Extendibility of Some P_{k} Sets

TL;DR

This paper investigates the extendibility of certain P_{k} sets to Diophantine quadruples, demonstrating non-extendibility for specific values of k and exploring properties of these sets.

Contribution

It establishes non-extendibility results for P_{k} sets with k=2 and k=-3 and provides new properties of P_{k} sets related to Diophantine m-tuples.

Findings

P_{k} sets with k=2 cannot be extended to Diophantine quadruples.

P_{k} sets with k=-3 cannot be extended to Diophantine quadruples.

The paper identifies specific properties of P_{k} sets relevant to their extendibility.

Abstract

A set of m distinct positive integers {a_{1},...a_{m}} is called a Diophantine m-tuple if a_{i}a_{j}+n is a square for each 1\leqi<j\leqm . The aim of this study is to show that some P_{k} sets can not be extendible to a Diophantine quadruple when k=2 and k=-3 and also to give some properties about P_{k} sets.

Equations

$$7m+2=x^2\left(1\right)$$

$$14m+2=y^2\left(2\right)$$

$$41m+2=z^2\left(3\right).$$

$$y^{2}-2x^{2}=-2\ \$$

$$2(x^2-1)=y^2\left(4\right).$$

$$x^2-1=2y_1^2.$$

$$2x_1\left(x_1+1\right)=y_1^2.$$

$$x^2-8y_2^2=1.$$

$$m+2=x^2\left(5\right)$$

$$7m+2=y^2\left(6\right)$$

$$14m+2=z^2\left(7\right).$$

$$z^{2}-2y^{2}=-2\ \ \ \$$

$$2(y^2-1)=z^2\left(8\right).$$

$$y^2-1=2z_1^2.$$

$$2y_1\left(y_1+1\right)=z_1^2.$$

$$y^2-8z_2^2=1.$$

$$41m+2=x^2\left(9\right)$$

$$239m+2=y^2\left(10\right)$$

$$478m+2=z^2\left(11\right).$$

$$2y^{2}-z^{2}=2\ \ \$$

$$2(y^2-1)=z^2.$$

$$3kt+2=x^2$$

$$x^2\equiv 2(\mathrm{mod}3).$$

$$3m-3=x^2\left(12\right)$$

$$4m-3=y^2\left(13\right)$$

$$13m-3=z^2\left(14\right)$$

$$3y^2-4x^2=3\left(15\right)$$

$$3(y^2-1)=4x^2\left(16\right)$$

$$y^2-1=12x_1^2\left(17\right)$$

$$y_1\left(y_1+1\right)=3x_1^2.\left(18\right)$$

Full text

INTRODUCTION

Let $n$ be an integer and $A$ be a set $\{a_{1},a_{2},a_{3},...,a_{m}\}$ with $m$ different positive integers. This set has the property $D\left(n\right)$ if $a_{i}a_{j}+n$ is a perfect square for all $1\leq i<j\leq m.$ Such a set is called as a Diophantine m-tuple (with the property $D\left(n\right)$) or a $P_{n}$ set of size $m$.

The problem extending of $P_{k}$ sets is an old one dating from the times of Diophantus $\left[3\right].$ So, this problem has been studied extensively. Diophantus studied finding four numbers such that the product of any two of them, increased by 1, is a perfect square. In $\left[2,4,9\right]$, the more general form of this problem was considered.The most famous result in this area is the one found by Baker and Davenport $\left[1\right]$ who proved for the four numbers $1,3,8,120$ the property that the product of any two, increased by $1$, is a perfect square. Dujella $\left[4\right]$ has a vast amount of literature about this interesting problem.

Filipin $\left[7\right]$ studied on the set $P_{-1}=\{1,10,c\}$ and proved that there do not exist different positive integers $c,d>1$ such that the product of any two distinct elements of the set $\{1,10,c,d\}$ diminished by $1$ is a perfect square.

In 1985, Brown $\left[2\right]$, Gupta and Singh $\left[8\right]$, Mohanty and Ramasamy $\left[11\right]$ proved independently that if  $n\equiv 2\left(\mathop{\mathrm{m}od}4\right)$ then there does not exist a $P_{n}$ set of size 4. Peker et al. $\left[16\right]$ showed that for an integer $k$ such that $\left|k\right|\geq 2,$ the $D\left(-2k+1\right)-$triple $\left\{1,k^{2},k^{2}+2k-1\right\}$ can not be extended to a $D\left(-2k+1\right)-$quadruple.

Mohanty and Ramasay $\left[10\right]$ showed non-extendibility of the set $P_{-1}=\{1,5,10\}$. Kedlaya $\left[9\right]$ gave a list of non-extendibity of $P_{-1}$   triples $\{1,2,45\}$, $\{1,2,4901\}$, $\{1,5,65\}$, $\{1,5,20737\}$, $\{1,10,17\}$ and $\{1,26,37\}$. Both Thamotherampillai $\left[17\right]$ and Mootha & Berzsenyi $\left[13\right]$ proved the non-extendibility of the set $P_{2}=\{1,2,7\}$ by u$\sin$g different methods. Ozer $\left[15\right]$ gave some $P_{11}$ and $P_{-11\text{ }}$ triples.

Diophantine triples are classified as regular or irregular according to whether they provide the condition below.

In this paper, we give some  non-extendable $P_{2}$ and $P_{-3}$ sets by using solutions of simultaneous Pell equations.

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By eliminating $m$ from the equations$\ \left(1\right)$ and $\left(2\right)$,* we get*

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and this gives

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In the last equation, since the left hand side is even, the right hand side must be even, too. So, we can write  $y=2y_{1}$ $\left(y_{1}\in\mathbb{Z}\right)$. Then the last equation becomes

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From this, we can conclude that $x$ must be odd, that is, $x=2x_{1}+1\ (x_{1}\in\mathbb{Z})$. Then, the last equation will be as follows

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Since the left hand side of this equation has multiple of $2$,* the right hand side must be even. So, we can write $y_{1}=2y_{2}\ \left(y_{2}\in\mathbb{Z}\right).$* Now, we have* y=4y_{2}.\If we write $y=4y_{2}$ in the equation $(4),$ we find*

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The fundamental solution of this Pell equation is $\left(x,y_{2}\right)=(3,1)\in\mathbb{Z}^{2}.$ So, all solutions of this Pell equation can be given in the form $x_{n}+\sqrt{8}(y_{2})_{n}=(3+\sqrt{8})^{n}$ by the usual methods. We get the general recurrence relation for the solutions of $x$ as $x_{n}=6x_{n-1}-x_{n-2}.$* Using this general recurrence relation, we can get some values of $m$ from the equation $\left(1\right)$. For these values of $m$, one can easily check that none of these values give any square of an integer for the equation $\left(3\right)$. So, we can not get an integer solution for $z.$* This means that the set $P_{2}=\{7,14,41\}$ is non-extendable.

Now, in the following theorem we show that another $P_{2}$ triple, namely $\{1,7,14\}$ can not be extended to a Diophantine quadruple by using above method.

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where $x,y,z$ are some integers satisfying the above equations.*      *

By eliminating $m$ from the equations $\left(6\right)$ and $\left(7\right)$ we get

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and this yields

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In the last equation, since the left side is even, the right side must be even, too. So, we can write $\mathit{z}=2z_{1}$ $\left(z_{1}\in\mathbb{Z}\right),$* then the equation $\left(8\right)$* becomes

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From this, we can conclude that $y$ must be odd, that is, $y=2y_{1}+1\ (y_{1}\in\mathbb{Z})$. Then, we find

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Since the left hand side of this equation has multiple of $2$,* the right hand side must be even. So, we can write $z_{1}=2z_{2}\ \left(z_{2}\in\mathbb{Z}\right).$*

Consequently, z=4z_{2}.\If we write $z=4z_{2}$ in the equation $(8),$ we find

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The fundamental solution of this Pell equation is $\left(y,z_{2}\right)=(3,1)\in\mathbb{Z}^{2}.$ So, all solutions of this Pell equation can be given in the form $y_{n}+\sqrt{8}(z_{2})_{n}=(3+\sqrt{8})^{n}$ by the usual methods. We get the general recurrence relation for the solutions of $y$ as $y_{n}=6y_{n-1}-y_{n-2}.$ Using this general recurrence relation, we can get some values of $m$ from the equation $\left(6\right)$. For these values of $m$,* one can easily check that none of these values give any square of an integer for the equation $\left(5\right)$. So, we can not get an integer solution for $x.$* This means that the set $P_{2}=\{1,7,14\}$ is non-extendable.

In addition, we can give another $P_{2}$ triple which can not be extended to a quadruple.

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where $x,y,z$ are some integers satisfying the above equations.*      *

By eliminating $m$ from the equations $\left(10\right)$ and $\left(11\right)$,* we get*

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and

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Since this equation is the same, equation $\left(4\right)$ and $\left(8\right),$ in the next steps, we will apply similar methods done in Theorem 1 and 2. Hence, this completes the proof.

Let’s assume that $3k$ $(k\in\mathbb{Z}^{+})$ is an element of the set $P_{2}.$ For any $t\epsilon P_{2}$,* the equation*

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must be satisfied. In modulo3,* the following equation is deduced,              *

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Since $3$ is an odd prime and $\left(2,3\right)=1,$ using Legendre Symbol and Euleur Criterion, we get $\left(\dfrac{2}{3}\right)=-1$ which means that the last equation  is unsolvable. Therefore, $3k$ $(k\in\mathbb{Z}^{+})$ can not be an element of the set $P_{2}.$

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where $x,y,z$ are some integers satisfying the above equations. Eliminating of $m$ between $\left(12\right)$, $\left(13\right)$ yields

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and then we have to abtain following equation

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If we consider equation $\left(16\right)$, since left hand side is divided 3, right hand side must be divided 3 too. So $3\mid x^{2}$ and consequently $3\mid x.$ And we write $x=3x_{1}$ $\left(x_{1}\in\mathbb{Z}\right)$. Byusing this relation if we rewrite equation $\left(16\right),$we get

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this equation gives us $y$ must be odd. Therefore $y=2y_{1}+1$ $\left(y_{1}\in\mathbb{Z}\right)$ and from last equation  we obtain the last equation,

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Left hand side of the equation  $\left(18\right)$ is the multiplication of two consecutive numbers. And we know that multiplication of two consecutive numbers is always even. So the right hand side must be even, that is* $x_{1}=2x_{2}\ \left(x_{2}\in\mathbb{Z}\right)$. We obtain from this x=6x_{2}\, and the equation $(18)$ * turn such Pell equation

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This Pell equations’ main solution is $\left(x_{2},y\right)=(1,7)\in\mathbb{Z}^{2}.$* All of the solutions of this Pell equation can be find such that $y_{n}+\sqrt{48}\left(x_{2}\right)_{n}=\left(7+\sqrt{48}\right)^{n},$* $n=0,1,2,...$ .For * $y$* solutions, the  general recurrence relation is * $y_{n+2}=14y_{n+1}-y_{n}.$ For considering equations $\left(12\right),\left(13\right),\left(14\right)$* we can not find a $m$ value, this gives our assumption is not true.  So the $P_{-3}$ set $\left\{3,4,13\right\}$ can not be extended. This completes the proof.

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By eliminating $m$ from the   equations $\left(21\right)$ and $\ \left(22\right)$ we get

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and

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Since this equation is the same equation with $\left(4\right)$ and $\left(8\right),$ the next steps are same in Theorem 1 and 2 . Therefore this completes the proof.

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where $x,y,z$ are some integers satisfying above equations. Eliminating $m$ from the equations$\ \left(23\right)$ and $\left(24\right)$ yieldst

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and then we obtain following equation

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In the last equation, since the left side is even, the right side must be even too. So we can write  $y=2y_{1}$ $\left(y_{1}\in\mathbb{Z}\right),$* last equation gives us*

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From this, we can conclude that $x$ must be odd. If we put $x=2x_{1}+1\ (x_{1}\in\mathbb{Z}),$ then we find

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The left hand side of this equation has multiple of $2,$ that is, the right hand side must be even. So we can put $y_{1}=2y_{2}\ \left(y_{2}\in\mathbb{Z}\right).$ Thus we can write y=4y_{2}.\Putting $y=4y_{2}$ in the equation (26)\ \yields

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Main solution of this Pell equation is $\left(x,y_{2}\right)=(1,7)\in\mathbb{Z}^{2}.$ All  of the solutions of this Pell equation are of the form $x_{n}+\sqrt{8}(y_{2})_{n}=(3+\sqrt{8})^{n}$, $n=0,1,2,3,...$. For $x$ * solutions * general recurrence relation is $x_{n}=6x_{n-1}-x_{n-2}.$ From the equations $\left(23\right),\left(24\right),\left(25\right)$ we can not find a $m$ value. Hence, our assumption is not true. This completes the proof.

Assume that $5k$ $\left(k\in\mathbb{Z}^{+}\right)$ is an element of both the sets $P_{2}$ and $P_{-3}$. For any $t_{1}\in P_{2}$ and $t_{2}\in P_{-3},$ we can write the following equations,

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From these equations, one can deduce in modulo 5 the followings

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Since $5$ is an odd prime and $\left(2,5\right)=1$, using from Legendre Symbol, we get $\left(\frac{2}{5}\right)=-1$ which means that the equation $\left(27\right)$ is unsolvable. Therefore, $5k$ $\left(k\in\mathbb{Z}^{+}\right)$ can not be an element of both the sets $P_{2}$ and $P_{-3}$.