# On extensions of $D(4)$-triples by adjoining smaller elements

Marija Bliznac Trebješanin and Pavao Radić

###### Abstract

This paper examines the problem of obtaining a $D(4)$-quadruple by adding a
smaller element to a $D(4)$-triple. We prove some relations between elements
of observed hypothetical $D(4)$-quadruples under which conjecture of the
uniqueness of such an extension holds. Also, it is shown that for any
$D(4)$-triple there are at most two extensions with a smaller element.

2020 Mathematics Subject Classification: 11D09, 11B37, 11J68.
Keywords: Diophantine $m$-tuples, Pellian equations, Hypergeometric method.

##  1 Introduction

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 $D(n)$, if the product of any
two of its distinct elements increased by $n$ is a perfect square.

One of the most frequently studied questions is how large these sets can be.
In the classical case $n=1$, first studied by Diophantus, Dujella proved in
[7] that a $D(1)$ sextuple doesn’t exist and that there are at most finitely
many quintuples. Over the years, many authors have improved the upper bound on
the number of $D(1)$-quintuples. Finally, in [13], He, Togbé and Ziegler
proved the nonexistence of $D(1)$-quintuples.

Similar conjectures and observations can be made for the case $n=4$ as for the
case $n=1$. In view of this observation, Filipin and the first author proved
in [4] that a $D(4)$-quintuple also doesn’t 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.

If we observe a $D(4)$-triple $\\{a,b,c\\}$, then it can be extended with an
element $d$ greater than $\max\\{a,b,c\\}$, namely, we take

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

A $D(4)$-quadruple $\\{a,b,c,d_{+}(a,b,c)\\}$ is called a regular quadruple.
If $\\{a,b,c,d\\}$ is a $D(4)$-quadruple such that $d>\max\\{a,b,c\\}$ and
$d\neq d_{+}(a,b,c)$ then it is called an irregular quadruple.

######  Conjecture 1.1.

Any $D(4)$-triple $\\{a,b,c\\}$ has a unique extension to a Diophantine
quadruple $\\{a,b,c,d\\}$ by an element $d>\max\\{a,b,c\\}$.

Results which support this conjecture can be found, for example, in [1], [3],
[8], [11] and [12].

Besides extending a $D(4)$-triple $\\{a,b,c\\}$, $a<b<c$, to a
$D(4)$-quadruple with an element $d_{+}$, we can define number

$d=d_{-}(a,b,c)=a+b+c+\frac{1}{2}(abc-\sqrt{(ab+4)(ac+4)(bc+4)}),$

which will give us another extension. If $c\neq a+b+2r$ then $0<d_{-}<c$ (but
$d_{-}$ is not smaller than $a$). Notice that $c=d_{+}(a,b,d_{-}(a,b,c))$.

In the case of $D(1)$-triples, Cipu, Dujella and Fujita in [5] studied
extensions of the triples by adjoining an element smaller than all elements of
the given triple. Similarly, the first author in [2] initiated a study of the
analogous problem in the case of $D(4)$-triples.

######  Conjecture 1.2.

Suppose that $\\{a_{1},b,c,d\\}$ is a $D(4)$-quadruple with $a_{1}<b<c<d$.
Then, $\\{a_{2},b,c,d\\}$ is not a $D(4)$-quadruple for any integer $a_{2}$
with $a_{1}\neq a_{2}<b$.

Its validity is checked in [2] for $c<0.25b^{3}$ and is also true for $c\geq
39247b^{4}$ as a consequence of [3, Theorem 1.6].

The main result of this paper is to give some conditions under which
hypothetical counterexamples of Conjecture 1.2 may exist.

######  Theorem 1.3.

Assume that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples
with $a_{1}<a_{2}<b<c<d$. Then the following hold:

  1. 1)

$a_{2}>4a_{1}$.

  2. 2)

$a_{2}>0.0251a_{1}^{2}$.

  3. 3)

If $a_{1}\geq 2$ then $b<a_{2}^{2}$ and $a_{2}\geq 317$.

  4. 4)

$0.25a_{1}^{2}b^{3}<c<0.25a_{2}^{2}b^{3}$.

If a $D(4)$-triple $\\{b,c,d\\}$ has two different extensions to a
$D(4)$-quadruple with an element smaller than $\min\\{b,c,d\\}$ then both of
those quadruples must be irregular.

######  Corollary 1.4.

If $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples,
$a_{1}<a_{2}<b<c<d$, then both of them are irregular $D(4)$-quadruples.

Also, it follows that there are at most two extensions of a $D(4)$-triple
$\\{b,c,d\\}$ with a smaller integer.

######  Corollary 1.5.

Let $\\{b,c,d\\}$ be a $D(4)$-triple. Then, there exist at most two positive
integers $a$ with $a<\min\\{b,c,d\\}$ such that $\\{a,b,c,d\\}$ is a
$D(4)$-quadruple.

By combining Theorem 1.3 with results from [3], [7] and [10] we will show that
there are at most finitely many $D(4)$-triples $\\{b,c,d\\}$ which can be
extended to two different $D(4)$-quadruples with smaller elements
$a_{1}<a_{2}<b$.

######  Corollary 1.6.

There are only finitely many quintuples $\\{a_{1},a_{2},b,c,d\\}$ with
$a_{1}<a_{2}<b<c<d$ such that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are
Diophantine $D(4)$-quadruples.

From Corollary 1.4 we see that any counterexample to Conjecture 1.2 gives rise
to two irregular $D(4)$-quadruples, which would be in contradiction to
Conjecture 1.1, which proves the following.

######  Corollary 1.7.

Conjecture 1.1 implies Conjecture 1.2.

To demonstrate these results, we use mainly the approach and concepts from [6]
to derive similar outcomes for extensions of a $D(4)$-triple. In Section 2 we
define the system of Pellian equations and introduce our main tools used in
search of its solution. More precisely, we prove lemmas which give us bounds
on the indices of possible solutions. In Section 3, we clarify the
circumstances under which the conditions of Lemma 2.5 are met, reinforcing its
applicability in the computer implementation of the algorithm crucial to
finishing the majority of our proofs. The remaining portion of the paper is
dedicated to the proof of each statement of Theorem 1.3 and its consequences.
However, the proof of Corollary 1.6 is addressed separately in the final
section, as it necessitated an auxiliary result that had not been previously
established for $D(4)$-quadruples.

These results improve [2] significantly by giving new relations between
elements of $D(4)$-quadruples $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$.

##  2 Pellian equations and bounds for indices of solutions

Assume that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples
with $a_{1}<a_{2}<b<c<d$. Let $r_{1},r_{2},s_{1},s_{2},t,x_{1},x_{2},y,z$ be
positive integers satisfying

$\begin{array}[]{lll}a_{1}b+4=r_{1}^{2},&\quad a_{1}c+4=s_{1}^{2},&\quad
a_{1}d+4=x_{1}^{2},\\\ a_{2}b+4=r_{2}^{2},&\quad a_{2}c+4=s_{2}^{2},&\quad
a_{2}d+4=x_{2}^{2},\\\ bc+4=t^{2},&\quad bd+4=y^{2},&\quad
cd+4=z^{2}.\end{array}$

Considering $x_{1},x_{2},y,z$ as unknowns, we obtain the following system of
Pellian equations:

$\displaystyle a_{1}z^{2}-cx_{1}^{2}$ $\displaystyle=4(a_{1}-c)$ (1)
$\displaystyle a_{2}z^{2}-cx_{2}^{2}$ $\displaystyle=4(a_{2}-c)$ (2)
$\displaystyle bz^{2}-cy^{2}$ $\displaystyle=4(b-c)$ (3)

As described in [10], any positive integer solution to (2) and (3) is given by
$z=v_{m}=w_{n}$ for some non-negative integers $m,n$, where $\\{v_{m}\\}$ and
$\\{w_{n}\\}$ are recurrent sequences

$\displaystyle v_{0}$ $\displaystyle=z_{(0)},\
v_{1}=\frac{1}{2}(s_{2}z_{(0)}+cx_{(0)}),\ v_{m+2}=s_{2}v_{m+1}-v_{m}$ (4)
$\displaystyle w_{0}$ $\displaystyle=z_{(1)},\
w_{1}=\frac{1}{2}(tz_{(1)}+cy_{(1)}),\ w_{n+2}=tw_{n+1}-w_{n}.$ (5)

Moreover, Theorem 1.3 and Lemma 2.14 in [3] imply that a solution
$z=v_{m}=w_{n}$ also satisfies

$\displaystyle m\equiv n\equiv 0\ (\bmod\ 2)\ \text{and}\
x_{(0)}=y_{(1)}=|z_{(0)}|=|z_{(1)}|=2\ \text{with}\ z_{(0)}z_{(1)}>0;$ (6)
$\displaystyle m\equiv n\equiv 1\ (\bmod\ 2)\ \text{and}\
x_{(0)}=y_{(1)}=r_{2},\ |z_{(0)}|=t,\ |z_{(1)}|=s_{2}\ \text{with}\
z_{(0)}z_{(1)}>0.$ (7)

It can easily be shown that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ cannot
both be regular, so from now on we can use the following lemma for lower bound
on $b$.

######  Lemma 2.1 ([3, Lemma 2.2]).

Let $\\{a,b,c,d\\}$ be an irregular $D(4)$-quadruple with $a<b<c<d$. Then
$b>10^{5}$.

Also, note that it holds $a_{2}-a_{1}\geq 2$ since [2, Theorem 1.1] assures
that $a_{2}\neq a_{1}+1$ for all $a_{1}\neq 3$. If $a_{1}=3$ and
$a_{2}=a_{1}+1=4$, then $\\{a_{1},a_{2},b,c,d\\}$ would be a $D(4)$-quintuple,
which is a contradiciton with [4, Theorem 1]. Hence, as $a_{1}\geq 1$, we have
$a_{2}\geq 3$.

We will now recall some useful tools, first introduced in [14], and establish
some improvements on bounds on indices $m$ and $n$ in terms of elements of an
irregular $D(4)$-quadruple.

######  Theorem 2.2.

Let $a_{1},a_{2}$ be integers such that $0<a_{1}\leq a_{2}-2,\ a_{2}\geq 3$
and let’s denote $a_{1}^{\prime}:=\max\\{4(a_{2}-a_{1}),4a_{1}\\}$. Also, let
$N$ be a multiple of $a_{1}a_{2}$ and assume that $N\geq
396.59a_{1}^{\prime}a_{2}^{2}(a_{2}-a_{1})^{2}$. Then the numbers
$\theta_{1}=\sqrt{1+\frac{4a_{2}}{N}}$ and
$\theta_{2}=\sqrt{1+\frac{4a_{1}}{N}}$ satisfy

$\left\\{\left|\theta_{1}-\frac{p_{1}}{q}\right|,\left|\theta_{1}-\frac{p_{2}}{q}\right|\right\\}>\frac{a_{1}}{3.53081\cdot
10^{27}a_{1}^{\prime}a_{2}N}q^{-\lambda}$

for all integers $p_{1},p_{2},q$, with $q>0$, where

$\lambda=1+\frac{\log(2.500788a_{1}^{-1}a_{1}^{\prime}a_{2}N)}{\log(0.04216a_{1}^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}N^{2})}<2.$

###### Proof.

The validity of this assertion is obvious from the proof of [4, Theorem 2]
with $g$ replaced by $1$. ∎

The proof of the following result can be derived as in [8, Lemma 6].

######  Lemma 2.3.

. Let $N=a_{1}a_{2}c$ and let $\theta_{1}$, $\theta_{2}$ be as in Theorem 2.2.
Then, all positive solutions to the system of Pellian equations (1) and (2)
satisfy

$\max\left\\{\left|\theta_{1}-\frac{s_{1}a_{2}x_{1}}{a_{1}a_{2}z}\right|,\left|\theta_{2}-\frac{s_{2}a_{1}x_{2}}{a_{1}a_{2}z}\right|\right\\}<\frac{2c}{a_{1}}z^{-2}.$

The next result is proven similarly as [4, Lemma 6] or [9, Lemma 8].

######  Lemma 2.4.

Assume that $c>4.1\cdot 10^{-5}a_{2}b^{3}$ and $b>10^{5}$. If $z=w_{n}$ with
$n\geq 4$, then

$\log z>\frac{n}{2}\log(bc).$

Combining these results yields a very useful lemma.

######  Lemma 2.5.

Let $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ be $D(4)$-quadruples such that
Theorem 2.2 and Lemma 2.4 hold. If $z=v_{m}=w_{n}$ has a solution for some
integers $m$ and $n$ with $n\geq 4$, then

$n<\frac{8\log(8.4034\cdot
10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c)\log(0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c)}{\log(bc)\log(0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c)}.$

###### Proof.

We apply Theorem 2.2 with

$q=a_{1}a_{2}z,\>\>p_{1}=s_{1}a_{2}x_{1},\>\>p_{2}=s_{2}a_{1}x_{2},\>\>N=a_{1}a_{2}c.$

Combining Theorem 2.2 with Lemma 2.3, we get

$z^{2-\lambda}<7.06162\cdot
10^{27}a_{1}a_{1}^{\prime}a_{2}^{4}c^{2}<(8.4034\cdot
10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c)^{2}$

and

$\displaystyle\frac{1}{2-\lambda}$
$\displaystyle=\frac{\log(0.04216a_{1}a_{2}(a_{2}-a_{1})^{-2}c^{2})}{\log(0.016858a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c)}$
$\displaystyle<\frac{2\log(0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c)}{\log(0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c)}.$

Our claim now results from the comparison of the above inequalities with those
in Lemma 2.4. ∎

From [3, Lemma 2.6] we know that a regular quadruple is obtained as a solution
of $z=v_{m}=w_{n}$ for $m,n\leq 2$. Moreover, [3, Lemma 2.10] together with
[3, Theorem 1.3] gives us the next lower bound for $n$ and $m$, which is
useful since we know that one of our $D(4)$-quadruples is irregular.

######  Lemma 2.6.

If $v_{m}=w_{n}$ has a solution for $m>2$, then $6\leq m\leq 2n+1$ and $n\geq
7$.

######  Lemma 2.7 ([3, Lemma 2.9]).

Assume that $z=v_{m}=w_{n}$ has a solution for some integers $m$ and $n$. If
$c>b^{\varepsilon}$, $b>10^{5}$, $1\leq\varepsilon<12$, then

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

The next lemma gives us a lower bound on index $n$ in terms of elements $b$
and $c$ and will be used throughout the paper in combination with Lemma 2.5.

######  Lemma 2.8.

Assume that $c>\max\\{0.56b^{3},0.001a_{2}^{2}b^{2}\\}$ and $b>10^{5}$. If
$z=v_{m}=w_{n}$ has a solution for some integers $m$ and $n$ with $n\geq 4$,
then $m\equiv n\ (\bmod\ 2)$ and $n>0.09226b^{-1/2}c^{1/4}$ for odd $n$ and
$n>0.340134b^{-1/2}c^{1/2}$ for even $n$.

###### Proof.

As in the proof of [3, Lemma 3.1] we know that $m\equiv n\ (\bmod\ 2)$.

First, if $m$ and $n$ are even we observe the case $z_{0}=z_{1}=\pm 2$. Since
$c>0.56b^{3}>b^{2.94}$, we have $\varepsilon=2.94$ in Lemma 2.7 which yields
$m<1.3415n+0.97.$ From Lemma 2.6 we have $n\geq 8$ so we can take $m<1.47n.$
These observations are applied to [3, Lemma 4.1] with parameters $L=1.47$,
$\rho=1$, $a_{0}=3$, $b_{0}=10^{5}$, $c_{0}=0.56\cdot 10^{15}$ (Note that
Lemma requires $\rho>1$, but the same holds for $\rho=1$). We get

$m>0.499997b^{-1/2}c^{1/2},$

hence

$n>0.340134b^{-1/2}c^{1/2}.$

Second, we observe the case when $m$ and $n$ are odd. From Lemma 2.6 we have
$n\leq m$ and $n\geq 7$ so we can, similarly as before, take $m<1.48n.$

From the proof of [10, Lemma 13] we know that congruences

$\displaystyle\pm 2t(a_{2}m(m+1)-bn(n+1))$ $\displaystyle\equiv
2r_{2}s_{2}(n-m)\ (\bmod\ c),$ (8) $\displaystyle\pm
2s_{2}(a_{2}m(m+1)-bn(n+1))$ $\displaystyle\equiv 2r_{2}t(n-m)\ (\bmod\ c),$

hold. Since $s_{2}t\equiv 16\ (\bmod\ c)$ after multiplying both sides of
these congruences we see that we have

$64(a_{2}m(m+1)-bn(n+1))^{2}\equiv 64r_{2}^{2}(m-n)^{2}\ (\bmod\ c).$ (9)

Assume the contrary, that $n\leq 0.09226b^{-1/2}c^{1/4}$.

If we show that

$|a_{2}m(m+1)-bn(n+1)|+r_{2}(m-n)<\frac{c^{1/2}}{64},$

then we know that (9) is an equality. First, let’s assume that $bn(n+1)\geq
a_{2}m(m+1)$. Then, since $r_{2}<b$ we have

$\displaystyle bn(n+1)-$ $\displaystyle a_{2}m(m+1)+r_{2}(m-n)$
$\displaystyle<bn^{2}+bn+0.48r_{2}n-a_{2}m(m+1)$
$\displaystyle<bn^{2}+0.96bn<1.5bn^{2}.$

And since we assumed $n\leq 0.09226b^{-1/2}c^{1/4}$ we have
$1.5n^{2}<0.0128c^{1/2}<1/64\cdot c^{1/2}$.

Now, assume that $bn(n+1)<a_{2}m(m+1)$. Then

$\displaystyle a_{2}m(m+1)-$ $\displaystyle bn(n+1)+r_{2}(m-n)$
$\displaystyle<1.48a_{2}n(1.48n+1)-bn(n+1)+r_{2}\cdot 0.48n$
$\displaystyle<n^{2}(2.1904a_{2}-b)+n(1.48a_{2}-b)+0.481n\sqrt{a_{2}b}.$

Since $a_{2}<b<c^{1/3}$ and $0.56\cdot 10^{15}<c$ and $n\leq
0.09226b^{-1/2}c^{1/4}$ we further have,

$\displaystyle a_{2}m(m+1)-$ $\displaystyle bn(n+1)+r_{2}(m-n)<$
$\displaystyle<(1.1904\cdot 0.09226^{2}+0.961\cdot 0.09226\cdot
10^{-15/12}\cdot 0.56^{-1/6})c^{1/2}$ $\displaystyle<1/64c^{1/2}.$

So, congruence (9) is an equality, which yields

$|a_{2}m(m+1)-bn(n+1)|=r_{2}(m-n).$ (10)

After inserting that in congruence (8) we get

$2r_{2}(m-n)(\pm t+s_{2})\equiv 0\ (\bmod\ c).$ (11)

Since $n\leq 0.09226b^{-1/2}c^{1/4}$ and $c^{1/4}>0.1778\sqrt{ab}$, inequality

$\displaystyle 2r_{2}(m-n)|\pm t+s_{2}|$ $\displaystyle<0.96nr_{2}\cdot 2t$
$\displaystyle<1.92n\cdot 1.0002\sqrt{a_{2}b}\cdot 1.00001\sqrt{bc}$
$\displaystyle<0.1772a^{1/2}b^{1/2}c^{3/4}<c,$

asserts that (11) is also an equality:

$2r_{2}(m-n)(\pm t+s_{2})=0.$

So, either $t=s_{2}$ or $m=n$. From $t=s_{2}$, we would have $b=a_{2}$, which
cannot hold. If $m=n$, from (10) we would also get $a_{2}=b$, a contradiction.

This implies that our assumption was wrong and $n>0.09226b^{-1/2}c^{1/4}$. ∎

Let’s observe the first step of constructing $D(4)$-pairs $\\{a_{1},b\\}$ and
$\\{a_{2},b\\}$ when $a_{1}$ and $a_{2}$ are given integers. We have that
$b=(r_{1}^{2}-4)/a_{1}$, where $r_{1}$ is a solution to equation

$r_{1}^{2}a_{2}-r_{2}^{2}a_{1}=4(a_{2}-a_{1}),$

with unknowns $(r_{1},r_{2})$. We can rewrite this equation and observe

$(r_{1}a_{2})^{2}-r_{2}^{2}(a_{1}a_{2})=4a_{2}(a_{2}-a_{1}).$ (12)

If $a_{1}a_{2}$ is a perfect square, this is the equation of a form

$(r_{1}a_{2})^{2}-(r_{2}\sqrt{a_{1}a_{2}})^{2}=4a_{2}(a_{2}-a_{1})$ (13)

which has only finitely many solutions for each pair $\\{a_{1},a_{2}\\}$.

On the other hand, if $a_{1}a_{2}$ isn’t a perfect square, (12) is a Pellian
equation whose solutions can be easily described for any fixed pair
$\\{a_{1},a_{2}\\}$.

If we want to extend the $D(4)$-pairs $\\{a_{1},b\\}$ and $\\{a_{2},b\\}$ to
$D(4)$-triples $\\{a_{1},b,c\\}$ and $\\{a_{2},b,c\\}$ we have a very similar
problem. The element $c$ is equal to $(s_{1}^{2}-4)/a_{1}$ where $s_{1}$ is
obtained from equation (12) or (13) in which $(r_{1},r_{2})$ is replaced by
$(s_{1},s_{2})$. Thus, both $b$ and $c$ are elements of the solution sequences
of the Pellian equation (12) or elements of the set of finitely many solutions
of equation (13).

In most of our proofs, we’ll be dealing with finitely many remaining pairs
$\\{a_{1},a_{2}\\}$ with known upper (and lower) bounds on the elements $b$
and $c$. This means that to complete the proof, it will be sufficient to
search for elements of solution sequences that also satisfy some given bounds.

##  3 Note on applying Lemma 2.5

Let $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ be such $D(4)$-quadruples that
Theorem 2.2 and Lemma 2.4 hold, i.e. that Lemma 2.5 holds. Let’s assume that
$a_{1},a_{2}$ and $b$ are fixed values and consider the function

$\varphi(c)=\frac{8\log(8.4034\cdot
10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c)\log(0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c)}{\log(bc)\log(0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c)}.$

In many cases, when using Lemma 2.5, it is useful to show that this function
is decreasing in $c$ for values of $c$ we observe, so that we obtain an upper
bound for $\varphi(c)$ when we replace $c$ with its lower bound (usually the
lower bound is expressed in terms of $a_{1}$, $a_{2}$ and/or $b$).

Since $a_{2}>a_{1}\geq 1$, it can easily be seen that

$0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}>0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}$

so

$\varphi_{1}(c)=\frac{\log(0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c)}{\log(0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c)}>1$

and $\varphi_{1}(c)$ is decreasing in $c$. Our goal is to reach the same
conclusion regarding

$\varphi_{2}(c)=\frac{\log(8.4034\cdot
10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c)}{\log(bc)},$

i.e. we need to show that $b<8.4034\cdot
10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}$. Hence, if both
$\varphi_{1}(c)$ and $\varphi_{2}(c)$ are decreasing and both
$\varphi_{1}(c)>0$ and $\varphi_{2}(c)>0$ it implies that $\varphi(c)$ is
decreasing in $c$.

Since [2, Theorem 1.2] states that $c>0.25b^{3}$ and $0.25b^{3}>0.25\cdot
10^{5}a_{2}^{2}$ it is easy to see that all arguments of logarithms in the
expression for $\varphi(c)$ are greater than $1$ except maybe
$0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$. So, we must
check that

$c>(0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2})^{-1}=0.01685^{-1}a_{1}^{-1}a_{1}^{\prime}a_{2}(a_{2}-a_{1})^{2}$
(14)

holds before concluding that $\varphi(c)$ is decreasing in $c$.

Let’s first prove that $b\geq 8.4034\cdot
10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}$ cannot hold.

######  Lemma 3.1.

Assume that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples
with $a_{1}<a_{2}<b<c<d$. Then

$b<8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}.$

###### Proof.

Let’s assume the contrary, $b>8.4034\cdot
10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}.$ Since
$c>0.25a_{1}^{2}b^{3}$, we also have

$c>1.186\cdot 10^{42}a_{1}^{5}a_{2}^{6}$ (15)

for both options $a_{1}^{\prime}\in\\{4a_{1},4(a_{2}-a_{1})\\}$. Observe that

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}$
$\displaystyle<1.681\cdot 10^{14}\cdot a_{2}^{3},$ $\displaystyle
0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}$ $\displaystyle<0.102665\cdot
a_{2},$ $\displaystyle b$ $\displaystyle>2.376\cdot 10^{14}\cdot a_{2}^{2},$
$\displaystyle 0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}$
$\displaystyle>0.0042125\cdot a_{2}^{-4}.$

Now, from Lemma 2.5 we have

$n<\frac{\log(1.681\cdot 10^{14}\cdot a_{2}^{3}c)\log(0.102665\cdot
a_{2}c)}{\log(2.376\cdot 10^{14}\cdot a_{2}^{2}c)\log(0.0042125\cdot
a_{2}^{-4}c)}.$

The right-hand side of this inequality is decreasing in $c$, so we can use
(15), $n>0.09226(c/b^{2})^{1/4}$ and $c/b^{2}>0.25a_{1}^{2}b\geq
0.25b>4.2\cdot 10^{13}\cdot a_{2}^{2}$ in the previous inequality and we get
that it cannot hold for $a_{2}>1$. ∎

We have the following corollary from these observations and Lemma 3.1.

######  Corollary 3.2.

Let $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ be $D(4)$-quadruples such that
Lemma 2.5 holds. If
$c>0.01685^{-1}a_{1}^{-1}a_{1}^{\prime}a_{2}(a_{2}-a_{1})^{2}$, then the
function $\varphi(c)$ is decreasing in $c$.

##  4 Upper bound on element $c$

The goal of this section is to prove the upper bound on $c$ in terms of
$a_{2}$ and $b$ and to prove one lower bound on $a_{2}$ in terms of $a_{1}$.

######  Theorem 4.1.

Assume that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples
with $a_{1}<a_{2}<b<c<d$. Then the following hold:

  1. 1)

$a_{2}>2a_{1}$;

  2. 2)

$0.25a_{1}^{2}b^{3}<c<0.25a_{2}^{2}b^{3}$.

We begin by improving [2, Theorem 1.2].

######  Theorem 4.2.

If $\\{a_{1},b,c\\}$ and $\\{a_{2},b,c\\}$ are $D(4)$-triples with
$a_{1}<a_{2}<b<c\leq 0.25a_{1}^{2}b^{3}$, then $\\{a_{1},a_{2},b,c\\}$ is a
$D(4)$-quadruple.

###### Proof.

This statement and the proof are the same as [2, Theorem 1.2] with the
assumption $c\leq 0.25b^{3}$ replaced by $c\leq 0.25a_{1}^{2}b^{3}$. ∎

Since [4, Theorem 1] holds, Theorem 4.2 has the following corollary, which
proves the first inequality of the second statement in Theorem 4.1.

######  Corollary 4.3.

If $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples with
$a_{1}<a_{2}<b<c<d$, then $c>0.25a_{1}^{2}b^{3}$ and $d>0.25a_{1}^{2}c^{3}$.

###### Proof.

First, assume that $c\leq 0.25a_{1}^{2}b^{3}$. By using the previous theorem
on $D(4)$-triples $\\{a_{1},b,c\\}$ and $\\{a_{2},b,c\\}$ we get that
$\\{a_{1},a_{2},b,c\\}$ is a $D(4)$-quadruple, so $\\{a_{1},a_{2},b,c,d\\}$ is
a $D(4)$-quintuple, which cannot hold. For the second statement assume that
$d\leq 0.25a_{1}^{2}c^{3}$. By using the previous theorem on $\\{a_{1},c,d\\}$
and $\\{a_{2},c,d\\}$ we get that $\\{a_{1},a_{2},c,d\\}$ is a
$D(4)$-quadruple, so $\\{a_{1},a_{2},b,c,d\\}$ is a $D(4)$-quintuple, which is
a contradiction. ∎

The proof of Corollary 1.4 now easily follows.

###### Proof of Corollary 1.4.

Notice that since $a_{i}bc+4c>r_{i}s_{i}t$ we have
$d_{+}(a_{i},b,c)<a_{i}bc+4c$. On the other hand, since
$c>0.25a_{1}^{2}b^{3}$, we have

$d>0.25a_{1}^{2}c^{3}>0.25a_{1}^{2}(0.25a_{1}^{2}b^{3})^{2}c>0.25^{3}a_{1}^{6}b^{4}a_{2}bc.$

So, by using the fact that $b>10^{5}$ it is easy to see that
$d>d_{+}(a_{2},b,c)>d_{+}(a_{1},b,c)$. ∎

###### Proof of Theorem 4.1.

_1)_ Assume that $a_{2}\leq 2a_{1}$. By Corollary 4.3 we know
$c>0.25a_{1}^{2}b^{3}$. We can use Theorem 2.2 since $a_{1}^{\prime}=4a_{1}$,
$b>10^{5}$, $b>a_{2}>a_{2}-a_{1}\geq 2$, hence

$N=a_{1}a_{2}c>\frac{0.25}{8}10^{5}a_{1}^{\prime}a_{2}^{2}(a_{2}-a_{1})^{2}>396.59a_{1}^{\prime}a_{2}^{2}(a_{2}-a_{1})^{2}.$

Also, it is easy to see that lower bounds on $c$ from the statements of Lemmas
2.4 and 2.8 are satisfied since $a_{2}-a_{1}\geq 2$ and $a_{2}\leq 2a_{1}$
imply $a_{1}\geq 2$.

Hence, we can combine lower bound $n>0.09226b^{-1/2}c^{1/4}$ and upper bound
on $n$ from Lemma 2.5.

Note again that $a_{1}<a_{2}\leq 2a_{1}$ implies
$a_{1}^{\prime}=\max\\{4a_{1},4(a_{2}-a_{1})\\}=4a_{1}$. Also, $2\leq
a_{2}-a_{1}$. So we have

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c$
$\displaystyle<1.68068\cdot 10^{14}a_{2}^{3}c$ $\displaystyle
0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c$
$\displaystyle<0.102665a_{2}c$ $\displaystyle
0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$
$\displaystyle>0.01685a_{2}^{-3}c.$

If we moreover have $b>ka_{2}$ for some $k\geq 1$, then

$n<\frac{8\log(1.68068\cdot
10^{14}a_{2}^{3}c)\log(0.102665a_{2}c)}{\log(ka_{2}c)\log(0.01685a_{2}^{-3}c)}.$
(16)

The right-hand side of the inequality (16) is decreasing in $c$ when
$c>238a_{2}^{3}$ which is true in our case. Since

$c>0.25a_{1}^{2}b^{3}>0.0625\cdot k^{3}a_{2}^{5}$

and $c/b^{2}>0.0625a_{2}^{2}b>6.25\cdot 10^{3}a_{2}^{2}$ we observe the
inequality

$0.09226\cdot(6.25\cdot 10^{3}a_{2}^{2})^{1/4}<\frac{8\log(1.050425\cdot
10^{13}a_{2}^{8}k^{3})\log(0.006417a_{2}^{6}k^{3})}{\log(0.0625a_{2}^{6}k^{4})\log(0.001053125a_{2}^{2}k^{3})}.$
(17)

We have $b>a_{2}$, so $k=1$ and we get from (17) that $a_{2}\leq 7401$. Now,
$b>10^{5}\geq\frac{10^{5}}{7401}a_{2}$, so we can take $k=\frac{10^{5}}{7401}$
and calculate that $a_{2}\leq 2860$. We repeat the process six more times and
get $a_{2}\leq 1976$.

Notice that
$(0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2})^{-1}<10^{15}<c$
so we can use Corollary 3.2 for any pair $\\{a_{1},a_{2}\\}$,
$a_{1}<a_{2}<1976$. For these finitely many pairs $\\{a_{1},a_{2}\\}$, we
wrote a program in Wolfram Mathematica where we observed the original
inequality from Lemma 2.5, and substituted $c$ with its lower bound
$0.25a_{1}^{2}b^{3}$. For all values $\\{a_{1},a_{2}\\}$ with $1406\leq
a_{2}\leq 1976$ this inequality yields a contradiction with $b>10^{5}.$ For
each of the remaining pairs $\\{a_{1},a_{2}\\}$, $a_{2}\leq 1405$ we searched
for solutions as described at the end of Section 2. To check the pairs where
$a_{1}a_{2}$ isn’t a perfect square, we used the inequality from Lemma 2.5 to
obtain an upper bound for $b$ and then searched for solutions of the equation
(12) where $b>10^{5}$ and $b$ is less than the calculated bound. It remained
to search for solutions of equation (12) which yield the element $c$ such that
$0.25a_{1}^{2}b^{3}<c<39247b^{4}$ and $\sqrt{bc+4}$ is an integer. This search
returned that none of such elements $b$ and $c$ exist within those bounds.

_2)_ Assume to contrary, that $c\geq 0.25a_{2}^{2}b^{3}$. It is easy to see
that the assumptions of Lemmas 2.5 and 2.8 are satisfied. We know from 1) that
$a_{2}>2a_{1}$, so $a_{1}^{\prime}=4(a_{2}-a_{1}).$ Also,
$\frac{a_{2}}{2}<a_{2}-a_{1}<a_{2}$. Then

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c$
$\displaystyle<1.1885\cdot 10^{14}a_{2}^{3}c$ $\displaystyle
0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c$ $\displaystyle<0.29039c$
$\displaystyle 0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$
$\displaystyle>0.0042125a_{2}^{-4}c.$

If we have $b>ka_{2}$ for some $k\geq 1$, then $c>0.25a_{2}^{5}k^{3}$ and
$c/b^{2}\geq 0.25\cdot 10^{5}a_{2}^{2}$. Since $0.01685^{-1}\cdot 4\cdot
a_{1}^{-1}a_{2}(a_{2}-a_{1})^{3}<238a_{2}^{4}$ and
$c>0.25bk^{2}a_{2}^{4}>25000a_{2}^{4}$, Corollary 3.2 implies that
$\varphi(c)$ is decreasing in $c$. Hence, we use $c>0.25a_{2}^{5}k^{3}$ and
observe the inequality

$0.09226(0.25\cdot 10^{5}a_{2}^{2})^{1/4}<\frac{8\log(6.2872\cdot
10^{16}a_{2}^{8}k^{3})\log(153.61632a_{2}^{5}k^{3})}{\log(0.25a_{2}^{6}k^{4})\log(2.2284a_{2}k^{3})}.$
(18)

We have $b>a_{2}$. So we can take $k=1$ in (18) which yields $a_{2}\leq
32499$. Now, $k=10^{5}/32499$ implies $a_{2}\leq 10741$, and after ten more
iterations we get $a_{2}\leq 2050$.

For those finitely many remaining pairs $\\{a_{1},a_{2}\\}$ we repeat the
process described in the first case of this proof. Again, it leads to the
conclusion that there are no quadruples $\\{a_{1},a_{2},b,c\\}$ which satisfy
all necessary conditions. ∎

###### Proof of Corollary 1.5.

Assume that there exist three positive integers $a_{1},a_{2},a_{3}$ with
$a_{1}<a_{2}<a_{3}<\min\\{b,c,d\\}$ such that $\\{a_{i},b,c,d\\}$
($i\in\\{1,2,3\\}$) are $D(4)$-quadruples. Applying Theorem 4.1 to
$\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ yields $c<0.25a_{2}^{2}b^{3}$.
Also, applying that same theorem to $\\{a_{2},b,c,d\\}$ and
$\\{a_{3},b,c,d\\}$ yields $0.25a_{2}^{2}b^{3}<c$, a contradiction. ∎

##  5 Lower bounds on element $a_{2}$

Proving the properties of the case where $a_{1}=1$ is sometimes difficult
since it doesn’t satisfy the conditions of Lemma 2.8, thus requiring a
different approach. For the proof of Proposition 5.2 we will first show
separately that $a_{1}=1$ implies $a_{2}\geq 5$.

######  Lemma 5.1.

Assume that $\\{1,b,c,d\\}$ is a $D(4)$-quadruple. Then $\\{3,b,c,d\\}$ and
$\\{4,b,c,d\\}$ are not $D(4)$-quadruples.

###### Proof.

Assume that $\\{1,b,c,d\\}$ and $\\{3,b,c,d\\}$ are $D(4)$-quadruples. As
explained at the end of Section 2, both $b$ and $c$ can be derived from a
sequence of solutions of equation

$x^{2}-3y^{2}=-8,$ (19)

since $b=r_{1}^{2}-4=y^{2}-4$ and $c=s_{1}^{2}-4=y^{2}-4$.

It is easy to prove that positive integer $y$ is given by

$y_{n}=\frac{3+\sqrt{3}}{3}(2+\sqrt{3})^{n}+\frac{3-\sqrt{3}}{3}(2-\sqrt{3})^{n},\
n\geq 0,$

n nonnegative integer.

From Theorem 4.1 we have $0.25b^{3}<c<0.25\cdot 9b^{3}=2.25b^{3}$. Let’s
denote $b=y_{j}^{2}-4$ and $c=y_{k}^{2}-4$ for some integers $j,k>0$. Then

$\displaystyle y_{k}^{2}$
$\displaystyle<2.25(y_{j}^{2}-4)^{3}+4<2.25y_{j}^{6},$ $\displaystyle
y_{k}^{2}$ $\displaystyle>0.25(y_{j}^{2}-4)^{3}+4>0.24y_{j}^{6}.$

For $j\leq 5$ we have the following values:

$j$ | $1$ | $2$ | $3$ | $4$ | $5$
---|---|---|---|---|---
$r_{j}$ | $6$ | $22$ | $82$ | $306$ | $1142$
$r_{j}^{2}-4$ | $32$ | $480$ | $6720$ | $93632$ | $1304160$

It is easy to see if $b=32$, there is no $c$ such that
$8192=0.25b^{3}<c<0.25\cdot 9b^{3}=73728$. The same is easy to check for
$b=r_{j}^{2}-4$, $j\leq 3$. So we can assume that $k>j>3$.

Since $1.57735(2+\sqrt{3})^{n}<y_{n}<1.57736(2+\sqrt{3})^{n}$ for $n\geq 6$,
we observe inequality

$1.57735^{2}(2+\sqrt{3})^{2k}<2.25\cdot 1.57736^{6}(2+\sqrt{3})^{6j},$

which implies $k\leq 3j+1$. On the other hand, we also observe

$0.24\cdot 1.57735^{6}(2+\sqrt{3})^{6j}<1.57736^{2}(2+\sqrt{3})^{2k},$

and get $k>3j+0.15$, i.e. $k=3j+1$. But,

$\displaystyle c-$ $\displaystyle 0.25\cdot
3^{2}b^{3}=r_{3j+1}^{2}-4-2.25(r_{j}^{2}-4)^{3}=$
$\displaystyle=8((2-\sqrt{3})^{4j+2}+(2+\sqrt{3})^{4j+2})-34((2-\sqrt{3})^{2j+1}+(2+\sqrt{3})^{2j+1})+\frac{128}{3}>0$

which is a contradiction with Theorem 4.1. So, $\\{3,b,c,d\\}$ cannot be a
$D(4)$-quadruple if $\\{1,b,c,d\\}$ is.

Now, assume that $\\{1,b,c,d\\}$ and $\\{4,b,c,d\\}$ are $D(4)$-quadruples.
Then there exist positive integers $r_{1},r_{2}$ such that $b+4=r_{1}^{2}$ and
$4b+4=r_{2}^{2}.$ It follows that $r_{1}$ and $r_{2}$ satisfy equation

$(2r_{1})^{2}-r_{2}^{2}=12,$

which has only the solution $r_{1}=r_{2}=2$. So there are no such
$D(4)$-quadruples. ∎

We are now ready to prove statement 1) of Theorem 1.3.

######  Proposition 5.2.

Assume that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples
with $a_{1}<a_{2}<b<c<d$. Then $a_{2}>4a_{1}$.

###### Proof.

Statement for $a_{1}=1$ follows from Lemma 5.1. So, we may assume that
$a_{1}\geq 2$. Also, let’s assume to the contrary, that $2a_{1}<a_{2}\leq
4a_{1}$. Since $a_{1}\geq 2$, it is easy to check that the assumptions of
Lemmas 2.5 and 2.8 are satisfied. As we know $a_{2}>2a_{1}$, it follows that
$a_{1}^{\prime}=4(a_{2}-a_{1})$. Also, from $a_{2}\leq 4a_{1}$ we obtain
$a_{2}-a_{1}\leq\frac{3}{4}a_{2}$. Then

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c$
$\displaystyle<1.02921\cdot 10^{14}a_{2}^{3}c$ $\displaystyle
0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c$ $\displaystyle<0.29039c$
$\displaystyle 0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$
$\displaystyle>0.002496a_{2}^{-3}c.$

On the other hand, if we write $b>ka_{2}$ for some $k\geq 1$, we have
$c>0.015625a_{2}^{5}k^{3}$ and $c/b^{2}\geq 1.5626\cdot 10^{3}a_{2}^{2}$.
Since $0.01685^{-1}\cdot 4\cdot a_{2}(a_{2}-a_{1})^{3}<238a_{2}^{4}$ and
$c>0.25bk^{2}a_{2}^{4}>25000a_{2}^{4}$, Corollary 3.2 implies that
$\varphi(c)$ is decreasing in $c$. So we can use lower bound
$c>0.015625a_{2}^{5}k^{3}$ and observe

$0.09226(1.5625\cdot 10^{3}a_{2}^{2})^{1/4}<\frac{8\log(1.02921\cdot
10^{14}\cdot\frac{1}{64}a_{2}^{8}k^{3})\log(0.29039\cdot\frac{1}{64}a_{2}^{5}k^{3})}{\log(\frac{1}{64}a_{2}^{6}k^{4})\log(0.002496\cdot\frac{1}{64}a_{2}^{2}k^{3})}.$
(20)

We have $b>a_{2}$. So we can take $k=1$ in (20) which yields $a_{2}\leq
15917$. Now, $k=10^{5}/15917$ implies $a_{2}\leq 7478$, and after eight more
iterations we get $a_{2}\leq 5179$.

For those finitely many remaining pairs $\\{a_{1},a_{2}\\}$ we repeat the
process described in the proof of Theorem 4.1 and conclude that there are no
quadruples $\\{a_{1},a_{2},b,c\\}$ which satisfy all necessary conditions. ∎

Unfortunately, some computer calculations took more than 24 hours to complete.
To reduce the remaining number of cases that need to be checked, we will
divide our next proof into several steps. First, we will establish a weaker
bound, and then we will use this bound to prove the final result.

######  Theorem 5.3.

Assume that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples
with $a_{1}<a_{2}<b<c<d$. Then

$a_{2}>0.0251a_{1}^{2}.$

###### Proof.

We will first prove that $a_{2}>0.01a_{1}^{2}$. Assume to the contrary, that
$a_{2}\leq 0.01a_{1}^{2}$. From Proposition 5.2 we have $a_{2}>4a_{1}$. This
implies $a_{1}\geq 401$ and $a_{2}\geq 1605.$ It is easy to see that the
assumptions of Lemmas 2.5 and 2.8 are now satisfied.

Notice that $3a_{1}<a_{2}-a_{1}<a_{2}\leq 0.01a_{1}^{2}$. We obtain the
following inequalities:

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c$
$\displaystyle<1.6807\cdot 10^{9}a_{1}^{11/2}c$ $\displaystyle
0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c$
$\displaystyle<0.00685a_{1}^{1/2}c$ $\displaystyle
0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$
$\displaystyle>421250a_{1}^{-7}c,$

so we observe an inequality

$n<\frac{8\log(1.6807\cdot 10^{9}\cdot
a_{1}^{11/2}c)\log(0.00685a_{1}^{1/2}c)}{\log(bc)\log(421250a_{1}^{-7}c)}.$
(21)

We will separate the proof in two cases depending on relations between $b$ and
$a_{1}^{2}$.

Case 1: $b\geq a_{1}^{2}$.

Here we have $c>0.25a_{1}^{8}$. The right-hand side of inequality (21) is
decreasing in $c$ for $c>421250^{-1}a_{1}^{7}$ which obviously holds. Also,
$n>0.09226\cdot 0.25^{1/4}a_{1}$. Applying all those inequalities to (21)
yields $a_{1}\leq 552$.

Now we can rewrite the inequality (21) and together with
$n>0.2465c^{1/12}$obtain numerical upper bound on $c$, i.e. $c<2.2666\cdot
10^{24}$. This implies $b<2.086\cdot 10^{8}\cdot a_{1}^{-2/3}$. We observe
these finitely many pairs $\\{a_{1},a_{2}\\}$, $a_{2}\leq 0.01a_{1}^{2}$ and
search for extensions within these bounds for $b$ and $c$ and computer search
returns that there are no such numbers.

Case 2: $b<a_{1}^{2}$.

Observe that, since $a_{1}<a_{2}/4<b/4$,
$\frac{3}{4}a_{2}<a_{2}-a_{1}<a_{2}\leq 0.01a_{1}^{2}$ and
$0.25b^{4}<c<0.25b^{5}$ we have

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c$
$\displaystyle<8.4034\cdot 10^{13}b^{3}c<2.101\cdot 10^{13}b^{8}$
$\displaystyle 0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c$
$\displaystyle<0.137c<0.03425b^{5}$ $\displaystyle
0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$
$\displaystyle>1.725\cdot 10^{7}b^{-7}c>4.3125\cdot 10^{6}b^{-3}.$

Also, $n>0.09226b^{-1/2}c^{1/4}>0.09226\cdot 0.25^{1/4}b^{1/2}$. Applying this
to inequalities from 2.5 and 2.8 yields $b\leq 162$ which is a contradiction.

Thus, we have proved that $a_{2}>0.01a_{1}^{2}$.

Assume now that $a_{2}\leq 0.0251a_{1}^{2}$. From Proposition 5.2 we have
$a_{2}>4a_{1}$ and get $a_{1}\geq 160$ and $a_{2}>4\cdot 160=640$. It is easy
to see that the assumptions of Lemmas 2.5 and 2.8 are satisfied.

We follow the steps from the first part of the proof. Notice that
$0.003a_{1}^{2}<a_{2}-a_{1}<a_{2}\leq 0.0251a_{1}^{2}$ holds. so we obtain an
inequality

$n<\frac{8\log(1.678\cdot
10^{10}a_{1}^{11/2}c)\log(0.3286a_{1}^{-1/2}c)}{\log(bc)\log(10613a_{1}^{-7}c)}.$
(22)

As before, we observe two cases.

Case 1: $b\geq a_{1}^{2}$. Since the same argument holds as before, we can use
$c>10613^{-1}a_{1}^{7}$ and $n>0.09226\cdot 0.25^{1/4}a_{1}$. Applying all
those inequalities yields $a_{1}\leq 682$.

Now we can rewrite the inequality (22) by inserting numerical bounds on
$a_{1}$ and together with $n>0.2465c^{1/12}$ obtain $c<3.06\cdot 10^{24}$ and
$b<2.205\cdot 10^{8}\cdot a_{1}^{-2/3}$. We observe these finitely many pairs
$\\{a_{1},a_{2}\\}$, $0.01a_{1}^{2}<a_{2}\leq 0.0251a_{1}^{2}$ and search for
extensions within these bounds for $b$ and $c$. Computer program described
before returns that there is no $c$ within the given bounds.

Case 2: $b<a_{1}^{2}$. Because of $b>10^{5}$, this assumption yields
$a_{1}\geq 317$.

Observe that, since $a_{1}<a_{2}/4<b/4$,
$0.007b<0.007a_{1}^{2}<a_{2}-a_{1}\leq 0.0251a_{1}^{2}$ for $a_{1}\geq 317$,
and $0.25b^{4}<c<0.25b^{5}$ we have

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c$
$\displaystyle<8.4034\cdot 10^{13}b^{3}c<2.101\cdot 10^{13}b^{8}$
$\displaystyle 0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c$
$\displaystyle<14.5785c<3.645b^{5}$ $\displaystyle
0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$
$\displaystyle>1.09\cdot 10^{6}b^{-7}c>272500b^{-3}.$

Also, $n>0.09226b^{-1/2}c^{1/4}>0.09226\cdot 0.25^{1/4}b^{1/2}$. Applying this
to inequalities from Lemmas 2.5 and 2.8 yields $b\leq 64$ which is a
contradiction. ∎

##  6 Upper bound on element $b$

Now we wish to establish further relations between elements $a_{2}$ and $b$.

When $a_{1}=1$, we have $c>0.25a_{1}^{2}b^{3}=0.25b^{3}$ and that lower bound
doesn’t satisfy Lemma 2.8, so we prove the next theorem only when $a_{1}\geq
2$.

######  Theorem 6.1.

Assume that $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ are $D(4)$-quadruples
with $a_{1}<a_{2}<b<c<d$. If $2\leq a_{1}$ then $b<a_{2}^{2}$.

###### Proof.

The proof can be split into two cases. In the first case, when $a_{1}\leq
159$, we observe that $a_{2}>4a_{1}$, which gives a better lower bound for
$a_{2}$, and if $a_{1}\geq 160$, a better lower bound for $a_{2}$ is given by
$a_{2}>0.0251a_{1}^{2}$.

Case $a_{1}\leq 159$: Assume to the contrary, that $b\geq a_{2}^{2}$. Then
$a_{2}\leq b^{1/2}$ and $a_{1}<0.25b^{1/2}$. Since $a_{1}\geq 2$ it is easy to
check that the assumptions of Lemmas 2.5 and 2.8 are satisfied. It holds

$\displaystyle 8.4034\cdot 10^{13}a_{1}^{1/2}(a_{1}^{\prime})^{1/2}a_{2}^{2}c$
$\displaystyle<8.4034\cdot 10^{13}b^{3/2}c,$ $\displaystyle
0.20533a_{1}^{1/2}a_{2}^{1/2}(a_{2}-a_{1})^{-1}c$
$\displaystyle<0.0153b^{1/2}c,$ $\displaystyle
0.01685a_{1}(a_{1}^{\prime})^{-1}a_{2}^{-1}(a_{2}-a_{1})^{-2}c$
$\displaystyle>0.0000115c,$

where we have used $a_{2}\geq 9$ and $\frac{3}{4}a_{2}<a_{2}-a_{1}<a_{2}$.
Also $n>0.09226\sqrt{0.5}b^{1/4}$. Now we observe

$0.09226\sqrt{0.5}b^{1/4}<\frac{8\log(8.4034\cdot
10^{13}b^{3/2}c)\log(0.0153b^{1/2}c)}{\log(bc)\log(0.0000115c)},$

and the right-hand side is a function decreasing in $c$ since
$c>0.0000115^{-1}$. So, we can use a lower bound $c>0.25a_{1}^{2}b^{3}\geq
b^{3}$ in this inequality and obtain $b<1.846\cdot 10^{9}$. This implies
$a_{2}\leq 42965$. We can improve this upper bound on $a_{2}$ furthermore by
observing for each fixed $a_{1}\in[8,159]$ a similar inequality with $a_{2}$.
However, for values $a_{1}\in[2,7]$, this approach yields a larger bound
compared to $42965$. We do a computer search and it returns there don’t exist
elements $b$ and $c$ which satisfy all of the conditions.

Case $a_{1}\geq 160$: If $a_{1}\geq 160$ then $a_{2}\geq 643$. We use
$a_{2}>0.0251a_{1}^{2}$, so $a_{1}<\frac{a_{2}^{1/2}}{\sqrt{0.0251}}$ and
$a_{2}-a_{1}>a_{2}\left(1-\frac{1}{\sqrt{0.0251a_{2}}}\right)>0.751a_{2}$.

We will first prove that $b<3a_{2}^{2}$ and then we will prove that
$b<a_{2}^{2}$. Thus, we will assume to the contrary that $b\geq 3a_{2}^{2}$.
Writing our steps more generally, let’s assume that $b\geq ka_{2}^{2}$. As
before, it is easy to see that Lemmas 2.5 and 2.8 hold. We have $c>0.25\cdot
160^{2}k^{3}a_{2}^{6}=6400k^{3}a_{2}^{6}$, $c/b^{2}>6400ka_{2}^{2}$ and
$n>0.09226\cdot(6400k)^{1/4}a_{2}^{1/2}$. It is easy to see from Corollary 3.2
that we can observe the inequality

$0.09226\cdot(6400k)^{1/4}a_{2}^{1/2}<\frac{8\log(2.7136\cdot 10^{18}\cdot
k^{3}\cdot a_{2}^{35/4})\log(4599.392\cdot k^{3}\cdot
a_{2}^{23/4})}{\log(6400\cdot k^{4}\cdot a_{2}^{8})\log(4313.6\cdot k^{3}\cdot
a_{2}^{2})},$

which yields $a_{2}\leq 533$ for $k=3$. This is a contradiction with
$a_{2}\geq 643$, so $b<3a_{2}^{2}$.

Now we assume that $b\geq a_{2}^{2}$ holds. Previous inequality for $k=1$
yields $a_{2}\leq 1140$ which implies $a_{1}\leq 231$. Then we also have
$b<3a_{2}^{2}\leq 3898800$.

Again, we observe finitely many pairs $\\{a_{1},a_{2}\\}$ and search for $b$
and $c$ such that $\max\\{10^{5}+1,a_{2}^{2}\\}\leq b<3a_{2}^{2}$ and
$0.25a_{1}^{2}b^{3}<c<0.25a_{2}^{2}b^{3}$, and computer search returns that
none such $D(4)$-pair $\\{b,c\\}$ exists. ∎

It is now easy to finish the proof of statement 3) of Theorem 1.3. Since
$10^{5}<b<a_{2}^{2}$, when $a_{1}\geq 2$, we have $a_{2}\geq 317.$

##  7 Proof of Corollary 1.6

First, we will prove a numerical upper bound on element $c$ in an irregular
$D(4)$-quadruple $\\{a,b,c,d\\}$. The proof follows [7, Proposition 4] and
[10, Lemma 13]. In this lemma, $(v_{m})$ and $(w_{n})$ denote sequences of
solutions described as in (4) and (5) but with $a$ instead of $a_{2}$, i.e. as
in [3].

######  Lemma 7.1.

Let $\\{a,b,c,d\\}$ be a $D(4)$-quadruple such that $a<b<c<d_{+}<d$,
$c>b^{2.8}$, $b>\max\\{4a,10^{5}\\}$. Then $c<10^{2157}$ and $m<3.65\cdot
10^{21}$.

###### Proof.

Assume that $d>d_{+}$. First, we will prove $n>c^{0.01}$. Assume to the
contrary, that $n\leq c^{0.01}$. Notice that for $c>b^{2.8}$ we have from
Lemma 2.7 that $m<1.36n+1$. Also, since $b>10^{5}$ we have $c>10^{13}$. From
Lemma 2.6 we know $n\geq 7$.

The only cases we have to observe are (6) and (7), where $m$ and $n$ are of
the same parity. Then from [10, Lemma 5] we have $n\leq m$.

1) Observe the case (6), where $v_{2m^{\prime}}=w_{2n^{\prime}}$ and
$|z_{0}|=2$, $z_{0}=z_{1}$.
We use $n^{\prime}<c^{0.01}$. As in [10, Lemma 13], congruence

$\pm am^{\prime 2}+sm^{\prime}\equiv\pm bn^{\prime 2}+tn^{\prime}\ (\bmod\ c)$

is equality since $\max\\{am^{\prime 2},sm^{\prime},bn^{\prime
2},tn^{\prime}\\}<0.01c$. Hence, $\pm am^{\prime 2}+sm^{\prime}=\pm bn^{\prime
2}+tn^{\prime}.$ As in [7, Lemma 11], by using $b>4a$ and $c>10^{13}$ it is
easy to see that

$\frac{m}{n}=\frac{2m^{\prime}}{2n^{\prime}}>\frac{t}{s}\cdot\frac{1-\frac{bn^{\prime
2}}{tn^{\prime}}}{1+\frac{am^{\prime
2}}{sm^{\prime}}}>\sqrt{\frac{b}{a}}\cdot\frac{0.999}{\sqrt{1.001}\cdot
1.001}>1.99,$

which is a contradiction with $m<1.36n+1$.

2) Observe the case (7), where $v_{2m^{\prime}+1}=w_{2n^{\prime}+1}$,
$|z_{0}|=t$, $|z_{1}|=s$ and $z_{0}z_{1}>0$.
As in [10, Lemma 13], we see that congruence

$\pm\frac{1}{2}(astm^{\prime}(m^{\prime}+1)-bstn^{\prime}(n^{\prime}+1))\equiv
2r(n^{\prime}-m^{\prime})\ (\bmod\ c)$

holds. If we square it and use $s^{2}\equiv t^{2}\equiv 4\ (\bmod\ c)$ we get

$(2am^{\prime}(m^{\prime}+1)-2bn^{\prime}(n^{\prime}+1))^{2}\equiv
4r^{2}(n^{\prime}-m^{\prime})^{2}\ (\bmod\ c).$

It is easy to see that both sides are less than $c$, so we actually have
equality, i.e.

$am^{\prime}(m^{\prime}+1)-bn^{\prime}(n^{\prime}+1)=\pm
r(n^{\prime}-m^{\prime}).$

As in [7, Lemma 11], we use it to show

$\frac{2n^{\prime}+1}{2m^{\prime}+1}-\sqrt{\frac{a}{b}}=\frac{\pm
4r(n^{\prime}-m^{\prime})+b-a}{\sqrt{b}(2m^{\prime}+1)(\sqrt{b}(2n^{\prime}+1)+\sqrt{a}(2m^{\prime}+1))}.$
(23)

Denote the expression on the right-hand side of (23) with $A$. Since
$m^{\prime}<1.36n^{\prime}+0.68$, we have

$A<\frac{0.36n^{\prime}+0.68}{2n^{\prime}+1}\cdot
2.01\cdot\frac{1}{2m^{\prime}+1}+\frac{1}{(2m^{\prime}+1)(2n^{\prime}+1)}.$

Inequality $m^{\prime}<1.36n^{\prime}+0.68$ also implies

$\frac{2n^{\prime}+1}{2m^{\prime}+1}>\frac{2n^{\prime}+1}{2.72n^{\prime}+2.36}.$

Denote the right-hand side of this inequality with $B$. Since $n\geq 7$
implies $n^{\prime}\geq 3$ then $B>0.65$ and $A<0.26\cdot
2.01\cdot\frac{1}{7}+\frac{1}{49}<0.1$ which yields

$0.65<\frac{2n^{\prime}+1}{2m^{\prime}+1}<0.5+0.1,$

a contradiction.

Hence, we have $n>c^{0.01}$ when $d>d_{+}$. From [3, Proposition 5.4] we have

$\frac{m}{\log(38.92(m+1))}<2.7717\cdot 10^{12}\log^{2}c.$ (24)

Since $c^{0.01}<n\leq m$, i.e. $c<m^{100}$ then

$\frac{m}{\log(38.92(m+1))\log^{2}(m)}<2.7717\cdot 10^{16}.$

This implies $m<3.65\cdot 10^{21}$ and $c<m^{100}<10^{2157}$. ∎

###### Proof of Corollary 1.6.

Let $\\{a_{1},b,c,d\\}$ and $\\{a_{2},b,c,d\\}$ be $D(4)$-quadruples. From
Corollary 1.4 we know both of them are irregular quadruples, so Lemma 7.1
implies $c<10^{2157}$ and $m<3.65\cdot 10^{21}.$ Since $b>10^{5}$ and
$b>a_{2}$, we have $c>0.25a_{1}^{2}b^{3}>b^{2.8}>a_{2}^{2.8}$ , so
$s_{2}=\sqrt{a_{2}c+4}<c^{7/10}$. From the proof of [10, Lemma 5] we see
$v_{m}<cx_{0}s_{2}^{m-1}$. Now [3, Lemma 2.5] implies

$\displaystyle v_{m}$
$\displaystyle<cx_{0}s_{2}^{m-1}<1.00317c\sqrt[4]{a_{2}c}s_{2}^{m-1}<1.00317c\sqrt[4]{c^{1/2.8}\cdot
c}\cdot c^{\frac{7}{10}m-\frac{7}{10}}$ $\displaystyle<1.00317c^{0.64}\cdot
c^{\frac{7}{10}m}<10^{10^{25}}.$

Since $v_{m}^{2}=cd+4>d$ we have $d<10^{10^{26}}$, which proves our statement
– there are only finitely many $D(4)$-triples $\\{b,c,d\\}$, $b<c<d$, which
can be extended to two different $D(4)$-quadruples with smaller elements
$a_{1}<a_{2}<b$. ∎

######  Remark.

For further research it would be interesting to establish some better bounds
on elements, similar to those presented in [6]. The first step would be to
prove a lower bound for $b$ in the form of $b>ka_{2}$, where $k>1$. The key
concept, observed in [6], is to exclude solutions $c=c_{\nu}^{\pm}$ when
$b\leq ka_{2}$. The sequence $\\{c_{\nu}^{\pm}\\}_{\nu}$ refers to a specific
sequence of solutions to a Pellian equation that emerges in the problem of
extending a $D(4)$-pair $\\{a,b\\}$ to a $D(4)$-triple $\\{a,b,c\\}$. An
explicit formula for this sequence can be found, for example, in [3]. However,
in the case of $D(4)$-$m$-tuples, a similar approach cannot be employed since
there exist multiple $c$ values of this form which fall within the bounds
established in this paper. Consequently, alternative methods need to be
devised to either eliminate the possibility of such $c$ values or demonstrate
that $b>ka_{2}$ for some $k>1$.

## References

  * [1] Lj. Baćić, A. Filipin, On the extensibility of $D(4)$-pair $\\{k-2,k+2\\}$, J. Comb. Number Theory 5 (2013), 181–197.
  * [2] M. Bliznac Trebješanin, $D(4)$-triples with two largest elements in common, Math. Slovaca, 73(2) (2023), 343–352.
  * [3] M. Bliznac Trebješanin, Extension of a Diophantine triple with the property $D(4)$, Acta Math. Hungar. 163 (2021), 213–-246.
  * [4] M. Bliznac Trebješanin, A. Filipin, Nonexistence of $D(4)$-quintuples, J. Number Theory, 194 (2019), 170–217.
  * [5] M. Cipu, A. Dujella and Y. Fujita, Diophantine triples with largest two elements in common, Period. Math. Hungar. 82 (2021), 56–68.
  * [6] M. Cipu, A. Dujella, Y. Fujita, Extensions of a Diophantine triple by adjoining smaller elements, Mediterr. J. Math. 19 (2022), Article 187, (20pp)
  * [7] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math. 566 (2004), 183–214.
  * [8] 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.
  * [9] A. Filipin, There does not exist a $D(4)$-sextuple, J. Number Theory 128 (2008), 1555–1565.
  * [10] A. Filipin, On the size of sets in which $xy+4$ is always a square, Rocky Mountain J. Math. 39 (2009), no. 4, 1195–1224.
  * [11] A. Filipin, The extension of some $D(4)$-pairs, Notes Number Theory Discrete Math. 23 (2017), 126–135.
  * [12] A. Filipin, Bo He, A. Togbé, On a family of two-parametric D(4)-triples, Glas. Mat. Ser. III 47 (2012), 31–51.
  * [13] B. He, A. Togbé, V. Ziegler, There is no Diophantine quintuple, Trans. Amer. Math. Soc. 371 (2019), 6665–6709.
  * [14] J. H. Rickert, Simultaneous rational approximations and related Diophantine equations, Proc. Cambridge Philos. Soc. 113 (1993) 461–472.

Marija Bliznac Trebješanin
University of Split, Faculty of Science
Ruđera Boškovića 33, 21000 Split, Croatia
Email<EMAIL_ADDRESS>

Pavao Radić
University of Split, Faculty of Science
Ruđera Boškovića 33, 21000 Split, Croatia
Email<EMAIL_ADDRESS>