Remark on a Paper by Izadi and Baghalaghdam about Cubes and Fifth Powers Sums

TL;DR

This paper refines a method to find integer solutions to a specific Diophantine equation involving fifth and third powers, demonstrating that there are infinitely many positive solutions.

Contribution

It improves upon previous techniques to prove the existence of infinitely many solutions to the equation involving cubes and fifth powers.

Findings

The equation has infinitely many positive solutions.

Refined method enhances solution search.

Supports conjecture of infinite solutions.

Abstract

In this paper, we refine the method introduced by Izadi and Baghalaghdam to search integer solutions to the Diophantine equation $X_1^5+X_2^5+X_3^5=Y_1^3+Y_2^3+Y_3^3$. We show that the Diophantine equation has infinitely many positive solutions.

Equations

$$a(X_1^{\prime 5}+X_2^{\prime 5})+\sum _{i=0}^n{a}_i{X}_i^5=b(Y_1^{\prime 3}+Y_2^{\prime 3})+\sum _{i=0}^m{b}_i{Y}_i^3$$

$$X_1^5+X_2^5+X_3^5=Y_1^3+Y_2^3+Y_3^3$$

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

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

$$\{\begin{array}{rl}{X}_1=t+x_1,& X_2=t-x_1,X_3=\alpha t,\\ Y_1=t+v,& Y_2=t-v,Y_3=\beta t.\end{array}$$

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

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

$$\alpha ,\beta >0,0\le F(t)<t^2,t>|x_1|$$

$$a=\frac{2+{\alpha }^5}{6},b=\frac{20x_1^2-8-{\beta }^3}{6},c=\frac{5}{3}{x}_1^4.$$

$$0<\alpha ,\beta ,b<0<b^2-4ac$$

$$C:v^2=\frac{1}{2}{t}^4-\frac{2009}{3}{t}^2+\frac{80}{3},$$

$$C^{\prime }:v^2=\frac{1}{2}{T}^4+88T^3+\frac{15415}{3}{T}^2+\frac{334312}{3}T+760^2$$

$$E:y^2+\frac{41789}{285}xy+133760y=x^3-\frac{76876021}{324900}{x}^2-1155200x+\frac{2460032672}{9},$$

$$T=\frac{2\cdot 760(x+\frac{15415}{3})-\frac{334312^2}{2\cdot 3^2\cdot 760}}{y},v=-760+\frac{T(Tx-\frac{334312}{3})}{2\cdot 760}.$$

$$P_1=\left(-\frac{1802189}{1521},\frac{5513659679}{417430}\right),P_2=\left(-\frac{351379}{363},\frac{47356344241}{2276010}\right).$$

$$C_0=\left\{(t,v)\in C\mid 0\le F(t)<t^2\right\}\subset C$$

$$F(t)=\frac{1}{2}{t}^4-\frac{2009}{3}{t}^2+\frac{80}{3}=0,\tilde{F}(t)=\frac{1}{2}{t}^4-\frac{2009}{3}{t}^2+\frac{80}{3}-t^2=0$$

$$t=\pm \frac{1}{3}\sqrt{6027\pm 3\sqrt{4035601}},t=\pm \frac{2}{3}\sqrt{1509\pm 3\sqrt{252979}}.$$

$$a_1$$

$$a_2$$

$$Q=2P_1-P_2=\left(\frac{304845381192111829037}{58470412871306667},-\frac{4767546475726965161322288395890039}{4652843756178203561643745770}\right)$$

$$(t_0,v_0)=\left(\frac{170815619844155909156204}{4664941095250009917983},-\frac{690740884062625663919872925291699877683029096}{21761675422152362106175457381859866386788289}\right)$$

$$X_1$$

$$\text{vspace}2pc{X}_2$$

$$X_3$$

$$Y_1$$

$$Y_2$$

$$Y_3$$

$$X_1^5+X_2^5=Y_1^3+Y_2^3+Y_3^3.$$

$$C:v^2=\frac{1}{3}{t}^4-639t^2+\frac{50000}{3}$$

$$E:y^2+\frac{1867}{9}xy-400y=x^3-\frac{3676525}{324}{x}^2-1200x+\frac{367652500}{27}.$$

$$x_0=\frac{9233921838917810856046138588468998730}{71226852166762122405616706766475947}$$

$$t_0=\frac{7869911761727476320751662986237524106650}{180965667579279848488380712753242417827}$$

Full text

The present paper is to appear in Math. J. Okayama University.

1. Introduction

In [2], Izadi and Baghalaghdam consider the Diophantine equation:

[TABLE]

where $n,m\in\mathbb{N}\cup\{0\},\ a,b\neq 0,\ a_{i},b_{i}$ are fixed arbitrary rational numbers. They use theory of elliptic curves to find nontrivial integer solutions to (1). In particular, they discuss the equation:

[TABLE]

and obtain integer solutions, for example:

[TABLE]

[TABLE]

However, no positive solutions are presented in their paper [2]. In this paper, we refine their method to find positive solutions to (2).

Consider the Diophantine equation (2). Let:

[TABLE]

Then we get a quartic curve:

[TABLE]

with parameters $x_{1},\ \alpha,\ \beta\in\mathbb{Q}$. If we get a rational point $(t,v)$ on $C$, we can compute a rational solution to (2) (see [2]).

Once we obtain rational solutions to (2), we can obtain integer solutions by multiplying an appropriate value to $X_{i},\ Y_{i}$. In the same way, in order to obtain solutions in positive integers, it suffices to search positive rational solutions to equation (2).

2. Additional Requirements for Positive Solutions

Suppose that a positive rational solution $(X_{i},Y_{i})_{1\leq i\leq 3}$ to (2) is obtained from a given point $(t,v)$ on the quartic $C$.

Proposition 2.1**.**

Let $\alpha,\beta,x_{1}\in\mathbb{Q}$ and

[TABLE]

A rational point $(t,v)$ on the curve $C:v^{2}=F(t)$ in (4) produces a positive rational solution to (2) by (3) if and only if

[TABLE]

hold.

Proof.

If $X_{i}$ and $Y_{i}$ are positive in the solution in the form (3), we have $t=(X_{1}+X_{2})/2>0$, $\alpha=2X_{3}/(X_{1}+X_{2})>0$ and $\beta=2Y_{3}/(Y_{1}+Y_{2})>0$. For $(t,v)\in C$, one has that $0\leq v^{2}=F(t)<v^{2}+Y_{1}Y_{2}=t^{2}$. It follows from $x_{1}^{2}<x_{1}^{2}+X_{1}X_{2}=t^{2}$ that $t>|x_{1}|$ for $t>0$. Conversely, suppose the inequalities in (5) hold. Then the given point $(t,v)$ on $C$ satisfies $v^{2}=F(t)<t^{2}$. This and (5) immediately imply $X_{i},Y_{i}>0$ in (3). ∎

Proposition 2.2**.**

Under the same assumption as Proposition 2.1, let

[TABLE]

Then $a,b,c$ satisfy $b^{2}-4ac>0$ and $b<0$ if and only if there exists a real number $t$ such that $F(t)<t^{2}$.

Proof.

Let $\tilde{F}(t)=F(t)-t^{2}$. Since $\tilde{F}(0)=5x_{1}^{4}/3\geq 0,$ and in this case $a>0,$ it is easy to see that the following conditions are equivalent to each others:

(i) There exists a real number $t$ such that $F(t)<t^{2}$.

(ii) The equation $\tilde{F}(t)=0$ has four distinct solutions.

(iii) The quadratic equation $ax^{2}+bx+c=0$ has two distinct non-negative solutions.

(iv) The discriminant $D=b^{2}-4ac$ of the quadratic function $f(x)=ax^{2}+bx+c$ is positive, and the axis of the quadratic function $-b/2a$ is positive, and $f(0)\geq 0$.

The condition (iv) holds if and only if “$b^{2}-4ac>0$ and $b<0$”, since $a>0$ and $f(0)=c=5x_{1}^{4}/3\geq 0$. ∎

3. Example for $X_{1}^{5}+X_{2}^{5}+X_{3}^{5}=Y_{1}^{3}+Y_{2}^{3}+Y_{3}^{3}$

Let us first search parameters $(x_{1},\alpha,\beta)$ such that

[TABLE]

with $a,b,c$ given by (6) and such that the quartic curve $C$ of (4) has at least one rational point. Note that these are necessary to satisfy conditions of Proposition 2.1, 2.2. Then, the curve $C$ is birationally equivalent to an elliptic curve $E$ over $\mathbb{Q}$. If $E$ has positive rank, then $C$ has infinitely many rational points.

Let $(x_{1},\alpha,\beta)=(2,1,16)$. Then the quartic:

[TABLE]

has a rational point $(t,v)=(44,760).$ By $T=t-44$, we transform $C$ into

[TABLE]

which is birationally equivalent over $\mathbb{Q}$ to the cubic elliptic curve (see [5, Theorem 2.17], [2]):

[TABLE]

where:

[TABLE]

Using the Sage software [3], we find that the cubic curve $E$ is an elliptic curve which has rank 2 and the generators of $E$ are:

[TABLE]

We now consider the subset

[TABLE]

whose points satisfy another condition (5) of Proposition 2.1. The two quartic equations:

[TABLE]

have respectively solutions:

[TABLE]

Let us take larger solutions as:

[TABLE]

If a point $(t_{0},v_{0})$ on $C$ satisfies $a_{1}\leq t_{0}\leq a_{2}$, then $(t_{0},v_{0})$ lies on $C_{0}$. We now make use of the composition law of points on the elliptic curve $E$. Since $E$ has positive rank, we can test infinitely many rational points of $E$ till finding a point $(t_{0},v_{0})$ on $C_{0}$. We find that the rational point

[TABLE]

on $E$ corresponds to

[TABLE]

on $C_{0}$, and creates a positive rational solution:

[TABLE]

Next we shall prove that the Diophantine equation (2) has infinitely many positive solutions. The real locus of elliptic curve $E(\mathbb{R})$ can be regarded as a compact topological subspace of complex projective variety $E$.

Lemma 3.1**.**

If the rank of elliptic curve $E$ over $\mathbb{Q}$ is positive, every point of $E(\mathbb{Q})$ is an accumulation point in $E(\mathbb{R})$.

Proof.

Since $E(\mathbb{R})$ is a compact topological group, and $E(\mathbb{Q})$ is an infinite subgroup of $E(\mathbb{R})$, there is at least one accumulation point of $E(\mathbb{Q})$ in $E(\mathbb{R})$. The group operations are homeomorphisms from $E(\mathbb{R})$ to itself. Therefore all points of $E(\mathbb{Q})$ are accumulation points of $E(\mathbb{R})$. ∎

Theorem 3.2**.**

The Diophantine equation (2) has infinitely many positive solutions.

Proof.

The part of $C_{0}$ has one rational point $(t_{0},v_{0})$ which corresponds to the above point $Q$. By Lemma 3.1, the point $Q$ is an accumulation point of $E(\mathbb{Q})$ in $E(\mathbb{R})$, and $(t_{0},v_{0})$ is that of $C(\mathbb{Q})$ in $C(\mathbb{R})$. Thus the part of $C_{0}$ includes infinitely many rational points. Since $2=|x_{1}|<a_{1}=36.59635926...$, they correspond to positive rational solutions to (2). ∎

4. Example for $X_{1}^{5}+X_{2}^{5}=Y_{1}^{3}+Y_{2}^{3}+Y_{3}^{3}$

Let $\alpha=0$. Then (2) gives another Diophantine equation:

[TABLE]

In the same way, we can obtain a rational or positive rational solutions of it. For example, let $x_{1}=10,\ \beta=18$. Then the quartic curve:

[TABLE]

has a rational point $(t,v)=(-5,30)$ and can be regarded as an elliptic curve over $\mathbb{Q}$ that has rank 2. It is birationally equivalent to:

[TABLE]

From this, we can compute positive rational solutions to (7). For example, there is a point $Q=(x_{0},y_{0})$ on $E$ with

[TABLE]

corresponding to $(t_{0},v_{0})$ on $C$ with

[TABLE]

which creates the following solution to (7):

[TABLE]

The case of $\beta=0$ will be discussed briefly in 5.2 below.

5. Parameters $(x_{1},\alpha,\beta)$ from Trivial Solutions

5.1.

There are several trivial solutions; for example:

[TABLE]

We call solutions to (2) which consist of $0,\ \pm 1$ trivial.　We are going to check some of them to search integer (or positive) solutions.

A solution to (2) may decide parameter. For example, when $X_{i}=Y_{i}=1$ ($i=1,2,3$), we get $(x_{1},\alpha,\beta)=(0,1,1)$. Then:

[TABLE]

has a singular point $(t,v)=(0,0)$　and can be parametrized by one parameter. Let us divide both sides of $C$ by $t^{4}$ and substitute $s,\ w$ for $1/t,\ v/t^{2}$ respectively. Then:

[TABLE]

has a rational point $(s,w)=(1,0)$. Hence we can parametrize rational points on $C^{\prime}$ and integer solutions to (2). That is to say we have:

[TABLE]

where $k\in\mathbb{Q}$. We can see that large enough $k$ give positive solutions to (2). For example:

[TABLE]

where $k=2$. Since $X_{1}=X_{2}=X_{3}=Y_{3}$, this solution also gives positive solution to another Diophantine equation $3X^{5}=Y_{1}^{3}+Y_{2}^{3}+X^{3}$. Moreover it satisfies $X_{1}+X_{2}+X_{3}=Y_{1}+Y_{2}+Y_{3}$ because $\alpha=\beta$.

5.2.

From another trivial solution:

[TABLE]

we can derive parameters $(x_{1},\alpha,\beta)=(\frac{1}{2},0,0)$. Then:

[TABLE]

is an elliptic curve defined over $\mathbb{Q}$ with rational point $(t,v)=(\frac{1}{2},\frac{1}{2})$. It is birationally equivalent to:

[TABLE]

over $\mathbb{Q}$ and has rank 1. Hence we can apply the method of Section 3 to compute positive solutions to

[TABLE]

as a special case of (2) with $X_{1},X_{2},Y_{1},Y_{2}>0$, $X_{3}=Y_{3}=0$ (where $\alpha=\beta=0$ in (3)). For example, a point

[TABLE]

on $E$ corresponding to the point

[TABLE]

on $C$ creates the positive solution to (8):

[TABLE]

5.3.

There exists one more parameter with $\beta=0$, $(x_{1},\alpha,\beta)=(0,0,0)$, which is derived from the trivial solution:

[TABLE]

Then the rational points on:

[TABLE]

can be parametrized. Thus we have:

[TABLE]

where $k\in\mathbb{Q}$. For example, substituting $2$ for $k$, we have:

[TABLE]

The solutions which are obtained in these way give solutions to another Diophantine equation $2X^{5}=Y_{1}^{3}+Y_{2}^{3}$.

5.4.

It is not simple to find parameters $(x_{1},\alpha,\beta)$ that produce elliptic curves for non-trivial solutions $(X_{i},Y_{i})_{1\leq i\leq 3}$. In particular, the author could not find a good parameter for $\beta=0,\ \alpha\neq 0$:

Question 5.1**.**

Find (a good method for) positive solutions to:

[TABLE]

Acknowledgement: The author would like to thank the referee for many valuable suggestions to improve this article.