On the Diophantine equations $z^2=f(x)^2 \pm g(y)^2$ concerning Laurent polynomials

TL;DR

This paper investigates rational solutions to specific Diophantine equations involving Laurent polynomials using elliptic curve theory, providing insights into their parametric solutions.

Contribution

It applies elliptic curve theory to analyze rational solutions of equations involving Laurent polynomials, a novel approach for these types of Diophantine equations.

Findings

Identification of conditions for rational solutions

Explicit parametric solutions for certain Laurent polynomials

Extension of elliptic curve methods to Laurent polynomial equations

Abstract

By the theory of elliptic curves, we study the nontrivial rational parametric solutions and rational solutions of the Diophantine equations $z^2=f(x)^2 \pm g(y)^2$ for some simple Laurent polynomials $f$ and $g$.

Equations

$$z^2=f(x{)}^2\pm f(y{)}^2,$$

$$f(x)f(y)=f(z{)}^n,$$

$$z^2=f(x{)}^2+g(y{)}^2$$

$$z^2=f(x{)}^2-g(y{)}^2$$

$$f(x)=\sum _k{a}_k{x}^k,a_k\in \mathbb{F},$$

$$f(x)=ax+b+\frac{c}{x},a{x}^2+bx+c+\frac{d}{x},ax+b+\frac{c}{x}+\frac{d}{x^2}$$

$$x=tT,y=T.$$

$$\begin{array}{rl}{T}^2{t}^2{z}^2=& T^4{t}^4+2T^{3}a{t}^3+[T^6+2c{T}^5+(c^2+2d)T^4+(2cd+2e)T^3\\ & +(a^2+2ce+d^2+2b)T^2+2Tde+e^2]t^2+2Tabt+b^2=:g_1(t).\end{array}$$

$$v=p{t}^2+qt+b,$$

$$v^2-g_1(t)=\sum _{i=1}^4{A}_i{t}^i,$$

$$\begin{array}{rl}{A}_1=& 2abT-2bq,\\ A_2=& T^6+2c{T}^5+(c^2+2d)T^4+(2cd+2e)T^3+(a^2+2ce+d^2+2b)T^2\\ & +2deT-2pb+e^2-q^2,\\ A_3=& 2a{T}^3-2pq,\\ A_4=& T^4-p^2.\end{array}$$

$$p=-T^2,q=-aT.$$

$$v^2-g_1(t)=\sum _{i=1}^4{A}_i{t}^i=0$$

$$t=-\frac{4abT}{T^6+2c{T}^5+(c^2+2d)T^4+(2cd+2e)T^3+(2ce+d^2+4b)T^2+2deT+e^2}.$$

$$\begin{array}{rl}{t}_1=& -\frac{4abT}{T^6+2T^{5}c+(c^2+2d)T^4+(2cd+2e)T^3+(2ce+d^2+4b)T^2+2Tde+e^2},\\ v_1=& \frac{{\phi }_1}{{\psi }_1},\end{array}$$

$$\begin{array}{rl}{\phi }_1=& b(T^{12}+4c{T}^{11}+(6c^2+4d)T^{10}+(4c^3+12cd+4e)T^9+(c^4+12c^{2}d+4a^2\\ & +12ce+6d^2+8b)T^8+(4c^{3}d+8a^{2}c+12c^{2}e+12c{d}^2+16bc+12de)T^7\\ & +(4a^2{c}^2+4c^{3}e+6c^2{d}^2+8a^{2}d+8b{c}^2+24cde+4d^3+16bd+6e^2)T^6\\ & +(8a^{2}cd+12c^{2}de+4c{d}^3+8a^{2}e+16bcd+12c{e}^2+12d^{2}e+16be)T^5+(8a^{2}ce\\ & +4a^2{d}^2+6c^2{e}^2+12c{d}^{2}e+d^4+16bce+8b{d}^2+12d{e}^2+16b^2)T^4+4e(2a^{2}d\\ & +3cde+d^3+4bd+e^2)T^3+2e^2(2a^2+2ce+3d^2+4b)T^2+4Td{e}^3+e^4),\\ {\psi }_1=& (T^6+2c{T}^5+(c^2+2d)T^4+(2cd+2e)T^3+(2ce+d^2+4b)T^2+2deT+e^2{)}^2.\end{array}$$

$$v=p(t+t_1{)}^2+q(t+t_1)+v_1,$$

$$x=tT,y=T.$$

$$\begin{array}{rl}{T}^4{t}^2{z}^2=& T^6{t}^4+2T^{5}a{t}^3+[T^6+2T^{5}c+(a^2+c^2+2b+2d)T^4+(2cd+2e)T^3\\ & +(2ce+d^2)T^2+2Tde+e^2]t^2+2T^{3}abt+b^2{T}^2=:g_2(t).\end{array}$$

$$v=p{t}^2+qt+bT,$$

$$v^2-g_2(t)=\sum _{i=1}^4{A}_i{t}^i,$$

$$\begin{array}{rl}{A}_1=& 2T^{3}ab-2Tbq,\\ A_2=& T^6+2c{T}^5+(a^2+c^2+2b+2d)T^4+(2cd+2e)T^3+(2ce+d^2)T^2\\ & +(2de-2pb)T+e^2-q^2,\\ A_3=& 2T^{5}a-2pq,\\ A_4=& T^6-p^2.\end{array}$$

$$p=-T^3,q=-a{T}^2.$$

$$v^2-g_2(t)=\sum _{i=1}^4{A}_i{t}^i=0$$

$$t=-\frac{4abT}{T^6+2c{T}^5+(c^2+2d)T^4+(2cd+2e)T^3+(2ce+d^2+4b)T^2+2deT+e^2}.$$

$$\begin{array}{rl}{t}_1=& -\frac{4ab{T}^3}{T^6+2c{T}^5+(c^2+4b+2d)T^4+(2cd+2e)T^3+(2ce+d^2)T^2+2Tde+e^2},\\ v_1=& \frac{{\phi }_1}{{\psi }_1},\end{array}$$

$$\begin{array}{rl}{\phi }_1=& bT(T^{12}+4c{T}^{11}+(4a^2+6c^2+8b+4d)T^{10}+(8a^{2}c+4c^3+16bc+12cd\\ & +4e)T^9+(4a^2{c}^2+c^4+8a^{2}d+8b{c}^2+12c^{2}d+16b^2+16bd+12ce+6d^2)T^8\\ & +(8a^{2}cd+4c^{3}d+8a^{2}e+16bcd+12c^{2}e+12c{d}^2+16be+12de)T^7\\ & +(8a^{2}ce+4a^2{d}^2+4c^{3}e+6c^2{d}^2+16bce+8b{d}^2+24cde+4d^3+6e^2)T^6\\ & +(8a^{2}de+12c^{2}de+4c{d}^3+16bde+12c{e}^2+12d^{2}e)T^5+(4a^2{e}^2+6c^2{e}^2\\ & +12c{d}^{2}e+d^4+8b{e}^2+12d{e}^2)T^4+4e(3cde+d^3+e^2)T^3\\ & +2e^2(2ce+3d^2)T^2+4Td{e}^3+e^4),\\ {\psi }_1=& (T^6+2c{T}^5+(c^2+4b+2d)T^4+(2cd+2e)T^3+(2ce+d^2)T^2+2Tde+e^2{)}^2.\end{array}$$

$$v=p(t+t_1{)}^2+q(t+t_1)+v_1,$$

$$x=T,y=\frac{d}{a}T.$$

$$a^4{T}^2{z}^2=(T+a{)}^2(T+c{)}^2((a^4+d^4)T^2+(2a^{4}b+2a{d}^{3}e)T+a^2(a^2{b}^2+d^2{e}^2)).$$

Full text

This research was supported by the National Natural Science Foundation of China (Grant No. 11501052).

1. Introduction

In 2010, A. Togbe and M. Ulas [10] considered the rational solutions of the Diophantine equations

[TABLE]

where $f$ being quadratic and cubic polynomials. At the same year, B. He, A. Togbe and M. Ulas [6] further investigated the integer solutions of Eq. (1.1) for some special polynomials $f$.

In 2016, Y. Zhang [11] studied the nontrivial rational parametric solutions of the Diophantine equations

[TABLE]

where $n=1,2,$ concerning the Laurent polynomials $f$. Moreover, we [12] considered the nontrivial rational parametric solutions and rational solutions of Eq. (1.1) for some simple Laurent polynomials $f$.

In 2017, Sz. Tengely and M. Ulas [9] investigated the integer solutions of the Diophantine equations

[TABLE]

and

[TABLE]

for different polynomials $f$ and $g$.

In this paper, we continue the study of [11] and [12], and consider the nontrivial rational parametric solutions and rational solutions of Eqs. (1.2) and (1.3) for some simple Laurent polynomials $f$ and $g$. The nontrivial solution $(x,y,z)$ of Eqs. (1.2) and (1.3) respectively means that $f(x)g(y)\neq 0$ and $f(x)^{2}\neq g(y)^{2}$.

Recall that a Laurent polynomial with coefficients in a field $\mathbb{F}$ is an expression of the form

[TABLE]

where $x$ is a formal variable, the summation index $k$ is an integer (not necessarily positive) and only finitely many coefficients $a_{k}$ are nonzero. Here we mainly care about the simple Laurent polynomials

[TABLE]

with nonzero integers $a,b,c,d$. In the sequel, without losing the generality we may assume that $a=1$, guaranteed by the shape of Eqs. (1.2) and (1.3).

By the theory of elliptic curves, we give a positive answer to the Question 3.3 of [12], and prove

Theorem 1.1**.**

For $f=x+a+b/x$ and $g=y^{2}+cy+d+e/y$ with nonzero integers $a,b,c,d,e$, Eq. (1.2) or Eq. (1.3) has infinitely many nontrivial rational parametric solutions.

Theorem 1.2**.**

For $f=x+a+b/x$ and $g=y+c+d/y+e/y^{2}$ with nonzero integers $a,b,c,d,e$, Eq. (1.2) or Eq. (1.3) has infinitely many nontrivial rational parametric solutions.

Theorem 1.3**.**

For $f=(x+a)(x+b)(x+c)/x$ and $g=(y+d)(y+e)(y+cd/a)/y$ with nonzero integers $a,b,c,d,e$, Eq. (1.2) or Eq. (1.3) has infinitely many nontrivial rational solutions.

Theorem 1.4**.**

For $f=(x+a)(x+b)(x+c)/x$ and $g=(y+d)(y+e)(y+cd/a)/y^{2}$ with nonzero integers $a,b,c,d,e$, Eq. (1.2) or Eq. (1.3) has infinitely many nontrivial rational solutions.

2. The proofs of Theorems

Proof of Theorem 1.1..

For $f=x+a+b/x$ and $g=y^{2}+cy+d+e/y$, let

[TABLE]

Then Eq. (1.2) equals

[TABLE]

Let $v=Ttz,$ and consider the curve $\mathcal{C}_{1}:~{}v^{2}=g_{1}(t)$. In order to prove Theorem 1.1 we must show that the curve $\mathcal{C}_{1}$ has infinitely many $Q(T)$-rational points. The curve $\mathcal{C}_{1}$ is a quartic curve with rational point $P=(0,b)$. By the method of Fermat [4, p. 639], using the point $P$ we can produce another point $P^{\prime}=(t_{1},v_{1})$, which satisfies the condition $t_{1}v_{1}\neq 0$. In order to construct a such point $P^{\prime}$, we put

[TABLE]

where $p,q$ are indeterminate variables. Then

[TABLE]

where the quantities $A_{i}=A_{i}(p,q)$ are given by

[TABLE]

The system of equations $A_{4}=A_{3}=0$ in $p,q$ has a solution given by

[TABLE]

This implies that the equation

[TABLE]

has the rational roots $t=0$ and

[TABLE]

Then we have the point $P^{\prime}=(t_{1},v_{1})$ with

[TABLE]

where

[TABLE]

Again, using the method of Fermat, we let

[TABLE]

and get another $P^{\prime\prime}=(t_{2},v_{2})$ on the curve $\mathcal{C}_{1}$. Repeating the above process any numbers of times completes the proof of Theorem 1.1 for Eq. (1.2). The same method can be used to give a proof for Eq. (1.3).

∎

Proof of Theorem 1.2..

For $f=x+a+b/x$ and $g=y+c+d/y+e/y^{2}$, let

[TABLE]

Then Eq. (1.2) leads to

[TABLE]

Let $v=T^{2}tz,$ and consider the curve $\mathcal{C}_{2}:~{}v^{2}=g_{2}(t)$. In order to prove Theorem 1.2 we must show that the curve $\mathcal{C}_{2}$ has infinitely many $Q(T)$-rational points. The curve $\mathcal{C}_{2}$ is a quartic curve with rational point $Q=(0,bT)$. By the method of Fermat [4, p. 639], and using the point $Q$ we can produce another point $Q^{\prime}=(t_{1},v_{1})$ satisfying the condition $t_{1}v_{1}\neq 0$. In order to construct a such point $Q^{\prime}$, we take

[TABLE]

where $p,q$ are indeterminate variables. Then

[TABLE]

where the quantities $A_{i}=A_{i}(p,q)$ are given by

[TABLE]

The system of equations $A_{4}=A_{3}=0$ in $p,q$ has a solution given by

[TABLE]

This implies that the equation

[TABLE]

has the rational roots $t=0$ and

[TABLE]

Then we have the point $P^{\prime}=(t_{1},v_{1})$ with

[TABLE]

where

[TABLE]

Applying the method of Fermat, we take

[TABLE]

and obtain another point $Q^{\prime\prime}=(t_{2},v_{2})$ on the curve $\mathcal{C}_{2}$. Repeating the above process any numbers of times completes the proof of Theorem 1.2 for Eq. (1.2). The same method can be used to give a proof for Eq. (1.3). ∎

Proof of Theorem 1.3..

For $f=(x+a)(x+b)(x+c)/x$ and $g=(y+d)(y+e)(y+cd/a)/y$, set

[TABLE]

Then Eq. (1.2) reduces to

[TABLE]

To prove the theorem, let us consider the curve

[TABLE]

Note that the point $(T,S)=(-b,ade-bd^{2})$ lies on $\mathcal{C}_{3}$, then it can be parameterized by

[TABLE]

where $r$ is a rational parameter. Hence,

[TABLE]

where

[TABLE]

This completes the proof of Theorem 1.3 for Eq. (1.2). A similar method for Eq. (1.3). ∎

Proof of Theorem 1.4..

For $f=(x+a)(x+b)(x+c)/x$ and $g=(y+d)(y+e)(y+cd/a)/y^{2}$, put

[TABLE]

Then Eq. (1.2) equals

[TABLE]

To prove the theorem, consider the curve

[TABLE]

The curve $\mathcal{C}_{4}$ is a quartic curve with rational point $R=(0,ae)$. By the method of Fermat [4, p. 639], and using the point $R$, we can produce another point $R^{\prime}=(T_{1},S_{1})$, which satisfies the condition $T_{1}S_{1}\neq 0$. In order to construct a such point $R^{\prime}$, we put

[TABLE]

where $p,q$ are indeterminate variables. Then

[TABLE]

where the quantities $A_{i}=A_{i}(p,q)$ are given by

[TABLE]

The system of equations $A_{4}=A_{3}=0$ in $p,q$ has a solution given by

[TABLE]

This implies the equation

[TABLE]

has the rational roots $t=0$ and

[TABLE]

Then we have the point $R^{\prime}=(T_{1},S_{1})$, where

[TABLE]

Using the method of Fermat, we let

[TABLE]

and find another point $R^{\prime\prime}=(T_{2},S_{2})$ on the curve $\mathcal{C}_{4}$. Repeat the above process any numbers of times, this completes the proof of Theorem 1.4 for Eq. (1.2). The same method for giving a proof for Eq. (1.3). ∎

3. Some related questions

We have studied the rational parametric solutions and rational solutions of Eqs. (1.2) and (1.3) for $f,g=x+a+b/x,x^{2}+ax+b+c/x$ and $x+a+b/x+c/x^{2}$, but we don't get the same results for other Laurent polynomials.

Question 3.1**.**

For $f=x^{3}+ax^{2}+bx+c+d/x$ or $x^{2}+ax+b+c/x+d/x^{2}$, do Eqs. (1.2) and (1.3) have rational solutions? If they have, are there infinitely many?

Finding the integer solutions of Eqs. (1.2) and (1.3) is also an interesting question.

Question 3.2**.**

Does there exist a Laurent polynomial such that Eqs. (1.2) and (1.3) have infinitely many integer solutions?