A note on the high power Diophantine equations

TL;DR

This paper investigates specific high-power Diophantine equations, providing parametric solutions, methods for generating new solutions, and proving the existence of infinitely many solutions under certain conditions.

Contribution

It offers new parametric solutions for particular high-power Diophantine equations and introduces methods to generate additional solutions, expanding understanding of their solution spaces.

Findings

Parametric solutions for the equations with u=1,3

Method to produce new solutions when u=3

Existence of infinitely many solutions under specified conditions

Abstract

In this paper, we solve the simultaneous Diophantine equations(SDE) x_1^u+...+x_n^u=k(y_1^u+...+y_{n/k}); u=1,3, where n >3, and k< n, is a divisor of n , and obtain nontrivial parametric solution for them. Furthermore we present a method for producing another solution for the above Diophantine equation (DE) for the case u = 3, when a solution is given. We work out some examples and find nontrivial parametric solutions for each case in nonzero integers. Also we prove that the other DE p_1x_1^{a_1}+....+p_nx_n^{a_n}=q_1y_1^{b_1}+...+q_my_m^{b_m} , has parametric solution and infinitely many solutions in nonzero integers with the condition that: there is a i such that p_i=1, and (a_i,a_1....a_{i-1}a_{i+1}...b_1...b_m)=1, or there is a j such that q_j=1, and (b_j,a_1...a_nb_1...b_{j-1}b_{j+1}..b_m)=1. Finally we study the DE x^a_y^b=z^c.

Equations

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

$$x_1^{\mu }+x_2^{\mu }+\cdots +x_n^{\mu }=k\cdot (y_1^{\mu }+y_2^{\mu }+\cdots +y_{\frac{n}{k}}^{\mu }),\mu =1,3,$$

$$\sum _{i=1}^n{p}_i.x_i^{a_i}=\sum _{j=1}^m{q}_j.y_j^{b_j}$$

$$t=\frac{x_1^2{X}_1+\cdots +x_n^2{X}_n-k{y}_1^2{Y}_1-\cdots -k{y}_{\frac{n}{k}}^2{Y}_{\frac{n}{k}}}{-x_1{X}_1^2-\cdots -x_n{X}_n^2+k{y}_1{Y}_1^2+\cdots +k{y}_{\frac{n}{k}}{Y}_{\frac{n}{k}}^2}:=\frac{A}{B}.$$

$$\sum _{i=1}^n{x}_i^3=k\cdot \sum _{j=1}^{\frac{n}{k}}{y}_j^3,$$

$$x_1^{a_1}=q_1{y}_1^{b_1}+\cdots +q_m{y}_m^{b_m}-p_2{x}_2^{a_2}-p_3{x}_3^{a_3}-\cdots -p_n{x}_n^{a_n}.$$

Full text

Also we prove that the other DE $\sum_{i=1}^{n}p_{i}\cdot x_{i}^{a_{i}}=\sum_{j=1}^{m}q_{j}\cdot y_{j}^{b_{j}}$, has parametric solution and infinitely many solutions in nonzero integers with the condition that: there is a $i$ such that $p_{i}=1$, and

$(a_{i},a_{1}\cdot a_{2}\cdots a_{i-1}\cdot a_{i+1}\cdots a_{n}\cdot b_{1}\cdot b_{2}\cdots b_{m})=1$, or there is a $j$ such that $q_{j}=1$, and $(b_{j},a_{1}\cdots a_{n}\cdot b_{1}\cdots b_{j-1}\cdot b_{j+1}\cdots b_{m})=1$. Finally we study the DE $x^{a}+y^{b}=z^{c}$.

1 Introduction

The cubic Diophantine equations has been studied by some mathematicians.

Gerardin gave partial solutions of the SDE

[TABLE]

in 1915-16 (as quoted by Dickson, pp. 565, 713 of 5 (5)) 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 7 (7). In 4 (4) Choudhry presented a complete four-parameter solution of (1) in terms of quadratic polynomials in which each parameter occurs only in the first degree.

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

[TABLE]

where $n\geq 3$, and $k\neq n$, is a divisor of $n$ ($\frac{n}{k}\geq 2$).

Also we study the other DE

[TABLE]

where $m,n,a_{i},b_{i}\in\mathbb{N}$ and $p_{i},q_{i}\in\mathbb{Z}$. This is a generalization of the Fermat’s equation.

A common generalization of Fermat’s equation is $Ax^{a}+By^{b}+Cz^{c}=0$, where $a$, $b$, $c$ $\in\mathbb{N}_{\geq 2}$ and $A,B,C\in\mathbb{Z}$ are given integers with $ABC\neq 0$ and $x,y,z\in\mathbb{Z}$ are variables.

In $1995$, Darmond and Granville (See 6 (6)) proved:

If $A,B,C\in\mathbb{Z}$ , $ABC\neq 0$ and $a,b,c\in\mathbb{N}_{\geq 2}$ be such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<1$, then the equation $Ax^{a}+By^{b}+Cz^{c}=0$ has only finitely many solutions $x,y,z\in\mathbb{Z}$ with $(x,y,z)=1$.

The following theorem is due to Beukers (See 3 (3)):

Let $A,B,C\in\mathbb{Z}$ , $ABC\neq 0$ and $a,b,c\in\mathbb{N}_{\geq 2}$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>1$. Then the equation $Ax^{a}+By^{b}+Cz^{c}=0$ has either zero or infinitely many solutions $x,y,z\in\mathbb{Z}$ with $(x,y,z)=1$.

To the best of our knowledge the SDE (2) and the DE (3) has not already been considered by any other authors.

We prove the following main theorems:

Main Theorem 1

Let $n\geq 3$, and $k\neq n$, be a divisor of $n$ ($\frac{n}{k}\geq 2$). Then the SDE (2) have infinitely many nontrivial parametric solutions in nonzero integers.

Main Theorem 2

Let $m,n,a_{i},b_{i}\in\mathbb{N}$ and $p_{i},q_{i}\in\mathbb{Z}$. Suppose that there is a $i$ such that $p_{i}=1$, and $(a_{i},a_{1}\cdot a_{2}\cdots a_{i-1}\cdot a_{i+1}\cdots a_{n}\cdot b_{1}\cdot b_{2}\cdots b_{m})=1$ or there is a $j$ such that $q_{j}=1$, and $(b_{j},a_{1}\cdots a_{n}\cdot b_{1}\cdots b_{j-1}\cdot b_{j+1}\cdots b_{m})=1$. Then the DE (3) has parametric solution and infinitely many solutions in nonzero integers. This solves the DE $x^{a}+y^{b}=z^{c}$ in the special cases of $(a,b,c)=1$, or $(a,b,c)=2$.

2 The SDE ${\displaystyle\sum_{i=1}^{n}x_{i}^{\mu}=k\cdot\sum_{j=1}^{\frac{n}{k}}y_{j}^{\mu}}$; $\mu=1,3$

In this section, we proof the first main theorem.

Proof

: Firstly, it is clear that if

$X=(x_{1},\cdots,x_{n},y_{1},\cdots,y_{\frac{n}{k}})$, and $Y=(X_{1},\cdots,X_{n},Y_{1},\cdots,Y_{\frac{n}{k}}),$

be two solutions for the SDE (2), then for any arbitrary rational number $t$, $X+tY$ is also a solution for $\mu=1$, i.e.,

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

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

We suppose that $X+tY$, is also a solution for the case $\mu=3$, where $t$ is a parameter, we wish to find $t$. By plugging

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

into the SDE (2), we get:

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

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

$3t^{2}(x_{1}X_{1}^{2}+x_{2}X_{2}^{2}+\cdots+x_{n}X_{n}^{2}-ky_{1}Y_{1}^{2}-ky_{2}Y_{2}^{2}-\cdots-ky_{\frac{n}{k}}Y_{\frac{n}{k}}^{2})+$

$3t(x_{1}^{2}X_{1}+x_{2}^{2}X_{2}+\cdots+x_{n}^{2}X_{n}-ky_{1}^{2}Y_{1}-ky_{2}^{2}Y_{2}-\cdots-ky_{\frac{n}{k}}^{2}Y_{\frac{n}{k}})=0.$

Therefore $t=0$ or

[TABLE]

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

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

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

If we pick up trivial parametric solutions $X$ and $Y$ properly, we will get a nontrivial parametric solution for the SDE (2). We should mention that not every trivial parametric solutions of $X$ and $Y$ necessarily give rise to a nontrivial parametric solution for the SDE (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 the SDE (2). It is clear that the trivial parametric solutions for the right hand side of the SDE (2) can be similarly found by the given trivial parametric solutions of the left hand side of the SDE (2).

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

A) The case $n\geq 3$ and $\frac{n}{k}\geq 3$:

There are $4$ different possible cases for $n$:

$1$. $n=2\alpha+1$, $\alpha$ is even.

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

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

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

$2$. $n=2\alpha+1$, $\alpha$ is odd.

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

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

$3$. $n=2\alpha$, $\alpha$ is even.

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

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

$4$. $n=2\alpha$, $\alpha$ is odd.

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

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

B) The case $n\geq 3$ and $\frac{n}{k}=2$:

$5$. $n=2k\geq 3$ and $\frac{n}{k}=2$.

$X=(p_{1},-p_{1},p_{2},-p_{2},\cdots,p_{k},-p_{k},q_{1},-q_{1})$,

$Y=(r_{1},r_{2},r_{1},r_{2},\cdots,r_{1},r_{2})$.

(In the final case we present the trivial parametric solutions X, and Y, completely.)

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 for the SDE (2) and by changing properly the fixed values, finally get infinitely many nontrivial one parameter parametric solutions for the SDE (2).

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

Now, we work out some examples.

Example 1

$a^{\mu}+b^{\mu}+c^{\mu}+d^{\mu}+e^{\mu}+f^{\mu}=3\cdot(g^{\mu}+h^{\mu})$*; $\mu=1,3$.

Trivial parametric solutions (case $5$):

$X=(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7},x_{8})=(p_{1},-p_{1},p_{2},-p_{2},p_{3},-p_{3},q_{1},-q_{1}),$

$Y=(y_{1},y_{2},y_{3},y_{4},y_{5},y_{6},y_{7},y_{8})=(r_{1},r_{2},r_{1},r_{2},r_{1},r_{2},r_{1},r_{2})$,

$A=x_{1}^{2}y_{1}+x_{2}^{2}y_{2}+x_{3}^{2}y_{3}+x_{4}^{2}y_{4}+x_{5}^{2}y_{5}+x_{6}^{2}y_{6}-3x_{7}^{2}y_{7}-3x_{8}^{2}y_{8}$,

$B=-x_{1}y_{1}^{2}-x_{2}y_{2}^{2}-x_{3}y_{3}^{2}-x_{4}y_{4}^{2}-x_{5}y_{5}^{2}-x_{6}y_{6}^{2}+3x_{7}y_{7}^{2}+3x_{8}y_{8}^{2}$,

nontrivial parametric solution:

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

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

$c=x_{3}B+Ay_{3}=p_{2}B+r_{1}A$,

$d=x_{4}B+Ay_{4}=-p_{2}B+r_{2}A$,

$e=x_{5}B+Ay_{5}=p_{3}B+r_{1}A$,

$f=x_{6}B+Ay_{6}=-p_{3}B+r_{2}A$,

$g=x_{7}B+Ay_{7}=q_{1}B+r_{1}A$,

$h=x_{8}B+Ay_{8}=-q_{1}B+r_{2}A$.

Example $1$:

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

Solution:

$371^{\mu}+756^{\mu}+476^{\mu}+651^{\mu}+581^{\mu}+546^{\mu}=3\cdot(686^{\mu}+441^{\mu})$; $\mu=1,3$..

Example $2$:

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

Solution:

$257^{\mu}+458^{\mu}+167^{\mu}+548^{\mu}+347^{\mu}+368^{\mu}=3\cdot(482^{\mu}+233^{\mu})$; $\mu=1,3$.

Example 2

$a^{\mu}+b^{\mu}+c^{\mu}+d^{\mu}+e^{\mu}+f^{\mu}+g^{\mu}+h^{\mu}=2\cdot(i^{\mu}+j^{\mu}+k^{\mu}+l^{\mu})$*; $\mu=1,3$.

Trivial parametric solutions (case $3$):

$X=(x_{1},x_{2},\cdots,x_{12})=$* $(p_{1},-p_{1},p_{2},-p_{2},p_{3},-p_{3},p_{4},-p_{4},q_{1},-q_{1},q_{2},-q_{2})$,*

$Y=(y_{1},y_{2},\cdots,y_{12})=$* $(r_{1},r_{2},-r_{1},-r_{2},r_{3},r_{4},-r_{3},-r_{4},s_{1},s_{2},-s_{1},-s_{2})$, *

Example:

$p_{1}=2$, $p_{2}=5$, $p_{3}=7$, $p_{4}=6$, $q_{1}=5$, $q_{2}=3$, $r_{1}=7$, $r_{2}=8$, $r_{3}=9$, $r_{4}=9$, $s_{1}=10$, $s_{2}=6$, $A=-593$, $B=1129$,

Solution:

$(-1893)^{\mu}+(-7002)^{\mu}+9796^{\mu}+(-901)^{\mu}+2566^{\mu}+(-13240)^{\mu}+12111^{\mu}+(-1437)^{\mu}=2\cdot[(-285)^{\mu}+(-9203)^{\mu}+9317^{\mu}+171^{\mu}]$; $\mu=1,3$

Example 3

$a^{\mu}+b^{\mu}+c^{\mu}+\cdots+m^{\mu}+n^{\mu}+o^{\mu}=5\cdot(p^{\mu}+q^{\mu}+r^{\mu})$*; $\mu=1,3$.

Trivial parametric solutions (case $2$):

$X=(x_{1},x_{2},\cdots,x_{18})=$**

$(p_{1},-p_{1},p_{2},-p_{2},p_{3},-p_{3},p_{4},-p_{4},p_{5},-p_{5},p_{6},-p_{6},p_{7},-p_{7},0,q_{1},-q_{1},0)$,

$Y=(y_{1},y_{2},\cdots,y_{18})=$

$(r_{1},r_{2},-r_{1},-r_{2},r_{3},r_{4},-r_{3},-r_{4},r_{5},r_{6},-r_{5},-r_{6},r_{7},0,-r_{7},s_{1},0,-s_{1})$,

Example:

$p_{1}=1$, $p_{2}=3$, $p_{3}=5$, $p_{4}=4$, $p_{5}=1$, $p_{6}=2$, $p_{7}=3$, $q_{1}=2$, $r_{1}=2$, $r_{2}=3$, $r_{3}=5$, $r_{4}=1$, $r_{5}=6$, $r_{6}=6$, , $r_{7}=7$, $s_{1}=3$, $A=-19$, $B=-253$,

Solution:

$(-291)^{\mu}+196^{\mu}+(-721)^{\mu}+816^{\mu}+(-1360)^{\mu}+1246^{\mu}+(-917)^{\mu}+1031^{\mu}+(-367)^{\mu}+139^{\mu}+(-392)^{\mu}+620^{\mu}+(-892)^{\mu}+759^{\mu}+133^{\mu}=5\cdot[(-563)^{\mu}+506^{\mu}+57^{\mu}]$; $\mu=1,3$

Example 4

$a^{\mu}+b^{\mu}+c^{\mu}+d^{\mu}=2\cdot(e^{\mu}+f^{\mu})$*; $\mu=1,3$.

Trivial parametric solutions(case $5$):

$X=(u,u,v,v,u,v)$,

$Y=(-v,1,-1,v,-u,u)$,

nontrivial parametric solution:

$a=-u^{2}-uv+uv^{3}+2u^{4}-vu^{2}+v^{3}-v^{4}$,

$b=-u^{2}v^{2}-uv-uv^{3}+2u^{4}+2u^{3}v-u^{2}v-v^{2}+v^{3}+2u^{3}-2uv^{2}$,

$c=-v^{4}+2u^{2}v^{2}-v^{3}-2u^{3}-u^{2}-uv-uv^{3}+2u^{3}v+u^{2}v+2uv^{2}$,

$d=u^{2}v^{2}-3uv^{3}+4u^{3}v-uv-v^{2}+vu^{2}-v^{3}$,

$e=u^{2}v^{2}-u^{2}-u^{3}-uv-2uv^{3}+3u^{3}v+uv^{2}$,

$f=-uv-v^{2}-v^{4}+u^{3}v+u^{3}-uv^{2}+2u^{4}$.

Example:

$u=2$, $v=1$, or $u=10$, $v=7$, or $u=100$, $v=11$, or $u=100$, $v=1$,

Solutions:

$1^{\mu}+4^{\mu}+5^{\mu}+8^{\mu}=2\cdot(2^{\mu}+7^{\mu})$; $\mu=1,3$.

$201^{\mu}+257^{\mu}+168^{\mu}+224^{\mu}=2\cdot(180^{\mu}+245^{\mu})$; $\mu=1,3$.

$900895^{\mu}+1002355^{\mu}+100874^{\mu}+202334^{\mu}=2\cdot(148400^{\mu}+954829^{\mu})$; $\mu=1,3$.

$1^{\mu}+201^{\mu}+10000^{\mu}+10200^{\mu}=2\cdot(100^{\mu}+10101^{\mu})$; $\mu=1,3$.

Example 5

$a^{\mu}+b^{\mu}+c^{\mu}+d^{\mu}+e^{\mu}+f^{\mu}=2\cdot(g^{\mu}+h^{\mu}+i^{\mu})$*; $\mu=1,3$.

Trivial parametric solutions (case: $4$, $1$):

$X=(1,v,u,v,-u,-1,v,u,-u)$

$Y=(v,v,u,1,-1,-u,-u,v,u)$,

nontrivial parametric solution:

$a=-3u^{3}+v^{4}-v+u+u^{2}+vu^{2}-uv+2v^{2}u-u^{3}v+2v^{3}u-2v^{2}u^{2}$,

$b=-4u^{3}v+4v^{3}u$,

$c=-4u^{3}+4v^{2}u^{2}$,

$d=-v^{4}+v-u^{3}-u^{2}-u+uv-3u^{3}v+2u^{2}v^{2}+2v^{3}u-u^{2}v+2v^{2}u$,

$e=3u^{4}-v-v^{3}-v^{2}+u+uv-2v^{2}u^{2}+v^{2}u+v^{3}u-2vu^{3}-2v^{2}u+2u^{2}v$,

$f=4u^{3}+u^{4}+v-u+v^{2}+v^{3}-3v^{2}u-2vu^{2}-uv-uv^{3}-2u^{2}v^{2}+2u^{3}v$,

$g=u^{4}+u^{3}+u^{2}-v^{3}-v^{4}-v^{2}-u^{3}v+v^{3}u+u^{2}v-uv^{2}$,

$h=-3u^{4}+u^{2}+u^{3}+v^{2}+v^{3}+v^{4}-2uv+u^{3}v+v^{3}u-v^{2}u-u^{2}v$,

$i=2u^{4}-2u^{3}-2u^{2}+2uv+2v^{3}u+2v^{2}u-4vu^{3}$.

Example:

$u=2$, $v=1$, or $u=2$, $v=10$, or $u=7$, $v=100$,

Solutions:

$(-8)^{\mu}+(-8)^{\mu}+(-16)^{\mu}+(-8)^{\mu}+11^{\mu}+13^{\mu}=2\cdot[7^{\mu}+(-11)^{\mu}+(-4)^{\mu}]$*; $\mu=1,3$. *

$563^{\mu}+320^{\mu}+64^{\mu}+(-211)^{\mu}+(-5)^{\mu}+(-91)^{\mu}=2\cdot[(-388)^{\mu}+536^{\mu}+172^{\mu}];$* $\mu=1,3$. *

$173777^{\mu}+42800^{\mu}+2996^{\mu}+(-130549)^{\mu}+7510^{\mu}+(-10934)^{\mu}=2\cdot[(-144557)^{\mu}+165839^{\mu}+21518^{\mu}]$; $\mu=1,3$.

Example 6

$a^{\mu}+b^{\mu}+c^{\mu}=d^{\mu}+e^{\mu}+f^{\mu}$*; $\mu=1,3$.

Trivial parametric solutions (case $2$):

$X=(u,v,1,u,v,1)$,

$Y=(v,-1,-u,-1,-u,v)$,

nontrivial parametric solution:

$a=-u^{3}-v^{3}+vu^{3}-v^{3}u-2uv+v^{2}u-vu^{2}+u^{2}-v^{2}$,

$b=-uv^{3}-vu^{2}+uv+v^{2}u^{2}+v^{3}-u^{2}v+u+u^{2}+v^{2}u+v$,

$c=-uv^{2}-v+u+vu^{2}+v^{2}-u^{3}v+v^{2}u+u^{3}+v^{2}u^{2}+uv$,

$d=-u^{2}v^{2}-uv-u^{3}+u^{2}+vu^{3}+2uv^{2}-u^{2}v+v^{2}+u+u^{2}+v$,

$e=-uv^{3}-v^{2}-vu^{2}+2uv+2u^{2}v^{2}+v^{3}-u^{3}v+uv^{2}+u^{2}+u^{3}$,

$f=-uv^{2}-v-u^{2}+u+u^{2}v^{2}-v^{3}-uv-v^{3}u$.

Example:

$u=10$, $v=3$,

Solution:

$762^{\mu}+145^{\mu}+251^{\mu}=195^{\mu}+377^{\mu}+586^{\mu}$; $\mu=1,3$.

Example 7

$a^{\mu}+b^{\mu}+c^{\mu}+d^{\mu}+e^{\mu}+f^{\mu}+g^{\mu}+h^{\mu}+i^{\mu}=3\cdot(j^{\mu}+k^{\mu}+l^{\mu})$*; $\mu=1,3$.

Trivial parametric solutions (case: $1$, $2$):

$X=(x_{1},x_{2},\cdots,x_{12})=$* $(u,v,-u,t,-s,-t,0,s,-v,t,0,-t)$,*

$Y=(y_{1},y_{2},\cdots,y_{12})=$* $(v,t,-s,u,-t,s,-v,-u,0,0,t,-t)$,*

Example $1$:

$u=1$, $v=2$, $s=3$, $t=10$,

Solution:

$1168^{\mu}+7346^{\mu}+(-2003)^{\mu}+(-4185)^{\mu}+(-6844)^{\mu}+7525^{\mu}+(-1670)^{\mu}+(-2341)^{\mu}+1004^{\mu}=3\cdot[(-5020)^{\mu}+8350^{\mu}+(-3330)^{\mu}]$*; $\mu=1,3$. *

Example $2$:

$u=1$, $v=10$, $s=20$, $t=30$,

Solution:

$83515^{\mu}+208720^{\mu}+(-173005)^{\mu}+(-170301)^{\mu}+(-148970)^{\mu}+358230^{\mu}+(-89490)^{\mu}+(-128449)^{\mu}+59750^{\mu}=3\cdot[(-179250)^{\mu}+268470^{\mu}+(-89220)^{\mu}]$; $\mu=1,3$.

3 Producing another solution of the DE ${\displaystyle\sum_{i=1}^{n}x_{i}^{3}=k\cdot\sum_{j=1}^{\frac{n}{k}}y_{j}^{3}}$, when a solution is given.

Let $X=(x_{1},x_{2},\cdots,x_{n},y_{1},y_{2},\cdots,y_{\frac{n}{k}})$ be a primitive solution for the DE

[TABLE]

under conditions that :

$x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}}\neq 0$, and

$-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+2ky_{2}^{2}+\cdots+2ky_{\frac{n}{k}}^{2}\neq 0$.

(These conditions will be necessary because we wish the other solution not to be trivial.)

We try to find another solution by using the first solution.

Define two variables function:

$F(x,y):=(2x_{1}x+y)^{3}+(2x_{2}x+y)^{3}+\cdots+(2x_{n}x+y)^{3}$

$-k(2y_{1}x+y)^{3}-k(2y_{2}x+y)^{3}-\cdots-k(2y_{\frac{n}{k}}x+y)^{3}$.

(Also, we may define: $F(x,y):=(tx_{1}x+y)^{3}+(tx_{2}x+y)^{3}+\cdots+(tx_{n}x+y)^{3}$

$-k(ty_{1}x+y)^{3}-k(ty_{2}x+y)^{3}-\cdots-k(ty_{\frac{n}{k}}x+y)^{3}$, where $t$ is an arbitrary integer.)

It is clear that if $F(x,y)=0$, then

$(2x_{1}x+y,2x_{2}x+y,\cdots,2x_{n}x+y,2y_{1}x+y,2y_{2}x+y,\cdots,2y_{\frac{n}{k}}x+y)$,

is another solution for the DE (4). We have

$F(x,y)=6xy[(2x_{1}^{2}+2x_{2}^{2}+\cdots+2x_{n}^{2}-2ky_{1}^{2}-\cdots-2ky_{\frac{n}{k}}^{2})x+\\ (x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}})y].$

Therefore if we define

$x:=x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}}\neq 0$,

and

$y:=-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+\cdots+2ky_{\frac{n}{k}}^{2}\neq 0$,

then we get $F(x,y)=0$, which gives rise to another solution for the DE (4):

$x^{\prime}_{1}=2x_{1}x+y=2x_{1}(x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}})+$

$(-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+\cdots+2ky_{\frac{n}{k}}^{2}),$

$x^{\prime}_{2}=2x_{2}x+y=2x_{2}(x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}})+$

$(-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+\cdots+2ky_{\frac{n}{k}}^{2}),$

⋮

$x^{\prime}_{n}=2x_{n}x+y=2x_{2}(x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}})+$

$(-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+\cdots+2ky_{\frac{n}{k}}^{2}),$

$y^{\prime}_{1}=2y_{1}x+y=2y_{1}(x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}})+$

$(-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+\cdots+2ky_{\frac{n}{k}}^{2}),$

$y^{\prime}_{2}=2y_{2}x+y=2y_{2}(x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}})+$

$(-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+\cdots+2ky_{\frac{n}{k}}^{2}),$

⋮

$y^{\prime}_{\frac{n}{k}}=2y_{\frac{n}{k}}x+y=2y_{2}(x_{1}+x_{2}+\cdots+x_{n}-ky_{1}-ky_{2}-\cdots-ky_{\frac{n}{k}})+$

$(-2x_{1}^{2}-2x_{2}^{2}-\cdots-2x_{n}^{2}+2ky_{1}^{2}+\cdots+2ky_{\frac{n}{k}}^{2})$.

By continuing this method for the new obtained solution, we get infinitely many solutions for the DE (4). This means that we may get infinitely many nontrivial solutions for the DE (4) by using a given solution.

If the primitive solution $X$ be a parametric solution, we get another parametric solution for the DE (4).

Example 8

$a^{3}+b^{3}+c^{3}=A^{3}+B^{3}+C^{3}$

Let $(a,b,c,A,B,C)$ be a solution for the above DE. We find another solution by using the first solution.

Example $1$: $(a,b,c,A,B,C)=(1,155,209,-41,227,107)$,

Another solution:

$125^{3}+169^{3}+250^{3}=62^{3}+277^{3}+97^{3}$*. *

*Example $2$: *$(a,b,c,A,B,C)=(-62,169,250,-125,277,97),$

Another solution:

$81^{3}+12555^{3}+16929^{3}=(-3321)^{3}+18387^{3}+8667^{3}.$

4 The DE ${\displaystyle\sum_{i=1}^{n}p_{i}\cdot x_{i}^{a_{i}}=\sum_{j=1}^{m}q_{j}\cdot y_{j}^{b_{j}}}$;

In this section, we prove the second main theorem. Let the above DE be in the form:

$x_{1}^{a_{1}}+p_{2}x_{2}^{a_{2}}+p_{3}x_{3}^{a_{3}}+\cdots+p_{n}x_{n}^{a_{n}}=q_{1}y_{1}^{b_{1}}+\cdots+q_{m}y_{m}^{b_{m}}$,

where $(a_{1},a_{2}\cdots a_{n}\cdot b_{1}\cdots b_{m})=1$.

Then we have:

[TABLE]

Define:

$m:=q_{1}s_{1}^{b_{1}}+\cdots+q_{m}s_{m}^{b_{m}}-p_{2}t_{2}^{a_{2}}-p_{3}t_{3}^{a_{3}}-\cdots-p_{n}t_{n}^{a_{n}}$,

where $t_{i}$ and $s_{i}$ are arbitrary integers. Now we introduce the our parametric solution:

$y_{1}=s_{1}\cdot m^{\frac{k}{b_{1}}}$,

$y_{2}=s_{2}\cdot m^{\frac{k}{b_{2}}}$,

$\vdots$

$y_{m}=s_{m}\cdot m^{\frac{k}{b_{m}}}$,

$x_{1}=m^{\frac{k+1}{a_{1}}}$,

$x_{2}=t_{1}\cdot m^{\frac{k}{a_{2}}}$,

$\vdots$

$x_{n}=t_{n}\cdot m^{\frac{k}{a_{n}}}$,

where $k$ is a natural number such that

$k\equiv 0\pmod{b_{1}}$,

$k\equiv 0\pmod{b_{2}}$,

$\vdots$

$k\equiv 0\pmod{b_{m}}$,

$k\equiv 0\pmod{a_{2}}$,

$\vdots$

$k\equiv 0\pmod{a_{n}}$,

$k\equiv-1\pmod{a_{1}}$.

From the Chinese remainder theorem, we know that there exists a solution for $k$, since $(a_{1},a_{2}\cdots a_{n}\cdot b_{1}\cdots b_{m})=1$, that is, the all exponents used in $x_{i}$ and $y_{j}$ are natural numbers. We claim that this is a parametric solution for the DE (5):

$q_{1}y_{1}^{b_{1}}+\cdots+q_{m}y_{m}^{b_{m}}-p_{2}x_{2}^{a_{2}}-p_{3}x_{3}^{a_{3}}-\cdots-p_{n}x_{n}^{a_{n}}=$

$q_{1}\cdot(s_{1}m^{\frac{k}{{b_{1}}}})^{b_{1}}+q_{2}\cdot(s_{2}m^{\frac{k}{b_{2}}})^{b_{2}}+\cdots+q_{m}\cdot(s_{m}m^{\frac{k}{b_{m}}})^{b_{m}}-p_{2}(t_{2}\cdot m^{\frac{k}{a_{2}}})^{a_{2}}-$

$\cdots-p_{n}(t_{n}\cdot m^{\frac{k}{a_{n}}})^{a_{n}}=$

$q_{1}\cdot s_{1}^{b_{1}}\cdot m^{k}+q_{2}\cdot s_{2}^{b_{2}}\cdot m^{k}+\cdots+q_{m}\cdot s_{m}^{b_{m}}\cdot m^{k}-p_{2}\cdot t_{2}^{a_{2}}\cdot m^{k}-\cdots-p_{n}\cdot t_{n}^{a_{n}}\cdot m^{k}=$

$m^{k}(q_{1}\cdot s_{1}^{b_{1}}+q_{2}\cdot s_{2}^{b_{2}}+\cdots+q_{m}\cdot s_{m}^{b_{m}}-p_{2}\cdot t_{2}^{a_{2}}-\cdots-p_{n}\cdot t_{n}^{a_{n}})=m^{k}\cdot m=$

$m^{k+1}=(m^{\frac{k+1}{a}})^{a}=x_{1}^{a}.$

Now the proof of the second main theorem is completed. Since $s_{i}$, and $t_{i}$ were arbitrary, we also obtain infinitely many solutions for the above DE.

Example 9

*The DE $x_{1}^{5}+x_{2}^{6}=y_{1}^{7}+y_{2}^{8}+y_{3}^{9}$, has infinitely many solutions in integers.

Since we have: $x_{1}^{5}=y_{1}^{7}+y_{2}^{8}+y_{3}^{9}-x_{2}^{6}$ and $(5,6\cdot 7\cdot 8\cdot 9)=1$, then by using the previous theorem, we conclude that the aforementioned DE has infinitely many solutions in integers. Put: $m:=r^{7}+s^{8}+t^{9}-w^{6}$, where $r$, $s$, $t$ and $w$ are arbitrary integers. As an example, if we let: $(r,s,t,w)=(3,2,1,3)$ and $k=504$, then we get:

$m=1715$,

$y_{1}=r\cdot m^{\frac{k}{7}}=3\cdot 1715^{72}$,

$y_{2}=s\cdot m^{\frac{k}{8}}=2\cdot 1715^{63}$,

$y_{3}=t\cdot m^{\frac{k}{9}}=1715^{56}$,

$x_{1}=m^{\frac{k+1}{5}}=1715^{101}$,

$x_{2}=w\cdot m^{\frac{k}{6}}=3\cdot 1715^{84}$.

Namely, we have:

$(1715^{101})^{5}+(3\cdot 1715^{84})^{6}=(3\cdot 1715^{72})^{7}+(2\cdot 1715^{63})^{8}+(1715^{56})^{9}.$* *

By letting $(r,s,t,w)=(3,2,2,2)$ and $k=504$, we get:

$(2891^{101})^{5}+(2\cdot 2891^{84})^{6}=(3\cdot 2891^{72})^{7}+(2\cdot 2891^{63})^{8}+(2\cdot 2891^{56})^{9}.$

By changing $(r,s,t,w)$, we obtain infinitely many solutions.

Example 10

*The DE $x^{5}=6(y_{1}^{7}+y_{2}^{7}+y_{3}^{7})$ has infinitely many solutions in integers. If we put: $m:=6r^{7}+6s^{7}+6t^{7}$, where $(r,s,t)=$ $(1,1,1)$ and $k=14$, we get:

$m=18$,

$y_{1}=r\cdot m^{\frac{k}{7}}=18^{2}$,

$y_{2}=s\cdot m^{\frac{k}{7}}=18^{2}$,

$y_{3}=t\cdot m^{\frac{k}{7}}=18^{2}$,

$x=m^{\frac{k+1}{5}}=18^{3}$.

*Namely, we have: $(18^{3})^{5}=6((18^{2})^{7}+(18^{2})^{7}+(18^{2})^{7})$. *

By letting: $(r,s,t)=(3,2,5)$ and $k=14$, we obtain:

$(482640^{3})^{5}=6((3\cdot 482640^{2})^{7}+(2\cdot 482640^{2})^{7}+(5\cdot 482640^{2})^{7})$.

Example 11

*The DE $x^{7}=13y_{1}^{11}+11y_{2}^{13}$ has infinitely many solutions in integers. If we put: $m:=13r^{11}+11s^{13}$, where $(r,s,t)=(1,1,1)$ and $k=286$, we get:

$m=24$,

$y_{1}=r\cdot m^{\frac{k}{11}}=24^{26}$,

$y_{2}=s\cdot m^{\frac{k}{13}}=24^{22}$,

$x=m^{\frac{k+1}{7}}=24^{41}$.

Namely, we have: $(24^{41})^{7}=13\cdot(24^{26})^{11}+11\cdot(24^{22})^{13}$.

Example 12

Is the DE $x_{1}^{5}+x_{2}^{5}=y_{1}^{6}+y_{2}^{6}$ solvable? Yes. To do this, let $y_{1}=y_{2}$. Then we have: $x_{1}^{5}+x_{2}^{5}=2y_{1}^{6}$. We may let $x_{1}=2u$, $x_{2}=2v$. So we get: $y_{1}^{6}=2^{4}(u^{5}+v^{5})$, that is solvable.

5 The DE $x^{a}+y^{b}=z^{c}$

In this part, we solve the above DE, where $a$, $b$, $c$ are fixed natural numbers and $x$, $y$, $z$ are variables. We have three cases: $(a,b,c)=1$ , $(a,b,c)\geq 3$ or $(a,b,c)=2$. If $(a,b,c)=1$, by using the second main theorem, we proved that the DE has infinitely many solutions in integers. If $(a,b,c)\geq 3$, the Fermat last theorem says that the DE dose not have any solutions in integers. Then it suffices to study the case $(a,b,c)=2$. However, we know that the Diophantine equations $x^{4}+y^{4}=z^{2}$ and $x^{4}-y^{4}=z^{2}$ have not any solutions in integers. So it suffices only to study the case where at most one of $a$, $b$, $c$ is divisible by $4$. In the sequel, we study this case by proving several theorems.

Theorem 5.1

The DE $x^{2}+y^{2B}=z^{2C}$, where $B$ and $C$ are both odd and $(B,C)=1$, has infinitely many solutions in integers.

Proof

: we have $x^{2}+(y^{B})^{2}=(z^{C})^{2}$, then we get:

$x=m^{2}-n^{2}$,

$y^{B}=2mn$,

$z^{C}=m^{2}+n^{2}$.

We may suppose that $m=2^{B-1}\cdot t_{1}^{B}$, $n=t_{2}^{B}$. By plugging these into the equations, we get :

$y=2t_{1}t_{2}$,

$x=2^{2(B-1)}\cdot t_{1}^{2B}-t_{2}^{2B}$,

$z^{C}=2^{2(B-1)}\cdot t_{1}^{2B}+t_{2}^{2B}$.

Since $(C,2B)=1$, the DE $z^{C}=2^{2(B-1)}\cdot t_{1}^{2B}+t_{2}^{2B}$ is solvable for $z$, $t_{1}$, $t_{2}$ from the second main theorem. Also $x$ and $y$ are computed from $y=2t_{1}t_{2}$, $x=2^{2(B-1)}\cdot t_{1}^{2B}-t_{2}^{2B}$, as well. The proof is completed.

Theorem 5.2

The DE $x^{2A}+y^{2B}=z^{2C}$, where $A$, $B$, $C$ are all odd and $(A,C)=1$ and $(AC,B)=1$ has infinitely many solutions in integers.

We note that in this case $(a,b,c)=(2A,2B,2C)=2$ and none of $a=2A$, $b=2B$ and $c=2C$ is divisible by $4$. We wish to solve $x^{a}+y^{b}=z^{c}$ in the case of $(a,b,c)=2$ and at most one of $a$, $b$, $c$ is divisible by $4$, this theorem is one of the desired cases where $(a,b,c)=2$ and none of $a$, $b$ and $c$ is divisable by $4$.

Proof

: We know that the DE $X^{A}+Y^{B}=Z^{C}$ with the condition $(AC,B)=1$ is solvable and its solution is:

$m:=r^{C}-s^{A}$,

$Z=m^{\frac{k}{C}}\cdot r$,

$X=m^{\frac{k}{A}}\cdot s$,

$Y=m^{\frac{k+1}{B}}$,

and

$k\equiv 0\pmod{AC}$,

$k\equiv-1\pmod{B}$,

where $r$ and $s$ are arbitrary integers. Now, if we obtain a solution for the DE $X^{A}+Y^{B}=Z^{C}$ from the solution just introduced, where $X$, $Y$ and $Z$ are squares, then we get a solution for the DE $x^{2A}+y^{2B}=z^{2C}$ by putting $X=x^{2}$, $Y=y^{2}$ and $Z=z^{2}$. From the above solution for $X$, $Y$ and $Z$ , we see that if $m$, $r$ and $s$ be squares, then $X$, $Y$ and $Z$ will be squares, as well. Then since $m$, $r$ and $s$ are related together with $m:=r^{C}-s^{A}$, we see that it suffices to solve the DE $M^{2}=R^{2C}-S^{2A}$, where we set:

$m=M^{2}$,

$r=R^{2}$,

$s=S^{2}$.

(we wanted $m$, $r$ and $s$ to be squares). Fortunately the DE $M^{2}=R^{2C}-S^{2A}$ is solvable from the previous theorem due to $(A,C)=1$. Then we obtain $m$, $r$, $s$, and in the end we get $x$, $y$, $z$, and the proof is completed.

Theorem 5.3

The DE $x^{4A}+y^{2B}=z^{2C}$, where $B$ and $C$ are both odd and $(A,C)=1$ and $(AC,B)=1$ has infinitely many solutions in integers.

Proof

: We know that the DE $X^{2A}+Y^{B}=Z^{C}$ with the condition $(2AC,B)=1$ is solvable and its solution is:

$m:=r^{C}-s^{2A}$,

$Z=m^{\frac{k}{C}}\cdot r$,

$X=m^{\frac{k}{2A}}\cdot s$,

$Y=m^{\frac{k+1}{B}}$,

and

$k\equiv 0\pmod{C}$,

$k\equiv 0\pmod{2A}$,

$k\equiv-1\pmod{B}$,

where $r$ and $s$ are arbitrary integers. Now, if we obtain a solution for the DE $X^{2A}+Y^{B}=Z^{C}$ from the solution just introduced, where $X$, $Y$ and $Z$ are squares, next we get a solution for the DE $x^{4A}+y^{2B}=z^{2C}$ by putting $x=X^{2}$, $y=Y^{2}$ and $z=Z^{2}$. From the above solution for $X$, $Y$, and $Z$, we see that if $m$, $r$ and $s$ be squares, then $X$, $Y$, and $Z$ will be squares, as well. Then since m, r and s are related together with $m:=r^{C}-s^{2A}$, we see that it suffices to solve the DE $M^{2}=R^{2C}-S^{4A}$, where we set:

$m=M^{2}$,

$r=R^{2}$,

$s=S^{2}$.

(We wanted $m$, $r$ and $s$ to be squares). Fortunately the DE $M^{2}=R^{2C}-S^{4A}$ is solvable from the previous theorems: we have $M^{2}+(S^{2A})^{2}=(R^{C})^{2}$, and then we get:

$M=t_{1}^{2}-t_{2}^{2}$,

$S^{2A}=2t_{1}t_{2}$,

$R^{C}=t_{1}^{2}+t_{2}^{2}$.

We may suppose that $t_{1}=2^{2A-1}\cdot p^{2A}$, $t_{2}=q^{2A}$. By substituting these in the above equations, we get :

$S=2pq$,

$M=2^{2(2A-1)}\cdot p^{4A}-q^{4A}$,

$R^{C}=2^{2(2A-1)}\cdot p^{4A}+q^{4A}$.

Since $(C,4A)=1$, the DE $R^{C}=2^{2(2A-1)}\cdot p^{4A}+q^{4A}$ is solvable for $R$, $p$ and $q$, from the second main theorem. And $M$, and $S$ are computed from $S=2pq$, $M=2^{2(2A-1)}\cdot p^{4A}-q^{4A}$, as well, the proof is completed.

We note that the previous theorem is the case that in the DE $x^{a}+y^{b}=z^{c}$, exactly one of $a$ or $b$ is divisible by $4$, and $(a,b,c)=2$.

Example 13

*We wish to solve the DE $x^{6}+y^{10}=z^{14}$. We know that the DE $x^{3}+y^{5}=z^{7}$ is solvable and its solution is:

$m:=r^{7}-s^{3}$,

$z=m^{\frac{k}{7}}\cdot r$,

$x=m^{\frac{k}{3}}\cdot s$,

$y=m^{\frac{k+1}{5}}$,

and

$k\equiv 0\pmod{21}$,

$k\equiv-1\pmod{5}$,

where $r$, and $s$ are arbitrary integers.

Now, if we obtain a solution for the DE $x^{3}+y^{5}=z^{7}$ from the solution just introduced, where $x$, $y$ and $z$ are squares, then we get a solution for the DE $x^{6}+y^{10}=z^{14}$. From the above solution for $x$, $y$ and $z$, we see that if $m$, $r$ and $s$ be squares, then $x$, $y$ and $z$ will be squares, as well. Since $m:=r^{7}-s^{3}$, we see that it suffices to solve the DE $M^{2}=R^{14}-S^{6}$, where:

$m=M^{2}$,

$r=R^{2}$,

$s=S^{2}$. Fortunately the DE $M^{2}=R^{14}-S^{6}$ is solvable from the previous theorems: we have $M^{2}+S^{6}=R^{14}$, and then we get:

$M=t_{1}^{2}-t_{2}^{2}$,

$S^{3}=2t_{1}t_{2}$,

$R^{7}=t_{1}^{2}+t_{2}^{2}$.

We may suppose that $t_{1}=4p^{3}$,$t_{2}=q^{3}$. By substituting these in the above equations, we get :

$S=2pq$,

$M=16p^{6}-q^{6}$,

$R^{7}=16p^{6}+q^{6}$.

Since $(7,6)=1$, the DE $R^{7}=16p^{6}+q^{6}$ is solvable for $R$, $p$ and $q$ from the previous theorems:

$M^{\prime}:=16u^{6}+v^{6}$,

$p=M^{\prime\frac{k}{6}}\cdot u$,

$q=M^{\prime\frac{k}{6}}\cdot v$,

$R=M^{\prime\frac{k+1}{7}}$,

and

$K\equiv 0\pmod{6}$ ,

$k\equiv-1\pmod{7}$.

By putting $u=v=1$, $k=6$, we get:

$(15^{28}\cdot 17^{170}\cdot 2)^{6}+(15^{17}\cdot 17^{102})^{10}=(15^{12}\cdot 17^{73})^{14}$.

Theorem 5.4

The DE $x^{2}+y^{2}=z^{2^{n}\cdot c}$, where $c$ is odd and $n\geq 2$, is solvable.

Proof

: We try to solve the DE by using induction. If $n=2$, we have

$x^{2}+y^{2}=z^{4c}$, then

$x=m^{2}-n^{2}$,

$y=2mn$,

$z^{2c}=m^{2}+n^{2}$.

Again we put:

$n=2t_{1}t_{2}$,

$m=2t_{1}^{2}-t_{2}^{2}$,

$z^{c}=t_{1}^{2}+t_{2}^{2}$.

Since $(2,c)=1$, the DE $z^{c}=t_{1}^{2}+t_{2}^{2}$ is solvable, then the main DE is solvable as well, and the proof for $n=2$ is complete. Next, suppose the above DE is solvable for $n$, then we solve it for $n+1$. We have:

$x^{2}+y^{2}=z^{2^{n+1}\cdot c}$, then

$x=m^{2}-n^{2}$,

$y=2mn$,

$z^{2^{n}\cdot c}=m^{2}+n^{2}$.

Since the DE $z^{2^{n}\cdot c}=m^{2}+n^{2}$, is solvable for $m$, $n$, and $z$ by our assumption, then the DE $x^{2}+y^{2}=z^{2^{n+1}\cdot c}$, is solvable as well, and the proof is completed.

Remark 1

*If in the previous theorem, we take $c=1$, we can solve it by another beautiful method. We start from the identity

$(m^{2}-n^{2})^{2}+(2mn)^{2}=(m^{2}+n^{2})^{2}$.

We see that: if in the above identity $m^{2}+n^{2}$, to be square, then we get a solution for the DE $x^{2}+y^{2}=z^{4}$. As an example, we can get $m$ and $n$ from the Pythagorean triples. For $m=4$, $n=3$, we obtain:

$7^{2}+24^{2}=(5^{2})^{2}=(3^{2}+4^{2})^{2}=5^{4}$.

By letting $n=7$, $m=24$, we obtain: $572^{2}+336^{2}=(7^{2}+24^{2})^{2}=(5^{4})^{2}=5^{8}$.

By continuing in this way, if we put $m=572$, $n=336$, we get a solution for the DE $x^{2}+y^{2}=z^{16}$. If we change $m$, $n$, we obtain another identity. If $m=120$, $n=119$, we have : $239^{2}+28560^{2}=13^{8}$. It is clear that by using this method we can find infinitely many solutions for the DE $x^{2}+y^{2}=z^{2^{n}\cdot c}$ with $(x,y,z)=1$, as mentioned in the Beukers theorem (see 3 (3)**) in the introduction.

In the end, it is clear that at some variables parametric solutions obtained for the Diophantine equations, we may get one variable parametric solutions for each case of the above Diophantine equations by fixing the other variables.

Acknowledgements

The authors would like to express their hearty thanks to the anonymous referee for a careful reading of the paper and for many careful comments and remarks which improved its quality.