Let $F(x, y)\in \mathbb{Z}[x, y]$ be an irreducible binary form of degree $n\geq3$. Thue [1] proved that the number of solutions of Thue equations $F(x, y)=k$ is finite. Furthermore, an explicit Thue equations can be solved by the method of A. Baker's [2] linear form in logarithms of algebraic numbers. Therefore, researchers focused on the parametrized Thue equations and inequalities, and abundant results are obtained. For cubic Thue equations and inequalities, see Thomas [3], Xia-Chen-Zhang [4] and Hoshi [5]. For quartic case, see Chen-Voutier [6] and Xia-Chen-Zhang [7]. While for the sextic case, we refer to Wakabayashi [8].

However, all of the above Thue equations and inequalities have at most two parameters. An interesting question is that whether or not there exist a solvable strategy for Thue equations with more than two parameters. In this research, we use the method proposed in [7] to study the Thue equations which consists of four integral parameters. Our main result is the following:

Theorem  Define

$ \begin{eqnarray}\label{equation} f(x, y)=sx^4+4tbdx^3y+6b^2dsx^2y^2+4b^3d^2txy^3+sb^4d^2y^4=N, \end{eqnarray} $

(1.1)

and let $\theta$ be the real root of $f(x, 1)=0$, then we have

$ |y|^{3-\lambda}\leq\frac{cN\varepsilon^2}{\rho^3b^3\sqrt d^3}, $

where $\lambda=\frac{\log(8\sqrt d\varepsilon)}{\log(\varepsilon/(8\sqrt ds^2))}$, $c=32.3144d\sqrt d\varepsilon(\frac{\sqrt d(b\sqrt d+1)\varepsilon}{4s^2})^\lambda$, $\varepsilon=t\sqrt d+\sqrt{t^2d-s^2}$, and $\rho=\sqrt{\varepsilon^2-s^2}$. If $\varepsilon>64ds^3$, this is an effective upper bound for $|y|$.

Let $n$ be a positive integer, suppose $\alpha, w \in(0, 1)$ are real numbers,

$ 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 $R_n(x)=x^\alpha q_n(x)-p_n(x).$

Lemma 1  Let $R_n(w)$ be defined above, we have $|R_n(w)|\leq\ (1-w^{\alpha})(1-\sqrt{w})^{2n}.$

proof  Put $f(t)=\frac{(1-t)(w-t)}{t}$. By Lemma 1 in [7], we get

$ \begin{eqnarray*} |R_n(w)|\leq|(n+1)\binom{n+\alpha}{n}\binom{\alpha}{n+1}|\int_1^w f(t)^n t^{\alpha-1} dt. \end{eqnarray*} $

Since $|f'(t)|=\frac{w-t^2}{t^2}.$ It is easy to obtain $|f(t)|<|f(\sqrt{w})|=(1-\sqrt w)^2.$

While

$ |(n+1)\binom{n+\alpha}{n}\binom{\alpha}{n+1}|= |\alpha \cdot \frac{(n^2-\alpha^2)((n-1)^2-\alpha^2)\cdots(1^2-\alpha^2)}{(n!)^2}|\leq \alpha, $

we obtain

$ |R_n(w)|\leq \alpha \int_1^w |1-\sqrt w|^{2n} t^{\alpha -1}dt=(1-w^{\alpha})(1-\sqrt{w})^{2n}. $

Thus Lemma 1 is proved.

Lemma 2  Let $q_n(w)$ be defined above, then we have

$ |q_n(w)|<\frac{(1+\sqrt w)^{2n}}{\sqrt w^{\alpha}}(1+\frac{1-\sqrt w}{\pi}). $

Proof  By Lemma 2 in [7], we get

$ q_n(w)=\frac{(-1)^n}{2\pi i}\oint_c(1-wt)^{n-\alpha}(1-t)^{n+\alpha}t^{-n-1}dt, $

where $c$ denotes the integration path encircles the origin once in the positive sense.

In order to get the estimation of $q_n(w)$, we cut the complex plain from 1 to $1/\sqrt w$, and consider the integration path $c=c_1+c_2+c_3$, where $c_1$ denotes the path that starts from 1 and proceeds along the positive real axis to $1/\sqrt w$, $c_2$ denotes the path that circles the origin in positive sense by a circle of radius $1/\sqrt w$, and $c_3$ denotes the path that starts from $1/\sqrt w$ back to 1 along the lower part of real axis.

While putting $f_1(t)=\frac{(1-wt)(1-t)}{t}$ and $g_1(t)=\frac{(1-wt)^{1-\alpha}(1-t)^{1+\alpha}}{t^2}, $ we have

$ \begin{eqnarray}\label{formula1} |q_n(w)|&=&\frac{1}{2\pi}|\oint_cf_1(t)^{n-1}g_1(t)dt|\\ \nonumber &\leq&\frac{1}{2\pi}(|\int_{c_1}f_1(t)^{n-1}g_1(t)dt|+|\int_{c_2}f_1(t)^{n-1}g_1(t)dt|+|\int_{c_3}f_1(t)^{n-1}g_1(t)dt|). \end{eqnarray} $

(2.1)

On the upper part of real axis, we have argt=0, so it leads to

$ |\int_{c_1}f_1(t)^{n-1}g_1(t)dt|=\int_1^{1/\sqrt w}|f_1(t)|^{n-1}|g_1(t)|dt. $

It's easy to get

$ |f_1(t)|<|f_1(\frac{1}{\sqrt w})|=(1-\sqrt w)^2, $

and

$ |g_1(t)|=(\frac{1}{t}-w)^{1-\alpha}(1-\frac{1}{t})^{1+\alpha}<g(\frac{1}{\sqrt w})=\sqrt w^{1-\alpha}(1-\sqrt w)^2. $

Straight forward computation shows that

$ |\int_{c_1}f_1(t)^{n-1}g_1(t)dt|\leq \sqrt w^{-\alpha}(1-\sqrt w)^{2n+1}. $

While on the lower part of real axis, we have argt= $2\pi$. Similarly, it leads to

$ |\int_{c_3}f_1(t)^{n-1}g_1(t)dt|=|\int_{1/\sqrt w}^1|f_1(te^{2\pi i})|^{n-1}|g_1(te^{2\pi i})|e^{2\pi i}dt|, $

and

$ |\int_{c_3}f_1(t)^{n-1}g_1(t)dt|\leq \sqrt w^{-\alpha}(1-\sqrt w)^{2n+1}. $

On the circle, after putting $t=\frac{e^{i\theta}}{\sqrt w}$, we have

$ \begin{eqnarray*} |\int_{c_2}f_1(t)^{n-1}g_1(t)dt|&=&|i\int_0^{2\pi}f_1( \frac{e^{i\theta}}{\sqrt w} )^{n-1}g_1(\frac{e^{i\theta}}{\sqrt w}) \frac{e^{i\theta}}{\sqrt w}d\theta|\\ &=&|\int_0^{2\pi}f_1(\frac{e^{i\theta}}{\sqrt w} )^{n-1}g_1(\frac{e^{i\theta}}{\sqrt w}) \frac{1}{\sqrt w}d\theta|. \end{eqnarray*} $

It is easy to get

$ |f_1(\frac{e^{i\theta}}{\sqrt w})|\leq |f_1(\frac{-1}{\sqrt w})|=(1+\sqrt w)^2 $

and

$ \begin{eqnarray*} |g_1(\frac{e^{i\theta}}{\sqrt w})|&=&|(\sqrt w e^{-i\theta}-w)^{1-\alpha}(\sqrt w e^{-i\theta}-1)^{1+\alpha}|\\ &=&\sqrt{(w-2w\sqrt w \cos\theta+w^2)^{1-\alpha}(w-2\sqrt w \cos\theta+1)^{1+\alpha}}\\ &=&\sqrt w ^{1-\alpha}(w+1-2\sqrt w\cos\theta)\\ &\leq&\sqrt w ^{1-\alpha}(1+\sqrt w)^2. \end{eqnarray*} $

So we have

$ |\int_{c_2}f_1(t)^{n-1}g_1(t)dt| \leq\frac{ 2\pi}{ \sqrt w^{\alpha}}(1+\sqrt w)^{2n}. $

By (2.1), we complete the proof of Lemma 2.

Lemma 3  Let $p_n(w)$ be denoted above, then we have

$ |p_n(w)|<|(1+\sqrt w)^{2n}\sqrt w^{\alpha}(1+\frac{1-\sqrt w}{\pi})|. $

Proof  Note that

$ p_n(w)=w^nq_n(\frac{1}{w}), $

and we know that the estimation of $|p_n(w)|$ is decided by that of $|q_n(\frac{1}{w})|$.

By Lemma 2 in [7], we have

$ |p_n(w)|=w^n|\frac{(-1)^n}{2\pi i}\oint_c(1-\frac{t}{w})^{n-\alpha}(1-t)^{n+\alpha}t^{-n-1}dt|, $

where $c=c_1+c_2+c_3$. In detail, $c_1$ denotes the path that starts from $w$ and proceeds along the positive real axis to $\sqrt w$, $c_2$ denotes the path that circles the origin in positive sense by a circle of radius $\sqrt w$, and $c_3$ denotes the path that starts from $\sqrt w$ and return to $w$ along the lower part of real axis.

Now we put

$ f_2(t)=\frac{(1-\frac{t}{w})(1-t)}{t}~\textrm{and}~g_2(t)=\frac{(1-\frac{t}{w})^{1-\alpha}(1-t)^{1+\alpha}}{t^2}, $

then we have

$ \begin{eqnarray}\label{formula2} &&|p_n(w)|=w^n\frac{1}{2\pi}|\oint_cf_2(t)^{n-1}g_2(t)dt|\\ &&\nonumber \leq w^n\frac{1}{2\pi}(|\int_{c_1}f_2(t)^{n-1}g_2(t)dt|+|\int_{c_2}f_2(t)^{n-1}g_2(t)dt|+\\ &&|\int_{c_3}f_2(t)^{n-1}g_2(t)dt|). \end{eqnarray} $

(2.2)

In a similar way as the proof in Lemma 2, we have

$ \begin{eqnarray*} |\int_{c_2}f_2(t)^{n-1}g_2(t)dt|&=&|\int_w^{\sqrt w}f_2(t)^{n-1}g_2(t)dt|\\ &\leq& |\int_w^{\sqrt w}(\frac{(1-\sqrt w)^2}{w})^{n-1}(\frac{1}{\sqrt w}-1)^2(\frac{1}{\sqrt w})^{1-\alpha}dt|\\ &\leq& \frac{(1-\sqrt w)^{2n}}{w^n}(1-\sqrt w)\sqrt w^\alpha, \end{eqnarray*} $

and also

$ |\int_{c_3}f_2(t)^{n-1}g_2(t)dt|\leq\frac{(1-\sqrt w)^{2n}}{w^n}(1-\sqrt w)\sqrt w^\alpha. $

On the circle, putting $t=\sqrt w e^{i \theta}$, we have

$ \begin{eqnarray*} |\int_{c_2}f_2(t)^{n-1}g_2(t)dt|&=&|i\int_0^{2\pi}f_2( \sqrt we^{i\theta} )^{n-1}g_2(\sqrt we^{i\theta})\sqrt w e^{i\theta} d\theta|\\ &=&|\int_0^{2\pi}f_2( \sqrt we^{i\theta} )^{n-1}g_2(\sqrt we^{i\theta})\sqrt w d\theta|\\ &\leq&|\int_0^{2\pi}f_2(- \sqrt w)^{n-1}g_2(-\sqrt w)\sqrt w d\theta|\\ &=&2\pi\frac{(1+\sqrt w)^{2n}}{w^n}\sqrt w^\alpha. \end{eqnarray*} $

From (2.2), it is direct to prove Lemma 3.

Lemma 4(see [6])  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 $y_n$ 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

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

Let $f(x, y), ~\varepsilon$ be as in Theorem and assume $\varepsilon>64dt^3, d>1$, where $d, s, t, b, d\in Z$ and $t>0$ without loss of generality. Then we have

$ f(x, 1)=sx^4+4tbdx^3+6b^2dsx^2+4b^3d^2tx+sb^4d^2 $

denoted by $f(x)$. If denoting the root of $f(x)=0$ as $\theta$, it is easy to show that $\theta$ satisfies

$ (\frac{\theta +b\sqrt d}{\theta-b\sqrt d})^4=\frac{t\sqrt d-s}{t\sqrt d+s}. $

For simplicity, we denote $z=t\sqrt d-s, u=t\sqrt d+s$, and $w=\frac{z}{u}.$

It is straightforward to get

$ \sqrt{w}=\frac{t\sqrt d+\sqrt{t^2d-s^2}-s}{t\sqrt d+\sqrt{t^2d-s^2}+s}=\frac{\varepsilon-s}{\varepsilon+s}. $

Putting $\rho=\sqrt{\varepsilon^2-s^2}$, we have $\sqrt[4]{w}=\pm\frac{\varepsilon-s}{\rho}, \pm\frac{\varepsilon-s}{\rho}i.$ Hence, the roots of $f(x)=0$ are

$ \begin{eqnarray*} \theta _0&=& b\sqrt d\frac{\frac{\varepsilon-s}{\rho}+1}{\frac{\varepsilon-s}{\rho}-1}=b\sqrt d ~\frac{\varepsilon-s+\rho}{\varepsilon-s-\rho}, \\ \theta _1&=& b\sqrt d\frac{-\frac{\varepsilon-s}{\rho}+1}{-\frac{\varepsilon-s}{\rho}-1}=b\sqrt d~ \frac{\varepsilon-s-\rho}{\varepsilon-s+\rho}, \\ \theta _2&=& b\sqrt d\frac{\frac{\varepsilon-s}{\rho}i+1}{\frac{\varepsilon-s}{\rho}i-1}=b\sqrt d ~\frac{s-\rho i}{\varepsilon}, \\ \theta_3 &=& b\sqrt d\frac{-\frac{\varepsilon-s}{\rho}i+1}{-\frac{\varepsilon-s}{\rho}i-1}=b\sqrt d ~\frac{s+\rho i}{\varepsilon}. \end{eqnarray*} $

Let $\delta_i=|x-y\theta_i|~~(i=0, 1, 2, 3), $ then we have

$ \delta_2=|x-\theta_2y|=|x-b\sqrt d\frac{s-\rho i}{\varepsilon}y|>\frac{\rho}{\varepsilon}b\sqrt d |y|. $

Similarly, we have $\delta_3>\frac{\rho}{\varepsilon}b\sqrt d |y|.$ If $\delta_0<\delta_1$, then we know that

$ \delta_1>\frac{\delta_0+\delta_1}{2}>\frac{|x-y\theta_0-(x-y\theta_1)|}{2}=|y|\frac{|\theta_0-\delta_1|}{2}=\frac{\rho b \sqrt d}{s}|y|. $

If $\delta_1<\delta_0 $, we similarly have $\delta_0>\frac{\rho b \sqrt d}{s}|y|.$ Since $|f(x, y)|=s\delta_0\delta_1\delta_2\delta_3, $ we obtain that

$ \begin{eqnarray}\label{formula3} \min\{\delta_0, \delta_1\}<\frac{N}{s(\frac{\rho}{\varepsilon}b\sqrt d|y|)^2~\frac{\rho b\sqrt d}{s}|y|}=\frac{N\varepsilon^2}{\rho^3b^3\sqrt d^3|y^3|}. \end{eqnarray} $

(3.1)

Thus we obtain an upper bound for $\delta_0$ or $\delta_1.$ Thereafter, we will get a lower bound for them by proving Theorem.

Proof of Theorem  Since

$ (\frac{\theta +b\sqrt d}{\theta-b\sqrt d})^4=\frac{t\sqrt d-s}{t\sqrt d+s}=w, $

it is easy to write $\theta_0, \theta_1$ as $\theta_i=b\sqrt d\frac{aw^{\frac{1}{4}}+a'}{aw^{\frac{1}{4}}-a'}, $ where $a=1$, if $i=0$; $a=\sqrt d$, if $i=1$, and' denote conjugate in $\mathbb{Q}[\sqrt d]$ that maps $Z+Z\sqrt d$ into $Z-Z\sqrt d$.

Since $0<w<1$, the lemmas can be applied into Theorem 1 after putting $\alpha=1/4 \in(0, 1)$. Now we use the Padè approximation method to formulate rational integers approximation so as to give an effective measure of the irrationality of $\theta$.

Let $p_n(w), q_n(w), R_n(w)$ be defined above and $\theta $ is one of $\theta_0, \theta_1 $. We know that

$ \begin{eqnarray*} \theta&=& b\sqrt d\frac{aw^\frac{1}{4}+a'}{aw^\frac{1}{4}-a'}\\ &=& b\sqrt d\frac{aw^\frac{1}{4}q_n(w)+a'q_n(w)}{aw^\frac{1}{4}q_n(w)-a'q_n(w)} \\ &=& \frac{b\sqrt d aR_n(w)+b\sqrt dap_n(w)+b\sqrt da'q_n(w)}{aR_n(w)+ap_n(w)-a'q_n(w)}\\ &=& \frac{4^n(\sqrt d)^{n+1}u^n b\sqrt d aR_n(w)+ 4^n(\sqrt d)^{n+1}u^n (b\sqrt dap_n(w)+b\sqrt da'q_n(w))}{4^n(\sqrt d)^{n+1}u^n aR_n(w)+4^n(\sqrt d)^{n+1}u^n (ap_n(w)-a'q_n(w))}. \end{eqnarray*} $

After putting

$ P_n=4^n(\sqrt d)^{n+1}u^n (b\sqrt dap_n(w)+b\sqrt da'q_n(w)) $

and

$ Q_n=4^n(\sqrt d)^{n+1}u^n (ap_n(w)-a'q_n(w)), $

we have

$ |P_n-Q_n\theta|=|4^n(\sqrt d)^{n+1}u^naR_n(w)(b\sqrt d-\theta)|, $

and denoted it as $R_n$.

Since $P_n, Q_n \in Z$, from the estimation of $p_n(w)$ and $q_n(w)$ in Lemma 2 and Lemma 3, we obtain

$ \begin{eqnarray*} |Q_n|&=&|4^n(\sqrt d)^{n+1}u^n (ap_n(w)-a'q_n(w))|\\ &\leq&4^n(\sqrt d)^{n+1}u^n|a|(|p_n(w)|+|q_n(w)|)\\ &\leq&4^n\sqrt d^{n+1}u^n(1+\sqrt w)^{2n}~|a||( \sqrt w^{\alpha}(1+\frac{1-\sqrt w}{\pi})+ \sqrt w^{-\alpha}(1+\frac{1-\sqrt w}{\pi}) )|\\ &\leq& 4^n\sqrt d^{n+1}u^n|a||(1+\sqrt w)^{2n}(\sqrt w^\alpha+\sqrt w^{-\alpha})(1+\frac{1-\sqrt w}{\pi})|\\ &=&C_Q(4\sqrt du(1+\sqrt w)^2)^n\\ &=&C_Q(8\sqrt d\varepsilon)^n, \end{eqnarray*} $

where

$ C_Q=\sqrt d|a|(\sqrt w^\alpha+\sqrt w^{-\alpha})(1+\frac{1-\sqrt w}{\pi}). $

Since $|a|\leq\sqrt d$ and $\varepsilon>64ds^3$, we have $\sqrt w=\frac{\varepsilon-s}{\varepsilon+s}>63/65$, so we can estimate that $\sqrt w^\alpha+\sqrt w^{-\alpha}<2.00006$, and $1+\frac{1-\sqrt w}{\pi}<1.00979$. Hence $C_Q<2.01965d$.

On the other hand, from Lemma 1, we have

$ \begin{eqnarray*} |R_n| &=&|4^n(\sqrt d)^{n+1}u^naR_n(w)(b\sqrt d-\theta)|\\ &<&4^n(\sqrt d)^{n+1}u^na|(1-w^\alpha)(1-\sqrt w)^{2n}(b\sqrt d-\theta)|\\ &=&C_R(4\sqrt du (1-\sqrt w)^2)^n =C_R(8\sqrt d s^2/\varepsilon)^n, \end{eqnarray*} $

where

$ C_R=\sqrt d |a(1-w^\alpha) (b\sqrt d-\theta)| \leq \sqrt d |a|(b\sqrt d+1)\leq d(b\sqrt d+1). $

Let $Q=8\sqrt d\varepsilon, k_0=2.01965b, E=\frac{\varepsilon}{8\sqrt ds^2}, l_0=d(b\sqrt d+1)$, result in Lemma 4 yields to

$ \begin{eqnarray}\label{formula4} \delta_0, \delta_1>\frac{1}{c|y|^\lambda}, \end{eqnarray} $

(3.2)

where $\lambda=\frac{\log(8\sqrt d\varepsilon)}{\log(\varepsilon/(8\sqrt ds^2))}$ and $c=32.3144d\sqrt d(\frac{\sqrt d(b\sqrt d+1)\varepsilon}{4s^2})^\lambda$.

By (3.1) and (3.2), we obtain $|y|^{3-\lambda}\leq\frac{cN\varepsilon^2}{\rho^3b^3\sqrt d^3}.$

Thus Theorem follows.

Actually, when $\varepsilon>64ds^3$, it directly leads to $\lambda<3$, so we are able to derived an effective upper bound for $y$.

In this research, a four-parametrized quartic Thue equations is solved by approximation certain crucial algebraic numbers in an elementary way. A computable upper bound for solutions is obtained as well, which is quite effective when $\varepsilon$ is much greater than $64ds^3$. In the mean time, the value of $\frac{1}{3-\lambda}$ decreases dramatically when $w$ approximate to 1.