A Thue equation is a Diophantine equation of the form

$ F(x, y)=k, $

where $F\in Z[x, y]$ is an irreducible binary form of degree $n\geq 3$ and $k$ is a non-zero rational integer. In 1909, Thue [1] proved that the thue equation has only finitely many solutions, however Thue didn't give a complete method. In 1986 Baker [2] used the theory of linear form in logarithms of algebraic numbers to solve the problem, in particular, he gave an effective upper bound for the solutions of Thue equation. In recent years, various families of Thue equations were studied (see [3-14]).

In 2012, Xia, Chen, Zhang[14] presented a new and simpler method to approximate certain algebraic numbers. Applying the method, authors derived an effective upper bound for the solutions $(x, y)$ of the two-parametric family of quartic Thue equation

$ tx^4-4sx^3y-6tx^2y^2+4sxy^3+ty^4=N $

for $s>32t^3$. The purpose of this research is to extend the method to solve the two-parametric family of sextic Thue equation

$ \begin{eqnarray} F_n(x, y)&=&x^6 - 2A_n x^5 y - 5(A_n + 3)x^4y^2 - 20x^3y^3 + 5A_nx^2y^4 + 2(A_n + 3)xy^5 + y^6\nonumber\\ &=&\pm 1, \end{eqnarray} $

(1.1)

where $A_n=54n^3+81n^2+54n+12$, $n>6.02\times 10^6$. In 1990, Alan Togbé proved that equation (1.1) has only trivial solutions, where $|n|\leq2.03\times10^6$. We extend the result of Alan Togbé, and consider n is any integer.

In this paper, we will show

Theorem 1   For any $n\in Z$, equation(1.1) has only trivial solutions,

$ (x, y)=(1, 0), (-1, 0), (0, 1), (0, -1), (1, -1), (-1, 1). $

Before the proof of the theorem, some lemmas are needed.

Lemma 2.1   (see [14]) Let $p_n(x)=\sum\limits_{k=0}^{n}\binom{n-\alpha}{n-k}\binom{n+\alpha}{k}x^k$, $q_n(x)=\sum\limits_{k=0}^{n}\binom{n-\alpha}{k}\binom{n+\alpha}{n-k}x^k$ and $R_n(x)=x^\alpha q_n(x)-p_n(x), $ where $n$ is a positive integer and $\alpha$ is a real number, then we have

(ⅰ) $q_n(x)=\sum\limits_{k=0}^{n}(-1)^{k}\binom{2n-k}{n-k}\binom{n-\alpha}{k}(1-x)^k; $

(ⅱ) Let$ x=w=\frac{si-t}{si+t}=e^{i\varphi}, 0 < \varphi < \frac{\pi}{2}, 0 < \alpha < 1, $ then $|q_n(w)|\leq4|1+\sqrt{w}|^{2(n-1)}; $

(ⅲ) Let $x=w=\frac{si-t}{si+t}=e^{i\varphi}, 0 < \varphi < \frac{\pi}{2}, 0 < \alpha < 1, $ then $|q_n(w)|\leq4|1+\sqrt{w}|^{2(n-1)}.$

Lemma 2.2   If $\alpha=1/6$, then $12^k3^{\lfloor k/2 \rfloor }\binom{n-\alpha}{k}$ is rational integer.

Proof  Let $k!=2^{s_2}3^{s_3}M$ with $(6, M)=1$. Since

$ \binom{n-\alpha}{k}=\frac{(6n-1)(6(n-1)-1)\cdots(6(n-k+1)-1)}{6^kk!}, $

then there exists $t$ such that $6t\equiv1({\rm mod}~ M)$ and

$ \begin{eqnarray*} && t^k(6n-1)(6(n-1)-1)...(6(n-k+1)-1)\\ &\equiv&(n-t)\cdots(n-k+1-t)\\ &\equiv&\binom{n-t}{k}k! \equiv 0~({\rm mod}~ M). \end{eqnarray*} $

Since $(t, M)=1$, we obtain

$ M|(6n-1)\cdots(6(n-k+1)-1). $

While

$ s_2=\lfloor\frac{k}{2}\rfloor+\lfloor \frac{k}{4}\rfloor+\lfloor\frac{k}{8}\rfloor+\cdots<k, $

and

$ s_3=\lfloor\frac{k}{3}\rfloor+\lfloor \frac{k}{3^2}\rfloor+\lfloor\frac{k}{3^3}\rfloor+\cdots<\lfloor \frac{k}{2} \rfloor. $

The lemma therefore follows:

Lemma 2.3    (see [11]) Let $\theta$ be an algebraic number. Suppose that there exists $k_0>0, l_0, Q>1, E>1 $ such that for all $n$ there are rational integers $P_n$ and $Q_n$ with $|Q_n| < k_0Q^n$ and $|Q_n\theta-P_n|\leq l_0E^{-n}$ and suppose further that $P_nQ_{n+1} \neq Q_nP_{n+1}$. Then, for any rational integers $x$ and $y$, $y\geq e/(2l_0)$, we have

$ |\theta y-x|>\frac{1}{cy^\lambda}, ~\textrm{where} ~c=2k_0Q(2l_0E)^\lambda, ~\lambda=\frac{\log Q}{\log E}. $

Consider the equation

$ \begin{eqnarray} f(x, y)&=&x^6-\frac{6A}{B}x^5y+(\frac{15A}{B}-15)x^4y^2+20x^3y^3- \frac{15A}{B}x^2y^4+(\frac{6A}{B}-6)xy^5 +y^6 \nonumber\\ &=&\pm 1, \end{eqnarray} $

(3.1)

where $A>0$, $A, B\in Z$. If we set $B=-n, A=-A_nB/3=18n^4+27n^3+18n^2+4n>0$, we have$f(x, y)=F_n(x, -y)$. Hence the solution of (1.1) could be deduced from the solution of (3.1).

Denote by $\theta$ as the root of $f(x, 1)=0$, straightforward computation shows that $\theta$ satisfies

$ \begin{eqnarray} (\frac{\theta+\rho}{\theta+\bar{\rho}})^6=\frac{A+B\bar{\rho}}{A+B\rho}, \end{eqnarray} $

(3.2)

where $\rho=-\frac{1}{2}+\frac{\sqrt{-3}}{2}$, $\bar{\rho}=-\frac{1}{2}-\frac{\sqrt{-3}}{2}$, and $-$ mean conjugate in $Z[\sqrt{-3}]$, so that $\overline{a+b\rho}=a+b\bar{\rho}$. Since$-3\equiv1(\mod 4)$, we know that $\{1, \frac{1+\sqrt {-3}}{2} \}$ are integral basis in $Z[\sqrt{-3}]$. Therefore, $\rho$, $\bar{\rho}$ are algebraic interger in $Z[\sqrt{-3}]$.

Putting $z=A+B\bar{\rho}, u=A+B\rho, w=z/u=e^{i\varphi}$, we have

$ \theta_k=\frac{\rho-\bar{\rho}w^{ \frac{1}{6} }e^{\frac{2k \pi }{6}}}{w^{\frac{1}{6}}e^{\frac{2k \pi }{6}}-1} =\frac{\rho e^{\frac{-k \pi }{6}}-\bar{\rho}w^{ \frac{1}{6} }e^{\frac{k \pi }{6}}}{w^{\frac{1}{6}}e^{\frac{k \pi }{6}}-e^{\frac{-k \pi }{6}}}, ~~k=0, 1, \cdots, 5. $

Note that $e^{\frac{-\pi }{6}}=i\bar{\rho}$, $e^{\frac{\pi }{6}}=-i\rho$, $e^{\frac{-\pi }{3}}=-\rho$, $e^{\frac{\pi }{3}}=-\bar{\rho}$, one can check that $\theta_k$ satisfies $\theta_k=\frac{v_k+\bar{v_k} w^{\frac{1}{6}}} {r_k w^{\frac{1}{6}}+\bar{r_k}}, $ where

$ \begin{equation} v_k=\left\{ \begin{array}{ll} \bar{\rho}-1, & k=0; \\ 1, & k=1; \\ \bar{\rho}+2, & k=2; \\ \rho, & k=3; \\ 2\rho+1, & k=4; \\ \rho+1, & k=5, \end{array} \right. ~~r_k=\left\{ \begin{array}{ll} 2\rho+1, & k=0; \\ -\rho, & k=1; \\ \bar{\rho}+2, & k=2; \\ -1, & k=3; \\ \bar{\rho}-1, & k=4; \\ \bar{\rho}, & k=5. \end{array} \right. \end{equation} $

(3.3)

Actually, this property of $\theta$ make possible for future proof. Furthermore, one can give explicit estimation for the roots as the lemma below.

Lemma 3.1   Denote by $\theta_k~(k=0, 1, \cdots, 5)$ as the root of $f(x, 1)=0$, then it satisfies

$ \begin{eqnarray} &&1+\frac{1}{2A_n}-\frac{5}{8A_n^2}-\frac{5}{16A_n^3}+\frac{9}{10A_n^4} \leq \theta_0 \leq 1+\frac{1}{2A_n}-\frac{5}{8A_n^2}-\frac{5}{16A_n^3}+\frac{1}{A_n^4}, \nonumber\\ && \frac{1}{2A_n}-\frac{7}{8A_n^2}+\frac{7}{16A_n^3}+\frac{89}{32A_n^4} \leq \theta_1 \leq \frac{1}{2A_n}-\frac{7}{8A_n^2}+\frac{7}{16A_n^3}+\frac{179}{64A_n^4}, \nonumber\\ && -1+\frac{3}{2A_n}-\frac{27}{8A_n^2}+\frac{69}{16A_n^3}+\frac{169}{8A_n^4} \leq \theta_2 \leq -1+\frac{3}{2A_n}-\frac{27}{8A_n^2}+\frac{69}{16A_n^3}+\frac{85}{4A_n^4}, \nonumber\\ && \frac{1}{2}+\frac{3}{8A_n}-\frac{9}{16A_n^2}+\frac{3}{128A_n^3}+\frac{155}{64A_n^4} \leq \theta_3 \leq \frac{1}{2}+\frac{3}{8A_n}-\frac{9}{16A_n^2}+\frac{3}{128A_n^3}+\frac{311}{128A_n^4}, \nonumber\\ && -2A_n-\frac{5}{2}-\frac{35}{8A_n}-\frac{105}{16A_n^2}-\frac{65}{32A_n^3} \leq \theta_4 \leq -2A_n-\frac{5}{2}-\frac{35}{8A_n}-\frac{105}{16A_n^2}-\frac{129}{64A_n^3}, \nonumber\\ && 2+\frac{3}{2A_n}-\frac{9}{8A_n^2}-\frac{39}{16A_n^3}+\frac{589}{64A_n^4} \leq \theta_5 \leq 2+\frac{3}{2A_n}-\frac{9}{8A_n^2}-\frac{39}{16A_n^3}+\frac{295}{32A_n^4}. \nonumber \end{eqnarray} $

Proof   One can check that $f(x, 1)$ change the sign between two formula in the inequalities.

Now we will give an effective approximation of $\theta$. Since

$ \begin{eqnarray*} \theta_k&=& \frac{v_k+\bar{v_k}w^\frac{1}{6}}{r_kw^\frac{1}{6}+\bar{r_k}}\\ &=&\frac{v_kq_n(w)+\bar{v_k}w^\frac{1}{6}q_n(w)}{r_kw^\frac{1}{6}q_n(w)+\bar{r_k}q_n(w)}\\ &=& \frac{v_kq_n(w)+\bar{v_k}p_n(w)+\bar{v_k}R_n(w)}{r_kp_n(w)+\bar{r_k}q_n(w)+r_kR_n(w)}\\ & = & \frac{S+\bar {v_k}R_n(w)}{T+r_kR_n(w)}, \end{eqnarray*} $

putting $S=v_kq_n(w)+\bar{v_k}p_n(w), ~T=r_kp_n(w)+\bar{r_k}q_n(w)$, we have

$ |T\theta_k-S|=\bar{v_k}R_n(w)-\theta R_n(w). $

Note that $1-w=1-\frac{A+B\bar{\rho}}{A+B\rho}=\frac{B\sqrt 3 i}{u}$, from Lemma 2.1, we have

$ \begin{eqnarray} q_n(w)=\sum\limits_{k=0}^{n}(-1)^{k}\binom{2n-k}{n-k}\binom{n-1/6}{k}(\frac{B\sqrt 3 i}{u})^k. \end{eqnarray} $

(3.4)

From Lemma 2.2, we get

$ \begin{eqnarray} 12^k3^{\lfloor \frac{k}{2} \rfloor }\binom{n-\frac{1}{6}}{k}\in Z. \end{eqnarray} $

(3.5)

Hence, if denote $Q_n=12^nu^nT$ and $P_n=12^nu^nS$, it's easy to prove that both are algebraic integer in $Q[\sqrt{-3}]$. Since $\bar{u}=z$, $\overline {u^np_n(w)}=u^nq_n(w)$. One can prove $\overline {Q_n}=Q_n$ and $\overline {P_n}=P_n$. Therefore we have $Q_n$, $P_n\in Z$. In another word, we get two arrays of rational integers $Q_n$, $P_n$ such that

$ |Q_n\theta_k-P_n|=|12^nu^nR_n(w)(\bar{v}-\theta r_k)|= R_n. $

In the following, we give estimation of upper bound for $Q_n, R_n$.

From Lemma 2.1, we have

$ |p_n(w)|=|q_n(w)| \leq 4|1+\sqrt{w}|^{2n-2} $

and $|R_n(w)| \leq \frac{\varphi}{3}|1-\sqrt{w}|^{2n}.$ So, for $A_n=54n^3+81n^2+54n+12$, and $|n| \geq 2.03\times10^6$, if denote $\varepsilon =2A-B+2\sqrt{A^2-AB+B^2}$, we obtain

$ \begin{eqnarray*} |Q_n|&&=&&|12^nu^n(r_kp_n(w)+\overline{r_k}q_n(w))|\\ &&\leq &&\frac{|6r_k|}{|1+\sqrt w|^2}|12u(1+\sqrt w)^2|^n\\ &&=&& \frac{|6r_k|}{|1+\sqrt w|^2}|12(2A-B+2\sqrt{A^2-AB+B^2})|^n\\ && = &&C_Q |12\varepsilon|^n, \end{eqnarray*} $

and

$ \begin{eqnarray*} |R_n|&&=&&|12^nu^nR_n(w)(\bar{v}-\theta r)|\\ &&\leq && \frac{|\overline{v_k}-\theta r_k|\cdot|\varphi|}{6}|12u(1-\sqrt w)^2|^n\\ &&=&& \frac{|\overline{v_k}-\theta r_k|\cdot|\varphi|}{6}|2A-B-2\sqrt{A^2-AB+B^2}|^n\\ &&= && C_R|\frac{\varepsilon}{36B^2}|^{-n}, \end{eqnarray*} $

where $C_Q=\frac{|6r_k|}{|1+\sqrt w|^2}$, $C_R=\frac{|\overline{v_k}-\theta r_k|\cdot|\varphi|}{6}$.

Now we will give estimation of $C_Q, C_R$. From(3.3) we have

$ \begin{eqnarray} \max\limits_{k=0, 1, \cdots, 5} {|r_k|}=\max\limits_{k=0, 1, \cdots, 5}{|v_k|}=3. \end{eqnarray} $

(3.6)

From definition, we have $w=\frac{A+B\bar{\rho}}{A+B\rho}=\frac{A-\frac{B}{2}-\frac{\sqrt 3}{2}Bi }{A-\frac{B}{2}+\frac{\sqrt 3}{2}Bi}=\frac{(A-\frac{B}{2}-\frac{\sqrt 3}{2}Bi)^2}{A^2-AB+B^2}$. Hence we get

$ |1+\sqrt w|^2=| 1+\frac{A-\frac{B}{2}-\frac{\sqrt 3}{2}Bi}{ \sqrt{A^2-AB+B^2} } |^2=2+2\frac{A-\frac{B}{2}}{\sqrt{A^2-AB+B^2}}. $

Note that $B=-n, ~A=18B^4-27B^3+18B^2-4B$ and $|B|>2.03\times10^6$, straightforward computation shows that

$ \begin{eqnarray} |1+\sqrt w|^2>3.999. \end{eqnarray} $

(3.7)

From (3.6) and (3.7), we have

$ \begin{eqnarray} C_Q=\frac{|6r_k|}{|1+\sqrt w|^2}&<\frac{18}{3.999}&<4.502. \end{eqnarray} $

(3.8)

On the other hand, from definition we have $e^{i\varphi}=w=\frac{(A-\frac{B}{2}-\frac{\sqrt 3}{2}Bi)^2}{A^2-AB+B^2}=\frac{A^2-AB-\frac{B^2}{2}}{A^2-AB+B^2}-\frac{\sqrt 3 B(A-\frac{B}{2})}{A^2-AB+B^2}$. So we have

$ \begin{eqnarray} |\varphi|\leq |\tan \varphi| =|\frac{\sqrt 3 B(A-\frac{B}{2})}{\frac{B^2}{2}} |&<0.0236. \end{eqnarray} $

(3.9)

If $\theta_k~(k=0, 1, \cdots, 5)$ is denoted as in Lemma 3.1, one can get estimation of root $\theta_k$:

$ \begin{eqnarray} \left\{ \begin{array}{ll} |\theta_k|&&<3, && \hbox{if~}k\neq 4, \nonumber\\ |\theta_k|&&<2.01|A_n|, && \hbox{if~}k=4. \end{array} \right. \end{eqnarray} $

Since $\varepsilon =2A-B+2\sqrt{A^2-AB+B^2} < 4.1A < 1.367|A_nB|$, from (3.6) and (3.9), we can get estimate of $C_R$.

For $\theta_k~~(k\neq4)$, we have

$ {C_R} = \frac{{|\overline {{v_k}} - \theta {r_k}| \cdot |\varphi |}}{6} < \frac{{3 + 9}}{6} \cdot 0.0236 < 0.0472. $

(3.10)

For $\theta_k~~(k=4)$, we also have

$ {C_R} = \frac{{|\overline {{v_k}} - \theta {r_k}| \cdot |\varphi |}}{6} < \frac{{3 + 3.01A}}{6} \cdot 0.0236 < 0.0357|{A_n}|. $

(3.11)

Now putting $Q=12\varepsilon, ~E=\frac{\varepsilon}{36B^2}$, $k_0=4.502$, then for $\theta_k$, where $k\neq4, k=4$, we set $l_0=0.0472$ or $l_0=0.0357|A_n|$ separately. From Lemma 2.2, computing

$ 2k_0Q(2l_0E)^{\lambda}=2\cdot4.502\cdot12\varepsilon (2\cdot0.0472\frac{\varepsilon}{36B^2})^{\lambda}=108.048\varepsilon(0.00263\frac{\varepsilon}{B^2})^{\lambda}, $

and

$ 2k_0Q(2l_0E)^{\lambda}=2\cdot4.502\cdot12\varepsilon(2\cdot0.0357\frac{|A_n|\varepsilon}{36B^2})=108.048\varepsilon (0.00199\frac{|A_n|\varepsilon}{B^2}), $

we have approximation to algebraic numbers as in followed lemma.

Lemma 3.2   If $\theta_k~(k=0, 1, \cdots, 5)$ is the root of

$ x^6-\frac{6A}{B}x^5+(\frac{15A}{B}-15)x^4+20x^3- \frac{15A}{B}x^2+(\frac{6A}{B}-6)x +1=0, $

then for any $x, y \in Z$, we have

$ |x-y\theta_k|>\frac{1}{c_ky^{\lambda}}, $

where

$ \begin{eqnarray*} \left\{ \begin{array}{ll} c_0=c_1=c_2=c_3=c_5=108.048\varepsilon(0.00263\frac{\varepsilon}{B^2})^{\lambda}, && \hbox{} \\ c_4=108.048\varepsilon(0.00199\frac{|A_n|\varepsilon}{B^2})^{\lambda}, && \hbox{} \\ \lambda=\frac{\log(12\varepsilon)}{\log{36B^2}-\log \varepsilon}. && \hbox{} \end{array} \right. \end{eqnarray*} $

Now we can prove the theorem. First, from Lemma 3.1, we have

$ \begin{eqnarray} &&&&\prod\limits_{j\neq0}|\theta_j-\theta_0|&<1.1\times2.1\times0.6\times2.01|A_n| \times1.1&<3.065|A_n|, \nonumber\\ &&&&\prod\limits_{j\neq0}|\theta_j-\theta_1|&<1.1\times1.1\times0.6\times2.01|A_n| \times2.1&<3.065|A_n|, \nonumber\\ &&&&\prod\limits_{j\neq0}|\theta_j-\theta_2|&<2.1\times1.1\times1.6\times2.01|A_n| \times3.1&<23.03|A_n|, \nonumber\\ &&&&\prod\limits_{j\neq0}|\theta_j-\theta_3|&<0.6\times0.6\times1.6\times2.01|A_n|\nonumber \times1.6&<1.853|A_n|, \\ &&&&\prod\limits_{j\neq0}|\theta_j-\theta_4|&<(2.01|A_n|)^5&<32.81|A_n|^5, \nonumber\nonumber\\ &&&&\prod\limits_{j\neq0}|\theta_j-\theta_5|&<1.1\times2.1\times3.1\times2.01|A_n| \times1.6&<23.03|A_n|. \end{eqnarray} $

(3.12)

Second, computation shows that when $n=|B|>2.03\times10^6$, we get $\lambda < 2.18$. It's easy to get

$ \begin{eqnarray*} |x-y\theta_k|&<\frac{2^5}{|y|^5\prod\limits_{j\neq k}|\theta_j-\theta_k|}. \end{eqnarray*} $

Hence

$ \begin{eqnarray} \textrm{If}~ k\neq4, ~~|x-y\theta_k|<\frac{2^5}{23.03|A_n||y|^5}, \\ \end{eqnarray} $

(3.13)

$ \begin{eqnarray} \textrm{If}~ k=4, ~~|x-y\theta_k|<\frac{2^5}{32.81|A_n|^5|y|^5}. \end{eqnarray} $

(3.14)

So we get an upper bound of $|x-\theta y|$. From Lemma 3.2 and (3.13), for $\theta_k~~(k\neq4)$, we have

$ \frac{1}{108.048\varepsilon(0.00263\frac{\varepsilon}{B^2})^{\lambda} |y|^{\lambda}}<|x-y\theta_k| < \frac{2^5}{23.03|A_n||y|^5}. $

Note that $\varepsilon < 4.1A < 1.367|A_nB|$, so if $(x, y)$ is type $k$ solution of (3.1), where $(k\neq4)$, we have

$ \begin{eqnarray} |y|<0.012^\frac{1}{5-\lambda}|A_n|^{\frac{\lambda}{5-\lambda}}|B|^{\frac{1-\lambda}{5-\lambda}} <0.01|A_n|^{0.7735}|B|^{-0.4185}. \end{eqnarray} $

(3.15)

For $\theta_4$, we have

$ \frac{1}{ 108.048\varepsilon(0.00199\frac{|A_n|\varepsilon}{B^2})^{\lambda} }<|x-y\theta_k| < \frac{2^5}{32.81|A_n|^5|y|^5}. $

So if $(x, y)$ is type $ 4$ solution of (3.1), we have

$ \begin{eqnarray} |y|<0.012^\frac{1}{5-\lambda}|A_n|^{\frac{\lambda}{5-\lambda}}|B|^{\frac{1-\lambda}{5-\lambda}} <0.0606|A_n|^{0.12766}|B|^{-1.19149}. \end{eqnarray} $

(3.16)

This is an upper bound for $|y|$. From well-known result in number theory, we know that when $|y|>1$, $x, y$ is partial quotient of $\theta$. In the following, we only need to verify whether $(p_n, q_n)$ is solution of (3.1) or not.

From Lemma 3.1, computation shows the continued fraction expansion of $\theta_k~~(k\neq4)$, the result is listed as below:

$ \begin{eqnarray*} &&\theta_0=[1, 2A_n+2, 1, 1, \lfloor \frac{2A_n}{25}\rfloor, \cdots], ~~\{ \frac{p_i}{q_i} \}=\{ \frac{1}{1}, \frac{2A_n+3}{2A_n+2}, \frac{2An+4}{2A_n+3}, \cdots\}, \\ &&\theta_1=[0, -2A_n-3, 1, 1, \lfloor \frac{4A_n}{70}\rfloor, \cdots], ~~ \{ \frac{p_i}{q_i} \}=\{ \frac{0}{1}, \frac{1}{-2A_n+3}, \frac{1}{-2A_n-2}, \cdots\}, \\ &&\theta_2=[-1, 2B_n+1, 1, 1, \lfloor \frac{B_n}{2}\rfloor, \cdots], ~~ \{ \frac{p_i}{q_i} \}=\{ \frac{-1}{1}, \frac{-2B_n}{2B_n+1}, \frac{-2B_n-1}{2B_n+2}, \cdots\}, \\ &&\theta_3=[0, 1, 1, 2B_n, 1, 1, \lfloor \frac{B_n}{2}\rfloor, \cdots], ~~ \{ \frac{p_i}{q_i} \}=\{ \frac{0}{1}, \frac{1}{1}, \frac{1}{2}, \frac{2B_n+1}{4B_n+1} \cdots\}, \\ &&\theta_5=[2, 2B_n, 1, 1, \lfloor \frac{B_n}{2}\rfloor, \cdots], ~~ \{ \frac{p_i}{q_i} \}=\{ \frac{2}{1}, \frac{4B_n+1}{2B_n}, \frac{4B_n+3}{2B_n+1}, \cdots\}, \end{eqnarray*} $

where $B_n=\frac{A_n}{3}=18n^3+27n^2+18n+4$.

One can observe that $q_2$ or $q_4$ has exceeded the upper bound of $|y|$, $0.01|A_n|^{0.7735}|B|^{-0.4185} $. Straight forward computation shows that it only exists trivial solution $\pm(x, y)=(0, 1), (1, -1)$.

One can also get the continued fraction expansion of $\theta_k~~(k\neq4)$ as below

$ \begin{eqnarray*} &&\theta_4=[-2A_n-3, 2, \lfloor \frac{2A_n}{35}\rfloor, \cdots], \\ &&\{ \frac{p_i}{q_i} \}=\{ \frac{-2A_n-3}{1}, \frac{-4A_n-5}{2}, \frac{ -\lfloor \frac{2A_n}{35} \rfloor (4A_n+5)-2A_n-3 }{ 2\lfloor \frac{2A_n}{35} \rfloor +1 }, \cdots\}.\\ \end{eqnarray*} $

One can observe that $q_2$ has exceeded the upper bound of $|y|$, $0.0606|A_n|^{0.12766}|B|^{-1.19149}$. Computation shows that it doesn't exist type $4$ solution.

Therefore, we know that (3.1) only has trivial solutions

$ \pm(x, y)=(1, 0), (0, 1)(1, -1). $

Since $F_n(x, y)=f(x, -y)$, so we proved that when $n>2.03\times10^6$, (2) only has trivial solutions. From the theorem developed by Alan Togbé [15], we prove the theorem.