Extension of a Diophantine triple with the property $D(4)$

TL;DR

This paper establishes an upper bound on how many ways a Diophantine triple with property D(4) can be extended to a quadruple and confirms the uniqueness of such an extension in specific cases.

Contribution

It provides new bounds on extensions of Diophantine triples with property D(4) and verifies the conjecture of uniqueness in certain instances.

Findings

Upper bound on the number of quadruple extensions

Confirmation of the uniqueness conjecture in special cases

Advancement in understanding Diophantine m-tuples with property D(4)

Abstract

In this paper, we give an upper bound on the number of extensions of a triple to a quadruple for the Diophantine $m$-tuples with the property $D(4)$. We also confirm the conjecture of the uniqueness of such an extension in some special cases.

Equations

$$b{s}^2-a{t}^2=4(b-a),$$

$$d_{\pm }=d_{\pm }(a,b,c)=a+b+c+\frac{1}{2}\left(abc\pm \sqrt{(ab+4)(ac+4)(bc+4)}\right).$$

$$ad+4=x^2,bd+4=y^2,cd+4=z^2.$$

$$c{x}^2-a{z}^2$$

$$c{y}^2-b{z}^2$$

$$v_0=z_0,v_1=\frac{1}{2}\left(s{z}_0+c{x}_0\right),v_{m+2}=s{v}_{m+1}-v_m,$$

$$w_0=z_1,w_1=\frac{1}{2}\left(t{z}_1+c{y}_1\right),w_{n+2}=t{w}_{n+1}-w_n.$$

$$c=c_{\nu }^{\pm }=\frac{4}{ab}\left\{{\left(\frac{\sqrt{b}\pm \sqrt{a}}{2}\right)}^2{\left(\frac{r+\sqrt{ab}}{2}\right)}^{2\nu }+{\left(\frac{\sqrt{b}\mp \sqrt{a}}{2}\right)}^2{\left(\frac{r-\sqrt{ab}}{2}\right)}^{2\nu }-\frac{a+b}{2}\right\}$$

$$1$$

$$z\sqrt{a}+x\sqrt{c}$$

$$z\sqrt{b}+y\sqrt{c}$$

$$z$$

$$>0.499b^{3/4}{c}^{7/4}>0.499b{c}^{6/4}>157bc$$

$$d>k\cdot b^{n-1.5}{c}^{n-0.5}$$

$$d>l\cdot a^{m-1.5}{c}^{n-0.5}$$

$$\frac{c}{2x_0}(s-1{)}^{m-1}$$

$$\frac{c}{2y_1}(t-1{)}^{n-1}$$

$$m<\frac{\epsilon +1}{0.999\epsilon }n+1.5-0.4\frac{\epsilon +1}{0.999\epsilon }.$$

$$2.00634^{-1}(ac{)}^{-1/4}(s-1{)}^{m-1}<1.000011(bc{)}^{1/4}{t}^{n-1}.$$

$$(s-1{)}^{m-3/2}<2.00956t^{n-0.5}<t^{n-0.4}.$$

$$s-1>0.9968c^{1/2}>{\left(0.99t\right)}^{\frac{\epsilon }{\epsilon +1}}>t^{0.999\frac{\epsilon }{\epsilon +1}}.$$

$$0.999(m-1.5)\frac{\epsilon }{\epsilon +1}<n-0.4$$

$$\tau =\frac{\sqrt{ab}}{r}\left(1-\frac{a+b+4/c}{c}\right),(<1).$$

$$w_3$$

$$w_5$$

$$w_7$$

$$v_5$$

$$v_7$$

$$v_9$$

$$v_{11}$$

$$v_{13}$$

$$cr(a^4{c}^4+9a^3{c}^3+27a^2{c}^2+30ac-b^3{c}^3-7b^2{c}^2-14bc+2)=$$

$$\mp st(a^4{c}^4+7a^3{c}^3+15a^2{c}^2+10ac-b^3{c}^3-5b^2{c}^2-6bc).$$

$$a^4{c}^4+9a^3{c}^3+27a^2{c}^2+30ac-b^3{c}^3-7b^2{c}^2-14bc+2<$$

Full text

2010 Mathematics Subject Classification: 11D09, 11D45, 11J86

Keywords: diophantine tuples, Pell equations, reduction method, linear forms in logarithms.

1 Introduction

Definition 1.1**.**

Let $n\neq 0$ be an integer. We call a set of $m$ distinct positive integers a $D(n)$-$m$-tuple, or $m$-tuple with the property $D(4)$, if the product of any two of its distinct elements increased by $n$ is a perfect square.

One of the most interesting and frequently studied questions is how large these sets can be. In the classical case (when $n=1$), first studied by Diophantus, Dujella proved in [10] that a $D(1)$-sextuple does not exist and that there are at most finitely many quintuples. Over the years many authors have improved the upper bound for the number of $D(1)$-quintuples. Finally, in [20], He, Togbé and Ziegler established the nonexistence of $D(1)$-quintuples. Details of the history of the problem with all references are available at this Web address [9].

Variants of the problem when $n=4$ or $n=-1$ are also studied frequently. In the case $n=4$, similar conjectures and observations can be made as in the $D(1)$ case. In the light of this observation, Filipin and the author have proven in [5] that a $D(4)$-quintuple also does not exist.

In both cases, $n=1$ and $n=4$, conjectures about the uniqueness of an extension of a triple to a quadruple with a larger element are still open. Moreover, in the case $n=-1$, a conjecture about the nonexistence of a quadruple is studied. For a survey of the latter case of the problem one can see [6].

A $D(4)$-pair can be extended with a larger element $c$ to form a $D(4)$-triple. The smallest such $c$ is $c=a+b+2r$, where $r=\sqrt{ab+4}$ and such a triple is often called a regular triple, or in the $D(1)$ case it is also called an Euler triple. There are infinitely many extensions of a pair to a triple and they can be studied by finding solutions to a Pellian equation

[TABLE]

where $s$ and $t$ are positive integers defined by $ac+4=s^{2}$ and $bc+4=t^{2}.$

For a $D(4)$-triple $\{a,b,c\}$, $a<b<c$, we define

[TABLE]

It is straightforward to check that $\{a,b,c,d_{+}\}$ is a $D(4)$-quadruple, which we will call a regular quadruple. If $d_{-}\neq 0$ then $\{a,b,c,d_{-}\}$ is also a regular $D(4)$-quadruple with $d_{-}<c$.

Conjecture 1.2**.**

Any $D(4)$-quadruple is regular.

Results which support this conjecture in some special cases can be found for example in [1], [13], [17] and [18]. Some of these results are stated in the next section and will be used in our proofs.

In [19], Fujita and Miyazaki approached this conjecture in the $D(1)$ case differently. They examined how many possibilities there are to extend a fixed Diophantine triple with a larger integer. They improved their result from [19] further in their joint work with Cipu [7], where they have shown that any triple can be extended to a quadruple in at most $8$ ways.

In this paper, we will follow the approach and ideas from [7] and [19] in order to deduce similar results for extensions of a $D(4)$-triple. Usually, the numerical bounds and coefficients are slightly better in the $D(1)$ case, which can be seen after comparing for instance Theorem 1.4 and [19, Theorem 1.5]. In order to overcome this problem, we deduce a better numerical lower bound on the element $b$ in an irregular $D(4)$-quadruple, and consider separately all special cases which appeared in the proof (similar to [5]).

Let $\{a,b,c\}$ be a $D(4)$-triple which can be extended to a quadruple with an element $d$. Then there exist positive integers $x,y,z$ such that

[TABLE]

By eliminating $d$ from these equations we get a system of generalized Pellian equations

[TABLE]

There exists only finitely many fundamental solutions $(z_{0},x_{0})$ and $(z_{1},y_{1})$ to these Pellian equations and any solution to the system satisfy $z=v_{m}=w_{n}$, where $m$ and $n$ are non-negative integers and ${v_{m}}$ and ${w_{n}}$ are recurrent sequences defined by

[TABLE]

The initial terms of these sequences were determined by Filipin in [15, Lemma 9]. Notably, one of the results of this paper is an improvement of that lemma by eliminating the case where $m$ and $n$ are even and $|z_{0}|$ is not explicitly determined.

Theorem 1.3**.**

Suppose that $\{a,b,c,d\}$ is a $D(4)$-quadruple with $a<b<c<d$ and that $w_{m}$ and $v_{n}$ are defined as before.

1. i)

If equation $v_{2m}=w_{2n}$ has a solution, then $z_{0}=z_{1}$ and $|z_{0}|=2$ or $|z_{0}|=\frac{1}{2}(cr-st)$. 2. ii)

If equation $v_{2m+1}=w_{2n}$ has a solution, then $|z_{0}|=t$, $|z_{1}|=\frac{1}{2}(cr-st)$ and $z_{0}z_{1}<0$. 3. iii)

If equation $v_{2m}=w_{2n+1}$ has a solution, then $|z_{1}|=s$, $|z_{0}|=\frac{1}{2}(cr-st)$ and $z_{0}z_{1}<0$. 4. iv)

If equation $v_{2m+1}=w_{2n+1}$ has a solution, then $|z_{0}|=t$, $|z_{1}|=s$ and $z_{0}z_{1}>0$.

Moreover, if $d>d_{+}$, case $ii)$ cannot occur.

Also, we improved a bound on $c$ in the terms of $b$ for which an irregular extension might exist.

Theorem 1.4**.**

Let $\{a,b,c,d\}$ be a $D(4)$-quadruple and $a<b<c<d$. If one of the following conditions hold

1. i)

if $b<2a$ and $c\geq 890b^{4}$ or 2. ii)

if $2a\leq b\leq 12a$ and $c\geq 1613b^{4}$ or 3. iii)

if $b>12a$ and $c\geq 39247b^{4}$

then we must have $d=d_{+}$.

Theorem 1.5**.**

Let $\{a,b,c,d\}$ be a $D(4)$-quadruple and $a<b<c<d_{+}<d$. Then any $D(4)$-quadruple $\{e,a,b,c\}$ with $e<c$ must be regular.

For a fixed $D(4)$-triple $\{a,b,c\}$, denote by $N$ the number of positive integers $d>d_{+}$ such that $\{a,b,c,d\}$ is a $D(4)$-quadruple. The next theorem is proven as in [7], and similar methods yielded analogous results.

Theorem 1.6**.**

Let $\{a,b,c\}$ be a $D(4)$-triple with $a<b<c$.

1. i)

If $c=a+b+2r$, then $N\leq 3$. 2. ii)

If $a+b+2r\neq c<b^{2}$, then $N\leq 7$. 3. iii)

If $b^{2}<c<39247b^{4}$, then $N\leq 6$. 4. iv)

If $c\geq 39247b^{4}$, then $N=0$.

This theorem implies the next corollary.

Corollary 1.7**.**

Any $D(4)$-triple can be extended to a $D(4)$-quadruple with $d>\max\{a,b,c\}$ in at most $8$ ways. A regular $D(4)$-triple $\{a,b,c\}$ can be extended to a $D(4)$-quadruple with $d>\max\{a,b,c\}$ in at most $4$ ways.

We remark that we can apply the previous results on a family of triples $c$ which arises from the Pellian equation (1) for the fundamental solutions $(t_{0},s_{0})=(\pm 2,2)$ which gives us a slightly better estimate on a number of extensions when $b<6.85a$.

Proposition 1.8**.**

Let $\{a,b\}$ be a $D(4)$-pair with $a<b$. Let $c=c^{\pm}_{\nu}$ be given by

[TABLE]

where $\nu\in\mathbb{N}$.

1. i)

If $c=c_{1}^{+}$ or $c=c_{1}^{-}$, then $N\leq 3$. 2. ii)

If $c_{2}^{+}\leq c\leq c_{4}^{+}$ then $N\leq 6$. 3. iii)

If $c=c_{2}^{-}$ and $a\geq 2$ then $N\leq 6$ and if $a=1$ then $N\leq 7$. 4. iv)

If $c\geq c_{5}^{-}$ or $c\geq c_{4}^{-}$ and $a\geq 35$ then $N=0$.

Corollary 1.9**.**

Let $\{a,b,c\}$ be a $D(4)$-triple. If $a<b\leq 6.85a$ then $N\leq 6$.

2 Preliminary results about elements of a $D(4)$-$m$-tuple

We begin this section by listing some known results.

Lemma 2.1**.**

Let $\{a,b,c\}$ be a $D(4)$-triple and $a<b<c$. Then $c=a+b+2r$ or $c>\max\{ab+a+b,4b\}$.

Proof.

This follows from [15, Lemma 3] and [11, Lemma 1]. ∎

The next lemma can be deduced similarly as [20, Lemma 2].

Lemma 2.2**.**

Let $\{a,b,c,d\}$ be a $D(4)$-quadruple such that $a<b<c<d_{+}<d$. Then $b>10^{5}$.

Proof.

This result extends the result from [4, Lemma 2.2] and [2, Lemma 3], and is established similarly by using the Baker-Davenport reduction method as described in [12]. For the computation we have used the Mathematica 11.1 software package on a computer with the following specifications; Intel(R) Core(TM) i7-4510U CPU @2.00-3.10 GHz processor. The computation took approximately 170 hours in order to check all of the possibilities. ∎

From [17, Theorem 1.1] we have a lower bound on $b$ in the terms of the element $a$.

Lemma 2.3**.**

If $\{a,b,c,d\}$ is a $D(4)$-quadruple such that $a<b<c<d_{+}<d$ then $b\geq a+57\sqrt{a}$.

The next lemma gives us possibilities for the initial terms of the sequences $v_{m}$ and $w_{n}$, and will be improved by Theorem 1.3.

Lemma 2.4**.**

[15, Lemma 9]** Suppose that $\{a,b,c,d\}$ is a $D(4)$-quadruple with $a<b<c<d$ and that $w_{m}$ and $v_{n}$ are defined as before.

1. i)

If equation $v_{2m}=w_{2n}$ has a solution, then $z_{0}=z_{1}$ and $|z_{0}|=2$ or $|z_{0}|=\frac{1}{2}(cr-st)$ or $z_{0}<1.608a^{-5/14}c^{9/14}$. 2. ii)

If equation $v_{2m+1}=w_{2n}$ has a solution, then $|z_{0}|=t$, $|z_{1}|=\frac{1}{2}(cr-st)$ and $z_{0}z_{1}<0$. 3. iii)

If equation $v_{2m}=w_{2n+1}$ has a solution, then $|z_{1}|=s$, $|z_{0}|=\frac{1}{2}(cr-st)$ and $z_{0}z_{1}<0$. 4. iv)

If equation $v_{2m+1}=w_{2n+1}$ has a solution, then $|z_{0}|=t$, $|z_{1}|=s$ and $z_{0}z_{1}>0$.

Remark**.**

From the proof of [15, Lemma 9] we see that the case where $v_{2m}=w_{2n}$ and $z_{0}<1.608a^{-5/14}c^{9/14}$ holds only when $d_{0}=\frac{z_{0}^{2}-4}{c}$, $0<d_{0}<c$, is such that $\{a,b,d_{0},c\}$ is an irregular $D(4)$-quadruple. As we can see from the statement of Theorem 1.3, this case will be proven impossible and only cases where $z_{0}$ is given in the terms of elements of a triple $\{a,b,c\}$ will remain.

Using the lower bound on $b$ in an irregular quadruple from Lemma 2.2 we can slightly improve [15, Lemma 1].

Lemma 2.5**.**

Let $(z,x)$ and $(z,y)$ be positive solutions of (2) and (3). Then there exist solutions $(z_{0},x_{0})$ of (2) and $(z_{1},y_{1})$ of (3) in the ranges

[TABLE]

such that

[TABLE]

The statement of the next lemma follows the notations from Theorem 1.3.

Lemma 2.6**.**

Let $\{a,b,c\}$ be a $D(4)$-triple such that $z=v_{m}=w_{n}$ has a solution $(m,n)$ for which $d=d_{+}=\frac{z^{2}-4}{c}$. Then only one of the following cases can occur:

1. i)

$(m,n)=(2,2)$* and $z_{0}=z_{1}=\frac{1}{2}(st-cr)$,* 2. ii)

$(m,n)=(1,2)$* and $z_{0}=t$, $z_{1}=\frac{1}{2}(st-cr)$,* 3. iii)

$(m,n)=(2,1)$* and $z_{0}=\frac{1}{2}(st-cr)$, $z_{1}=s$,* 4. iv)

$(m,n)=(1,1)$* and $z_{0}=t$, $z_{1}=s$.*

Proof.

If $n\geq 3$ then

[TABLE]

where we have used Lemma 2.5 and $bc>10^{10}$ from Lemma 2.2. On the other hand, when $d=d_{+}$ we have $z=\frac{1}{2}(cr+st)<cr<cb$, which is an obvious contradiction. So, when $d=d_{+}$, we must have $n\leq 2$. Also, since $a<b<c<d$, from the proof of [15, Lemma 6] we see that when $n\leq 2$, the only possibility is $d=d_{+}$ when $(m,n;z_{0},z_{1})\in\{(1,1;t,s),(1,2;t,\frac{1}{2}(st-cr)),(2,1;\frac{1}{2}(st-cr),s),(2,2;\frac{1}{2}(st-cr),\frac{1}{2}(st-cr))\}$. ∎

This lemma can now be used to get a lower bound on $d$ in terms of the elements of a triple $\{a,b,c\}$.

Lemma 2.7**.**

Suppose that $\{a,b,c,d\}$ is a $D(4)$-quadruple with $a<b<c<d_{+}<d$ and that $v_{m}$ and $w_{n}$ are defined as before. If $z\geq w_{n}$, $n=4,5,6,7$, then

[TABLE]

where $k=0.249979,0.249974,0.249969,0.249965$ respectively.

If $z\geq v_{m}$, $n=4,5,6$, then

[TABLE]

where $l=0.243775,0.242245,0.240725$ respectively.

Proof.

In the proof of [15, Lemma 5] it has been shown that

[TABLE]

We use $bc>10^{10}$ for $d>d_{+}$ and $d=\frac{z^{2}-4}{c}$ to obtain the desired inequalities. ∎

We know a relation between $m$ and $n$ if $v_{m}=w_{n}$.

Lemma 2.8**.**

[15, Lemma 5]** Let $\{a,b,c,d\}$ be $D(4)$-quadruple. If $v_{m}=w_{n}$ then $n-1\leq m\leq 2n+1$.

We remark that we can also prove that a better upper bound holds using a more precise argument and the fact that $c<7b^{11}$ (which is proven in [14, Lemma 8]).

Lemma 2.9**.**

If $c>b^{\varepsilon}$, $1\leq\varepsilon<12$, then

[TABLE]

Proof.

If $v_{m}=w_{n}$ then

[TABLE]

Since $c>b>10^{5}$ we can easily check that $s-1>0.9968a^{1/2}c^{1/2}$ and $b^{1/2}c^{1/2}<t<1.0001b^{1/2}c^{1/2}$, so the previous inequality implies

[TABLE]

On the other hand, the assumption $b<c^{1/\varepsilon}$ yields $t<1.0001c^{\frac{\varepsilon+1}{2\varepsilon}}$ and

[TABLE]

So we observe that an inequality

[TABLE]

must hold, which proves our statement. ∎

The next lemma follows from the results of [16, Lemma 5] and [14, Lemma 5].

Lemma 2.10**.**

If $v_{m}$=$w_{n}$ has a solution for $m>2$, then $6\leq m\leq 2n+1$ and $n\geq 7$ or the case $|z_{0}|<1.608a^{-5/14}c^{9/14}$ from Lemma 2.4 holds and either $6\leq m\leq 2n+1$ or $m=n=4$.

Lemma 2.11**.**

Assume that $c\leq 0.243775a^{2.5}b^{3.5}$. If $z=v_{m}=w_{n}$ for some even $m$ and $n$ then $d>0.240725a^{4.5}c^{5.5}$.

Proof.

Let us assume that $z_{0}\notin\left\{2,\frac{1}{2}(cr-st)\right\}$, i.e. there exists an irregular $D(4)$-quadruple $\{a,b,d_{0},c\}$, $c>d_{0}$ but then Lemma 2.7 implies $c>0.243775a^{2.5}b^{3.5}$, a contradiction. So, we must have $z_{0}\in\left\{2,\frac{1}{2}(cr-st)\right\}$ and by Lemma 2.10 we know that $\max\{m,n\}\geq 6$. The statement now follows from Lemma 2.7. ∎

By using the improved lower bound on $d$ in an irregular quadruple from Lemma 2.7 we can prove the next result in the same way as [15, Lemma 9] is proved.

Lemma 2.12**.**

If $v_{m}=w_{n}$ for some even $m$ and $n$ and $|z_{0}|\notin\left\{2,\frac{1}{2}(cr-st)\right\}$ then $|z_{0}|<1.2197b^{-5/14}c^{9/14}$.

We can prove some upper and lower bounds on $c$ in the terms of smaller elements, depending on the value of $z_{0}$, similarly as in [19], by using Lemma 2.5

Lemma 2.13**.**

Set

[TABLE]

We have that

1. i)

$|z_{0}|=\frac{1}{2}(cr-st)$* implies $c<ab^{2}\tau^{-4}$,* 2. ii)

$|z_{1}|=\frac{1}{2}(cr-st)$* implies $c<a^{2}b\tau^{-4}$,* 3. iii)

$|z_{0}|=t$* implies $c>ab^{2}$,* 4. iv)

$|z_{1}|=s$* implies $c>a^{2}b$,*

and ii) and iii) cannot occur simultaneously when $d>d_{+}$.

The next lemma follows easily by induction.

Lemma 2.14**.**

Let $\{v_{z_{0},m}\}$ denote a sequence $\{v_{m}\}$ with an initial value $z_{0}$ and $\{w_{z_{1},n}\}$ denote a sequence $\{w_{n}\}$ with an initial value $z_{1}$. It holds that $v_{\frac{1}{2}(cr-st),m}=v_{-t,m+1}$, $v_{-\frac{1}{2}(cr-st),m+1}=v_{t,m}$ for each $m\geq 0$ and $w_{\frac{1}{2}(cr-st),n}=w_{-s,n+1}$, $w_{-\frac{1}{2}(cr-st),n+1}=w_{s,n}$ for each $n\geq 0$.

For the proof of Theorem 1.6 we will use the previous lemma. It is obvious that if we "shift" a sequence as in Lemma 2.14 the initial term of the new sequence would not necessarily satisfy the bounds from Lemma 2.5. In the next lemma we will prove some new lower bounds on $m$ and $n$ when $|z_{0}|=t$ and $|z_{1}|=s$, but without assuming the bounds on $z_{0}$ and $z_{1}$ from Lemma 2.5. Since Filipin has deduced in [16] that $n\geq 7$ in any case $(z_{0},z_{1})$ which appears when "shifting" sequences as in Lemma 2.14, we can consider that bound already established. Even though the following proof is analogous to the proof of [7, Lemma 2.6], there is more to consider in the $D(4)$ case which is why we present some details of the proof.

Proposition 2.15**.**

Let $\{a,b,c\}$ be a $D(4)$-triple, $a<b<c$, $b>10^{5}$ and $c>ab+a+b$. Let us assume that the equation $v_{m}=w_{n}$ has a solution for $m>2$ such that $|z_{0}|=t$, $|z_{1}|=s$, $z_{0}z_{1}>0$ and $m\equiv n\equiv 1\ (\bmod\ 2)$. Then $\min\{m,n\}\geq 9$.

Proof.

It is easy to see that $v_{1}=w_{1}<v_{3}<w_{3}$. If we show that

1. i)

$w_{3}<v_{5}<w_{5}<v_{9}$ and $v_{7}\neq w_{5}$, 2. ii)

$v_{7}<w_{7}<v_{13}$ and $v_{9}\neq w_{7}\neq v_{11}$,

from Lemma 2.8 we see that it leads to a conclusion that $\min\{m,n\}\geq 9$.

Let $(z_{0},z_{1})=(\pm t,\pm s)$. We derive that

[TABLE]

and

[TABLE]

[TABLE]

Since $a<b$, $c>ab$, and the sequences $v_{m}$ and $w_{n}$ are increasing, it is easy to see that $w_{3}<v_{5}<v_{7}$, $v_{5}<w_{5}<v_{7}<v_{9}<v_{11}$, $v_{7}<w_{7}$, $w_{5}<v_{9}$ and $w_{7}<v_{13}$. It remains to prove that $v_{7}\neq w_{5}$, $v_{9}\neq w_{7}$ and $v_{11}\neq w_{7}$.

From the Lemma 2.10 and the explanation before this lemma, we consider that $v_{7}\neq w_{5}$ is already proven but it is not hard to follow the next proof in order to deduce this case also. We will show only that $v_{9}\neq w_{7}$, and the case $v_{11}\neq w_{7}$ can be shown similarly with only some technical details changed.

Let us assume to the contrary, that $v_{9}=w_{7}$. We have for $(z_{0},z_{1})=(\pm t,\pm s)$ that

[TABLE]

1. a)

**Case $z_{0}=-t$

**Since $cr>st$ we have

[TABLE]

which leads to

[TABLE]

Since $c>10^{5}$ we have $b>316a$. On the other hand, we easily see that

[TABLE]

which can be used to prove

[TABLE]

So

[TABLE]

By combining equations (4) and (5) we see that $b<2.006a$, which is in a contradiction to $b>316a$. 2. b)

**Case $z_{0}=t$

**Since $cr>st$ we have

[TABLE]

which leads to

[TABLE]

If $16a^{3}c^{3}<12b^{2}c^{2}$ then $c<0.75\frac{b^{2}}{a^{3}}$. On the other hand, if $16a^{3}c^{3}\geq 12b^{2}c^{2}$ then $2a^{4}c^{4}<2b^{3}c^{3}$. In each case, inequality

[TABLE]

holds. Since $c>10^{5}$, $b>46a$ and since $c>ab$ we have $b>a^{5/2}$ and $c>a^{7/2}$. It is easily shown that $cr+st>632r^{2}$. We can use this to see that

[TABLE]

from which we get an upper bound on $v_{9}$. Now, from $v_{9}=w_{7}$ we have

[TABLE]

Notice that $c>b^{3}1.001^{-1}a^{-4}>0.999b^{7/5}>9.99\cdot 10^{6}$.

If we consider $v_{9}=w_{7}$ modulo $c^{2}$ and use the fact that $st(cr-st)\equiv 16\ (\bmod\ c)$ it yields a congruence

[TABLE]

Since $2r(cr-st)<4.01c$ and $16(6b-10a)<96b<96\cdot(1.001a^{4}c)^{1/3}<97c^{5/7}<c$, we have that one of the equalities

[TABLE]

must hold.

If $k=0$, we have $2r(cr-st)=16(6b-10a)$. The inequality $2r(cr-st)>3.8r^{2}>3.8ab$ implies $96b>16(6b-10a)>3.8ab$ so $a\leq 25$. Now, we have

[TABLE]

so $16(6b-10a)>2\sqrt{ab+4}\cdot 0.0000005b^{5/2}$. For each $a\in[1,25]$, we get from this inequality a numerical upper bound on $b$ which is in a contradiction to $b>10^{5}$.

If $k\neq 0$, we have a quadratic equation in $c$ with possible solutions

[TABLE]

where $A=r^{2}(16-4k)+k^{2}>0$, $B=-(32(2r^{2}-6)(6b-10a)+16r^{2}(a+b))<0$ and $E=64(4(6b-10a)^{2}-r^{2})>0$.

If $k\leq 3$, $A>4r^{2}$ and $c_{\pm}<\frac{-B}{A}<\frac{32\cdot 2r^{2}+6b+16r\cdot 32b}{4r^{2}}=100b$. Since $c>ab+a+b$, we have $a\leq 98$ and from $b^{3}<1.001a^{4}c<100.1a^{4}b$ we have $b<100.1^{1/2}a^{2}<96089$ which is in a contradiction to $b>10^{5}$.

In only remains to check if $k=4$. But in this case we express $b$ in the terms of $a$ and $c$ and get

[TABLE]

where $A=400ac-9216>0$, $B=c(16a^{2}-640a+832)+64a+30720$ and $E=16c^{2}+1216ac+25600a^{2}-256>0$.

We have $B^{2}+4AE>4AE>25536ac^{3}$, so it is not hard to see that $b_{-}<0$. Also, $B<0$ when $a\in[2,38]$, and $B>0$ otherwise. When $B>0$, $b_{+}<\frac{\sqrt{B^{2}+4AE}}{2A}<\frac{\sqrt{256a^{4}c^{2}+400ac^{2}(16c+2000a)}}{798ac}$ and since $a<c^{2/7}$, we get $b_{+}<\frac{84a^{1/2}c^{3/2}}{798ac}<0.106c^{1/2}a^{-1/2}$. On the other hand, $b_{+}>\frac{\sqrt{25536ac^{3}}-16a^{2}c}{800ac}>0.18c^{1/2}a^{-1/2}$, which is a contradiction to the previous inequality. In the last case, when $B<0$, we have $|B|<7232c$ and get a similar contradiction.∎

3 Proof of Theorem 1.3

Lemma 3.1**.**

Let us assume that $c>0.243775a^{2}b^{3.5}$ and that $z=v_{m}=w_{n}$ has a solution for $m$ and $n$, $n\geq 4$. Then $m\equiv n\ (\bmod\ 2)$ and

1. i)

if $m$ and $n$ are even and $b\geq 2.21a$, then

[TABLE]

and if $b<2.21a$ then

[TABLE] 2. ii)

if $m$ and $n$ are odd, then

[TABLE]

Proof.

Since from $b>10^{5}$ and $c>0.243775a^{2}b^{3.5}$ we have $c>ab^{2}$ and $\tau^{-4}<b/a$, and cases $i)$ and $ii)$ from Lemma 2.13 cannot hold. So, we see that the only options from Lemma 2.4 are $i)$, when $|z_{0}|=2$ or $|z_{0}|<1.219b^{-5/14}c^{9/14}$, and $iv)$. In each case we have $m\equiv n\ (\bmod\ 2)$.

First, let us consider the case $|z_{0}|<1.219b^{-5/14}c^{9/14}$, $z_{0}=z_{1}$ and $m$ and $n$ even. Since $(z_{0},x_{0})$ satisfies an equality

[TABLE]

we have

[TABLE]

where we have used the estimate $c^{2}b^{-5}>0.243775^{2}b^{2}>5.9426\cdot 10^{8}$. So,

[TABLE]

Similarly, we get

[TABLE]

From [15, Lemma 12] we have that the next congruence holds

[TABLE]

From $b>10^{5}$ and $c>0.243775b^{3.5}$ we have $c>b^{3.377}$, so we can use $\varepsilon=3.377$ in the inequality from Lemma 2.9 and get

[TABLE]

This implies that the inequality $m<1.34n$ holds for every possibility $m$ and $n$ even, except for $(m,n)=(6,4)$, which we will study separately.

Now we study the case where $b\geq 2.21a$ and we assume to the contrary, that $n\geq 0.45273b^{-9/28}c^{5/28}$. Then from $c>b>10^{5}$ we have

[TABLE]

[TABLE]

and from inequalities (9) and (10) we also have

[TABLE]

and similarly

[TABLE]

In the case $(m,n)=(6,4)$ we can prove the same final inequalities. So, from congruence (11) we see that the equation

[TABLE]

must hold.

On the other hand, from equation $az_{0}^{2}-cx_{0}^{2}=4(a-c)$, since $|z_{0}|\neq 2$ in this case, and $c\mid(z_{0}^{2}-4)$ we have $z_{0}^{2}\geq c+4$. Let us assume that $z_{0}^{2}<\frac{5c}{a}$, then we would have $c(x_{0}^{2}-9)+4a<0$ and since $c>4a$ we must have $x_{0}=2$ and $|z_{0}|=2$, which is not our case. So, here we have $z_{0}^{2}\geq\max\{c+4,\frac{5c}{a}\}$.

Now,

[TABLE]

and similarly

[TABLE]

When $z_{0}>0$, i.e. $z_{0}>2$, we have

[TABLE]

Since $m\geq 4$ we have from the previous inequality that $2.21n(n+1)<1.35001m^{2}<1.35001(1.2975n+0.9811)$ must hold. But then we get $n<1$, an obvious contradiction.

On the other hand, if $z_{0}<0$, i.e. $z_{0}<-2$, we similarly get

[TABLE]

Since, $m<1.2975n+0.9811$ and $b>2.21a$ we have $n\leq 6$. If $b\geq 2.67a$, we would have $n<4$. So, it only remains to study the case $2.21a\leq b<2.67a$ and $n\leq 6$. In that case we have $c>0.243775a^{2}c^{3.5}>0.03419b^{5.5}$, so we can put $\varepsilon=5.2$ in Lemma 2.9 and get $m<1.1936n+1.0226$. Inserting in the inequality (13) yields that only the case $(m,n)=(4,4)$ remains, but it doesn’t satisfy equation (13).

Now, let us study the case where $b<2.21a$. We again consider the congruence (11), which after squaring and using $z_{0}^{2}\equiv t^{2}\equiv s^{2}\equiv 4\ (\bmod\ c)$ yields a congruence

[TABLE]

Let us denote $C=4((am^{2}-bn^{2})^{2}-y_{1}^{2}n^{2}-x_{0}^{2}m^{2})$, and (14) multiplied by $s$ and by $t$ respectively shows that

[TABLE]

Now, assume that $n\leq\min\{0.35496a^{-1/2}b^{-1/8}c^{1/8},0.177063b^{-11/28}c^{3/28}\}.$ Then also $n\leq 0.45273b^{-9/28}c^{5/28}$ holds, so we again have an equality in the congruence (11), i.e.

[TABLE]

We have that $x_{0}^{2}<y_{1}^{2}<1.4877b^{2/7}c^{2/7}.$ Since $b<2.21a$ we have $c>0.04991b^{5.5}$so we can take $\varepsilon=5.239$ in Lemma 2.9 and we get $m<1.1921n+1.0232$, and since we know $m,n\geq 4$ and $m$ and $n$ even, we also have $m<1.34n$ from this inequality. This yields

[TABLE]

On the other hand, we have from our assumption on $n$ that

[TABLE]

and

[TABLE]

so, $|Cs|<|Ct|<c$. On the other hand, $8sx_{0}y_{1}mn<8tx_{0}y_{1}mn$ and

[TABLE]

So, in congruences (15) and (16) we can only have

[TABLE]

If $k=l=0$, we would have $s=t$, which is not possible. When $k=l=1$, we get $c=8(t+s)x_{0}y_{1}mn<0.5c+0.5c<c$, a contradiction. In the case $k=0$ and $l=1$ we get $cs=8(s^{2}-t^{2})x_{0}y_{1}mn<0$, and in the case $k=1$ and $l=0$, we have $ct=8(t^{2}-s^{2})x_{0}y_{1}mn$, so $c(t-s)<8(t^{2}-s^{2})x_{0}y_{1}mn$, which leads to a contradiction as in the case $k=l=1$.

So, $n>\min\{0.35496a^{-1/2}b^{-1/8}c^{1/8},0.177063b^{-11/28}c^{3/28}\}$ must hold.

It remains to consider the case when $m$ and $n$ are odd. In this case, a congruence from [15, Lemma 3] holds,

[TABLE]

Let us assume that $n\leq 0.30921b^{-3/4}c^{1/4}$. In this case we have $m<1.2975n+0.9811$ and since $m\geq n\geq 5$ are both odd we also have $m<1.47n$. Notice that $2t(am(m+1)-bn(n+1))<2tbm(m+1)$ holds. Also, from $c>0.243775b^{3.5}>7.7\cdot 10^{16}$ we have $b<0.243775^{-2/7}c^{2/7}$. So it suffice to observe that

[TABLE]

which means that we have equalities in congruences (17) and (18) and implies

[TABLE]

Since $s\neq t$, the only possibility is $n=m$, but then $2t(am(m+1)-bn(n+1))=0$ implies $a=b$, a contradiction. So, our assumption for $n$ was wrong and $n>0.30921b^{-3/4}c^{1/4}$ when $n$ and $m$ are odd. ∎

Various versions of Rickert’s theorem from [23] and results derived from them are often used when considering problems of $D(1)$-$m$-tuples and $D(4)$-$m$-tuples. In this paper we will use a lemma from [2] and give a new version of that result, which improves it in some special cases.

Lemma 3.2**.**

[2, Lemmas 6 and 7]** Let $\{a,b,c,d\}$ be a $D(4)$-quadruple, $a<b<c<d$ and $c>308.07a^{\prime}b(b-a)^{2}a^{-1}$, where $a^{\prime}=\max\{4a,4(b-a)\}$. Then

[TABLE]

By combining results and proofs of [2, Lemma 7] and [8, Lemma 3.3], the next generalization of [5, Lemma 7] can be proved.

Lemma 3.3**.**

Let $\{a,b,c,d\}$ be a $D(4)$-quadruple, $a<b<c<d$, $b>10^{5}$ and $c>59.488a^{\prime}b(b-a)^{2}a^{-1}$, where $a^{\prime}=\max\{4a,4(b-a)\}$. If $z=v_{m}=w_{n}$ for some $m$ and $n$ then

[TABLE]

Proposition 3.4**.**

Let $\{a,b,c,d\}$ be a $D(4)$-quadruple such that $a<b<c<d$ and that equation $z=v_{m}=w_{n}$ has a solution for some $m$ and $n$. If $b\geq 2.21a$ and $c>\max\{0.24377a^{2}b^{3.5},2.3b^{5}\}$ or $b<2.21a$ and $c>1.1b^{7.5}$ then $d=d_{+}$.

Proof.

Let us assume that $d>d_{+}$. Since $c>0.24377a^{2}b^{3.5}$ in both cases, we can use Lemma 3.1, and as in the proof of that lemma, $m$ and $n$ have the same parity. Also, since in any case $c>2.3b^{5}>308.07a^{\prime}b(b-a)^{2}a^{-1}$, we can use Lemmas 3.2 and 3.3 (note Lemma 3.2 will give better results in these cases).

Let us assume that $m$ and $n$ are even and $b\geq 2.21a$. Then we have $n>0.45273b^{-9/28}c^{5/28}$, $a^{\prime}=\max\{4a,4(b-a)\}=4(b-a)$, $a\leq\frac{b}{2.21}$ and $b>b-a>\frac{1.21}{2.21}b$. We estimate

[TABLE]

Now, from Lemma 3.2 we have an inequality

[TABLE]

where the right-hand side is decreasing in $c$ for $b>0$, $c>2.3b^{5}$ so we can observe the inequality in which we have replaced $c$ with $2.3b^{5}$ which gives us $b<19289$, a contradiction to $b>10^{5}.$

Let us consider the case when $m$ and $n$ are even and $2a<b<2.21a$. Then $a^{\prime}=4(b-a)$, $\frac{b}{2}<b-a<\frac{1.21}{2.21}b$. Denote

[TABLE]

Then by Lemma 3.2

[TABLE]

The right-hand side of this inequality is decreasing in $c$ for $b>0$, $c>1.1b^{7.5}$, and for each possibility for $F$ we get $b<722$ and $b<81874$ respectively, a contradiction in either case.

In the case when $m$ and $n$ are even and $b<2a$, we have $a^{\prime}=4a$ and $57<b-a<\frac{b}{2}$. With $F$ defined as before, we have $F>\{0.35921b^{9/16},0.17888b^{23/56}\}$ and

[TABLE]

Again, the right-hand side is decreasing in $c$ for $c>1.1b^{7.5}$. We get $b<1396$ for the first choice for $F$, and $b<98413$ for the second, again, a contradiction.

It remains to consider the case when $m$ and $n$ are odd. If $b\geq 2a$ we have $a^{\prime}=4(b-a)$ and $c>2.3b^{5}$, so similarly as before

[TABLE]

In this case we have $n>0.30921b^{-3/4}c^{1/4}$ so by Lemma 3.2 we observe an inequality

[TABLE]

Since the right-hand side is decreasing in $c$ for $c>2.3b^{5}$ we get $b<97144$, a contradiction.

If $b<2a$ and $c>1.1b^{7.5}$ we observe

[TABLE]

and since the right-hand side is decreasing in $c$ for $c>1.1b^{7.5}$ we get $b<48$, a contradiction. ∎

Proof of Theorem 1.3.

Let us assume that $d=d_{+}$. Then Lemma 2.6 gives us all of the possible fundamental solutions and indices.

Let us assume that $d>d_{+}$. Then from Lemma 2.10 we have $n\geq 4$. Here we only consider a possibility when $m$ and $n$ are even and $z_{0}=z_{1}\notin\left\{2,\frac{1}{2}(cr-st)\right\}$. Then from [15] we know that for $d_{0}=\frac{z_{0}^{2}-4}{c}<c$, $D(4)$-quadruple $\{a,b,d_{0},c\}$ is irregular.

Denote $\{a_{1},a_{2},a_{3}\}=\{a,b,d_{0}\}$ such that $a_{1}<a_{2}<a_{3}$.

If $a_{3}\leq 0.234775a_{1}^{2.5}a_{2}^{3.5}$ holds then by Lemma 2.11 we also have an inequality $c>0.240725a_{1}^{4.5}a_{3}^{5.5}\geq 0.240725a^{4.5}b^{5.5}$. If $b>2.21a$, since $b>10^{5}$, this inequality implies $c>76a^{4.5}b^{5.5}$. On the other hand, if $b\leq 2.21a$, we have $a\geq 45249$ and $c>0.240725\cdot 2.21^{-2}a^{2.5}b^{7.5}>10a^{2}b^{7.5}$. We see that in either case we can use Proposition 3.4 and conclude $d=d_{+}$, i.e. we have a contradiction to the assumption $d>d_{+}$.

It remains to consider the case $a_{3}>0.234775a_{1}^{2.5}a_{2}^{3.5}$. If $b=a_{2}$ then from Lemma 2.7 we get

[TABLE]

It is easy to see that we again have conditions of Proposition 3.4 satisfied and can conclude that $d=d_{+}$ is the only possibility. On the other hand, if $b=a_{3}$, then $b>0.243775a_{1}^{2.5}a_{2}^{3.5}>0.243775\cdot 10^{17.5}$ since $a_{2}>10^{5}$. As such, by Lemma 3.1 we need to consider two cases. First, when $a_{2}\geq 2.21a_{1}$ then $n^{\prime}>0.45273a_{2}^{-9/28}b^{5/28}>0.45273(0.243775^{-2/7}b^{2/7})^{-9/28}b^{5/28}>0.3976b^{17/196}>11$. By Lemma 2.7 it follows that $z^{\prime}>w_{6}$ and $c>0.249969a_{1}^{4.5}b^{5.5}$ which, as before, yields a contradiction when Proposition 3.4 is applied. The second case is when $a_{2}<2.21a_{1}$. If $n^{\prime}\geq 6$, we have $c>0.249969a_{1}^{4.5}b^{5.5}>b^{7.5}$, since $a_{1}>\frac{a_{2}}{2.21}>\frac{10^{5}}{2.21}$ and get the same conclusion as before. If $n<6$, by Lemma 2.10, we see that we have $m^{\prime}$ and $n^{\prime}$ even, so $n^{\prime}=4$. Since $b>0.0335a_{2}^{6}>a_{2}^{5.7}$ from Lemma 2.9 we have that $m^{\prime}<1.1766n^{\prime}+1.0294$, i.e. $m^{\prime}=4$. From the proof of [14, Lemma 5] we know that $m^{\prime}=n^{\prime}=4$ can hold only when $|z_{0}^{\prime}|<1.2197(b^{\prime})^{-5/14}(c^{\prime})^{9/14}$. As such, we have a $0<d_{0}^{\prime}<b$ such that $\{a,d_{0}^{\prime},d_{0},b\}$ is an irregular $D(4)$-quadruple and we can use the same arguments to prove that such a quadruple cannot exist by Proposition 3.4, or we have a new quadruple with $0<d_{0}^{\prime\prime}<d_{0}^{\prime}<b$. Since this process cannot be repeated infinitely, for some of those quadruples in the finite process we must have $n>6$, a contradiction to Proposition 3.4.

The last assertion of Theorem 1.3 follows from Lemma 2.13. ∎

4 Proofs of Theorems 1.4 and 1.5

A more general result for the lower bound on $m$ can be established by following an analogous argument as in [4, Proposition 3.1] and [5, Lemma 16].

Lemma 4.1**.**

Let $\{a,b,c,d\}$ be a $D(4)$-quadruple with $a<b<c<d$ for which $v_{2m}=w_{2n}$ has a solution such that $z_{0}=z_{1}=\pm 2$ and $Ln\geq m\geq n\geq 2$ and $m\geq 3$, for some real number $L>1$.

Suppose that $a\geq a_{0}$, $b\geq b_{0}$, $c\geq c_{0}$ and $b\geq\rho a$ for some positive integers $a_{0}$, $b_{0}$, $c_{0}$ and a real number $\rho>1$. Then

[TABLE]

where $\alpha$ is any real number satisfying both inequalities

[TABLE]

[TABLE]

*with $\lambda=(a_{0}+4)^{1/2}(\rho a_{0}+4)^{-1/2}$.

Moreover, if $c^{\tau}\geq\beta b$ for some positive real numbers $\beta$ and $\tau$ then*

[TABLE]

Lemma 4.2**.**

Let us assume that $c>b^{4}$ and $z=v_{m}=w_{n}$ has a solution for some positive integers $m$ and $n$. Then $m\equiv n\ (\bmod\ 2)$ and $n>0.5348b^{-3/4}c^{1/4}$.

Proof.

As in the proof of Lemma 3.1, we can see that $m\equiv n\ (\bmod\ 2)$ when $c>b^{4}$, and when $m$ and $n$ are even, the only possibility is $|z_{0}|=2$. By Theorem 1.3 we need to consider two cases.

Case $m$ and $n$ are even, $|z_{0}|=|z_{1}|=2$.

From Lemma 2.9 we have $m<1.252n+0.9995$. Since from Lemma 2.10 we have $m\geq 6$ we also have $n\geq 6$ and $m<1.4n$. Using Lemma 4.1 yields $m>0.4999998b^{-1/2}c^{1/2}$, and finally

[TABLE] 2. 2)

Case $m$ and $n$ odd, $|z_{0}|=t$, $|z_{1}|=s$.

Congruences (17) and (18) from the proof of Lemma 3.1 hold. Let us assume to the contrary, that $n<0.5348b^{-3/4}c^{1/4}$. Then

[TABLE]

which means that we have equality in those congruences. This implies that a contradiction can be established as in the proof of Lemma 3.1.∎

Proof of Theorem 1.4.

Let us assume that $d>d_{+}$. We prove this as in Proposition 3.4

1. i)

Case $b<2a$ and $c\geq 890b^{4}$.

Since $a^{\prime}=4a$ and $b>10^{5}$, inequality

[TABLE]

holds. We can use Lemma 3.2 and Lemma 4.2 and observe inequality

[TABLE]

and since the right-hand side is decreasing in $c$ for $c>890b^{4}$ we get $b<99887$ a contradiction to $b>10^{5}$. 2. ii)

Case $2a\leq b\leq 12a$ and $c\geq 1613b^{4}$.

Here we have $a^{\prime}=4(b-a)$ and $\frac{b}{2}\leq b-a\leq\frac{11}{12}b$, so

[TABLE]

We observe an inequality

[TABLE]

After noting that the right-hand side is decreasing in $c$ for $c>1613b^{4}$, we deduce $b<99949$, a contradiction. 3. iii)

Case $b>12a$ and $c\geq 52761b^{4}$.

Let us first assume that $c\geq 52761b^{4}$. Here we have $a^{\prime}=4(b-a)$ and $\frac{b}{12}<b-a<b$, so

[TABLE]

We use Lemma 3.3 to obtain inequality

[TABLE]

Similarly as before, after noting that the right-hand side is decreasing in $c$ for $c>52761b^{4}$, we deduce $b<99998$, a contradiction.

Now, we assume that $39247b^{4}<c<52761b^{4}$. We can modify the method in the following way. For $a=1$, we only have to notice that the right-hand side in the inequality in the Lemma 3.3 is decreasing in $c$, insert our lower bound on $c$, and calculate an upper bound on $b$ from the inequality. We get $b<999994$, a contradiction. For $a\geq 2$, we modify estimate $0.016858a(a^{\prime})^{-1}b^{-1}(b-a)^{-2}c>0.0042145b^{-4}ca_{0}$, where $a_{0}=2$ and get $b<73454$, a contradiction. ∎

Lemma 4.3**.**

If $\{a,b,c,d\}$ is a $D(4)$-quadruple with $a<b<c<d_{+}<d$ then

[TABLE]

Proof.

From [16, Lemma 5] and Theorem 1.3 we have that $m\geq 6$ or $n\geq 7$ so Lemma 2.7 implies $d>0.249965b^{5.5}c^{6.5}$ or $d>0.240725a^{4.5}c^{5.5}$. ∎

Proof of Theorem 1.5.

Let $\{a,b,c,d\}$ be a $D(4)$-quadruple such that $a<b<c<d_{+}<d$ and let us assume to the contrary, that there exists $e<c$ such that $\{e,a,b,c\}$ is an irregular $D(4)$-quadruple. Then by Lemma 4.3 $c>0.240725a^{\prime 4.5}b^{5.5}>0.240725\cdot(10^{5})^{1.5}b^{4}>52761b^{4}$, where $a^{\prime}=\min\{a,e\}\geq 1$ or $c>0.249965a^{5.5}c^{6.5}>52761b^{4}$. Theorem 1.4 implies that $\{a,b,c,d\}$ must be a regular quadruple; a contradiction. ∎

5 Proof of Theorem 1.6

In this section, we aim to split our problem into several parts. We will consider separately the case when a triple $\{a,b,c\}$ is regular ($c=a+b+2r$), and when it is not regular ($c>ab+a+b$). In the latter case, we will consider solutions of the equation $z=v_{m}=w_{n}$ without assuming that the inequalities from Lemma 2.5 hold when $|z_{0}|\neq 2$. Lemmas from this section will usually also address separately the case when $c>ab+a+b$. More specifically, only this case when $(|z_{0}|,|z_{1}|)=(t,s)$ and $z_{0}z_{1}>0$, which can then be used to prove the results to all the other cases (except the case $|z_{0}|=2$) by using Lemma 2.14.

In what follows, we will introduce results concerning linear forms in three logarithms. These results will establish that there are at most $3$ possible extensions of a triple to a quadruple for a fixed fundamental solution. Then, we will use a result of Laurent from [21] in order to establish our final technical tools and finish the proof of Theorem 1.6.

First, we prove that the inequality from Lemma 2.8 holds also for the case $c>ab+a+b$ without assuming that the inequalities from Lemma 2.5 hold.

Lemma 5.1**.**

Let $\{a,b,c\}$ be a Diophantine triple with $b>\max\{a,10^{5}\}$ and $c>ab+a+b$. If $v_{m}=w_{n}$ has a solution with $m\geq 1$ such that $(|z_{0}|,x_{0};|z_{1}|,y_{1})=(t,r;s,r)$ and $z_{0}z_{1}>0$, then $n-1\leq m$.

Proof.

Notice that $c>\max\{4b,ab+a+b\}$ and $c>ab+a+b>r^{2}$. The statement follows similarly to the argument for [7, Lemma 2.9], and so we opt to omit the proof. ∎

Let $\{a,b,c\}$ be a $D(4)$ triple. We define and observe

[TABLE]

a linear form in three logarithms, where $\xi=\frac{s+\sqrt{ac}}{2}$, $\eta=\frac{t+\sqrt{bc}}{2}$ and $\mu=\frac{\sqrt{b}(x_{0}\sqrt{c}+z_{0}\sqrt{a})}{\sqrt{a}(y_{1}\sqrt{c}+z_{1}\sqrt{b})}$. This linear form and its variations were already studied before, for example in [15]. From [15, Lemma 10] we know that

[TABLE]

where $\kappa_{0}$ is a coefficient which is defined in the proof of the lemma with

[TABLE]

Lemma 5.2**.**

Let $(m,n)$ be a solution of the equation $z=v_{m}=w_{n}$ and assume that $m>0$ and $n>0$. Then

[TABLE]

where

1. i)

$(\kappa,\delta)=(2.7\sqrt{ac},0)$* if the inequalities from Lemma 2.5 hold,* 2. ii)

$(\kappa,\delta)=(6,0)$* if $|z_{0}|=2$,* 3. iii)

$(\kappa,\delta)=\left(1/(2ab),0\right)$* if $z_{0}=t$,* 4. iv)

$(\kappa,\delta)=\left(2.0001b/c,1\right)$* if $z_{0}=-t$, $b>10^{5}$ and $c>ab+a+b$.*

Proof.

From Lemma 2.5 we get

[TABLE]

Inserting in the expression (24) yields $\kappa_{0}<2.7\sqrt{ac}$.

If $|z_{0}|=2$, equation (2) yields $x_{0}=2$. Using $c>4a$ gives us the desired estimate.

If $c>a+b+2r$ and $z_{0}=t$ then $x_{0}=r$ and $\kappa_{0}=\frac{8(c-a)}{(t\sqrt{a}+r\sqrt{c})^{2}}<\frac{1}{2ab}$.

In the last case, we observe that

[TABLE]

where we have used that $c>ab>10^{5}a$. ∎

The next result is due to Matveev [22] and can be used to get a better lower bound on the linear form (22) than (23).

Theorem 5.3** (Matveev).**

Let $\alpha_{1},\alpha_{2},\alpha_{3}$ be a positive, totally real algebraic numbers such that they are multiplicatively independent. Let $b_{1},b_{2},b_{3}$ be rational integers with $b_{3}\neq 0$. Consider the following linear form $\Lambda$ in the three logarithms:

[TABLE]

Define real numbers $A_{1},A_{2},A_{3}$ by

[TABLE]

where $D=[\mathbb{Q}(\alpha_{1},\alpha_{2},\alpha_{3}):\mathbb{Q}]$. Put

[TABLE]

Then we have

[TABLE]

with

[TABLE]

Put $\alpha_{1}=\xi=\frac{s+\sqrt{ac}}{2}$, $\alpha_{2}=\mu=\frac{\sqrt{b}(x_{0}\sqrt{c}+z_{0}\sqrt{a})}{\sqrt{a}(y_{1}\sqrt{c}+z_{1}\sqrt{b})}$ and $\alpha_{3}=\eta=\frac{t+\sqrt{bc}}{2}$. We can easily show, similarly as in [14] or [5], that

[TABLE]

It follows by similar arguments as in [19] that $h(\alpha_{2})<\frac{1}{4}\log(P_{2})$ where

[TABLE]

First, we consider the case $c=a+b+2r<4b$, so that the case $iii)$ in Lemma 2.13 cannot hold. Also, the case where $|z_{1}|=s$ and $|z_{0}|=\frac{1}{2}(cr-st)$ cannot occur since the same lemma implies $c=a+b+2r>a^{2}b$ ($a=1$) and this case can be eliminated with the same arguments as in [18]. Also, since $s=a+r$ and $t=b+r$, we have $\frac{1}{2}(cr-st)=2$, so the only case is $|z_{0}|=|z_{1}|=x_{0}=y_{1}=2$. As in [19] we easily get

[TABLE]

Now we will study the case when $c>ab+a+b$ and $(|z_{0}|,x_{0},|z_{1}|,y_{1})=(t,r;s,r)$ without assuming the inequalities from the Lemma 2.5. Here we have

[TABLE]

and

[TABLE]

so in this case

[TABLE]

Theorem 1.4 implies that $c<39247b^{4}$, from which it follows that $A_{3}=2\log\eta>2\log(0.071c^{5/8})>\frac{5}{4}\log(0.014c)$. This together with Lemma 5.1 implies

[TABLE]

Since by Lemma 2.10 and Theorem 1.3 we know that $m\geq 6$, we can use $B<m+1$. The numbers $\alpha_{1}$, $\alpha_{2}$ and $\alpha_{3}$ are multiplicatively independent (this can be shown similarly as in [20, Lemma 19]). We can now apply Theorem 5.3 which proves the next result.

Proposition 5.4**.**

For $m\geq 6$ we have

[TABLE]

If we have a $D(4)$-quadruple $\{a,b,c,d\}$ then $z=\sqrt{cd+4}$ is a solution of the equation $z=v_{m}=w_{n}$ for some $m$, $n$ and fundamental solutions $(z_{0},z_{1})$.

Let $\{a,b,c\}$ be a $D(4)$-triple and let us assume that there are $3$ solutions to the equation $z=v_{m}=w_{n}$ which belong to the same fundamental solution. We denote them with $(m_{i},n_{i})$, $i=0,1,2$. Let us assume that $m_{0}<m_{1}<m_{2}$ and $m_{1}\geq 4$. Denote

[TABLE]

As in [19], we borrow an idea of Okazaki from [3] in order to find a lower bound on $m_{2}-m_{1}$ in the terms of $m_{0}$. We omit the proof since it is analogous to [19, Lemma 7.1].

Lemma 5.5**.**

Assume that $v_{m_{0}}$ is positive. Then

[TABLE]

where

[TABLE]

In particular, if $m_{0}>0$ and $n_{0}>0$ then

[TABLE]

Proposition 5.6**.**

Suppose that there exist $3$ positive solutions $(x_{(i)},y_{(i)},z_{(i)})$, $i=0,1,2$, of the system of Pellian equations (2) and (3) with $z_{(0)}<z_{(1)}<z_{(2)}$ belonging to the same class of solutions and $c>b>10^{5}$. Put $z_{(i)}=v_{m_{i}}=w_{n_{i}}$. Then $m_{0}\leq 2$.

Proof.

Let us assume to the contrary, that $m_{0}>2$. From Lemmas 2.10 and 2.6 we know that $m_{0}\geq 6$ and $m_{2}>\kappa^{-1}(ac)^{6-\delta}\log\eta>(2.7\sqrt{ac})^{-1}(ac)^{6}\log\sqrt{bc}>c^{5}$ by Lemmas 5.5 and 5.2. After observing that the left-hand side in the inequality of Proposition 5.4 is increasing in $m$ we get

[TABLE]

This inequality cannot hold for $c>10^{5}$. We conclude that $m_{0}\leq 2$. ∎

The next result is deduced following the logic in [5] (with only some technical details changed), and so we omit the proof.

Proposition 5.7**.**

Let $a,b,c$ be integers with $0<a<b<c$ and let $a_{1}=4a(c-b),$ $a_{2}=4b(c-a)$, $N=abz^{2}$, where $z$ is a solution of the system od Pellian equations (2) and (3). Put $u=c-b$, $v=c-a$ and $w=b-a$. Assume that $N\geq 10^{5}a_{2}$. Then the numbers

[TABLE]

satisfy

[TABLE]

for all integers $p_{1},p_{2},q$ with $q>0$, where $a_{1}^{\prime}=\max\{a_{1},a_{2}-a_{1}\}$ and

[TABLE]

Lemma 5.8**.**

Let $(x_{(i)},y_{(i)},z_{(i)})$ be positive solutions to the system of Pellian equations (2) and (3) for $i\in\{1,2\}$, and let $\theta_{1}$, $\theta_{2}$ be as in Proposition 5.7 with $z=z_{1}$. Then we have

[TABLE]

Proof.

It is not hard to see that from Proposition 5.7 we have

[TABLE]

and similarly

[TABLE]

∎

Proposition 5.9**.**

Suppose that $\{a,b,c,d_{i}\}$ are $D(4)$-quadruples with $a<b<c<d_{1}<d_{2}$ and $x_{(i)},y_{(i)},z_{(i)}$ are positive integers such that $ad_{i}+1=x_{(i)}^{2}$, $bd_{i}+1=y_{(i)}^{2}$ and $cd_{i}=z_{(i)}^{2}$ for $i\in\{1,2\}$.

1. i)

If $n_{1}\geq 8$, then

[TABLE]

More specifically,if $n_{1}=8$, $n_{2}<2628n_{1}$, and if $n_{1}\geq 9$ then $n_{2}<83n_{1}$. 2. ii)

If $c>ab+a+b$ and $(z_{0},z_{1})=(t,s)$, $z_{0}z_{1}>0$ and $n_{1}\geq 9$ then

[TABLE]

Proof.

Put $N=abz^{2}$, $p_{1}=acy_{(1)}y_{(2)}$, $p_{2}=bcx_{(1)}x_{(2)}$ and $q=abz_{(1)}z_{(2)}$ in Proposition 5.7 and Lemma 5.8. We get

[TABLE]

We use estimates for fundamental solutions from Lemma 2.5 and the inequality from the proof of Lemma 2.7 which gives us

[TABLE]

Since $z_{(1)}=w_{n_{1}}$, we use this inequality to show that

[TABLE]

and

[TABLE]

where we have used the assumption $n_{1}\geq 8$. So

[TABLE]

Now we can show that

[TABLE]

On the other hand

[TABLE]

By combining these inequalities we get

[TABLE]

where $\sigma=\frac{3.5205n_{1}+4.75675}{0.4795n_{1}-3.82175}$. If $n_{2}\geq n_{1}\sigma+1.1(\sigma-1)$, similarly as in [19], we would get a contradiction from

[TABLE]

where $A=\frac{y_{(1)}\sqrt{c}-z_{(1)}\sqrt{b}}{y_{(1)}\sqrt{c}+z_{(1)}\sqrt{b}}$ and $\xi=\frac{t+\sqrt{bc}}{2}$ as before. So, $n_{2}<n_{1}\sigma+1.1(\sigma-1)$ must hold, which proves the first statement.

The second statement is proven analogously by using the inequality

[TABLE]

Notice that in this case we didn’t use Lemma 2.5 since we know explicitly values $(z_{0},z_{1})=(t,s)$. ∎

We now consider a linear form in two logarithms

[TABLE]

for which we know that $\Gamma\neq 0$ since it is not hard to show that $\xi$ and $\eta$ are multiplicatively independent.

From Lemma 5.2 we have that $0<\Lambda_{i}<\kappa\xi^{-2m_{1}}$ for $i=1,2$, so

[TABLE]

We can now use Laurent’s theorem from [21] to find a lower bound on $|\Gamma|$, similarly as in [5] and [19].

Theorem 5.10** (Laurent).**

Let $\gamma_{1}$ and $\gamma_{2}$ be multiplicatively independent algebraic numbers with $|\gamma_{1}|\geq 1$ and $|\gamma_{2}|\geq 1$. Let $b_{1}$ and $b_{2}$ be positive integers. Consider the linear form in two logarithms

[TABLE]

where $\log\gamma_{1}$ and $\log\gamma_{2}$ are any determinations of the logarithms of $\gamma_{1}$ and $\gamma_{2}$ respectively. Let $\rho$ and $\mu$ be real numbers with $\rho>1$ and $1/3\leq\mu\leq 1$. Set

[TABLE]

Let $a_{1}$ and $a_{2}$ be real numbers such that

[TABLE]

where $D=[\mathbb{Q}(\gamma_{1},\gamma_{2}):\mathbb{Q}]/[\mathbb{R}(\gamma_{1},\gamma_{2}):\mathbb{R}]$. Let $h$ be a real number such that

[TABLE]

Put

[TABLE]

Then

[TABLE]

with $C=C_{0}\mu/(\lambda^{3}\sigma)$ and $C^{\prime}=\sqrt{C_{0}\omega\theta/\lambda^{6}}$

[TABLE]

Proposition 5.11**.**

If $z=v_{m_{i}}=w_{n_{i}}$ $(i\in\{1,2\})$ has a solution, then

[TABLE]

where $\rho=8.2$, $\mu=0.48$

[TABLE]

$h=4\log\left(\frac{2j}{\log\eta}+1\right)+4\log\left(\frac{\lambda}{\rho+3}\right)+7.06+\log\rho$* and $\sigma,\lambda,H,\omega,\theta$ are as in Theorem 5.10.*

Proof.

Similarly as in [5] we can take $a_{i}=(\rho+3)\log\gamma_{i}$, $i=1,2$, $h=4\log\left(\frac{2j}{\log\gamma_{1}}\right)+4\log\left(\frac{\lambda}{\rho+3}\right)+7.06+\log\rho$ which yields $C_{0}<C_{0}^{\prime}$ as defined in the statement of the proposition. Now, we finish the proof by combining Theorem 5.10 and Lemma 5.2. ∎

Proof of Theorem 1.6.

Let $\{a,b,c\}$ be a $D(4)$-triple and let $N(z_{0},z_{1})$ denote the number of nonregular solutions of the system (2) and (3), i.e. the number of integers $d>d_{+}$ which extend that triple to a quadruple and which correspond to the same fundamental solution $(z_{0},z_{1})$. From Proposition 5.6 we know that if we have $3$ possible solutions $m_{0}$, $m_{1}$, $m_{2}$ with the same fundamental solution $(z_{0},z_{1})$ then $m_{0}\leq 2$, so from Lemma 2.10 we know that $N(z_{0},z_{1})\leq 2$ for each possible pair $(z_{0},z_{1})$ in Theorem 1.3. Also, from the same theorem, we know that the case when $m$ is odd and $n$ is even cannot occur when $d>d_{+}$. So, if we denote with $N_{eo}$ the number of solutions $d>d_{+}$ when $m$ is even and $n$ is odd, and similarly for other cases, the number of extensions of a $D(4)$-triple to a $D(4)$-quadruple with $d>d_{+}$ is equal to

[TABLE]

Case $iv)$

This follows from Theorem 1.4.

Case $i)$

Since $c=a+b+2r<4b$, only the case $|z_{0}|=|z_{1}|=2$ can hold as explained before Proposition 5.4. This implies

[TABLE]

We now show that $N(2,2)\leq 1$. Assume to the contrary, that $N(2,2)=2$. Since $\frac{1}{2}(st-cr)=2$, then $(m_{0},n_{0})=(2,2)$ is a solution in this case. Beside that solution, there exist two more solutions $z=v_{m_{i}}=w_{n_{i}}$, $(m_{i},n_{i})$, $i=1,2$, such that $2=m_{0}<m_{1}<m_{2}$ and $m_{1}\geq 8$ and $n_{1}\geq 8$ by Lemma 2.10. From Lemmas 5.2 and 5.5 we have $\kappa=6$, $\Delta\geq 4$ and $m_{2}-m_{1}>\kappa^{-1}(ac)^{m_{0}}\delta\log\eta$, i.e.

[TABLE]

On the other hand, $n_{1}\geq 8$ and Proposition 5.9 gives us

[TABLE]

so $j=m_{2}-m_{1}\leq 5912m_{1}$. Using Proposition 5.11 yields $\frac{j}{\log\eta}<5.71\cdot 10^{8}$, which is a contradiction.

This proves that $N=N\left(2,2\right)+N\left(-2,-2\right)\leq 3$.

Cases $ii)$ and $iii)$

Since $c\neq a+b+2r$, we have $c>ab+a+b$. Let $N^{\prime}(z_{0},z_{1})$ denote a number of solutions of equation $v_{m}=w_{n}$, with $m>2$ and $b>10^{5}$, but without assuming that inequalities from Lemma 2.5 hold. Then, by Lemma 2.14 and Proposition 2.15 we have

[TABLE]

where $(z_{0}^{+},z_{1}^{+})\in\left\{\left(\frac{1}{2}(st-cr),\frac{1}{2}(st-cr)\right),\left(t,\frac{1}{2}(st-cr)\right),\right.$ $\left.\left(\frac{1}{2}(st-cr),s\right),\right.$ $\left.\left(t,s\right)\right\}$ and $(z_{0}^{-},z_{1}^{-})=(-z_{0}^{+},-z_{1}^{+})$.

It is not hard to see that the previous proof for $N\left(2,2\right)\leq 1$ didn’t depend for the element $c$, so it holds in this case too. We will now show that $N^{\prime}(-t,-s)\leq 2$ which by Lemma 2.14 implies $N^{\prime}(z_{0}^{-},z_{1}^{-})\leq 2$, and $N^{\prime}(t,s)\leq 2$ for $c<b^{2}$ and $N^{\prime}(t,s)\leq 1$ for $c>b^{2}$, which will prove our statements in these last two cases.

Let us assume the contrary, that $N^{\prime}(-t,-s)\geq 3$ (there exist $m_{0},m_{1},m_{2}$, $2<m_{0}<m_{1}<m_{2}$). By Proposition 2.15 we know that $m_{0}\geq 9$. From Lemma 5.2 we have $\Lambda<2.0001\frac{b}{c}\xi^{-2(m-1)}$ which can be used in Lemma 5.5 to get

[TABLE]

Now, we can use Proposition 5.4 with $B(m_{2})=m_{2}+1$, $\log\eta<\log(c/2)$. This gives us an inequality

[TABLE]

The left-hand side is increasing in $m_{2}$ so we can solve the inequality in $c$ which yields $c<56$, a contradiction to $c>b>10^{5}$.

Now, let us prove that $N^{\prime}(t,s)\leq 1$. Assume to the contrary that $N^{\prime}(t,s)\geq 2$ and for some $3$ solutions $1\leq m_{0}<m_{1}<m_{2}$ we also have $2<m_{1}$ ($m_{0}$ is associated with a regular solution). Then by Lemma 5.2, since $\Delta\geq 1$, $c>4b$ and $b>10^{5}$, we get

[TABLE]

On the other hand, as in the proof of [19, Lemma 7.1], it can be shown that $\frac{n_{2}-n_{1}}{m_{2}-m_{1}}>\frac{\log\xi}{\log\eta}$, which together with Proposition 5.9 implies

[TABLE]

where $f(n_{1})=\frac{2.0294n_{1}+8.96759}{0.4853n_{1}-3.85292}\left(1+\frac{1}{n_{1}}\right)$. These two inequalities yield $n_{1}>9.34047\cdot 10^{11}$ and $f(n_{1})\leq 4.1818$.

From $0<\Lambda_{1}<m_{1}\log\xi-n_{1}\log\eta+\log\mu$ we have

[TABLE]

So, we can use Proposition 5.11 and the inequality

[TABLE]

yields $\frac{m_{2}-m_{1}}{\log\eta}<152184$ which is in a contradiction to (25). So we must have $N^{\prime}(t,s)\leq 1$.

It remains to prove that $N^{\prime}(-t,-s)\leq 1$ for $c>b^{2}$. Again, let us assume to the contrary, that there are at least $2$ solutions $m_{1}<m_{2}$ besides a solution $(m_{0},n_{0})=(1,1)$ (which gives $d=d_{-}(a,b,c)$). Then

[TABLE]

After repeating steps as in the previous case, we get $f(n_{1})<4.1819$ and Proposition 5.11 yields $\frac{m_{2}-m_{1}}{\log\eta}<152184$, a contradiction.

So, when $c>b^{2}$ we have $N\leq 2+1+2+2=7$ and when $a+b+2r\neq c<b^{2}$ we have $N\leq 2+1+2+1=6$. ∎

6 Extension of a pair

For completeness, to give all possible results similar to the ones in [19] and [7], we have also considered extensions of a pair to a triple and estimated the number of extensions to a quadruple in such cases. Extensions of a pair to a triple were considered in [2] for the $D(4)$ case. Baćić and Filipin have shown that a pair $\{a,b\}$ can be extended to a triple with a $c$ given by

[TABLE]

where $\nu\in\mathbb{N}$. These extensions are derived from the fundamental solution $(t_{0},s_{0})=(\pm 2,2)$ of the Pell equation

[TABLE]

associated with the problem of an extension of a pair to a triple. Under some conditions for the pair $\{a,b\}$ we can prove that these fundamental solutions are the only ones.

The next result is an improvement of [2, Lemma 1].

Lemma 6.1**.**

Let $\{a,b,c\}$ be a $D(4)$-triple. If $a<b<6.85a$ then $c=c_{\nu}^{\pm}$ for some $\nu$.

Proof.

We follow the proof of [2, Lemma 1]. Define $s^{\prime}=\frac{rs-at}{2}$, $t^{\prime}=\frac{rt-bs}{2}$ and $c^{\prime}=\frac{(s^{\prime})^{2}-4}{a}$. The cases $c^{\prime}>b$, $c^{\prime}=b$ and $c^{\prime}=0$ are the same as in the [2, Lemma 1] and yield $c=c_{\nu}^{\pm}$. It is only left to consider the case $0<c^{\prime}<b$. Here we define $r^{\prime}=\frac{s^{\prime}r-at^{\prime}}{2}$ and $b^{\prime}=((r^{\prime})^{2}-4)/a$. If $b^{\prime}=0$ then it can be shown that $c^{\prime}=c_{1}^{-}$ and $c=c_{\nu}^{-}$ for some $\nu$. We observe that $b^{\prime}=d_{-}(a,b,c^{\prime})$ so

[TABLE]

This implies $b^{\prime}c^{\prime}\leq 6$. If $c^{\prime}=1$, since $b^{\prime}>0$ and $b^{\prime}c^{\prime}+4$ is a square, the only possibility is $b^{\prime}=5$. In that case, $a$ and $b$ extend a pair $\{1,5\}$. Then

[TABLE]

and $b=d_{+}(1,5,a)=a_{\nu+1}^{\pm}$ for the same choice of $\pm$. Define $k:=\frac{b}{a}=\frac{a_{\nu+1}^{\pm}}{a_{\nu}^{\pm}}$. It can be easily shown that $k\leq 8$ and decreasing as $\nu$ increases. Also,

[TABLE]

which gives us a contradiction to the assumption $b<6.85a$.

If $c^{\prime}\in\{2,3,4,6\}$ there is no $b^{\prime}$ which satisfies the necessary conditions. The case $c^{\prime}=5$ gives $b^{\prime}=1$ which is analogous to the previous case. ∎

Remark**.**

For a pair $(a,b)=(4620,31680)$, where $b>6.85a$, we have a solution $(s_{0},t_{0})=(68,178)$ of the Pellian equation (1), so it can be extended with a greater element to a triple with $c\neq c_{\nu}^{\pm}$. For example $c=146434197$. So, this result cannot be improved further more.

Proof of Proposition 1.8.

We have

[TABLE]

The aim is to use Theorem 1.6 and since $N=0$ for $b<10^{5}$ we can use the lower bound on $b$ (more precisely, on $b$ if $b<c$ and on $c$ otherwise).

Case $c_{1}^{\pm}$ implies that $\{a,b,c\}$ is a regular triple so $N\leq 3$.

If $a=1$ then $c_{2}^{-}<b^{2}$ so the best conclusion is $N\leq 7$. On the other hand, if $c=c_{2}^{-}$ and $a^{2}\geq b$ we have $c>b^{2}$ since $a+b-2r\geq 1$ so $N=0$. It remains to consider the case when $b>a^{2}$. Then it can be shown that $r^{2}<0.004b^{2}$ and $c_{2}^{-}>0.872ab^{2}$, so if $a\geq 2$ we can again conclude $N\leq 6$. Also if $c\geq c_{2}^{+}=(a+b)(ab+4)+2r(ab+2)>b^{2}$ we have $N\leq 6$. Observe that $c\geq c_{5}^{-}>a^{4}b^{4}$. If $b\leq 7104a$ then $c>\frac{10^{20}}{7104^{4}}b^{4}>39263b^{4}$. It follows that $N=0$ by Theorem 1.6. If $b>7104a$ then $a+b-2r>0.97b$ so $c_{5}^{-}>0.97ba^{4}b^{5}>97000b^{4}$, and again $N=0$.

Similarly, we observe $c_{4}^{-}>a^{3}b^{3}$, and if $b\leq 63a$ we get $N=0$. If $b>63a$ we have $c+b-2r>0.76b$ which will lead to $N=0$ when $a\geq 38$. Cases $a\leq 37$ can be studied separately. We remark that only $a\geq 35$ led to $N=0$, and others to $N\leq 6$. ∎

From Proposition 1.8 and Lemma 6.1 we can conclude the result of Corollary 1.9 after observing that $10^{5}<b<6.85a$ implies $a\geq 14599$.

Acknowledgement: The author was supported by the Croatian Science Foundation under the project no. IP-2018-01-1313.