}x^4+42p^{3/2}x+2dp^{1/2}\overline{x^2}\\&&+(d^2+1)p^{1/2}x)d_2\\ &&+(10x^4+(32p+d^2)\overline{x^2}+(4dp+p)x)e_2\\&&+(10\overline{x^4}+(32p+d^2)x^2+(4dp+p)\overline{x})f_2\\ & = &5pa_4+10dpa_3+(46p^2+5d^2p+dp)a_2\\&&+(2p^2+120dp^2+3d^3p+d^2p+dp)a_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p)a_0. \end{eqnarray}$

So we complete the proof of this lemma.

Lemma 9. Let $a_n$ is defined as above, then for any integer $n\geq0$ , we have

$\begin{eqnarray} a_{n+6}& = &5pa_{n+4}+10dpa_{n+3}+(46p^2+5d^2p+dp)a_{n+2}\\&&+(2p^2+120dp^2+3d^3p+d^2p+dp)a_{n+1}\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p)a_n. \end{eqnarray}$

Proof. By (2.4) and Lemma 8, we have

$\begin{eqnarray} G^{6}(m, 6;p)& = &5pG^{4}(m, 6;p)+10dpG^{3}(m, 6;p)+(46p^2+5d^2p+dp)G^{2}(m, 6;p)\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)G(m, 6;p)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p). \end{eqnarray}$

We multiple $G^{n}(m, 6;p)$ to the both side of the above formula, we have

$\begin{eqnarray} G^{n+6}(m, 6;p)& = &5pG^{n+4}(m, 6;p)+10dpG^{n+3}(m, 6;p)\\&&+(46p^2+5d^2p+dp)G^{n+2}(m, 6;p)\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)G^{n+1}(m, 6;p)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p)G^n(m, 6;p). \end{eqnarray}$

By Lemma 5, we can compare the coefficient of the above equation, we have

3. Proof of the theorem 1

In the formula below, we always let $k\geq1$ . By the following formula,

$\begin{eqnarray} \sum\limits_{a = 0}^{p-1}e_p\left(ma\right) = \left\{ \begin{array}{ll} p, & \text{ if }\ p\mid m; \\ 0, & \text{ otherwise, }\end{array}\right. \end{eqnarray}$

we have

$\begin{eqnarray} A(k, p)& = &\frac{1}{p}\sum\limits_{m = 0}^{p-1}\sum\limits_{x_1 = 0, x_2 = 0, \cdot\cdot\cdot, x_k = 0}^{p-1}e_p(m(x_1^6+x_2^6+\cdot\cdot\cdot+x_k^6))\\& = &\frac{1}{p}\sum\limits_{m = 0}^{p-1}G^k(m, 6;p). \end{eqnarray}$

(3.1)

By (8), we have

$\begin{eqnarray} A(k, p)& = &\frac{1}{p}\sum\limits_{m = 0}^{p-1}G^k(m, 6;p)\\ & = &\frac{1}{p}(\sum\limits_{m = 1}^{p-1}G^k(m, 6;p)+p^k)\\ & = &\frac{1}{p}((p-1)a_k+p^k) = \frac{p-1}{p}a_k+p^{k-1}. \end{eqnarray}$

(3.2)

So by Lemma 9, we have

$\begin{eqnarray} A(k+6, p)-p^{k+5}& = &5p(A(k+4, p)-p^{k+3})+10dp(A(k+3, p)-p^{k+2})\\ &&+(46p^2+5d^2p+dp)(A(k+2, p)-p^{k+1})\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)(A(k+1, p)-p^k)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\ &&+(129d^2+11d+6)p^2+d^4p)(A(k, p)-p^{k-1}). \end{eqnarray}$

So we have

And by Lemma 7 and (3.2), we have the initial conditions

$\begin{eqnarray} &&A(1, p) = 1, A(2, p) = 4(p-1)+p, A(3, p) = 10d(p-1)+p^2, \\ &&A(4, p) = 56p(p-1)+10d^2(p-1)+p^3, \\ &&A(5, p) = 188dp(p-1)+5d^3(p-1)+16dC(p)(p-1)+p^4.\\ &&A(6, p) = p^5+1400p^2(p-1)+(388d^2+8d-576)p(p-1)+d^2p-d^2. \end{eqnarray}$

So we complete the proof of the theorem.

4. Conclusion

The main purpose of this paper is using analytic methods to give a recurrence formula of the number of solutions of an equation over finite field. And we give an expression of the number of solutions of the above equation by the root of sixth degree polynomial. We use analytic methods to give a recurrence formula for the number of solutions of the above equation. And our method is based on the properties of the Gauss sum. It is worth noting that we used a novel method to simplify the steps and avoid complicated calculations.

Acknowledgements

The author thanks to referees for very important recommendations and warnings which improved the paper.

Conflict of interest

The author declares that there is no competing interest.

References

[1]	H. Agarwal, F. Husain, Development of payload capacity enhanced robust video watermarking scheme based on symmetry of circle using lifting wavelet transform and SURF, J. Inf. Secur. Appl., 59 (2021), 102846. https://doi.org/10.1016/j.jisa.2021.102846 doi: 10.1016/j.jisa.2021.102846
[2]	S. Xuecheng, L. Zheming, W. Zhe, L. Yongliang, A geometrically robust multi-bit video watermarking algorithm based on 2-D DFT, Multimed. Tools Appl., 80 (2021), 13491-13511. https://doi.org/10.1007/s11042-020-10392-9 doi: 10.1007/s11042-020-10392-9
[3]	G. Sandaruwan, L. Ranathunga, Robust and adaptive watermarking technique for digital images. IEEE International Conference on Industrial & Information Systems. IEEE, 2017, 1-6. https://doi.org/10.1109/ICⅡNFS.2017.8300387
[4]	M. Reza Keyvanpour, N. Khanbani, M. Boreiry, A secure method in digital video watermarking with transform domain algorithms, Multimed. Tools Appl., 80 (2021), 20449-20476. https://doi.org/10.1007/s11042-021-10730-5 doi: 10.1007/s11042-021-10730-5
[5]	D. Ariatmanto, F. Ernawan, An improved robust image watermarking by using different embedding strengths, Multimed. Tools Appl., 79 (2020), 12041-12067. https://doi.org/10.1007/s11042-019-08338-x doi: 10.1007/s11042-019-08338-x
[6]	J. Abraham, V. Paul, An imperceptible spatial domain color image watermarking scheme, J. King Saud Univ.-Comput. Inf. Sci., 31 (2019), 125-133. https://doi.org/10.1016/j.jksuci.2016.12.004 doi: 10.1016/j.jksuci.2016.12.004
[7]	W. Zhou, G. Jiang, M. Yu, F. Shao, Z. Peng, Reduced-reference stereoscopic image quality assessment based on view and disparity zero-watermarks, Signal Process. Image Commun., 29 (2014), 167-176. https://doi.org/10.1016/j.image.2013.10.005 doi: 10.1016/j.image.2013.10.005
[8]	D. Fan, Y. Li, S. Gao, A novel zero watermark optimization algorithm based on Gabor transform and discrete cosine transform, Concurr. Comput. Pract. Exp., 2 (2020). https://doi.org/10.1002/cpe.5689 doi: 10.1002/cpe.5689
[9]	C. Kumar, A. K. Singh, P. Kumar, et al, SPHIT-based multiple image watermarking in NSCT domain, Concurr. Comput. Pract. Exp., 32 (2020), e4912.1-e4912.9. https://doi.org/10.1002/cpe.4912 doi: 10.1002/cpe.4912
[10]	M. Moosazadeh, G. Ekbatanifard, An improved robust image watermarking method using DCT and YCoCg-R color space, Optik, 140 (2017), 975-988. https://doi.org/10.1016/j.ijleo.2017.05.011 doi: 10.1016/j.ijleo.2017.05.011
[11]	C. Zhu, Y. Li, W. Chi, S. Gao, D. Fan, Zero-watermarking algorithm for color image in contourlet domain based on Schur decomposition, Inf. Technol. Inf. Technol., 227 (2019), 94-98.
[12]	D. Wei, Y. Li, Convolution and Multichannel Sampling for the Offset Linear Canonical Transform and Their Applications, IEEE T. Signal PR., 67 (2019), 6009-6024. https://doi.org/10.1109/TSP.2019.2951191 doi: 10.1109/TSP.2019.2951191
[13]	D. Wei, M. Jiang, A fast image encryption algorithm based on parallel compressive sensing and DNA sequence, Optik, 238 (2021), 166748. https://doi.org/10.1016/j.ijleo.2021.166748 doi: 10.1016/j.ijleo.2021.166748
[14]	Y. Li, D. Wei, L. Zhang, Double-encrypted watermarking algorithm based on cosine transform and fractional Fourier transform in invariant wavelet domain, Inf. Sci., 551 (2021), 205-227. https://doi.org/10.1016/j.ins.2020.11.020 doi: 10.1016/j.ins.2020.11.020
[15]	W. Zhou, C. Liu, J. Lei, L. Yu, T. Luo, HFNet: Hierarchical feedback network with multilevel atrous spatial pyramid pooling for RGB-D saliency detection, Neurocomputing, 2021. https://doi.org/10.1016/j.neucom.2021.11.100 doi: 10.1016/j.neucom.2021.11.100
[16]	W. Zhou, J. Liu, J. Lei, L. Yu, J. Hwang, GMNet: Graded-feature multilabel-learning network for RGB-thermal urban scene semantic segmentation, IEEE T. Image Process., 30 (2021), 7790-7802. https://doi.org/10.1109/TIP.2021.3109518 doi: 10.1109/TIP.2021.3109518
[17]	M. Tancik, B. Mildenhall, R. NG, Invisible hyperlinks in physical photographs, Proceedings of the IEEE Computer Society Conference on Computer Vision and Pattern Recognition, 2020, 2117-2126. https://doi.org/10.1109/CVPR42600.2020.00219 doi: 10.1109/CVPR42600.2020.00219
[18]	S. Anguraj, P. S. Shantharajah, E J. Jeba, A steganographic method based on optimized audio embedding technique for secure data communication in the Internet of Things, Comput. Intell., 36 (2019), 557-573. https://doi.org/10.1111/coin.12253 doi: 10.1111/coin.12253
[19]	P. Hu, D. Peng, Z. Yi. Y. Xiang, Robust time-spread echo watermarking using characteristics of host signals, Electron. Lett., 52 (2016), 5-6. https://doi.org/10.1049/el.2015.1508 doi: 10.1049/el.2015.1508
[20]	M. Mosleh, S. Setayeshi, B. Barekatain, M. Mohammad, A novel audio watermarking scheme based on fuzzy inference system in DCT domain, Multimed. Tools Appl., 80 (2021), 20423-20447. https://doi.org/10.1007/s11042-021-10686-6 doi: 10.1007/s11042-021-10686-6
[21]	M. Abdelwahab Khaled, M. Abd El-atty Saied, Wi. El-Shafa, S. El-Rabaie, F. E. Abd El-Samie, Efficient SVD-based audio watermarking technique in FRT domain, Multimed. Tools Appl., 79 (2020), 5617-5648. https://doi.org/10.1007/s11042-019-08023-z doi: 10.1007/s11042-019-08023-z
[22]	H. Karajeh, T. Khatib, L. Rajab, M. Maqableh, A robust digital audio watermarking scheme based on DWT and Schur decomposition, Multimed. Tools Appl., 78 (2019), 18395-18418. https://doi.org/10.1007/s11042-019-7214-3 doi: 10.1007/s11042-019-7214-3
[23]	S. Bhargavi Latha, D. Venkata Reddy, A. Damodaram, Video watermarking using neural networks, Int. J. Inf. Comput. Secur., 14 (2021), 40-59. https://doi.org/10.1504/IJICS.2021.112207 doi: 10.1504/IJICS.2021.112207
[24]	J. Sang, Q. Liu, C. Song, Robust video watermarking using a hybrid DCT-DWT approach, J. Electron. Sci. Technol., 18 (2020), 179-189. https://doi.org/10.1016/j.jnlest.2020.100052 doi: 10.1016/j.jnlest.2020.100052
[25]	Kh. Manglem Singh, A robust rotation resilient video watermarking scheme based on the SIFT, Multimed. Tools Appl., 77 (2018), 16419-16444. https://doi.org/10.1007/s11042-017-5213-9 doi: 10.1007/s11042-017-5213-9
[26]	A. Rakesh, B. Sarabjeet Singh, Video watermarking scheme based on IDR frames using MPEG-2 structure, Int. J. Inf. Comput. Secur., 11 (2019), 585-603. https://doi.org/10.1504/IJICS.2019.103065 doi: 10.1504/IJICS.2019.103065
[27]	C. Li, Y. Yang, K. Liu, L. Tian, A Semi-Fragile video watermarking algorithm based on H.264/AVC, Wirel. Commun. Mob. Comput., 2020. https://doi.org/10.1155/2020/8848553 doi: 10.1155/2020/8848553
[28]	F. Madine, M. A. Akhaee, N. Zarmehi, A multiplicative video watermarking robust to H.264/AVC compression standard, Signal Process. Image Commun., 68 (2018), 229-240. https://doi.org/10.1016/j.image.2018.06.015 doi: 10.1016/j.image.2018.06.015
[29]	S. Gaj, A. Sur, PK. Bora, Prediction mode based H.265/HEVC video watermarking resisting re-compression attack, Multimed. Tools Appl., 79 (2020), 18089-18119. https://doi.org/10.1007/s11042-019-08301-w doi: 10.1007/s11042-019-08301-w
[30]	J. Dittmann, A. Steinmetz, R. Steinmetz, Content-based Digital Signature for Motion Pictures Authentication and Content-Fragile Watermarking, IEEE International Conference of the Multimedia Systems, 1 (1999), 574-579. https://doi.org/10.1109/MMCS.1999.779264 doi: 10.1109/MMCS.1999.779264
[31]	P. Agrawal, A. Khurshid, DWT and GA-PSO Based Novel Watermarking for Videos Using Audio Watermark, International conference on swarm intelligence. ICSI, 2014,212-220. https://doi.org/10.1007/978-3-319-11897-0_25
[32]	M. Sun dararajan, G. Yamuna, CWT and CS algorithm based video watermarking using audio watermark, Procedia Comput. Sci., 87 (2016), 93-98. https://doi.org/10.1016/j.procs.2016.05.132 doi: 10.1016/j.procs.2016.05.132
[33]	X. Wang, Y. Pan, Audio and video cross watermarking algorithm based-on visual saliency model, Electron. Meas. Technol., 40 (2017), 112-115.
[34]	Z. Esmaeilbeig, S. Ghaemmaghami, Compressed Video Watermarking for Authentication and Reconstruction of the Audio Part, 2018 15th International ISC (Iranian Society of Cryptology) Conference on Information Security and Cryptology. ISCISC. 2018, 1-6. https://doi.org/10.1109/ISCISC.2018.8546897
[35]	G. Sucharitha, R. Kumar Senapati, Biomedical image retrieval by using local directional edge binary patterns and Zernike moments, Multimed. Tools Appl., 79 (2020), 1847-1864. https://doi.org/10.1007/s11042-019-08215-7 doi: 10.1007/s11042-019-08215-7
[36]	J. Qu, Image encryption algorithm based on Logistic chaotic scrambling, Science and Technology Innovation, 2020, 2.
[37]	Y. Jiang, M. Cai, C. Song, SIFT based video watermarking algorithm against manifold attacks in contourlet domain, Comput. Simul., 35 (2018), 314-320.
[38]	C. Maiti, B. C. Dhara, Robust non-blind video watermarking using DWT and QR decomposition, Adv. Intel. Syst. Comput., 999 (2020), 333-343. https://doi.org/10.1007/978-981-13-9042-5_28 doi: 10.1007/978-981-13-9042-5_28
[39]	R. Singh, A. Ashok, M. Saraswat, Robust Video Watermarking in Frequency Domain for Copyright Protection, ACM International Conference Proceeding Series, AICPS, 2021,174-178. https://doi.org/10.1145/3474124.3474148

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