On the simultaneous Diophantine equations m.(x_1^k+....+x_{t_1}^k)=n.(y_1^k+....+y_{t_2}^k); k=1,3

TL;DR

This paper finds infinitely many solutions to specific simultaneous Diophantine equations involving sums of powers for k=1,3, by constructing parametric solutions and illustrating with examples.

Contribution

It introduces a method to generate infinitely many solutions for these equations using trivial parametric solutions and applies it to particular cases.

Findings

Established parametric solutions for the equations

Generated infinitely many solutions for specific cases

Provided explicit examples of solutions

Abstract

In this paper, we solve the simultaneous Diophantine equations m.(x_1^k+....+x_{t_1}^k)=n.(y_1^k+....+y_{t_2}^k); k=1,3, where t_1, t_2>3, and m, n are fixed arbitrary and relatively prime positive integers. This is done by choosing two appropriate trivial parametric solutions and obtaining infinitely many nontrivial parametric solutions. Also we work out some examples, in particular the Diophantine systems of A^k+B^k+C^k=D^k+E^4; k=1,3.

Equations

$$x^k+y^k+z^k=u^k+v^k+w^k;k=1,3$$

$$\begin{gathered} m\cdot (x_1^k+x_2^k+\cdots +x_{t_1}^k)=n\cdot (y_1^k+y_2^k+\cdots +y_{t_2}^k), \\ k=1,3 \end{gathered}$$

$$t=\frac{m{x}_1^2{X}_1+\cdots +m{x}_{t_1}^2{X}_{t_1}-n{y}_1^2{Y}_1-\cdots -n{y}_{t_2}^2{Y}_{t_2}}{-m{x}_1{X}_1^2-\cdots -m{x}_{t_1}{X}_{t_1}^2+n{y}_1{Y}_1^2+\cdots +n{y}_{t_2}{Y}_{t_2}}:=\frac{A}{B}.$$

Full text

1 Introduction

The cubic Diophantine equations has been studied by some mathematicians. The Diophantine equation $x^{3}+y^{3}=z^{3}$, which is a special case of Fermat’s last theorem, has no solution in integers.

Gerardin gave partial solutions of the simultaneous Diophantine equation (SDE)

[TABLE]

in 1915-16 (as quoted by Dickson, pp. 565, 713 of 4 (4)) and additional partial solutions were given by Bremner 1 (1). Subsequently, complete solutions were given in terms of cubic polynomials in four variables by Bremner and Brudno 2 (2), as well as by Labarthe 6 (6).

In 3 (3) Choudhry presented a complete four-parameter solution of ($1$) in terms of quadratic polynomials in which each parameter occurs only in the first degree.

The authors in a different paper used two different methods to solve the cubic Diophantine equation $x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=k\cdot(y_{1}^{3}+y_{2}^{3}+\cdots+y_{\frac{n}{k}}^{3})$, where $n\geq 3$, and $k\neq n$, is a divisor of $n$ ($\frac{n}{k}\geq 2$), and obtained infinitely many nontrivial parametric solutions (see 5 (5)).

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

[TABLE]

where $t_{1},t_{2}\geq 3$, $n$, $m$ are fixed arbitrary and relatively prime positive integers. To the best of our knowledge ($2$) has not already been considered by any other authors.

2 Main result

We prove the following main theorem:

Theorem 2.1

Let $t_{1},t_{2}\geq 3$, $n$, $m$ be fixed arbitrary and relatively prime positive integers. Then ($2$) has infinitely many nontrivial parametric solutions in non-zero integers $x_{1}$, $\cdots$, $x_{t_{1}}$, $y_{1}$, $\cdots$, $y_{t_{2}}$.

Proof

: Firstly, it is clear that if

$X=(x_{1},\cdots,x_{t_{1}},y_{1},\cdots,y_{t_{2}})$, and $Y=(X_{1},\cdots,X_{t_{1}},Y_{1},\cdots,Y_{t_{2}}),$

are two solutions of ($2$), then for any arbitrary rational numbers $t$, $X+tY$ is also a solution for $k=1$, i.e.,

$m\cdot[(x_{1}+tX_{1})+(x_{2}+tX_{2})+\cdots+(x_{t_{1}}+tX_{t_{1}})]=\\ n\cdot[(y_{1}+tY_{1})+(y_{2}+tY_{2})+\cdots+(y_{t_{2}}+tY_{t_{2}})]$.

We say that $X$ is a trivial parametric solution of ($2$), if it is nonzero and satisfies the system trivially. Let $X$, and $Y$ be two proper trivial parametric solutions of ($2$). (we will introduce them later.)

We suppose that $X+tY$, is also a solution for the case of $k=3$, where $t$ is a parameter. We wish to find $t$.

By plugging

$X+tY=(x_{1}+tX_{1},x_{2}+tX_{2},\cdots,x_{t_{1}}+tX_{t_{1}},y_{1}+tY_{1},y_{2}+tY_{2},\cdots,y_{t_{2}}+tY_{t_{2}})$,

into ($2$), we get:

$m\cdot[(x_{1}+tX_{1})^{3}+(x_{2}+tX_{2})^{3}+\cdots+(x_{t_{1}}+tX_{t_{1}})^{3}]=\\ n\cdot[(y_{1}+tY_{1})^{3}+(y_{2}+tY_{2})^{3}+\cdots+(y_{t_{2}}+tY_{t_{2}})^{3}]$.

Since $X$ and $Y$ are solutions of ($2$), after some simplifications, we obtain:

$3t^{2}(mx_{1}X_{1}^{2}+mx_{2}X_{2}^{2}+\cdots+mx_{t_{1}}X_{t_{1}}^{2}-ny_{1}Y_{1}^{2}-ny_{2}Y_{2}^{2}-\cdots-ny_{t_{2}}Y_{t_{2}}^{2})+$

$3t(mx_{1}^{2}X_{1}+mx_{2}^{2}X_{2}+\cdots+mx_{t_{1}}^{2}X_{t_{1}}-ny_{1}^{2}Y_{1}-ny_{2}^{2}Y_{2}-\cdots-ny_{t_{2}}^{2}Y_{t_{2}})=0.$

Therefore $t=0$ or

[TABLE]

By substituting $t$ in the above expressions, and clearing the denominator $B^{3}$, we get an integer solution of ($2$) as follows:

$(x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{t_{1}},y^{\prime}_{1},y^{\prime}_{2},\cdots,y^{\prime}_{t_{2}})=$

$(x_{1}B+AX_{1},x_{2}B+AX_{2},\cdots,x_{t_{1}}B+AX_{t_{1}},y_{1}B+AY_{1},y_{2}B+AY_{2},\cdots,y_{t_{2}}B+AY_{t_{2}})$.

If we pick up trivial parametric solutions $X$ and $Y$ properly, we will get a nontrivial parametric solution of ($2$). We should mention that not every trivial parametric solutions of $X$ and $Y$ necessarily give rise to a nontrivial parametric solution of ($2$). So the trivial parametric solutions must be chosen properly.

Now we introduce the proper trivial parametric solutions.

For the sake of simplicity, we only write down the trivial parametric solutions for the left hand side of ($2$). It is clear that the trivial parametric solutions for the right hand side of ($2$) can be similarly found by the given trivial parametric solutions of the left hand side of ($2$).

Let $p_{i}$, $q_{i}$, $s_{i}$, r_{i}$$\in\mathbb{Z}.

There are $4$ different possible cases for $t_{1}$:

$1$. $t_{1}=2\alpha+1$, $\alpha$ is even.

$X_{left}=(x_{1},x_{2},\cdots,x_{t_{1}})=(p_{1},-p_{1},p_{2},-p_{2},\cdots,p_{\alpha},-p_{\alpha},0)$, and

$Y_{left}=(X_{1},X_{2},\cdots,X_{t_{1}})=$

$(r_{1},r_{2},-r_{1},-r_{2},r_{3},r_{4},-r_{3},-r_{4},\cdots,r_{\alpha-1},r_{\alpha},-r_{\alpha-1},0,-r_{\alpha}).$

$2$. $t_{1}=2\alpha+1$, $\alpha$ is odd.

$X_{left}=(p_{1},-p_{1},p_{2},-p_{2},\cdots,p_{\alpha},-p_{\alpha},0)$, and

$Y_{left}=(r_{1},r_{2},-r_{1},-r_{2},\cdots,-r_{\alpha-2},-r_{\alpha-1},r_{\alpha},0,-r_{\alpha})$.

$3$. $t_{1}=2\alpha$, $\alpha$ is even.

$X_{left}=(p_{1},-p_{1},p_{2},-p_{2},\cdots,p_{\alpha},-p_{\alpha})$, and

$Y_{left}=(r_{1},r_{2},-r_{1},-r_{2},\cdots,r_{\alpha-1},r_{\alpha},-r_{\alpha-1},-r_{\alpha}).$

$4$. $t_{1}=2\alpha$, $\alpha$ is odd.

$X_{left}=(p_{1},-p_{1},p_{2},-p_{2},\cdots,p_{\alpha},-p_{\alpha})$, and

$Y_{left}=(r_{1},r_{2},-r_{1},-r_{2},\cdots,r_{\alpha-2},r_{\alpha-1},-r_{\alpha-2},r_{\alpha},-r_{\alpha-1},-r_{\alpha}).$

Finally, it can be easily shown that for every $i\neq j$, we have:

$x_{i}B+AX_{i}$ $\neq\pm$ $(x_{j}B+AX_{j})$, $y_{i}B+AY_{i}$ $\neq\pm$ $(y_{j}B+AY_{j})$ and

x_{i}B+AX_{i}$$\neq\pm $(y_{j}B+AY_{j})$,

i.e., it is really a nontrivial parametric solution in terms of the parameters $p_{i}$, $q_{i}$, $r_{i}$ and $s_{i}$.

It is clear that by fixing all of the parameters $p_{i}$, $q_{i}$, $r_{i}$ and $s_{i}$, but one parameter, we can obtain a nontrivial one parameter parametric solution of ($2$) and by changing properly the fixed values, finally get infinitely many nontrivial one parameter parametric solutions of ($2$).

Now, the proof of the theorem is completed. $\spadesuit$

In the remaining part, we worked out some examples.

3 Application to examples

Example 1

$m\cdot(x_{1}^{k}+x_{2}^{k}+x_{3}^{k})=n\cdot(y_{1}^{k}+y_{2}^{k}+y_{3}^{k});$* $k=1,3$.

Trivial parametric solutions (case $2$):

$X=(x_{1},x_{2},x_{3},y_{1},y_{2},y_{3})=$* $(p_{1},-p_{1},0,q_{1},-q_{1},0)$,*

$Y=(X_{1},X_{2},X_{3},Y_{1},Y_{2},Y_{3})=$* $(r_{1},0,-r_{1},s_{1},0,-s_{1})$,*

$A=mp_{1}^{2}r_{1}-nq_{1}^{2}s_{1}$, $B=-mp_{1}r_{1}^{2}+nq_{1}s_{1}^{2}$,

nontrivial parametric solution:

$x_{1}=p_{1}B+r_{1}A,$

$x_{2}=-p_{1}B,$

$x_{3}=-r_{1}A,$

$y_{1}=q_{1}B+s_{1}A,$

$y_{2}=-q_{1}B,$

$y_{3}=-s_{1}A.$

Numerical example:

$p_{1}=4$, $q_{1}=1$, $r_{1}=2$, $s_{1}=3$,

Solution:

$m\cdot[(30n)^{k}+(64m-36n)^{k}+(-64m+6n)^{k}]=\\ n\cdot[(80m)^{k}+(16m-9n)^{k}+(-96m+9n)^{k}];$* $k=1,3$.*

Example 2

$m\cdot(x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k})=n\cdot(y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k});$* $k=1,3$.

Trivial parametric solutions (case $3$):

$X=(x_{1},x_{2},x_{3},x_{4},y_{1},y_{2},y_{3},y_{4})=$* $(p_{1},-p_{1},p_{2},-p_{2},q_{1},-q_{1},q_{2},-q_{2})$,*

$Y=(X_{1},X_{2},X_{3},X_{4},Y_{1},Y_{2},Y_{3},Y_{4})=$* $(r_{1},r_{2},-r_{1},-r_{2},s_{1},s_{2},-s_{1},-s_{2})$,*

Numerical example:

$p_{1}=2$, $p_{2}=5$, $q_{1}=1$, $q_{2}=3$, $r_{1}=6$, $r_{2}=7$, $s_{1}=4$, $s_{2}=9$,

Solution:

$m\cdot[(-1456m+104n)^{k}+(-2093m+1248n)^{k}+(2093m-1924n)^{k}+(1456m+572n)^{k}]=n\cdot[(-1001m+156n)^{k}+(-2548m+1196n)^{k}+(1365m-1196n)^{k}+(2184m-156n)^{k}];$* $k=1,3$.*

Example 3

$x_{1}^{k}+x_{2}^{k}+\cdots+x_{5}^{k}=n\cdot(y_{1}^{k}+y_{2}^{k}+\cdots+y_{5}^{k});\ k=1,3.$*

*Trivial parametric solution (case $2$): *

$X=(x_{1},x_{2},x_{3},x_{4},x_{5},y_{1},y_{2},y_{3},y_{4},y_{5})=$* $(p_{1},-p_{1},p_{2},-p_{2},0,q_{1},-q_{1},q_{2},-q_{2},0)$, *

$Y=(X_{1},X_{2},X_{3},X_{4},X_{5},Y_{1},Y_{2},Y_{3},Y_{4},Y_{5})=$* $(r_{1},r_{2},-r_{1},0,-r_{2},s_{1},s_{2},-s_{1},0,-s_{2})$, *

$A=p_{1}^{2}r_{1}+p_{1}^{2}r_{2}-p_{2}^{2}r_{1}-nq_{1}^{2}s_{1}-nq_{1}^{2}s_{2}+nq_{2}^{2}s_{1}$*, *

$B=-p_{1}r_{1}^{2}+p_{1}r_{2}^{2}-p_{2}r_{1}^{2}+nq_{1}s_{1}^{2}-nq_{1}s_{2}^{2}+nq_{2}s_{1}^{2},$

nontrivial parametric solution:

$x_{1}=x_{1}B+X_{1}A=p_{1}B+r_{1}A$,

$x_{2}=x_{2}B+X_{2}A=-p_{1}B+r_{2}A$,

$x_{3}=x_{3}B+X_{3}A=p_{2}B-r_{1}A$,

$x_{4}=x_{4}B+X_{4}A=-p_{2}B$,

$x_{5}=x_{5}B+X_{5}A=-r_{2}A$,

$y_{1}=y_{1}B+Y_{1}A=q_{1}B+s_{1}A$,

$y_{2}=y_{2}B+Y_{2}A=-q_{1}B+s_{2}A$,

$y_{3}=y_{3}B+Y_{3}A=q_{2}B-s_{1}A$,

$y_{4}=y_{4}B+Y_{4}A=-q_{2}B$,

$y_{5}=y_{5}B+Y_{5}A=-s_{2}A$.

Numerical example $1$:

$p_{1}=5$, $p_{2}=6$, $q_{1}=7$, $q_{2}=8$, $r_{1}=1$, $r_{2}=2$, $s_{1}=3$, $s_{2}=4$,

Solution:

$(84-36n)^{k}+(33-417n)^{k}+(15+289n)^{k}+(-54-138n)^{k}+(-78+302n)^{k}=n\cdot[(180-292n)^{k}+(93-765n)^{k}+(-45+637n)^{k}+(-72-184n)^{k}+(-156+604n)^{k}];$* $k=1,3$. *

$n=0$* $\Longrightarrow$ $5^{3}+11^{3}+28^{3}=18^{3}+26^{3}$.*

It is interesting to see that $5+11+28=18+26$, too.

Remark 1

By fixing the values $p_{2}$, $r_{1}$, $r_{2}$, $n=0$, and letting $p_{1}=:p$, as a parameter, we can obtain infinitely many nontrivial parametric solutions for the simultaneous Diophantine equations of the form $A^{k}+B^{k}+C^{k}=D^{k}+E^{k}$, $k=1,3$.

Numerical example $2$:

$n=0$, $r_{1}=1$, $r_{2}=3$, $p_{2}=5$, $p_{1}=:p$,

Solution:

$(12p^{2}-5p-25)^{k}+(4p^{2}+5p-75)^{k}+(-4p^{2}+40p)^{k}=\\ (40p-25)^{k}+(12p^{2}-75)^{k};k=1,3$.

Acknowledgements.

We are very grateful to the referee for the careful reading of the paper and giving several useful comments which improved the quality of the paper.