On the Diophantine equations X^3+Y^3+Z^3+aU^k=a_0U_0^{t_0}+...+a_nU_n^{t_n}, k=3,4

TL;DR

This paper employs elliptic curve theory to solve complex Diophantine equations involving sums of cubes and higher powers, transforming them into elliptic curves to find infinitely many solutions and explore representations of powers as sums of various powers.

Contribution

The paper introduces a method to convert specific Diophantine equations into elliptic curves of positive rank, enabling the derivation of infinitely many solutions and representations of powers as sums of other powers.

Findings

Infinite solutions for equations with k=3,4.

Representation of sums of powers as sums of other powers.

Application of elliptic curves to solve complex Diophantine equations.

Abstract

In this paper, elliptic curves theory is used for solving the Diophantine equations X^3+Y^3+Z^3+aU^k=a_0U_0^{t_0}+...+a_nU_n^{t_n}, k=3,4 where n, ti are natural numbers and a, a_i are fixed arbitrary rational numbers. We try to transform each case of the above Diophantine equations to a cubic elliptic curve of positive rank, then get infinitely many integer solutions for each case. We also solve these Diophantine equations for some values of n, a, a_i, t_i, and obtain infinitely many solutions for each case, and show among the other things that how sums of four, five, or more cubics can be written as sums of four, five, or more biquadrates as well as sums of 5th powers, 6th powers and so on.

Equations

$$X^4+Y^4=2U^4+\sum _{i=1}^n{T}_i{U}_i^{{\alpha }_i},$$

$$\sum _{i=1}^n{a}_i{x}_i^6+\sum _{i=1}^m{b}_i{y}_i^3=\sum _{i=1}^n{a}_i{X}_i^6\pm \sum _{i=1}^m{b}_i{Y}_i^3,$$

$$X^3+Y^3+Z^3+a{U}^k=\sum _{i=0}^n{a}_i{U}_i^{t_i};k=3,4$$

$$t^2=(\frac{a}{6Z})U^3+(\frac{-Z^2}{6}-\frac{{\sum }_{i=0}^n{a}_i{U}_i^{t_i}}{6Z}).$$

$$Y^{\prime }=t.(\frac{-a}{6Z}),$$

$$X^{\prime }=U.(\frac{a}{6Z}),$$

$$Y^{\prime 2}=X^{\prime 3}+(-\frac{a^2}{216}-(\frac{a^2}{216Z^3}).\sum _{i=0}^n{a}_i{U}_i^{t_i}).$$

$$t^2=(\frac{a}{6Z})U^4+(\frac{-Z^2}{6}-\frac{{\sum }_{i=0}^n{a}_i{U}_i^{t_i}}{6Z}).$$

$$Q:=(\frac{-Z^2}{6}-\frac{{\sum }_{i=0}^n{a}_i{U}_i^{t_i}}{6Z}),$$

$$q^2=(\frac{-Z^2}{6}-\frac{{\sum }_{i=0}^n{a}_i{U}_i^{t_i}}{6Z}).$$

$$t^2=(\frac{a}{6Z})U^4+q^2.$$

$$U=\frac{2q{X}^{\prime }}{Y^{\prime }},$$

$$t=-q+\frac{U^2{X}^{\prime }}{2q},$$

$$Y^{\prime 2}=X^{\prime 3}+(\frac{aZ}{9}+\frac{a.({\sum }_{i=0}^n{a}_i{U}_i^{t_i})}{9Z^2})X^{\prime }.$$

$$t^2=(\frac{-1}{6Z})U^4+(\frac{-Z^2}{6}-\frac{a_0{U}_0^{t_0}+a_1}{6Z}).$$

$$q^2=(\frac{-Z^2}{6}-\frac{a_0{U}_0^{t_0}+a_1}{6Z}).$$

$$U=\frac{2q{X}^{\prime }}{Y^{\prime }},$$

$$t=-q+\frac{U^2{X}^{\prime }}{2q},$$

$$Y^{\prime 2}=X^{\prime 3}+(\frac{2q^2}{3Z})X^{\prime }.$$

$$(2t^3+6t{v}^2)+{\beta }^3{t}^3=(2t^5+10x_1^{4}t+20x_1^2{t}^3)+{\alpha }^5{t}^5.$$

$$v^2=\frac{2+{\alpha }^5}{6}{t}^4+\frac{20x_1^2-2-{\beta }^3}{6}{t}^2+(\frac{5}{3})x_1^4.$$

$$v^2=\frac{17}{3}{t}^4+\frac{5}{3}{t}^2+\frac{5}{3}.$$

$$v^2=\frac{17}{3}{T}^4+\frac{68}{3}{T}^3+\frac{107}{3}{T}^2+26T+9.$$

$$T=\frac{6(X+\frac{107}{3})-\frac{26^2}{6}}{Y},$$

$$v=-3+\frac{T(XT-26)}{6},$$

$$Y^2+\frac{26}{3}XY+136Y=X^3+\frac{152}{9}{X}^2-204X-\frac{10336}{3}.$$

$$M^2=X^3+\frac{107}{3}{X}^2+\frac{1156}{3}X+\frac{3536}{3}.$$

Full text

1. Introduction

The authors in two different papers, used elliptic curves to solve two Diophantine equations

[TABLE]

where $n,\alpha_{i}\in\mathbb{N}$, and $T_{i}$, are appropriate rational numbers, and

[TABLE]

where $n,m\in\mathbb{N}$ and $a_{i},b_{i}\neq 0\in\mathbb{Q}$. (see [1], [2])

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

[TABLE]

where $n,t_{i}\in\mathbb{N}\cup\{0\}$, and $a\neq 0$, $a_{i}$, are fixed arbitrary rational numbers.

This shows that how sums of four, five, or more cubics can be written as sums of four, five, or more biquadrates as well as sums of 5th powers, 6th powers and so on.

We conclude this introduction with a standard fact which is needed in section $3$. (see [4])

Lemma 1.1**.**

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 $(x,y)=(-a_{2},a_{1}a_{2}-a_{3})$.*

2. The Diophantine equation (DE) $X^{3}+Y^{3}+Z^{3}+aU^{3}=\sum_{i=0}^{n}a_{i}U_{i}^{t_{i}}$

** Main Theorem 2.1****.**

*Consider the DE (1.3) for the case $k=3$, where $n,t_{i}\in\mathbb{N}\cup\{0\}$, $a\neq 0$, $a_{i}$, are fixed arbitrary rational numbers.

We try to transform this DE to a cubic elliptic curve of positive rank.

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 $a$, $a_{i}$, $t_{i}$ and the other rational parameters $P_{i}$, $t$, $s$, yet to be found later. If the elliptic curve has positive rank, depending on the values of $P_{i}$, $s$, the DE (1.3) has infinitely many integer solutions.

Proof 2.1**.**

Firstly, it is clear that if we find rational solutions for each case of the Diophantine equations (1.3), then by canceling the denominators of $X$, $Y$, $Z$, $U$, $U_{i}$, and by multiplying both sides of these Diophantine equations by appropriate number $M$, we may obtain integer solutions for each case.

Let $X=-Z+t$, $Y=-Z-t$, where $Z$, $t$ are rational variables. By substituting these variables in the DE (1.3), and some simplifications, we get:

[TABLE]

By multiplying both sides of the Eq. (2.1), by $(\frac{a}{6Z})^{2}$, and letting

[TABLE]

and

[TABLE]

we get:

[TABLE]

Now by choosing appropriate values for $Z$ (equal to $s$), and $U_{i}$ (equal to $P_{i}$), such that the rank of the elliptic curve (2.4) to be positive, and by calculating $t$, $U$, $X$, $Y$, from the relations (2.2), (2.3), $X=-Z+t$ and $Y=-Z-t$, some simplifications and canceling the denominators of $X,Y,Z,U,U_{i}$, we obtain infinitely many integer solutions for the DE (1.3). The proof is completed.

Now we work out some examples.

2.1. Application to examples

Example 1**.**

$X^{3}+Y^{3}+Z^{3}+U^{3}=U_{0}^{3}+U_{1}^{3}+U_{2}^{3}$

i.e., the sum of $4$ cubics can be written as the sum of $3$ cubics.

The cubic elliptic curve (2.4) is $Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{U_{0}^{3}+U_{1}^{3}+U_{2}^{3}}{216Z^{3}})$.

By taking $Z=5$, $U_{0}=1$, $U_{1}=2$, $U_{2}=3$, the above elliptic curve becomes

$Y^{\prime 2}=X^{\prime 3}-\frac{161}{27000}$.

Rank=1.

Generator: $P=(X^{\prime},Y^{\prime})=(\frac{643}{90},\frac{2578}{135})$.

$(t,U)=(\frac{-5156}{9},\frac{643}{3})$.

Solution:

$9^{3}+18^{3}+27^{3}+5201^{3}=45^{3}+1929^{3}+5111^{3}$.

Example 2**.**

$X^{3}+Y^{3}+Z^{3}+U^{3}=U_{0}^{4}$

i.e., fourth power of an integer is written as a sum of four cubics.

The cubic elliptic curve (2.4) is $Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{U_{0}^{4}}{216Z^{3}})$.

Now letting $Z=-2$, and $U_{0}=7$, yields

$Y^{\prime 2}=X^{\prime 3}+\frac{2393}{1728}$.

Rank=2.

Generators: $P_{1}=(X^{\prime},Y^{\prime})=(\frac{115}{48},\frac{249}{64})$, $P_{2}=(X^{\prime\prime},Y^{\prime\prime})=(\frac{1320481}{291600},\frac{1528666129}{157464000})$.

$(t,U)=(\frac{747}{16},\frac{-115}{4})$, for the point $P_{1}$,

$(t^{\prime},U^{\prime})=(\frac{1528666129}{13122000},-\frac{1320481}{24300})$, for the point $P_{2}$.

Solutions:

$779^{3}+(-715)^{3}+(-32)^{3}+(-460)^{3}=56^{4}$.

$7774550645^{3}+(-7512110645)^{3}+(-131220000)^{3}+(-3565298700)^{3}=5103000^{4}$.

Example 3**.**

$X^{3}+Y^{3}+Z^{3}+U^{3}=U_{0}^{4}+U_{1}^{4}+U_{2}^{4}+U_{3}^{4}$

i.e., the sum of $4$ cubics is written as the sum of $4$ biquadrates.

The cubic elliptic curve (2.4) is $Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{U_{0}^{4}+U_{1}^{4}+U_{2}^{4}+U_{3}^{4}}{216Z^{3}})$.

Now letting $Z=-5$, $U_{0}=1$, $U_{1}=2$, $U_{2}=3$, $U_{3}=4$, yields

$Y^{\prime 2}=X^{\prime 3}+\frac{229}{27000}$.

Rank=1.

Generator: $P=(X^{\prime},Y^{\prime})=(\frac{367}{7260},\frac{14821}{159720})$.

$(t,U)=(\frac{14821}{5324},-\frac{367}{242})$.

Solution:

$1823404^{3}+519156^{3}+(-1171280)^{3}+(-355256)^{3}=10648^{4}+21296^{4}+31944^{4}+42592^{4}$.

Example 4**.**

$X^{3}+Y^{3}+Z^{3}+U^{3}=U_{0}^{5}+U_{1}^{5}+U_{2}^{5}+U_{3}^{5}$

i.e., the sum of $4$ cubics is written as the sum of $4$ fifth powers.

The cubic elliptic curve (2.4) is $Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{U_{0}^{5}+U_{1}^{5}+U_{2}^{5}+U_{3}^{5}}{216Z^{3}})$.

Now letting $Z=-6$, $U_{0}=1$, $U_{1}=2$, $U_{2}=3$, $U_{3}=4$, yields

$Y^{\prime 2}=X^{\prime 3}+\frac{271}{11664}$.

Rank=1.

Generator: $P=(X^{\prime},Y^{\prime})=(\frac{37}{324},\frac{917}{5832})$.

$(t,U)=(\frac{917}{162},-\frac{37}{9})$.

Solution:

$90672^{3}+2640^{3}+(-46656)^{3}+(-31968)^{3}=216^{5}+432^{5}+648^{5}+864^{5}$.

Example 5**.**

$X^{3}+Y^{3}+Z^{3}+U^{3}=U_{0}^{6}+U_{1}^{6}+U_{2}^{6}+U_{3}^{6}$

i.e., the sum of $4$ cubics is written as the sum of $4$ sixth powers.

The cubic elliptic curve (2.4) is $Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{U_{0}^{6}+U_{1}^{6}+U_{2}^{6}+U_{3}^{6}}{216Z^{3}})$.

Now letting $Z=-7$, $U_{0}=1$, $U_{1}=2$, $U_{2}=3$, $U_{3}=4$, yields

$Y^{\prime 2}=X^{\prime 3}+\frac{4547}{74088}$.

Rank=2.

Generators:

$P_{1}=(X^{\prime},Y^{\prime})=(\frac{284747}{318402},\frac{111956735}{127042398})$, $P_{2}=(X^{\prime\prime},Y^{\prime\prime})=(\frac{-162932615}{499254336},\frac{1819882008883}{11155338883584})$.

$(t,U)=(\frac{111956735}{3024819},-\frac{284747}{7581})$, $(t^{\prime},U^{\prime})=(\frac{1819882008883}{265603306752},\frac{162932615}{11887008})$.

Solutions:

$2529478892^{3}+(-1724877038)^{3}+(-402300927)^{3}+(-2158667007)^{3}=\\ 7581^{6}+15162^{6}+22743^{6}+30324^{6}$.

$489320985767551^{3}+5232371404673^{3}+484195324491480^{3}+\\ (-247276678586112)^{3}=$ $5943504^{6}+11887008^{6}+17830512^{6}+23774016^{6}$.

If we take $Z=-7$, $U_{0}=U_{1}=3$, $U_{2}=U_{3}=4$, the above elliptic curve becomes

$Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{3^{6}+3^{6}+4^{6}+4^{6}}{216.(-7)^{3}})=X^{3}+\frac{9307}{74088}$.

Rank=2.

Generators:

$P_{1}=(X^{\prime},Y^{\prime})=(\frac{1654081}{3572100},\frac{3201764021}{6751269000})$, $P_{2}=(X^{\prime\prime},Y^{\prime\prime})=(\frac{1526057}{71442},\frac{-666521155}{6751269})$.

$(t,U)=(\frac{3201764021}{160744500},-\frac{1654081}{85050})$, $(t^{\prime},U^{\prime})=(\frac{-1333042310}{321489},-\frac{1526057}{1701})$.

Solutions: (for the DE $X^{3}+Y^{3}+Z^{3}+U^{3}=2U_{0}^{6}+2U_{2}^{6}$):

$21634877605^{3}+(-10382762605)^{3}+(-5626057500)^{3}+(-15631065450)^{3}=\\$ $2.(85050)^{6}+2.(113400)^{6}$.

$(-1330791887)^{3}+1335292733^{3}+(-2250423)^{3}+(-288424773)^{3}=$

$2.(1701)^{6}+2.(2268)^{6}$.

Example 6**.**

$X^{3}+Y^{3}+Z^{3}+U^{3}=U_{0}^{7}+U_{1}^{7}+U_{2}^{7}+U_{3}^{7}$

i.e., the sum of $4$ cubics is written as the sum of $4$ seventh powers.

The cubic elliptic curve (2.4) is $Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{U_{0}^{7}+U_{1}^{7}+U_{2}^{7}+U_{3}^{7}}{216Z^{3}})$.

Now letting $Z=2$, $U_{0}=1$, $U_{1}=2$, $U_{2}=3$, $U_{3}=4$, yields

$Y^{\prime 2}=X^{\prime 3}-\frac{1559}{144}$.

Rank=1.

Generator: $P=(X^{\prime},Y^{\prime})=(\frac{55825}{1521},\frac{-52754017}{237276})$.

$(t,U)=(\frac{52754017}{19773},\frac{223300}{507})$.

Solution:

$365855455514133^{3}+(-366404379540849)^{3}+274462013358^{3}$ $+\\ 60441190910100^{3}=59319^{7}+118638^{7}+177957^{7}+237276^{7}$.

Example 7**.**

$X^{3}+Y^{3}+Z^{3}+U^{3}+U_{0}^{3}=U_{1}^{8}+U_{2}^{8}+U_{3}^{8}+U_{4}^{8}+U_{5}^{8}$

i.e., the sum of $5$ cubics is written as the sum of $5$ eighth powers.

The cubic elliptic curve (2.4) is $Y^{\prime 2}=X^{\prime 3}-\frac{1}{216}-(\frac{-U_{0}^{3}+U_{1}^{8}+U_{2}^{8}+U_{3}^{8}+U_{4}^{8}+U_{5}^{8}}{216Z^{3}})$.

Now letting $Z=2$, $U_{0}=1$, $U_{1}=\frac{2}{3}$, $U_{2}=2$, $U_{3}=\frac{1}{3}$, $U_{4}=3$, $U_{5}=\frac{4}{3}$, yields

$Y^{\prime 2}=X^{\prime 3}-\frac{14946019}{3779136}$.

Rank=1.

Generator: $P(X^{\prime},Y^{\prime})=(\frac{4905677718694537513}{2497113793896353424},\frac{-7515135371775048887228789899}{3946003812454638191475172032})$.

$(t,U)=(\frac{7515135371775048887228789899}{328833651037886515956264336},\frac{4905677718694537513}{208092816158029452})$.

Solution:

$X=\frac{6857468069699275855316261227}{328833651037886515956264336}$,

$Y=\frac{-8172802673850821919141318571}{328833651037886515956264336},$

$U=\frac{4905677718694537513}{208092816158029452}$,

$Z=2$, $U_{0}=1$, $U_{1}=\frac{2}{3}$, $U_{2}=2$, $U_{3}=\frac{1}{3}$, $U_{4}=3$, $U_{5}=\frac{4}{3}$.

Remark 2.2**.**

By choosing the other points on the above elliptic curves such as nP ($n=3,4$, $\cdots$,), P is one of the elliptic curves generators, we obtain infinitely many solutions for each case of the above Diophantine equations.

3. The DE $X^{3}+Y^{3}+Z^{3}+aU^{4}=\sum_{i=0}^{n}a_{i}U_{i}^{t_{i}}$

** Main Theorem 3.1****.**

*Consider the DE (1.3) for the case $k=4$, where $n,t_{i}\in\mathbb{N}\cup\{0\}$, and $a\neq 0$, $a_{i}$, are fixed arbitrary rational numbers.

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 $a$, $a_{i}$, $t_{i}$ and the other rational parameters $P_{i}$, $t$, $s$, yet to be found later. If the elliptic curve has positive rank, depending on the values of $P_{i}$, $s$, the DE (1.3) has infinitely many integer solutions.

Proof 3.1**.**

Let $X=-Z+t$, $Y=-Z-t$, where $Z$, $t$ are rational variables. By substituting these variables in the DE (1.3), and some simplifications, we get:

[TABLE]

Now by choosing appropriate values for $U_{i}$ (equal to $P_{i}$), and $Z$ (equal to $s$), such that the rank of the quartic elliptic curve (3.1) to be positive (Then we get infinitely many rational solutions for $U$, $t$.), by calculating $U$, $t$, $X$, $Y$, $U_{i}$, $Z$, from the relations (3.1), $X=-Z+t$, $Y=-Z-t$, $U_{i}=P_{i}$, $Z=s$, after some simplifications and canceling the denominators of the values obtained for variables, we obtain infinitely many integer solutions for the DE (1.3). The proof is completed.

Remark 3.2**.**

(If in the quartic elliptic curve (3.1),

[TABLE]

to be square (It is done by choosing appropriate values for $Z$, and $U_{i}$.), say $q^{2}$, we may use the lemma $1.1$ for transforming this quartic to a cubic 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 $a_{i}\in\mathbb{Q}$.

Then we solve the cubic elliptic curve just obtained of the rank$\geq 1$, and get infinitely many solutions for the above DE.

Generally, it is not essential that $Q$ to be square, because we may transform the quartic (3.1) to a quartic that its constant number is square if the rank of the quartic (3.1) is positive. The only important thing is that the rank of the quartic elliptic curve (3.1), to be positive for getting infinitely many nontrivial solutions for the above DE. See the example $8$.)

Now let us take

[TABLE]

Then for the quaratic elliptic curve (3.1), we have

[TABLE]

With the inverse transformation

[TABLE]

and

[TABLE]

the corresponding cubic elliptic curve is

[TABLE]

Then if the elliptic curve (3.7) has positive rank ( This is done by choosing appropriate values for $U_{i}$, and $Z$.), by calculating $U$, $t$, $X$, $Y$, from the relations (3.5), (3.6), $X=-Z+t$, and $Y=-Z-t$, after some simplifications and canceling the denominators of values obtained for variables, we obtain infinitely many integer solutions for the DE (1.3). Thus we conclude that we must choose appropriate values for $U_{i}$, and $Z$, such that the rank of the elliptic curve (3.7) to be positive, then obtain infinitely many solutions for the DE (1.3).

3.1. Application to examples

Example 8**.**

$X^{3}+Y^{3}+Z^{3}=U^{4}+a_{0}U_{0}^{t_{0}}+a_{1}U_{1}^{t_{1}}$

First of all, we solve the DE $X^{3}+Y^{3}+Z^{3}=U^{4}+a_{0}U_{0}^{t_{0}}+a_{1}$.

Let $X=-Z+t$ and $Y=-Z-t$. By substituting these variables in the above DE, and some simplifications, we get

[TABLE]

Let us transform the quaratic (3.8) to a cubic elliptic curve.

Let

[TABLE]

Now with the inverse transformation

[TABLE]

and

[TABLE]

the quaratic (3.8), maps to the cubic elliptic curve

[TABLE]

Then if the above elliptic curve has positive rank (This is done by choosing appropriate arbitrary values for $q$, and $Z$, such as the rank of the above elliptic curve to be $\geq 1$.), we obtain infinitely many rational solutions for the DE $X^{3}+Y^{3}+Z^{3}=U^{4}+a_{0}U_{0}^{t_{0}}+a_{1}$.

After multiplying both sides of the above DE by appropriate number $M$, we obtain integer solutions for the main DE $X^{3}+Y^{3}+Z^{3}=U^{4}+a_{0}U_{0}^{t_{0}}+a_{1}U_{1}^{t_{1}}$.

As an example, if we take $Z=-2$, $q=7$, $a_{0}=1$, $U_{0}=2$, $t_{0}=t_{1}=4$, ($a_{1}=580$),

the elliptic curve (3.12) becomes

$Y^{\prime 2}=X^{\prime 3}-\frac{49}{3}X^{\prime}$.

Rank=1.

Generator: $P=(X^{\prime},Y^{\prime})=(\frac{-300}{361},\frac{24730}{6859})$.

$(t,U)=(\frac{-46590103}{6115729},\frac{-7980}{2473})$.

Solution:

$(-210128161627205)^{3}+359736726432969^{3}+(-74804282402882)^{3}=\\ 48803517420^{4}+580.(15124197817)^{4}+30248395634^{4}$.

By taking $Z=-1$, $q=12$, $a_{0}=1$, $U_{0}=3$, $t_{0}=t_{1}=6$, ($a_{1}=136$), the elliptic curve (3.12) becomes $Y^{\prime 2}=X^{\prime 3}-96X^{\prime}$.

Rank=1.

Generator: $P=(X^{\prime},Y^{\prime})=(-8,16)$.

$2P=(25,-115)$.

$(t,U)=(\frac{8652}{529},\frac{-120}{23})$, (for the point 2P).

Solution:

$4856749^{3}+(-4297067)^{3}+(-279841)^{3}=63480^{4}+1587^{6}+136.(529)^{6}$.

Letting $Z=-1$, $q=9$, $a_{0}=1$, $U_{0}=2$, $t_{0}=t_{1}=8$, ($a_{1}=231$), in the elliptic curve (3.12), yields

$Y^{\prime 2}=X^{\prime 3}-54X^{\prime}$.

Rank=1.

Generator: $P=(X^{\prime},Y^{\prime})=(-2,10)$.

$(t,U)=(\frac{-261}{25},\frac{-18}{5})$.

Solution:

$(-3687500)^{3}+4468750^{3}+(-390625)^{3}=56250^{4}+250^{8}+231.(125)^{8}$.

Example 9**.**

$Y_{1}^{3}+Y_{2}^{3}+Y_{3}^{3}=X_{1}^{5}+X_{2}^{5}+X_{3}^{5}$ ($a=0$)

i.e., the sum of $3$ cubics can be written as the sum of $3$ fifth powers.

We solve this DE with another new method.

Let: $Y_{1}=t+v$, $Y_{2}=t-v$, $Y_{3}=\beta t$, $X_{1}=t+x_{1}$, $X_{2}=t-x_{1}$, $X_{3}=\alpha t$.

By substituting these variables in the above DE, we get

[TABLE]

Then after some simplifications and clearing the case of $t=0$, we obtain

[TABLE]

Note that $(\frac{5}{3})x_{1}^{4}$, is not a square. Then we may not use from the lemma $1.1$ for transforming this quartic to a cubic elliptic curve, but we do this work by another method. Let us take $x_{1}=1$, $\alpha=\beta=2$. Then the quartic (3.14) becomes

[TABLE]

By searching, we see that the above quartic has two rational points

$P_{1}=(1,3)$, and $P_{2}=(7,117)$, among others. Let us put $T=t-1$. Then we get

[TABLE]

Now with the inverse transformation

[TABLE]

and

[TABLE]

the quaratic (3.16), maps to the cubic elliptic curve

[TABLE]

Rank=2.

Generators: $P_{1}=(X^{\prime},Y^{\prime})=(\frac{-152}{9},\frac{280}{27})$, $P_{2}=(X^{\prime\prime},Y^{\prime\prime})=(\frac{-44}{3},\frac{20}{9})$.

To square the left hand of (3.19), let us put $M=Y+\frac{13}{3}X+68$. Then the cubic elliptic curve (3.19) transforms to the Weierstrass form

[TABLE]

Rank=2.

Generators: $G_{1}=(X^{\prime},M^{\prime})=(\frac{-44}{3},\frac{20}{3})$, $G_{2}=(X^{\prime\prime},M^{\prime\prime})=(\frac{-152}{9},\frac{140}{27})$.

Thus we conclude that we could transform the main quartic (3.15) to the cubic elliptic curve (3.20) of the rank equal to $2$.

Because of this, the above cubic elliptic curve has infinitely many rational points and we may obtain infinitely many solutions for the above DE too.

Since $G_{1}=(X^{\prime},M^{\prime})=(\frac{-44}{3},\frac{20}{3})$, we get $(t,v)=(7,-117)$, that is on the (3.15), by calculating $X_{i}$, $Y_{i}$, from the above relations and after some simplifications and canceling the denominators of $X_{i}$, $Y_{i}$, we obtain a solution for the above DE:

$(-110)^{3}+124^{3}+14^{3}=8^{5}+6^{5}+14^{5}$.

It is iteresting to see that $(-110)+124+14=8+6+14$, too.

Also we have $2G_{2}=(\frac{373}{36},\frac{-21721}{216})$.

By using this new point $2G_{2}=(\frac{373}{36},\frac{-21721}{216})$, we get $(t,v)=(\frac{11}{47},\frac{2943}{2209})$.

Solution:

$359227580^{3}+(-251874598)^{3}+107352982^{3}=128122^{5}+(-79524)^{5}+48598^{5}$.

The Sage software has been used for calculating the rank of the elliptic curves. (see [3])