On the Diophantine equations $ \sum_{i=1}^n a_ix_{i} ^6+\sum_{i=1}^m b_iy_{i} ^3= \sum_{i=1}^na_iX_{i}^6\pm\sum_{i=1}^m b_iY_{i} ^3 $

TL;DR

This paper employs elliptic curve theory to find infinitely many positive and parametric solutions to complex Diophantine equations involving sixth and third powers, applicable for arbitrary fixed parameters.

Contribution

It introduces a novel method using elliptic curves to solve a broad class of Diophantine equations with arbitrary parameters, yielding infinite solutions.

Findings

Infinitely many positive solutions are found.

Parametric solutions are explicitly constructed.

Method applies to any fixed nonzero integers $a_i$, $b_i$, $n$, and $m$.

Abstract

In this paper, the elliptic curves theory is used for solving the Diophantine equations $\sum_{i=1}^n a_ix_{i} ^6+\sum_{i=1}^m b_iy_{i} ^3= \sum_{i=1}^na_iX_{i}^6\pm\sum_{i=1}^m b_iY_{i} ^3$, where $n$, $m$ $\geq 1$ and $a_i$, $b_i$, are fixed arbitrary nonzero integers. By our method, we may find infinitely many nontrivial positive solutions and also obtain infinitely many nontrivial parametric solutions for the Diophantine equations for every arbitrary integers $n$, $m$, $a_i$ and $b_i$.

Equations

$$v^2=\frac{(2h{\sum }_{i=1}^n{a}_i{D}_i^5)}{({\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}{u}^4+\frac{(20h^3{\sum }_{i=1}^n{a}_i{D}_i^3-{\sum }_{i=1}^m{b}_i{B}_i^3)}{(3{\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}{u}^2+\frac{(2h^5{\sum }_{i=1}^n{a}_i{D}_i)}{({\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}.$$

$$L:=\frac{(2h^5{\sum }_{i=1}^n{a}_i{D}_i)}{({\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}$$

$$v^2=\frac{(-2h{\sum }_{i=1}^n{a}_i{D}_i^5)}{({\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}{u}^4-\frac{(20h^3{\sum }_{i=1}^n{a}_i{D}_i^3+{\sum }_{i=1}^m{b}_i{B}_i^3)}{(3{\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}{u}^2-\frac{(2h^5{\sum }_{i=1}^n{a}_i{D}_i)}{({\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}.$$

$$L^{\prime }:=\frac{(-2h^5{\sum }_{i=1}^n{a}_i{D}_i)}{({\sum }_{i=1}^m{b}_i{B}_i{C}_i^2)}$$

Full text

1. Introduction

Number theory is a vast and fascinating field of mathematics, sometimes called ”higher arithmetic”, consisting of the study of the properties of whole numbers. In mathematics, a Diophantine equations is a polynomial equation, usually in two or more unknowns, such that only the integer solutions are studied. The word Diophantine refers to the Hellenistic mathematician of the 3rd century, Diophantus of Alexanderia, who made a study of such equations. While individual equation present a kind of puzzle and have been considered throughout the history, the formulation of general theories of Diophantine equations (beyond the theory of quadratic forms) was an achievement of the twentieth century. As a history work, Euler conjectured in $1969$ that the Diophantine equation $A^{4}+B^{4}+C^{4}=D^{4}$, or more generally $A_{1}^{N}+A_{2}^{N}+\cdots+A_{N-1}^{N}=A_{N}^{N}$, ($N\geq 4$), has no solution in positive integers ( see [1]). Nearly two centuries later, a computer search (see [3]) found the first counterexample to the general conjecture (for $N=5$): $27^{5}+84^{5}+110^{5}+133^{5}=144^{5}$.

In $1986$, Noam Elkies found a method to construct an infinite series counterexamples for the $K=4$ case (see [2]). His smallest counterexample was:

$2682440^{4}+15365639^{4}+18796760^{4}=20615673^{4}$.

In this paper, we are interested in the study of Diophantine equations:

$\sum_{i=1}^{n}a_{i}x_{i}^{6}+\sum_{i=1}^{m}b_{i}y_{i}^{3}=\sum_{i=1}^{n}a_{i}X_{i}^{6}\pm\sum_{i=1}^{m}b_{i}Y_{i}^{3}$,

where $n$, $m$ $\geq 1$ and $a_{i}$, $b_{i}$, are fixed arbitrary nonzero integers.

Our main results are the following theorems:

** Main Theorem 1.1****.**

*Consider the Diophantine equation :

$\sum_{i=1}^{n}a_{i}x_{i}^{6}+\sum_{i=1}^{m}b_{i}y_{i}^{3}=\sum_{i=1}^{n}a_{i}X_{i}^{6}+\sum_{i=1}^{m}b_{i}Y_{i}^{3}$*,

*where $n$, $m$ $\geq 1$ and $a_{i}$, $b_{i}$, are fixed arbitrary nonzero integers. Let $Y^{2}=X^{3}+FX^{2}+GX+H$, be an elliptic curve in which the coefficients $F$, $G$, and $H$ are all functions of $a_{i}$, $b_{i}$, and the other rational parameters $B_{i}$, $C_{i}$, $D_{i}$, and $h$, yet to be found later. If the elliptic curve has positive rank, depending on the values of $B_{i}$, $C_{i}$, $D_{i}$, and $h$, the Diophantine equation has infinitely many integer solutions.

Proof. Let: $x_{i}=-D_{i}u+h$, $y_{i}=B_{i}u+C_{i}v$, $X_{i}=D_{i}u+h$, and

$Y_{i}=-B_{i}u+C_{i}v$*, where all variables are rational numbers. By substituting these variables in the above Diophantine equation, and after some simplifications, we get:

[TABLE]

*To complete the proof we use the following theorem for transforming this quartic to an elliptic curve of the form $y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}$, where $a_{i}\in\mathbb{Q}$. (see [5])

THEOREM. Let K be a field of characteristic not equal to $2$. Consider the equation

$v^{2}=au^{4}+bu^{3}+cu^{2}+du+q^{2}$*, with $a$, $b$, $c$, $d$ $\in K$.

*Let $x=\frac{2q(v+q)+du}{u^{2}}$, $y=\frac{4q^{2}(v+q)+2q(du+cu^{2})-(\frac{d^{2}u^{2}}{2q})}{u^{3}}$.

*Define $a_{1}=\frac{d}{q}$, $a_{2}=c-(\frac{d^{2}}{4q^{2}})$, $a_{3}=2qb$, $a_{4}=-4q^{2}a$, $a_{6}=a_{2}a_{4}$.

*Then $y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}$.

*The inverse transformation is $u=\frac{2q(x+c)-(\frac{d^{2}}{2q})}{y}$, $v=-q+\frac{u(ux-d)}{2q}$.

The point $(u,v)=(0,q)$ corresponds to the point $(x,y)=\infty$ and

$(u,v)=(0,−q)$* corresponds to $(x,y)=(−a_{2},a_{1}a_{2}-a_{3})$.

We see that if

[TABLE]

to be square, say $q^{2}$, (This is possible by fixing the other parameters and choosing appropriate values for $h$.), we may use the above theorem and transform the quartic elliptic curve to an elliptic curve of the form

$y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}$*, where $a_{i}\in\mathbb{Q}$.

If the above elliptic curve has positive rank (For every arbitrary nonzero integers $a_{i}$ and $b_{i}$, this is done by choosing appropriate values for $B_{i}$, $C_{i}$, $D_{i}$, $h$.), by calculating $x_{i}$, $y_{i}$, $X_{i}$, $Y_{i}$, from relations $x_{i}=-D_{i}u+h$, $y_{i}=B_{i}u+C_{i}v$, $X_{i}=D_{i}u+h$, and $Y_{i}=-B_{i}u+C_{i}v$, after some simplifications and canceling the denominators of $x_{i}=-D_{i}u+h$, $y_{i}=B_{i}u+C_{i}v$, $X_{i}=D_{i}u+h$, and $Y_{i}=-B_{i}u+C_{i}v$, we may obtain infinitely many integer solutions for the Diophantine equation. The proof is complete.

Now we are going to solve some couple of examples:

Example 1. We wish to solve the Diophantine equation:

$x_{1}^{6}+y_{1}^{3}=X_{1}^{6}+Y_{1}^{3}$,

we may assume that in the corresponding the elliptic curve (1.1), $B_{1}=C_{1}=D_{1}=1$,

$h=2$. This means that we let

$x_{1}=-u+2$, $y_{1}=u+v$, $X_{1}=u+2$, $Y_{1}=-u+v$.

( We may choose the other appropriate values for $B_{1}$, $C_{1}$, $D_{1}$, and $h$, so that the rank of the corresponding elliptic curve to be $\geq 1$ and $L$ to be square.

Then we get the elliptic curve:

$v^{2}=4u^{4}+53u^{2}+64$.

With the inverse transformation $u=\frac{16(X+53)}{Y}$ and $v=-8+\frac{u^{2}X}{16}$, this maps to the new elliptic curve

$Y^{2}=X^{3}+53X^{2}-1024X-54272$.

The rank of this elliptic curve is $1$ and its generator is the point

$P=(X,Y)=(-48,80)$. Because of this, the above elliptic curve has infinitely many rational points and we may obtain infinitely many solutions for the Diophantine equation too. Since $P=(X,Y)=(-48,80)$, we get $(u,v)=(1,-11)$, by calculating $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, from the above relations and after some simplifications and canceling the denominators of $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, we get the identity:

$1^{6}+12^{3}=3^{6}+10^{3}$.

Also we have: $2P=(X^{\prime},Y^{\prime})=(68,-660)$,

$3P=(X^{\prime\prime},Y^{\prime\prime})=(\frac{-27168}{841},\frac{490880}{24389})$ and $4P=(X^{\prime\prime\prime},Y^{\prime\prime\prime})=(\frac{1139524}{27225},\frac{1180380068}{4492125})$.

By using these three new points, we obtain the other solutions for the Diophantine equation, respectively as

$74^{6}+5768^{3}=14^{6}+7088^{3}$,

$1503^{6}+6175210^{3}=1919^{6}+5819322^{3}$,

$693358^{6}+2867762572208^{3}=1427882^{6}+2478237149768^{3}$.

By choosing the other points on the elliptic curve such as

$nP$ ($n=5,6$, $\cdots$ ) we obtain infinitely many solutions for the Diophantine equation.

Example 2. Let us solve the Diophantine equation:

$2x_{1}^{6}+y_{1}^{3}=2X_{1}^{6}+Y_{1}^{3}$,

we may assume that in the elliptic curve (1.1),

$B_{1}=3$, $D_{1}=C_{1}=1$, $h=\frac{4}{3}$. This means that we let

$x_{1}=-u+\frac{4}{3}$, $y_{1}=3u+v$, $X_{1}=u+\frac{4}{3}$, $Y_{1}=-3u+v$.

( We may choose the other appropriate values for $B_{1}$, $C_{1}$, $D_{1}$, and $h$, so that rank of the corresponding elliptic curve to be $\geq 1$ and $L$ to be square.

Then we get the elliptic curve:

$v^{2}=\frac{16}{9}u^{4}+\frac{1831}{243}u^{2}+(\frac{64}{27})^{2}$.

With the inverse transformation $u=\frac{\frac{128}{27}(X+\frac{1831}{243})}{Y}$ and $v=-\frac{64}{27}+\frac{u^{2}X}{\frac{128}{27}}$, we get the cubic elliptic curve

$Y^{2}=X^{3}+\frac{1831}{243}X^{2}-\frac{262144}{6561}X-\frac{479985664}{1594323}$.

The rank of this elliptic curve is $2$ and its generators are the points $P_{1}=(X,Y)=(\frac{-1792}{243},\frac{3328}{2187})$ and $P_{2}=(X^{\prime},Y^{\prime})=(\frac{-2411}{324},\frac{7007}{5832})$. Because of this, the above elliptic curve has infinitely many rational points and we may obtain infinitely many solutions for the Diophantine equation too. By using this point $P_{1}=(X,Y)=(\frac{-1792}{243},\frac{3328}{2187})$, we get $(u,v)=(\frac{1}{2},\frac{-149}{54})$, and by calculating $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, from the above relations and after some simplifications and canceling the denominators of $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, we get the identity:

$2.(15)^{6}+1380^{3}=2.(33)^{6}+408^{3}$.

Also we have: $2P_{1}=(\frac{16708}{729},\frac{-2392940}{19683})$.

By using these two new points $2P_{1}$, $P_{2}$, we obtain the other solutions for the Diophantine equation, respectively as

$2.(22773)^{6}+72216552^{3}=2.(1317)^{6}+653700972^{3}$,

$2.(501)^{6}+997572^{3}=2.(885)^{6}+398820^{3}$.

By choosing the other points on the elliptic curve such as

$nP_{1}$ or $nP_{2}$ ($n=3,4$, $\cdots$ ) we obtain infinitely many solutions for the Diophantine equation.

Example 3. We wish to solve the Diophantine equation:

$x_{1}^{6}+7y_{1}^{3}=X_{1}^{6}+7Y_{1}^{3}$,

we may assume that in the elliptic curve (1.1),

$B_{1}=C_{1}=D_{1}=1$, $h=\frac{2}{7}$.

This means that we let $x_{1}=-u+\frac{2}{7}$, $y_{1}=u+v$, $X_{1}=u+\frac{2}{7}$, $Y_{1}=-u+v$.

Then we get the elliptic curve:

$v^{2}=\frac{4}{49}u^{4}-\frac{747}{2401}u^{2}+(\frac{8}{343})^{2}$.

With the inverse transformation $u=\frac{\frac{16}{343}(X-\frac{747}{2401})}{Y}$ and $v=-\frac{8}{343}+\frac{u^{2}X}{\frac{16}{343}}$,

the cubic elliptic curve is

$Y^{2}=X^{3}-\frac{747}{2401}X^{2}-\frac{1024}{5764801}X+\frac{764928}{13841287201}$.

The rank of this elliptic curve is $1$ and its generator is the point

$P=(X,Y)=(\frac{752}{2401},\frac{240}{16807})$. Because of this, the above elliptic curve has infinitely many rational points and we may obtain infinitely many solutions for the Diophantine equation too. By using this point $P=(X,Y)=(\frac{752}{2401},\frac{240}{16807})$, we get $(u,v)=(\frac{1}{147},\frac{-3481}{151263})$, and by calculating $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, from the above relations and after some simplifications and canceling the denominators of $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, we get the identity:

$287^{6}+7.(31570)^{3}=301^{6}+7.(17164)^{3}$.

If we take $B_{1}=C_{1}=D_{1}=1$, $h=\frac{7}{2}$,

this means that in the above Diphantine equation, we have $x_{1}=-u+\frac{7}{2}$,

$y_{1}=u+v$, $X_{1}=u+\frac{7}{2}$, $Y_{1}=-u+v$,

this in turn results to

$v^{2}=u^{4}-\frac{81}{2}u^{2}+(\frac{49}{4})^{2}$.

With the inverse transformation $u=\frac{\frac{49}{2}(X+\frac{81}{2})}{Y}$ and $v=\frac{-49}{4}+\frac{u^{2}X}{\frac{49}{2}}$,

we get the new elliptic curve

$Y^{2}=X^{3}+\frac{81}{2}X^{2}-\frac{2401}{4}X-\frac{194481}{8}$.

The rank of this elliptic curve is $1$ and its generator is the point

$P=(X,Y)=(\frac{-77}{2},42)$. Because of this, the above elliptic curve has infinitely many rational points and we may obtain infinitely many solutions for the Diophantine equation too. By using this point $P=(X,Y)=(\frac{-77}{2},42)$, we get $(u,v)=(\frac{7}{6},\frac{-259}{18})$, and by calculating $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, from the above relations and after some simplifications and canceling the denominators of $x_{1}$, $y_{1}$, $X_{1}$, $Y_{1}$, we get the identity:

$7^{6}+7.(140)^{3}=14^{6}+7.(119)^{3}$.

Also we have: $2P=(X^{\prime},Y^{\prime})=(\frac{2009}{18},\frac{-36260}{27})$.

By using this new point $2P$, we obtain another solution for the Diophantine equation:

$251^{6}+7.(32089)^{3}=29^{6}+7.(40969)^{3}$

By choosing the other points on the elliptic curve such as

$nP$ ($n=3,4$, $\cdots$ ) we obtain infinitely many solutions for the Diophantine equation.

** Main Theorem 1.2****.**

*Consider the Diophantine equation:

$\sum_{i=1}^{n}a_{i}x_{i}^{6}+\sum_{i=1}^{m}b_{i}y_{i}^{3}=\sum_{i=1}^{n}a_{i}X_{i}^{6}-\sum_{i=1}^{m}b_{i}Y_{i}^{3}$*,

*where $n$, $m$ $\geq 1$ and $a_{i}$, $b_{i}$, are fixed arbitrary nonzero integers. Let $Y^{2}=X^{3}+FX^{2}+GX+H$, be an elliptic curve in which the coefficients $F$, $G$, and $H$ are all functions of $a_{i}$, $b_{i}$, and the other rational parameters $B_{i}$, $C_{i}$, $D_{i}$, and $h$, yet to be found later. If the elliptic curve has positive rank, depending on the values of $B_{i}$, $C_{i}$, $D_{i}$, and $h$, the Diophantine equation has infinitely many integer solutions.

Proof. Let: $x_{i}=D_{i}u+h$, $y_{i}=B_{i}u-C_{i}v$, $X_{i}=-D_{i}u+h$, and

$Y_{i}=B_{i}u+C_{i}v$*, where all variables are rational numbers. By substituting these variables in the above Diophantine equation, and after some simplifications, we get:

[TABLE]

We see that if

[TABLE]

to be square, say $q^{2}$, (This is possible by fixing the other parameters and choosing appropriate values for $h$.), we may transform (similar to the previous theorem) the above elliptic curve to an elliptic curve of the form

$y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}$*, where $a_{i}\in\mathbb{Q}$.

If the above elliptic curve has positive rank (For every arbitrary nonzero integers $a_{i}$ and $b_{i}$, this is done by choosing appropriate values for $B_{i}$, $C_{i}$, $D_{i}$, $h$.), by calculating $x_{i}$, $y_{i}$, $X_{i}$, $Y_{i}$, from the relations $x_{i}=D_{i}u+h$, $y_{i}=B_{i}u-C_{i}v$, $X_{i}=-D_{i}u+h$, and $Y_{i}=B_{i}u+C_{i}v$, and after some simplifications and canceling the denominators of $x_{i}=D_{i}u+h$, $y_{i}=B_{i}u-C_{i}v$, $X_{i}=-D_{i}u+h$, and $Y_{i}=B_{i}u+C_{i}v$, we may obtain infinitely many integer solutions for the Diophantine equation. Now the proof is complete.

Also we know that if $(x_{1},\cdots,x_{n},y_{1},\cdots,y_{m},X_{1},\cdots,X_{n},Y_{1},\cdots,Y_{m})$ is a solution for the Diophantine equations, then for every arbitrary $t$,

$(tx_{1},\cdots,tx_{n},t^{2}y_{1},\cdots,t^{2}y_{m},tX_{1},\cdots,tX_{n},t^{2}Y_{1},\cdots,t^{2}Y_{m})$ is a solution too.

Then if we obtain a rational solution, we may get an integer solution for the Diophantine equations by multiplying the both sides of the Diophantine equations by an appropriate $t$.

Finally we mention that each point on the elliptic curve can be represented in the form $(\frac{r}{s^{2}},\frac{t}{s^{3}})$, where $r$, $s$, $t$ $\in\mathbb{Z}$.

So if we put $nP=(\frac{r_{n}}{s_{n}^{2}},\frac{t_{n}}{s_{n}^{3}})$, that the point P is one of the elliptic curve generators, we may obtain a parametric solution for each case of Diophantine equations by using this new point $P^{\prime}=nP=(\frac{r_{n}}{s_{n}^{2}},\frac{t_{n}}{s_{n}^{3}})$. Also by choosing the other appropriate values of $B_{i}$, $C_{i}$, $D_{i}$, $h$, and getting the new elliptic curve of rank $\geq 1$( and repeating the above process) , we may obtain infinitely many nontrivial parametric solutions for each case of the above Diophantine equations.

We use the Sage software for calculating the rank of the elliptic curves. (see [4])