# On the $D(4)$-pairs $\\{a,ka\\}$ with $k\in\\{2,3,6\\}$

Kouèssi Norbert Adédji, Marija Bliznac Trebješanin, Alan Filipin and Alain
Togbé

(Date: August 27, 2024)

###### Abstract.

Let $a$ and $b=ka$ be positive integers with $k\in\\{2,3,6\\},$ such that
$ab+4$ is a perfect square. In this paper, we study the extensibility of the
$D(4)$-pairs $\\{a,ka\\}.$ More precisely, we prove that by considering three
families of positive integers $c$ depending on $a,$ if $\\{a,b,c,d\\}$ is the
set of positive integers which has the property that the product of any two of
its elements increased by $4$ is a perfect square, then $d$ in given by

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

As a corollary, we prove that any $D(4)$-quadruple which contains the pair
$\\{a,ka\\}$ is regular.

Keywords: Diophantine $m$-tuples, Pellian equations, Linear form in
logarithms, Reduction method.

##  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 an $m$-tuple with the property $D(n)$, if the product of
any two of its distinct elements increased by $n$ is a perfect square.

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

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

It is straightforward to check that $\\{a,b,c,d_{+}\\}$ is a $D(4)$-quadruple,
which we will call a regular quadruple. A quadruple which is not regular is
called an irregular quadruple. If $d_{-}\neq 0$ then $\\{a,b,c,d_{-}\\}$ is
also a regular $D(4)$-quadruple with $d_{-}<c$. It is conjectured that an
irregular quadruple doesn’t exist.

######  Conjecture 1.1.

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

In this paper, we consider extensions of the $D(4)$-pairs $\\{a,ka\\}$,
$k=2,3,6$ to a $D(4)$-quadruple $\\{a,ka,c,d\\}$ by following the method in
[2], where extensions of Diophantine pairs $\\{a,ka\\}$, with $k=3,8$, were
studied. We conjecture that $d=d_{+}(a,ka,c)$, i.e., that there is no
irregular quadruple of this form. The validity of the conjecture is shown for
some special pairs and triples. One of the results of interest for our case is
the following lemma, which gives us a lower bound for the second element of
the pair $b=ka$.

######  Lemma 1.2.

[5, Lemma 2.2] Let $\\{a,b,c,d\\}$ be a $D(4)$-quadruple such that
$a<b<c<d_{+}<d$. Then $b>10^{5}$.

If $\\{a,ka\\}$ is a $D(4)$-pair, then there exists $r\in\mathbb{N}$ such that

(1.1) $ka^{2}+4=r^{2}.$

Rewriting (1.1) as a Pellian equation, yields

(1.2) $r^{2}-ka^{2}=4.$

The theory of Pellian equations guarantees that there is only one fundamental
solution $(r_{1},a_{1})$ of (1.2), for any $k=2,3,6$, namely
$(r_{1},a_{1})\in\\{(6,4),(4,2),(10,4)\\}$ (in that order). All solutions
$(r_{p},a_{p})$ of the equation (1.2) are given by

(1.3)
$\displaystyle\frac{r_{p}+a_{p}\sqrt{k}}{2}=\left(\frac{r_{1}+a_{1}\sqrt{k}}{2}\right)^{p},\quad
p\in\mathbb{N}.$

It is easy to see that $\gcd(r,a)=2$ holds in every case. And since
$b=ka>10^{5}$ we can also deduce a lower bound for $a$. For $k=2$, we have
$a\geq a_{7}=161564$, which also gives us $b=2a\geq 323128$. In the case
$k=3$, we have $a\geq a_{9}=81090$ and $b=3a\geq 3a_{9}\geq 243270.$ Finally,
for $k=6$, the lower bounds $a\geq a_{5}=38804$, $b\geq 232824$ hold.

In general, if we extend a $D(4)$-pair $\\{a,b\\}$ to a $D(4)$-triple
$\\{a,b,c\\}$ then there exist $s,t\in\mathbb{N}$ such that

$\displaystyle ac+4$ $\displaystyle=s^{2},$ $\displaystyle bc+4$
$\displaystyle=t^{2}.$

Combining these two equalities yields a Pellian equation

(1.4) $at^{2}-bs^{2}=4(a-b).$

Its solutions $(t,s)$ are given by

$(t_{\nu}+s_{\nu}\sqrt{k})=(t_{0}+s_{0}\sqrt{k})\left(\frac{r+\sqrt{ab}}{2}\right)^{\nu},\quad\nu\geq
0,$

where $(t_{0},s_{0})$ is a fundamental solution of the equation (1.4) and
$\nu$ is a nonnegative integer.

In [5, Lemma 6.1], it has been shown that $(t_{0},s_{0})=(\pm 2,2)$ are the
only fundamental solutions when $b\leq 6.85a$, which is our case. We can
represent the solutions $(t_{\nu},s_{\nu})$ as pair of binary recurrence
sequences

(1.5) $\displaystyle t_{0}=\pm 2,\ t_{1}=b\pm r,\
t_{\nu+2}=rt_{\nu+1}-t_{\nu},$ (1.6) $\displaystyle s_{0}=2,\ s_{1}=r\pm a,\
s_{\nu+2}=rs_{\nu+1}-s_{\nu},\ \nu\geq 0.$

Since $c=\frac{s^{2}-4}{a}$, we give an explicit expression for the third
element $c$ in the terms of $a$ and $b$ by

(1.7)

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

where $\nu\geq 0$ is an integer. From [5, Proposition 1.8], if $a\geq 35$ then
$c\geq c_{4}^{-}$ cannot hold, i.e., it remains to observe the cases
$c\in\\{c_{1}^{\pm},c_{2}^{\pm},c_{3}^{\pm}\\}$. Let us list them in a more
suitable form

$\displaystyle c_{1}^{\pm}$ $\displaystyle=a+b\pm 2r,$ $\displaystyle
c_{2}^{\pm}$ $\displaystyle=(ab+4)(a+b\pm 2r)\mp 4r,$ $\displaystyle
c_{3}^{\pm}$ $\displaystyle=(a^{2}b^{2}+6ab+9)(a+b\pm 2r)\mp 4r(ab+3).$

Let us mention some observations in the case $c_{1}^{-}.$ Note that if
$k\in\\{2,3\\},$ then $c_{1}^{-}<a<b$ and in case $k=6$ we have
$a<c_{1}^{-}<b.$ So, in these cases, we consider the $D(4)$-triple
$\\{a,b,c\\}$ of the form $\\{c_{1}^{-},a,b\\}$ or $\\{a,c_{1}^{-},b\\}.$

The present paper deals with two closely related families, viz. those of
$D(4)$-triples mentioned in this section. The outcome of our study is the
theorem below, showing that each of the triples under scrutiny has a unique
extension to quadruple. In particular, the next result shows that the above
mentioned conjecture is true for the families of $c$ examined in this paper.

######  Theorem 1.3.

Let $k$ and $\nu$ be positive integers such that $k\in\\{2,3,6\\}.$ If
$\\{a,b,c_{\nu}^{\pm},d\\}$ is a $D(4)$ quadruple with $b=ka$, then it is
regular. In other words, we have $d=d_{\pm}.$

The organization of this paper is as follows. In Sections 2 and 3 of the
paper, we will essentially prove the useful results to achieve our main goal.
We devote Section 4 to the proof of Theorem 1.3 and we will end in Section 5
with some clarifications.

##  2\. Pellian equations and linear form in three logarithms

The goal of this section is to provide and prove the main technical tools used
in our proof of Theorem 1.3. These tools are related to the search for the
intersection of linear recurrent sequences and to linear form in logarithms.

###  2.1. System of simultaneous Pellian equations

Let us observe an extension of a triple $\\{a,b,c\\}$ to a quadruple
$\\{a,b,c,d\\}$

$\displaystyle ad+4$ $\displaystyle=x^{2},$ $\displaystyle bd+4$
$\displaystyle=y^{2},$ $\displaystyle cd+4$ $\displaystyle=z^{2}.$

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

(2.1) $\displaystyle az^{2}-cx^{2}$ $\displaystyle=4(a-c),$ (2.2)
$\displaystyle bz^{2}-cy^{2}$ $\displaystyle=4(b-c),$ (2.3) $\displaystyle
ay^{2}-bx^{2}$ $\displaystyle=4(a-b).$

Its solutions $(z,x)$, $(z,y)$ and $(y,x)$ satisfy

(2.4) $\displaystyle z\sqrt{a}+x\sqrt{c}$
$\displaystyle=(z_{0}\sqrt{a}+x_{0}\sqrt{c})\left(\frac{s+\sqrt{ac}}{2}\right)^{m},$
(2.5) $\displaystyle z\sqrt{b}+y\sqrt{c}$
$\displaystyle=(z_{1}\sqrt{a}+y_{1}\sqrt{c})\left(\frac{t+\sqrt{bc}}{2}\right)^{n},$
(2.6) $\displaystyle y\sqrt{a}+x\sqrt{b}$
$\displaystyle=(y_{2}\sqrt{a}+x_{2}\sqrt{b})\left(\frac{r+\sqrt{ab}}{2}\right)^{l},$

where $m,n,l$ are nonnegative integers and $(z_{0},x_{0})$, $(z_{1},y_{1})$
and $(y_{2},x_{2})$ are fundamental solutions of these equations.

Any solution to the system satisfies $z=v_{m}=w_{n}$, where ${v_{m}}$ and
${w_{n}}$ are recurrent sequences defined by

$\displaystyle v_{0}=z_{0},\ v_{1}=\frac{1}{2}\left(sz_{0}+cx_{0}\right),\
v_{m+2}=sv_{m+1}-v_{m},$ $\displaystyle w_{0}=z_{1},\
w_{1}=\frac{1}{2}\left(tz_{1}+cy_{1}\right),\ w_{n+2}=tw_{n+1}-w_{n}.$

The initial terms of these sequences are described in the next theorem.

######  Theorem 2.1.

[5, 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 the 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 the 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 the 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 the 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.

The next result will also be useful in our case.

######  Lemma 2.2.

[5, Lemma 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 $n\geq 0$.

These results prove the next lemma.

######  Lemma 2.3.

Assume that $\\{a,b,c,c^{\prime}\\}$ is not a $D(4)$-quadruple for any
$c^{\prime}$ with $0<c^{\prime}<c_{\nu-1}^{\pm}$. We have

  1. i)

If the equation $v_{2m}=w_{2n}$ has a solution, then $z_{0}=z_{1}=\pm 2$ and
$x_{0}=y_{1}=2$.

  2. ii)

If the equation $v_{2m+1}=w_{2n+1}$ has a solution, then $z_{0}=\pm t$,
$z_{1}=\pm s$, $x_{0}=y_{1}=r$ and $z_{0}z_{1}>0$.

######  Remark 2.4.

If $c=c_{1}^{\pm}=a+b\pm 2r,$ then it is enough to observe the case
$v_{2m}=w_{2n}$.

Now, we observe the solutions of the system of equations (2.2) and (2.6). More
precisely, we will determine the intersections $y=A_{n}=B_{l}$ of sequences
$(A_{n})_{n}$ and $(B_{l})_{l}$ defined by

(2.7) $\displaystyle A_{0}=y_{1},\ A_{1}=\frac{1}{2}(ty_{1}+bz_{1}),\
A_{n+2}=tA_{n+1}-A_{n},$ (2.8) $\displaystyle B_{0}=y_{2},\
B_{1}=\frac{1}{2}(ry_{2}+bx_{2}),\ B_{l+2}=rB_{l+1}-B_{l},\ n,j\geq 0.$

The next lemma, which is a part of Lemma 2 in [7], gives us a description of
the solutions of Pell equations.

######  Lemma 2.5.

[7, Lemma 2] If $(X,Y)$ is a positive integer solution to a generalized Pell
equation

$aY^{2}-bX^{2}=4(a-b),$

with $ab+4=r^{2}$, we have

$Y\sqrt{a}+X\sqrt{b}=(y_{0}\sqrt{a}+x_{0}\sqrt{b})\left(\frac{r+\sqrt{ab}}{2}\right)^{n},$

where $n\geq 0$ is an integer and $(x_{0},y_{0})$ is integer solution of the
equation such that

$1\leq x_{0}\leq\sqrt{\frac{a(b-a)}{r-2}}\quad\textit{and}\quad
1\leq|y_{0}|\leq\sqrt{\frac{(r-2)(b-a)}{a}}.$

The next lemma is proved as [2, Lemma 4].

######  Lemma 2.6.

Assume that $\\{a,b,c^{\prime},c\\}$ is not a $D(4)$-quadruple for any
$c^{\prime}$ with $0<c^{\prime}<c^{\pm}_{\nu-1}$ and $b\geq 832824$. Then,
$A_{2n}=B_{2l+1}$ has no solution. Moreover, if $A_{2n}=B_{2l}$ then
$y_{2}=2$. In other cases, we have $y_{2}=\pm 2.$

###### Proof.

It is straightforward to check by induction that

$\displaystyle A_{2n}\equiv y_{1}\ (\bmod\ b),\quad$ $\displaystyle
A_{2n+1}\equiv\frac{1}{2}(ty_{1}+bz_{1})\ (\bmod\ b),$ $\displaystyle
B_{2l}\equiv y_{2}\ (\bmod\ b),\quad$ $\displaystyle
B_{2l+1}\equiv\frac{1}{2}(ry_{2}+bx_{2})\ (\bmod\ b).$

From Lemma 2.5, we have

$|y_{2}|\leq\sqrt{\frac{(r-2)(b-a)}{a}}=\sqrt{\frac{(r-2)(k-1)a}{a}}=\sqrt{(k-1)(r-2)},$

where we have used that $b=2k$, $k\in\\{2,3,6\\}$. This implies

$|y_{2}|\leq\sqrt{(k-1)\sqrt{b^{2}/k+4}}\leq\begin{cases}0.85\sqrt{b},&\
k=2,\\\ 1.075\sqrt{b},&\ k=3,\\\ 1.43\sqrt{b},&\ k=6.\end{cases}$

Case 1:  If $A_{2n}=B_{2l}$ then $y_{1}\equiv y_{2}\pmod{b}$. From Lemma 2.3,
we have $y_{1}=2$, so $y_{2}\equiv 2\ (\bmod\ b)$. On the other hand,
$y_{2}<1.43\sqrt{b}<0.5b$, for $b\geq 9$ so $y_{2}=2.$

Case 2:  If $A_{2n}=B_{2l+1}$ we have $y_{1}=2$ and

(2.9) $2\equiv\frac{1}{2}(ry_{2}+bx_{2})\ (\bmod\ b).$

Since $ab+4=r^{2}$, we know that $g=\gcd(b,r)\in\\{1,2,4\\}$. After
multiplying congruence (2.9) by $2$ and using the fact that $r^{2}\equiv 4\
(\bmod\ b)$, we can divide the final congruence by $r$ and we get

(2.10) $r\equiv y_{2}\ \left(\bmod\ \frac{b}{g}\right).$

We can rewrite the upper bound on $|y_{2}|$ and get

$|y_{2}|\leq\begin{cases}0.002b,&\ k=2,\\\ 0.0022b,&\ k=3,\\\ 0.003b,&\
k=6,\end{cases}$

where we have used a lower bound on $b$ for each value of $k$. Also,
$r=\sqrt{ab+4}=\sqrt{b^{2}/k+4}=\sqrt{1/k+4/b^{2}}b$, so we can use lower
bounds of $b$ to get those of $r$ in the terms of $b$. More precisely,
$R_{0}\cdot b<r<R_{0}+0.01\cdot b$, where

$\begin{cases}R_{0}=0.7,&\ k=2,\\\ R_{0}=0.57,&\ k=3,\\\ R_{0}=0.4,&\
k=6.\end{cases}$

If $g=\gcd(b,r)=1$, for each $k$, we have $||y_{2}|-r|0.4b$, which is a
contradiction with a lower bound for $|y_{2}|$. On the other hand, if
$g=\gcd(b,r)\in\\{2,4\\}$ we have that $y_{2}=r+\frac{b}{4}\cdot p$, for some
$p\in\mathbb{Z}$. If $p\geq 0$, we have $|y_{2}|\geq r$ and a contradiction as
in the previous case. If $p\leq-4$, we have $|y_{2}|>0.29b$. For the remaining
cases, $p=-1,-2,-3$ we have $|y_{2}|>0.13b,0.09b,0.04b$, which is a
contradiction in each case.

Other possibilities for the parities of the indices of the sequences
$\\{A_{n}\\}$ and $\\{B_{l}\\}$ are solved similarly to [2, Lemma 4]. So, we
omit the details. ∎

Therefore, the fundamental solutions of equation (2.3) are $(y_{2},x_{2})=(\pm
2,2).$ Finally, we need to look at $x=Q_{m}=P_{l}$, for some non-negative
integers $m$ and $l,$ where the sequences $(Q_{m})_{m\geq 0}$ and
$(P_{l})_{l\geq 0}$ are obtained using (2.4) and (2.6) and given by

(2.11) $\displaystyle P_{0}=x_{2},\
P_{1}=\frac{1}{2}\left(rx_{2}+ay_{2}\right),\ P_{l+2}=rP_{l+1}-P_{l},$ (2.12)
$\displaystyle Q_{0}=x_{0},\ Q_{1}=\frac{1}{2}\left(sx_{0}+az_{0}\right),\
Q_{m+2}=sQ_{m+1}-Q_{m}.$

From the above, we conclude that for the equation $x=P_{l}=Qm,$ the following
two possibilities exist:
Type $a:$ If $l\equiv m\equiv 0\pmod{2},$ then $z_{0}=\pm 2,$ $x_{0}=y_{2}=2$
and $x_{2}=2.$
Type $b:$ If $m\equiv 1\pmod{2},$ then $z_{0}=\pm t,$ $x_{0}=r,$ $y_{2}=\pm 2$
and $x_{2}=2.$

For the rest of this study, we will carefully examine the following equation

(2.13) $x=Q_{m}=P_{l},$

while using the fundamental solutions of Types $a$ and $b.$ As we mentioned in
Remark 2.4, we only need to consider solutions in Type $a$ if $c=c_{1}^{\pm}$
since $\dfrac{1}{2}(cr-st)=\dfrac{1}{2}\left((a+b\pm 2r)r-(r\pm a)(b+\pm
r)\right)=\pm 2.$

###  2.2. A linear form in three logarithms

Solving recurrences (2.11) and (2.12), we obtain

$\displaystyle P_{l}$
$\displaystyle=\frac{1}{2\sqrt{b}}\left((y_{2}\sqrt{a}+x_{2}\sqrt{b})\alpha^{l}-(y_{2}\sqrt{a}-x_{2}\sqrt{b})\alpha^{-l}\right),$
$\displaystyle Q_{m}$
$\displaystyle=\frac{1}{2\sqrt{c}}\left((z_{0}\sqrt{a}+x_{0}\sqrt{c})\beta^{m}-(z_{0}\sqrt{a}-x_{0}\sqrt{c})\beta^{-m}\right),$

where

(2.14)
$\alpha=\dfrac{r+\sqrt{ab}}{2}\quad\mbox{and}\quad\beta=\dfrac{s+\sqrt{ac}}{2}.$

Let us define

(2.15)
$\gamma=\frac{\sqrt{c}(y_{2}\sqrt{a}+x_{2}\sqrt{b})}{\sqrt{b}(z_{0}\sqrt{a}+x_{0}\sqrt{c})}.$

We follow the strategy used in [2] with some improvements and define the
following linear form in three logarithms

(2.16) $\Lambda=l\log\alpha-m\log\beta+\log\gamma.$

######  Lemma 2.7.

  1. 1)

If the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ of Type a with $m\geq 1$,
then

$0<\Lambda<11.7\beta^{-2m}.$

  2. 2)

If the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ of Type b with $m\geq 1$,
then

$0<\Lambda<4.4a^{2}\beta^{-2m}.$

###### Proof.

We define

$E=\frac{y_{2}\sqrt{a}+x_{2}\sqrt{b}}{\sqrt{b}}\alpha^{l}\quad\mbox{and}\quad
F=\frac{z_{0}\sqrt{a}+x_{0}\sqrt{c}}{\sqrt{c}}\beta^{m}.$

Easily $E,\,F>1$ if $l,m\geq 1$. Then, the equation $P_{l}=Q_{m}$ becomes

(2.17) $\displaystyle
E+4\left(\frac{b-a}{b}\right)E^{-1}=F+4\left(\frac{c-a}{c}\right)F^{-1}.$

Because $c>b>10^{5}$, we have $\frac{c-a}{c}>\frac{b-a}{b}$. It follows that

(2.18)
$E+4\left(\frac{b-a}{b}\right)E^{-1}>F+4\left(\frac{b-a}{b}\right)F^{-1}$

and hence

$(E-F)\left(EF-4\left(\frac{b-a}{b}\right)\right)>0.$

Therefore, $E>F$. Moreover, by (2.17) we have

$0<E-F<4\left(\frac{c-a}{c}\right)F^{-1}<4F^{-1}.$

It follows that $\Lambda>0$, with

$\Lambda=\log\frac{E}{F}=\log\left(1+\frac{E-F}{F}\right)<\frac{E-F}{F}<4F^{-2}.$

Type $a$:

$\displaystyle\Lambda$ $\displaystyle<4\frac{c}{(\pm
2\sqrt{a}+2\sqrt{c})^{2}}\beta^{-2m}=\frac{c}{(\pm\sqrt{a}+\sqrt{c})^{2}}\beta^{-2m}$
$\displaystyle<\dfrac{k}{(\sqrt{k}-1)^{2}}\beta^{-2m}<11.7\beta^{-2m},\;\text{for}\;c>b=ka>10^{5}.$

Note that the above bound also holds for the $D(4)$-triple of the form
$\\{c_{1}^{-},a,b\\}$ and $\\{a,c_{1}^{-},b\\}.$

Type $b$:
$\bullet$ The case $z_{0}=t.$ We have

$\Lambda<4\frac{c}{(t\sqrt{a}+r\sqrt{c})^{2}}\beta^{-2m}<\frac{4}{r^{2}}\frac{c}{(\sqrt{a}+\sqrt{c})^{2}}\beta^{-2m}<\beta^{-2m},$

for $c>b>10^{5}$ and $r>100$.
$\bullet$ The case $z_{0}=-t.$ From (2.17), we get

$\displaystyle
F=E+4\left(\frac{b-a}{b}\right)E^{-1}-4\left(\frac{c-a}{c}\right)F^{-1}$
$\displaystyle>E-4\left(\frac{c-a}{c}\right)F^{-1}$
$\displaystyle>E-4\left(\dfrac{c-a}{c}\right)>0.$

The above inequality comes from the fact that $F>1.$ Thus, we obtain

$F^{-1}<\left(E-4\left(\dfrac{c-a}{c}\right)\right)^{-1}.$

Therefore, we have

$\displaystyle E-F$
$\displaystyle=4\left(\frac{c-a}{c}\right)F^{-1}-4\left(\frac{b-a}{b}\right)E^{-1}$
(2.19)
$\displaystyle<4\left(\dfrac{c-a}{c}\right)\left(E-4\left(\dfrac{c-a}{c}\right)\right)^{-1}-4\left(\dfrac{b-a}{b}\right)E^{-1}.$

Moreover, in Type b and for $m\geq 3,$ we have

$\displaystyle F$
$\displaystyle\geq\dfrac{r\sqrt{c}-t\sqrt{a}}{\sqrt{c}}\beta^{3}=\dfrac{4(c-a)}{\sqrt{c}(r\sqrt{c}+t\sqrt{a})}\cdot\left(\dfrac{s+\sqrt{ac}}{2}\right)^{3}$
(2.20) $\displaystyle>\dfrac{4(c-a)}{\sqrt{c}\cdot
2r\sqrt{c}}\cdot(\sqrt{ac})^{3}>4(c-a),$

which implies $E>F>4(c-a)$ and then

(2.21)
$\displaystyle\dfrac{c-a}{c}\left(E-4\left(\dfrac{c-a}{c}\right)\right)^{-1}<E^{-1}.$

When $m=1,$ we can easily see that $P_{l}\neq Q_{1}$ by following what is done
in the proof of [1, Lemma 3.3]. So, combining (2.2) and (2.21), we obtain

$E-F<4E^{-1}-4\left(\dfrac{b-a}{b}\right)E^{-1}=\dfrac{4a}{b}E^{-1}<\dfrac{4}{k}F^{-1}.$

Therefore, one can see that

$\Lambda=\log\dfrac{E}{F}=\log\left(1+\dfrac{E-F}{F}\right)<\dfrac{E-F}{F}<\dfrac{4}{k}F^{-2}$

and

$\displaystyle\dfrac{4}{k}F^{-2}=\dfrac{4}{k}\cdot\dfrac{c}{(r\sqrt{c}-t\sqrt{a})^{2}}\beta^{-2m}<\dfrac{c^{2}r^{2}}{k(c-a)^{2}}\beta^{-2m}.$

Using the fact that $c>b=ka>10^{5},$ we get

$r^{2}=ka^{2}+4<1.1ka^{2}\quad\text{and}\quad\dfrac{c^{2}}{k(c-a)^{2}}<\dfrac{k}{(k-1)^{2}}.$

Hence, $\Lambda=\log\dfrac{E}{F}<4.4a^{2}\beta^{-2m}.$ Considering all cases
in Type $b,$ we have $\Lambda<4.4a^{2}\beta^{-2m}$. This completes the proof
of Lemma 2.7. ∎

We now have the next result whose proof is similar to that of part 2) in [2,
Lemma 8].

######  Lemma 2.8.

If the equation $x=Q_{m}=P_{l}$ has a solution $(l,m)$ with $m\geq 3,$ then
$m\leq l.$

For any nonzero algebraic number $\alpha$ of degree $d$ over $\mathbb{Q}$,
whose minimal polynomial over $\mathbb{Z}$ is
$a_{0}\prod_{j=1}^{d}(X-\alpha^{(j)}),$ we denote by

$h(\alpha)=\frac{1}{d}\left(\log|a_{0}|+\sum_{j=1}^{d}\log\max\left(1,\left|\alpha^{(j)}\right|\right)\right)$

its absolute logarithmic height. We recall the following result due to Matveev
[11].

######  Lemma 2.9.

Denote by $\alpha_{1},\ldots,\alpha_{j}$ algebraic numbers, not $0$ or $1,$ by
$\log\alpha_{1},\ldots,$ $\log\alpha_{j}$ determinations of their logarithms,
by $D$ the degree over $\mathbb{Q}$ of the number field
$\mathbb{K}=\mathbb{Q}(\alpha_{1},\ldots,\alpha_{j}),$ and by
$b_{1},\ldots,b_{j}$ integers. Define $B=\max\\{|b_{1}|,\ldots,|b_{j}|\\},$
and $A_{i}=\max\\{Dh(\alpha_{i}),|\log\alpha_{i}|,0.16\\}$ ($1\leq i\leq j$),
where $h(\alpha)$ denotes the absolute logarithmic Weil height of $\alpha.$
Assume that the number

$\Lambda=b_{1}\log\alpha_{1}+\cdots+b_{n}\log\alpha_{j}$

does not vanish; then

$|\Lambda|\geq\exp\\{-C(j,\chi)D^{2}A_{1}\cdots A_{j}\log(eD)\log(eB)\\},$

where $\chi=1$ if $\mathbb{K}\subset\mathbb{R}$ and $\chi=2$ otherwise and

$C(j,\chi)=\min\left\\{\frac{1}{\chi}\left(\frac{1}{2}ej\right)^{\chi}30^{j+3}j^{3.5},2^{6j+20}\right\\}.$

######  Proposition 2.10.

Assume that $c\in\\{c_{1}^{+},c_{2}^{\pm}\\}.$ If $Q_{m}=P_{l},$ then

$\displaystyle\frac{l}{\log(el)}$ $\displaystyle<3.36\cdot
10^{13}\cdot\log^{2}(10.7c^{2}),\quad\text{with solutions of Type a},\;$
$\displaystyle\frac{l}{\log(el)}$ $\displaystyle<6.44\cdot
10^{13}\cdot\log^{2}(10.7c^{2}),\quad\text{with solutions of Type b}.$

If $Q_{m}=P_{l},$ with $c=c_{1}^{-},$ then we get

$\displaystyle\frac{l}{\log(el)}<4.5\cdot
10^{13}\cdot\log^{2}(72a),\quad\text{if}\;k=2,3$

and

$\displaystyle\frac{l}{\log(el)}<9\cdot
10^{13}\cdot\log^{2}(27a),\quad\text{if}\;k=6.$

###### Proof.

We apply Lemma 2.9 with $j=3$ and $\chi=1$ to the form (2.16). Here, we take

$D=4,b_{1}=l,b_{2}=-m,b_{3}=1,\alpha_{1}=\alpha,\alpha_{2}=\beta\;\text{and}\;\alpha_{3}=\gamma.$

Since $m\leq l,$ we can take $B=l.$ Also, we have

$h(\alpha_{1})=\frac{1}{2}\log\alpha\quad\text{and}\quad
h(\alpha_{2})=\frac{1}{2}\log\beta.$

$\bullet$ In the case $c\in\\{c_{1}^{+},c_{2}^{\pm}\\}$ with $b=ka,$ we have
$c-a>\left(1-\dfrac{1}{k}\right)c$ and $r<\dfrac{1}{3}c.$ Moreover, the
conjugates of $\alpha_{3}$ are

$\frac{\sqrt{c}(y_{2}\sqrt{a}\pm x_{2}\sqrt{b})}{\sqrt{b}(z_{0}\sqrt{a}\pm
x_{0}\sqrt{c})},$

and the leading coefficient of the minimal polynomial of $\alpha_{3}$ is
$a_{0}=16k^{2}(c-a)^{2}.$ We proceed with the following estimates

$\displaystyle h(\alpha_{3})$
$\displaystyle\leq\frac{1}{4}\left[\log(16k^{2}(c-a)^{2})+4\log\frac{\max\\{|\sqrt{c}(y_{2}\sqrt{a}\pm
x_{2}\sqrt{b})|\\}}{\min\\{|\sqrt{b}(z_{0}\sqrt{a}\pm
x_{0}\sqrt{c})|\\}}\right]$
$\displaystyle=\frac{1}{4}\left[\log(16k^{2}(c-a)^{2})+4\log\frac{2\sqrt{c}(1+\sqrt{k})}{\sqrt{k}(-t\sqrt{a}+r\sqrt{c})}\right]$
$\displaystyle<\frac{1}{4}\log\left[\frac{2^{4}r^{4}c^{4}(1+\sqrt{k})^{4}}{(c-a)^{2}}\right]<\frac{1}{4}\log\left(\frac{k^{2}(1+\sqrt{k})^{2}\cdot
c^{6}}{(\sqrt{k}-1)^{2}}\right)$
$\displaystyle<\frac{1}{4}\log(1223c^{6})<\dfrac{3}{4}\log(10.7c^{2}).$

Thus, we can take

$A_{1}=2\log\alpha,\quad A_{2}=2\log\beta,\quad A_{3}=3\log(10.7c^{2}).$

Applying Lemma 2.9, we get

(2.22) $\displaystyle\log|\Lambda|>-1.3901\cdot 10^{11}\cdot 16\cdot
12\cdot\log\alpha\cdot\log\beta\cdot\log(10.7c^{2})\cdot\log(4e)\cdot\log(el).$

From each inequality of Lemma 2.7, we have $l\log\alpha<2m\log\beta$ and also

(2.23) $\displaystyle\log|\Lambda|<-1.9m\log\beta\quad\text{in Type
a},\;\text{with}\;m\geq 1$

and

(2.24) $\displaystyle\log|\Lambda|<-0.99m\log\beta\quad\text{in Type
b},\;\text{with}\;m\geq 2.$

Moreover,

(2.25) $\displaystyle\log\beta<\frac{1}{2}\log(10.7c^{2}).$

Combining (2.22), (2.23), (2.24), and (2.25), we respectively get according to
Type a and Type b the following inequalities

$\displaystyle\frac{l}{\log(el)}<3.36\cdot
10^{13}\cdot\log^{2}(10.7c^{2})\;\text{and}\;\frac{l}{\log(el)}<6.44\cdot
10^{13}\cdot\log^{2}(10.7c^{2}).$

$\bullet$ In the case $c=c_{1}^{-}$ with solutions of Type a, we similarly get
the following inequalities

$\displaystyle\frac{l}{\log(el)}<4.5\cdot
10^{13}\cdot\log^{2}(72a),\quad\text{if}\;k=2,3$

and

$\displaystyle\frac{l}{\log(el)}<9\cdot
10^{13}\cdot\log^{2}(27a),\quad\text{if}\;k=6.$

This ends the proof. ∎

##  3\. Lower bounds of $m$ and $l$ in terms of $a$

In this section we will apply congruence relations to obtain some lower bounds
for the indices $m$ and $l$ satisfying the equation $x=Q_{m}=P_{l}.$

######  Lemma 3.1.

If $a$ is odd, then

(3.1) $\displaystyle Q_{2m}$ $\displaystyle\equiv
x_{0}+\dfrac{1}{2}a(cx_{0}m^{2}+sz_{0}m)\pmod{a^{2}},$ (3.2) $\displaystyle
P_{2l}$ $\displaystyle\equiv
x_{2}+\dfrac{1}{2}a(bx_{2}l^{2}+ry_{2}l)\pmod{a^{2}}.$

If $a$ is even, then

(3.3) $\displaystyle Q_{2m}$ $\displaystyle\equiv
x_{0}+\dfrac{1}{2}a(cx_{0}m^{2}+sz_{0}m)\pmod{\dfrac{1}{2}a^{2}},$ (3.4)
$\displaystyle P_{2l}$ $\displaystyle\equiv
x_{2}+\dfrac{1}{2}a(bx_{2}l^{2}+ry_{2}l)\pmod{\dfrac{1}{2}a^{2}}.$

###### Proof.

The proof is similar to that of [2, Lemma 15]. ∎

We now consider the following result.

######  Lemma 3.2.

If the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ of Type $a$, then we have

$l\geq\frac{1}{12}\left(-2+\sqrt{4+3\sqrt{a}}\right).$

###### Proof.

Let $\lambda\in\\{1,2,3\\}$ and $\nu=1,2,3,$ we have $c=c_{\nu}^{\pm}\equiv\pm
2\lambda r\pmod{a}.$ From (1.6), we see that $s\equiv 2,r\pmod{a}.$ We also
get that $b=ak\equiv 0\pmod{a}.$ Using Lemma 3.1, we have in all cases with
solutions in Type a

$cx_{0}m^{2}+sz_{0}m\equiv bx_{2}l^{2}+ry_{2}l\pmod{a}.$

Hence, we get

$\pm 4\lambda rm^{2}\pm 4m\equiv 2rl\pmod{a},\quad\mbox{if}\quad s\equiv
2\pmod{a}$

and

$\pm 4\lambda m^{2}\pm 2m-2l\equiv
0\pmod{\dfrac{a}{\gcd(a,r)}},\quad\mbox{if}\quad s\equiv r\pmod{a}.$

The case $s\equiv r\pmod{a}.$ Recall that in our case $\gcd(a,r)=2.$ Thus,

$\left|\pm 4\lambda m^{2}\pm
2m-2l\right|\geq\dfrac{a}{\gcd(a,r)}=\dfrac{a}{2}.$

Note also that $m\leq l$ and $\lambda\leq 3.$ So, we get
$12l^{2}+4l\geq\dfrac{a}{2},$ which implies

(3.5) $\displaystyle l\geq\dfrac{1}{12}\left(-2+\sqrt{4+6a}\right).$

The case $s\equiv 2\pmod{a}.$ In this case, by multiplying the congruence
obtained by $r$ and using the fact that $r^{2}\equiv 4\pmod{a},$ we get

(3.6) $\displaystyle\pm 16\lambda m^{2}\pm 4mr-8l\equiv 0\pmod{a}.$

Because $r^{2}\equiv 4\pmod{a},$ we conclude that $r\equiv\pm
2\pmod{a^{\prime}}$ for some $a^{\prime}$ which is a divisor of $a$ and
$a^{\prime}\geq\sqrt{a}.$ It follows that

$\displaystyle\pm 16\lambda m^{2}\pm 8m-8l\equiv 0\pmod{a^{\prime}}.$

Therefore, we deduce that

$\left|\pm 16\lambda m^{2}\pm 8m-8l\right|\geq a^{\prime}\geq\sqrt{a}.$

Using again $m\leq l$ and $\lambda\leq 3,$ we get

$48l^{2}+16l\geq\sqrt{a},$

which implies

(3.7) $\displaystyle l\geq\dfrac{1}{12}\left(-2+\sqrt{4+3\sqrt{a}}\right).$

Combining the inequalities (3.5) and (3.7), we obtain the desired inequality.
This completes the proof. ∎

Let us define

(3.8)
$\alpha^{\nu}=\left(\dfrac{r+\sqrt{ab}}{2}\right)^{\nu}=T_{\nu}+U_{\nu}\sqrt{ab},$

where $(T_{\nu},U_{\nu})$ is the $\nu$-th positive rational solution to the
Pell equation

$T^{2}-abU^{2}=T^{2}-(r^{2}-4)U^{2}=1.$

It is easy to show by induction that

(3.9) $\displaystyle T_{0}$
$\displaystyle=1,\;T_{1}=\dfrac{r}{2},\;T_{\nu+2}=rT_{\nu+1}-T_{\nu},$ (3.10)
$\displaystyle U_{0}$
$\displaystyle=0,\;U_{1}=\dfrac{1}{2},\;U_{\nu+2}=rU_{\nu+1}-U_{\nu},$

for $\nu\geq 0.$ From (1.5), (1.6), (3.9), and (3.10), we have

(3.11) $s=s_{\nu}^{\pm}=2T_{\nu}\pm 2aU_{\nu}\mbox{ and }t=t_{\nu}^{\pm}=\pm
2T_{\nu}+2bU_{\nu}.$

Considering congruences modulo $r,$ we get

$2T_{\nu}\equiv\pm 2\pmod{r},\;2U_{\nu}\equiv
0\pmod{r}\quad\mbox{if}\quad\nu\equiv 0\pmod{2},$

and

$2T_{\nu}\equiv 0\pmod{r},\;2U_{\nu}\equiv\pm
1\pmod{r}\quad\mbox{if}\quad\nu\equiv 1\pmod{2},$

which implies that $s\equiv\pm 2,\pm a\pmod{r}.$ The case $s\equiv\pm
a\pmod{r}$ leads to a contradiction if $c=c_{2}^{\pm}.$ Let us prove it. In
this case, we have

$s^{2}=ac_{2}^{\pm}+4\equiv a^{2}\pmod{r}.$

Multiplying the above congruence by $k$ and using the fact that
$ka^{2}=r^{2}-4\equiv-4\pmod{r},$ we get

(3.12) $\displaystyle kac_{2}^{\pm}\equiv-4k-4\pmod{r}.$

Furthermore, from $c=c_{2}^{\pm}$ defined in this paper, we have

(3.13) $\displaystyle kac_{2}^{\pm}\equiv 0\pmod{r}.$

So, combining (3.12) and (3.13), we get $4k+4\equiv 0\pmod{r}$ and therefore

$r\mid 4k+4\in\\{12,16,28\\},$

which is not possible in our case since $r>28$, i.e., $b>10^{5}.$ In
conclusion, we summarize what we have proved by the following result.

######  Proposition 3.3.

Let $c=c_{2}^{\pm}=(ab+4)(a+b\pm 2r)\mp 4r.$ There is no $D(4)$-triple
$\\{a,b,c\\}$ if $s\equiv\pm a\pmod{r}.$

Therefore, for solutions of Type $b$ with $s\equiv\pm 2\pmod{r},$ we get the
following key result.

######  Lemma 3.4.

Assume that $c=c_{2}^{\pm}.$ If the equation $P_{l}=Q_{m}$ has a solution
$(l,m)$ of Type $b$, then we have

$m>\begin{cases}(3\sqrt{2}-4)a/4,&\ k=2,\\\ (2\sqrt{3}-3)a/2,&\ k=3,\\\
(5\sqrt{6}-12)a/4,&\ k=6.\end{cases}$

###### Proof.

By induction, we obtain

$P_{l}\equiv\pm 2,\pm a\pmod{r}\quad\text{and}\quad
Q_{m}\equiv\dfrac{1}{2}amz_{0}\pmod{r}.$

Because, $t=t_{\nu}^{\pm}=\pm 2T_{\nu}+2bU_{\nu}\equiv\pm 2,\pm b\pmod{r}$ and
$z_{0}=\pm t,$ we would get

$Q_{m}\equiv\pm am,\pm 2m\pmod{r}.$

$\bullet$ The case $Q_{m}\equiv\pm 2m\pmod{r}.$ Thus, first the equation
$Q_{m}=P_{l}$ implies

$\pm 2m\equiv\pm 2\pmod{r},$

which becomes $\pm m\pm 1\equiv 0\pmod{r/2}$ and then $m\geq r/2-1$.
Secondarily, we see that

$\pm 2m\equiv\pm a\pmod{r},$

which leads to $2m+a\geq r.$ It follows that

$m\geq\dfrac{1}{2}(r-a)>\dfrac{1}{2}(\sqrt{k}-1)a,\quad\text{for}\;k\in\\{2,3,6\\}.$

$\bullet$ The case $Q_{m}\equiv\pm am\pmod{r}.$ Combining this with
$P_{l}\equiv\pm a\pmod{r},$ the equation $Q_{m}=P_{l}$ implies $am\equiv\pm
a\pmod{r},$ which, using $\gcd(a,r)=2$, gives $m\equiv\pm 1\pmod{r}$ and
$m\geq r/2-1.$ Now, we use the fact that $P_{l}\equiv\pm 2\pmod{r}$ to see
that the equation $Q_{m}=P_{l}$ implies

(3.14) $\displaystyle ma\equiv\pm 2\pmod{r}.$

Multiplying the above congruence by $ka$ and adding $4m,$ we would get

$m(ka^{2}+4)\equiv 4m\pm 2ka\pmod{r},$

which gives

(3.15) $\displaystyle 2ka\pm 4m\equiv 0\pmod{r}.$

Then, $m\geq\dfrac{1}{2}ka$ or the left hand side is positive and we get for
$k=2$ the possibilities:

(3.16) $\displaystyle 4a+4m$ $\displaystyle=3r,4r,5r,\ldots$ (3.17)
$\displaystyle 4a-4m$ $\displaystyle=r,2r.$

From (3.16), we have $4a+4m\geq 3r$ and using $r=\sqrt{2a^{2}+4}>a\sqrt{2}$ we
get

(3.18) $\displaystyle m>\dfrac{1}{4}(3\sqrt{2}-4)a.$

In the case (3.17), we have $4a-4m\leq 2r.$ Since $r<a\sqrt{2.1},$ we obtain

(3.19) $\displaystyle m>\dfrac{1}{2}(2-\sqrt{2.1})a.$

We conclude that for $k=2,$ the relation (3.18) holds in all cases. The other
cases $(k=3,6)$ can be treated in the same way. So, we omit them. ∎

##  4\. Proof of Theorem 1.3

In this section, we completely give the proof of Theorem 1.3 in two
subsections according to the values of $c.$

###  4.1. Proof of Theorem 1.3 with $c=c_{1}^{\pm},c_{2}^{\pm}$

We start by what follows.

$\bullet$ The case $b=2a.$ From (1.3), we easily get the following relation

(4.1) $\displaystyle
a=a_{p}=\dfrac{1}{\sqrt{2}}\left((3+2\sqrt{2})^{p}-(3-2\sqrt{2})^{p}\right),$

which gives using Proposition 2.10, Lemmas 3.2 and 3.4 the following result.

######  Lemma 4.1.

  1. 1)

For a $D(4)$-triple $\\{a,2a,c_{1}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1),$ if
the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ with $m\geq 3,$ then $p\leq
111$ and $l\leq 2.54\cdot 10^{20}.$

  2. 2)

For a $D(4)$-triple $\\{a,2a,c_{2}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1),$ if
the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ in Type $a$ with $m\geq 3,$
then $p\leq 116$ and $l\leq 2.57\cdot 10^{21}.$ If the equation $P_{l}=Q_{m}$
has a solution $(l,m)$ in Type $b$ with $m\geq 3,$ then $p\leq 28$ and $l\leq
2.82\cdot 10^{20}.$

$\bullet$ The case $b=3a.$ Using (1.3), we get

(4.2) $\displaystyle
a=a_{p}=\dfrac{1}{\sqrt{3}}\left((2+\sqrt{3})^{p}-(2-\sqrt{3})^{p}\right).$

Combining this with Proposition 2.10, Lemmas 3.2 and 3.4, we obtain the
following result.

######  Lemma 4.2.

  1. 1)

For a $D(4)$-triple $\\{a,3a,c_{1}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1),$ if
the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ with $m\geq 3,$ then $p\leq
149$ and $l\leq 2.56\cdot 10^{20}.$

  2. 2)

For a $D(4)$-triple $\\{a,3a,c_{2}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1),$ if
the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ in Type $a$ with $m\geq 3,$
then $p\leq 156$ and $l\leq 2.6\cdot 10^{21}.$ If the equation $P_{l}=Q_{m}$
has a solution $(l,m)$ in Type $b$ with $m\geq 3,$ then $p\leq 37$ and $l\leq
2.75\cdot 10^{20}.$

$\bullet$ The case $b=6a.$ By (1.3), we have

(4.3) $\displaystyle
a=a_{p}=\dfrac{1}{\sqrt{6}}\left((5+2\sqrt{6})^{p}-(5-2\sqrt{6})^{p}\right).$

Combining this with Proposition 2.10, Lemmas 3.2 and 3.4 we obtain the
following result.

######  Lemma 4.3.

  1. 1)

For a $D(4)$-triple $\\{a,6a,c_{1}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1),$ if
the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ with $m\geq 3,$ then $p\leq
85$ and $l\leq 2.52\cdot 10^{20}.$

  2. 2)

For a $D(4)$-triple $\\{a,6a,c_{2}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1),$ if
the equation $P_{l}=Q_{m}$ has a solution $(l,m)$ in Type $a$ with $m\geq 3,$
then $p\leq 89$ and $l\leq 2.56\cdot 10^{21}.$ If the equation $P_{l}=Q_{m}$
has a solution $(l,m)$ in Type $b$ with $m\geq 3,$ then $p\leq 22$ and $l\leq
3\cdot 10^{20}.$

For the remaining cases, we will use the following lemma which is a slight
modification of the original version of Baker-Davenport reduction method (see
[6, Lemma 5a]).

######  Lemma 4.4.

Assume that $M$ is a positive integer. Let $p/q$ be a convergent of the
continued fraction expansion of $\kappa$ such that $q>6M$ and let

$\eta=\parallel\mu q\parallel-M\cdot\parallel\kappa q\parallel,$

where $\parallel\cdot\parallel$ denotes the distance from the nearest integer.
If $\eta>0,$ then there is no solution of the inequality

$0<l\kappa-m+\mu<AB^{-l}$

in integers $l$ and $m$ with

$\frac{\log(Aq/\eta)}{\log(B)}\leq l\leq M.$

Dividing $0<\Lambda<11.7\beta^{-2m}$ and $0<\Lambda<4.4a^{2}\beta^{-2m}$ by
$\log\beta$ and using the fact that we have $\beta^{-2m}<\alpha^{-l}$ leads to
an inequality of the form

(4.4) $0<l\kappa-m+\mu<AB^{-l},$

where we consider for solutions of Type $a$

$\kappa:=\frac{\log\alpha}{\log\beta},\quad\mu:=\frac{\log\gamma}{\log\beta},\quad
A:=\frac{11.7}{\log\beta},\quad B:=\alpha,$

and for solutions of Type $b$

$\kappa:=\frac{\log\alpha}{\log\beta},\quad\mu:=\frac{\log\gamma}{\log\beta},\quad
A:=\frac{4.4a^{2}}{\log\beta},\quad B:=\alpha.$

For the remaining proof, we use Mathematica to apply Lemma 4.4. For the
computations, if the first convergent such that $q>6M$ does not satisfy the
condition $\varepsilon>0,$ then we use the next convergent until we find the
one that satisfies the conditions. The following table provides the
information on the results obtained.

$c$ | $b=2a$ | $b=3a$ | $b=6a$
---|---|---|---
$c_{1}^{\pm}$ | $l\leq 4$ | $l\leq 4$ | $l\leq 4$
$c_{2}^{\pm}$ | $l\leq 5$ | $l\leq 5$ | $l\leq 4$

Therefore, in all cases, we can conclude that

(4.5) $\displaystyle 3\leq m\leq l\leq 5.$

Combining this with Lemma 3.2 and the relations (4.1), (4.2), and (4.3), for
solutions in Type $a$, we get

$p\leq\begin{cases}8,&\ k=2,\\\ 11,&\ k=3,\\\ 6,&\ k=6,\end{cases}$

and therefore the equation $Q_{m}=P_{l}$ has no solutions in this range.
Combining now (4.5) with Lemma 3.4, in all cases, we get $a\leq 98$, which
contradicts the fact that $b=ka>10^{5}$, with $k=2,3,6.$ Finally, we need to
see what happens when $m\in\\{0,1,2\\}.$ For $m=0,$ we get $x=Q_{0}=P_{0}=2$,
which gives $d=0.$ By following what is done in the proof of [1, Lemma 3.3],
one can easily see that the only solution of equation (2.13) if
$m\in\\{1,2\\}$ is $(l,m)=(\nu,1)$ for $z_{0}=t$ i.e. $x=Q_{1}=P_{\nu+1}.$ In
this case we have

$x=Q_{1}=P_{\nu+1}=r(T_{\nu}\pm aU_{\nu})+a(bU_{\nu}\pm T_{\nu}),$

which implies $d=(x^{2}-4)/a=d_{\pm}.$

###  4.2. Proof of Theorem 1.3 with $c=c_{3}^{\pm}$

In the case $c=c_{3}^{\pm}$, recall that the problem is more complicated to
solve if $s\equiv\pm a\pmod{r}$ by considering the equation (2.13). The
difficulty lies in the fact that it is not easy to find a lower bound for $l$
and $m$ in term of $a$ in the equation (2.13) if $s\equiv\pm a\pmod{r}.$ To
overcome this situation, we will deal with this case by examining the equation
$z=v_{m}=w_{n}$ using Lemma 2.3. Now, we will give the lower bounds of the
indices $m$ and $n$ in the equation $z=v_{m}=w_{n}$, for $2<n<m<2n$ (where the
relationship between $m$ and $n$ follows from [9, Lemma 5] if $m$ and $n$ have
the same parity). First, we have the following result.

######  Lemma 4.5.

  1. i)

If the equation $z=v_{2m}=w_{2n}$ has a solution $(m,n)$ with $n>1,$ then
$m>0.495b^{-0.5}c^{0.5}.$

  2. ii)

If the equation $z=v_{2m+1}=w_{2n+1}$ has a solution $(m,n)$ with $n>1,$ then
$m^{2}>0.0625b^{-1}c^{0.5}.$

###### Proof.

$i)$ The statement follows from the proof of [3, Proposition 2.3]. Here we
only have to use that $b>10^{5}>10^{4}.$

$ii)$ Using Lemma 2.3 in the case of odd indices and from [9, Lemma 12], we
get what follows

(4.6)
$\pm\dfrac{1}{2}astm(m+1)+r(2m+1)\equiv\pm\dfrac{1}{2}bstn(n+1)+r(2n+1)\pmod{c}.$

Because $(st)^{2}\equiv 16\pmod{c},$ we conclude that $st\equiv\pm
4\pmod{c^{\prime}}$, for some $c^{\prime}$ which is a divisor of $c$ and
$c^{\prime}\geq\sqrt{c}.$ Also the $\pm$ sign means that one of the
congruences is true. Hence, we get

(4.7) $\pm 2am(m+1)+r(2m+1)\equiv\pm 2bn(n+1)+r(2n+1)\pmod{c^{\prime}}.$

Let us now assume the opposite i.e., $m^{2}\leq 0.0625b^{-1}c^{0.5}.$ Then, it
is easy to see that both sides of the congruence relation $(\ref{eq-26})$ are
less than $c^{\prime}$ and they have the same sign. More precisely, we have

$\max\bigl{(}2am(m+1),r(2m+1),2bn(n+1),r(2n+1)\bigr{)}\leq 2bm(m+1)$

and

$2bm(m+1)<4bm^{2}\leq\frac{c^{\prime}}{4}.$

Therefore, we get

$|\pm 2am(m+1)+r(2m+1)|<\dfrac{c^{\prime}}{2}\quad\mbox{and}\quad|\pm
2bn(n+1)+r(2n+1)|<\dfrac{c^{\prime}}{2}.$

Note that in the case of the sign “-”, the two quantities $\pm
2am(m+1)+r(2m+1)$ and $\pm 2bn(n+1)+r(2n+1)$ are negative and in the case of
the sign “+” they are positive. Thus, we actually have the equations

(4.8) $\displaystyle\pm 2am(m+1)+r(2m+1)=\pm 2bn(n+1)+r(2n+1).$

instead of a congruence. Notice that in above $b=ka$ with $k=2,3,6.$
Considering now congruence modulo $a,$ we get

$r(2m+1)\equiv r(2n+1)\pmod{a}$

which implies

(4.9) $\displaystyle 2m\equiv 2n\pmod{\dfrac{a}{2}}.$

Since $c_{3}^{\pm}<519a^{5},$ we get

$2n<2m\leq 2\cdot 0.0625^{0.5}\cdot
b^{-0.5}\cdot\left(519a^{5}\right)^{0.25}<\dfrac{a}{4}.$

Therefore, the congruence (4.9) gives an equation of the form $2m=2n.$ So,
from (4.8), we get $m=n=0$, which is a contradiction. This completes the
proof. ∎

Now, we will combine the lower bounds for indices $m$ and $n$ together with
the result obtained using Baker’s theory of linear forms in logarithms to
prove the main Theorem for large values of $p.$ Using the main result in [4]
the third author proved in [8] that $z=v_{m}=w_{n},$ for $n>2,$ implies

(4.10) $\displaystyle\dfrac{m}{\log(m+1)}<6.543\cdot 10^{15}\log^{2}c.$

Combining this with Lemma 4.5, in the case of even indices, we get

(4.11) $\frac{2\cdot 0.495b^{-0.5}c^{0.5}}{\log(2\cdot
0.495b^{-0.5}c^{0.5}+1)}<6.543\cdot 10^{15}\log^{2}c,$

and in the case of odd indices, we get the inequality

(4.12) $\frac{2\cdot 0.0625^{0.5}b^{-0.5}c^{0.25}+1}{\log(2\cdot
0.0625^{0.5}b^{-0.5}c^{0.25}+2)}<6.543\cdot 10^{15}\log^{2}c.$

Therefore, using Maple, the solutions obtained for inequalities (4.11) and
(4.12) are summarized in the following lemma.

######  Lemma 4.6.

  1. 1)

For the $D(4)$-triples $\\{a,2a,c_{3}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1)$
defined in (4.1), if $z=v_{2m}=w_{2n}$ has a solution $(m,n),$ then $p\leq 14$
and $m\leq 2.6\cdot 10^{21}$ but if $z=v_{2m+1}=w_{2n+1}$ has a solution
$(m,n),$ then $p\leq 40$ and $m\leq 2.1\cdot 10^{22}.$

  2. 2)

For the $D(4)$-triples $\\{a,3a,c_{3}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1)$
defined in (4.2), if $z=v_{2m}=w_{2n}$ has a solution $(m,n),$ then $p\leq 19$
and $m\leq 3\cdot 10^{21}$ but if $z=v_{2m+1}=w_{2n+1}$ has a solution
$(m,n),$ then $p\leq 54$ and $m\leq 2.2\cdot 10^{22}.$

  3. 3)

For the $D(4)$-triples $\\{a,6a,c_{3}^{\pm}\\}$ with $a=a_{p}$ $(p\geq 1)$
defined in (4.3), if $z=v_{2m}=w_{2n}$ has a solution $(m,n),$ then $p\leq 11$
and $m\leq 2.6\cdot 10^{21}$ but if $z=v_{2m+1}=w_{2n+1}$ has a solution
$(m,n),$ then $p\leq 30$ and $m\leq 2.1\cdot 10^{22}.$

Now, it remains to see what is happening for small values of $p$ by applying
Lemma 4.4. For this, we also need the inequality which follows from
$v_{m}=w_{n},n>2$ (that is [10, Lemma 9]),

$\displaystyle 0$
$\displaystyle<m\log\left(\dfrac{s+\sqrt{ac}}{2}\right)-n\log\left(\dfrac{t+\sqrt{bc}}{2}\right)+\log\dfrac{\sqrt{b}(x_{0}\sqrt{c}+z_{0}\sqrt{a})}{\sqrt{a}(y_{1}\sqrt{c}+z_{1}\sqrt{b})}$
$\displaystyle<2ac\left(\dfrac{s+\sqrt{ac}}{2}\right)^{-2m}.$

Using Lemma 4.6, we apply Lemma 4.4 considering $c=c_{3}^{\pm}$. Note that in
the case of even indices we have $z_{0}=z_{1}=\pm 2,\;x_{0}=y_{1}=2$ and in
the case of odd indices we have $x_{0}=y_{1}=r,\;z_{0}=\pm t,\;z_{1}=\pm s$
and $z_{0}z_{1}>0.$ We have done the reduction using Mathematica. In all cases
according to $b=ka$ with $k=2,3,6$, after at most $2$ steps of reduction, we
see that $z=w_{m}=w_{n}$ implies $n\leq m\leq 2.$ In these small ranges, it is
not diﬃcult to check that all solutions of $z=v_{m}=w_{n}$ will give the
extension of $D(4)$-triple $\\{a,b,c\\}$ to a quadruple with $d=d_{-}$ or
$d=d_{+}$. This completes the proof of Theorem 1.3.

##  5\. Final conclusion

Let us mention in this section that the choice of the $D(4)$-pairs
$\\{a,ak\\}$ with $k\in\\{2,3,6\\}$ is not random. In fact, the idea is to
investigate irregular $D(4)$-quadruples, which contain the pairs $\\{a,ka\\}$,
for any positive integer $k.$ In view of the recent work done by the second
author in [5], it appears that if $b=ka$ we must have $a<b\leq 6.85a$ i.e.
$k\in\\{2,3,4,5,6\\}.$ For $k=4,$ it is easy to check from equation (1.1) that
there is no $D(4)$-pair of the form $\\{a,4a\\}$. On the other hand, for
$k=5$, we obtain an equation of the type

$r^{2}-5a^{2}=4,$

whose positive solutions are defined by $(r,a)=(L_{2n},F_{2n})$, where $F_{n}$
and $L_{n}$ denote the $n$-th Fibonacci and Lucas numbers respectively. Here
also the extension of the $D(4)$-pair $\\{F_{2n},5F_{2n}\\}$ should be skipped
since it has already been studied (see [7], [10]). In short, it remains to
study $k=2,3,6$, which is the aim of this paper.

## Acknowledgements

The first author is supported by IMSP, Institut de Mathématiques et de
Sciences Physiques de l’Université d’Abomey-Calavi. The second and the third
authors are supported by the Croatian Science Foundation, grant HRZZ-
IP-2018-01-1313. The fourth author is partially supported by Purdue University
Northwest.

## References

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  * [2] K. N. Adédji, B. He, A. Pintér, A. Togbé, On the Diophantine pair $\\{a,3a\\}$, J. Number Theory 227 (2021), 330–351.
  * [3] Lj. Baćić and A. Filipin, A note on the number of $D(4)$-quintuples, Rad Hrvat. Akad. Znan. Umjet. Mat. Znan., 18 (2014), 7–13.
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  * [5] M. Bliznac Trebješanin, Extension of a Diophantine triple with the property $D(4)$, Acta Math. Hungar. 163 (2021), 213–-246.
  * [6] A. Dujella and A. Pethö, _A generalization of a theorem of Baker and Davenport_ , Quart. J. Math. Oxford Ser. (2) 49 (1998), 291–306.
  * [7] A. Dujella, A. M. S. Ramasamy, Fibonacci numbers and sets with the property $D(4)$, Bull. Belg. Math. Soc. Simon Stevin, 12(3) (2005), 401–412.
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Institut de Mathématiques et de Sciences Physiques, Université d’Abomey-
Calavi, Bénin
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University of Split, Faculty of Science, Ruđera Boškovića 33, 21000 Split,
Croatia
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Faculty of Civil Engineering, University of Zagreb, Fra Andrije Kačića-Miošića
26, 10000 Zagreb, Croatia
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Department of Mathematics and Statistics, Purdue University Northwest, 1401 S,
U.S. 421, Westville IN 46391 USA
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