# On a conjecture of Franušić and Jadrijević: Counter-examples

Kalyan Chakraborty, Shubham Gupta and Azizul Hoque  K. Chakraborty @Kerala
School of Mathematics, Kozhikode-673571, Kerala, India<EMAIL_ADDRESS>S. Gupta @Harish-Chandra Research Institute, A CI of Homi Bhabha National
Institute, Chhatnag Road, Jhunsi, Prayagraj - 211019<EMAIL_ADDRESS>A. Hoque @Department of Mathematics, Rangapara College, Rangapara,
Sonitpur-784505, Assam, India<EMAIL_ADDRESS>
###### Abstract.

Let $d\equiv 2\pmod{4}$ be a square-free integer such that $x^{2}-dy^{2}=-1$
and $x^{2}-dy^{2}=6$ are solvable in integers. We prove the existence of
infinitely many quadruples in $\mathbb{Z}[\sqrt{d}]$ with the property $D(n)$
when
$n\in\\{(4m+1)+4k\sqrt{d},(4m+1)+(4k+2)\sqrt{d},(4m+3)+4k\sqrt{d},(4m+3)+(4k+2)\sqrt{d},(4m+2)+(4k+2)\sqrt{d}\\}$
for $m,k\in\mathbb{Z}$. As a consequence, we provide few counter examples to a
conjecture of Franušić and Jadrijević (see Conjecture 1.1).

###### Key words and phrases:

Counter-example; Diophantine quadruples; Pellian equations; Quadratic fields

###### 2020 Mathematics Subject Classification:

11D09; 11R11

##  1\. Introduction

Given a non-zero integer $n$, a set of $m$ distinct positive integers
$\\{a_{1},a_{2},\cdots,a_{m}\\}$ such that $a_{i}a_{j}+n$ is a perfect square
for all $1\leq i<j\leq m$, is called a Diophantine $m$-tuple with the property
$D(n)$ or simply $D(n)$-$m$-set. The $D(1)$-$m$-sets are known as Diophantine
$m$-tuples, and have been studied since the time of Diophantus. The first
Diophantine quadruple, viz. $\\{1,3,8,120\\}$ was found by Fermat. It follows
from the work of Baker and Davenport [2] that the above quadruple cannot be
extended to a Diophantine quintuple. In 2004, Dujella [9] proved that there
are only finitely many Diophantine quintuples. However, there was a ‘folklore’
conjecture which states that there does not exist Diophantine quintuples.
Recently this conjecture has been settled by He, Togbé and Ziegler in [20]. On
the other hand, Bonciocat et al. [3] settled the conjecture of Dujella [7]
which states that there is no Diophantine $D(-1)$-quadruple. We direct the
readers to [4, 6, 11, 12, 19] for more information about $D(n)$-$m$-sets.

A general notion is obtained by considering elements of any commutative ring
instead of positive integers. However, many demanding and interesting problems
already occur in the setting fixed by the above definition. For a given non-
zero element $n$ of a commutative ring $\mathcal{R}$ with unity, a set
$\\{a_{1},a_{2},\cdots,a_{m}\\}\subset\mathcal{R}\setminus\\{0\\}$ is called a
Diophantine $m$-tuple with the property $D(n)$ or simply $D(n)$-$m$-set in
$\mathcal{R}$ if $a_{i}a_{j}+n$ is a perfect square in $\mathcal{R}$ for all
$1\leq i<j\leq m$.

In [7, 25], it was established that there exists a quadruple with the property
$D(n)$ if and only if $n$ can be expressed as the difference of two squares,
up to finitely many exceptions. Later, Dujella [8] proved that this fact also
true in case of Gaussian integers. Besides this, it was proved that for
certain integers $d$, the above fact is true in the ring of integers of
$\mathbb{Q}(\sqrt{d})$ for certain $d$’s (see, [1, 13, 14, 15, 18, 26]). These
results motivate Franušić and Jadrijević to post the following conjecture.

######  Conjecture 1.1 ([17, Conjecture 1]).

Let $\mathcal{R}$ be a commutative ring with unity $1$ and
$n\in\mathcal{R}\setminus\\{0\\}$. Then there exists a quadruple with the
property $D(n)$ if and only if $n$ can be expressed as a difference of two
squares in $\mathcal{R}$, up to finitely many exceptions.

This conjecture was verified successfully for ring of integers of certain
quadratic fields (see, [1, 13, 14, 15, 18, 26]). It follows from [16] and [22]
that Conjecture 1.1 is true for the ring of integers of
$\mathbb{Q}(\sqrt[3]{2})$. Recently Franušić and Jadrijević [17] confirmed
this conjecture for the ring of integers of bicyclic biquadratic field
$\mathbb{Q}[\sqrt{2},\sqrt{3}]$.

In this paper, we investigate the existence of infinitely many quadruples in
$\mathbb{Z}[\sqrt{d}]$ with the property $D(n)$ for certain
$n\in\mathbb{Z}[\sqrt{d}]$ under the assumptions that the equations

$x^{2}-dy^{2}=-1$ (1.1)

and

$x^{2}-dy^{2}=6$ (1.2)

are solvable in integers. Precisely, we prove the following:

######  Theorem 1.1.

Assume that $d\equiv 2\pmod{4}$ is a square-free positive integer and the
equations (1.1) and (1.2) are solvable. Then there exist infinity many
quadruples in $\mathbb{Z}[\sqrt{d}]$ with the property $D(n)$ when
$n\in\\{(4m+1)+4k\sqrt{d},(4m+1)+(4k+2)\sqrt{d},(4m+3)+4k\sqrt{d},(4m+3)+(4k+2)\sqrt{d},(4m+2)+(4k+2)\sqrt{d}\\}$
with $m,k\in\mathbb{Z}$.

As a consequence of Theorem 1.1, we construct some counter-examples of
Conjecture 1.1. In fact, under the assumption of Bunyakovsky’s conjecture, we
see that there are infinitely many such rings in which Conjecture 1.1 does not
hold.

##  2\. Preliminary Descent

We first fix up the following notations:

  * •

$(a,b):=a+b\sqrt{d}$,

  * •

$Nm(\alpha):=(a,b)(a,-b)$ for a given $\alpha=(a,b)$,

  * •

$(x,y)\equiv(a,b)\text{~{}(mod~{}}(c,d))$ denotes $x\equiv
a\pmod{c}\text{~{}~{}~{}and~{}~{}~{}}y\equiv b\pmod{d}.$

For any square-free integer $d\equiv 2\pmod{4}$ and $m,k\in\mathbb{Z}$, we
define:

$\displaystyle\mathcal{S}:=$
$\displaystyle\\{(4m,4k+1),(4m,4k+2),(4m,4k+3),(4m+1,4k+1),(4m+1,4k+3),(4m+2,$
$\displaystyle 4k+1),(4m+2,4k+3),(4m+3,4k+1),(4m+3,4k+3)\\},$
$\displaystyle\mathcal{T}:=$
$\displaystyle\\{(4m,4k),(4m+1,4k),(4m+1,4k+2),(4m+2,4k),(4m+2,4k+2),(4m+3,4k),$
$\displaystyle(4m+3,4k+2)\\}.$

Clearly, if $n\in\mathbb{Z}[\sqrt{d}]$ then $n\in\mathcal{S}\cup\mathcal{T}$.
We now extract the following lemma from [13, Proposition 5].

######  Lemma 2.1.

Let $d\in\mathbb{Z}$ such that $d\equiv 2\pmod{4}$. Then for
$n\in\mathcal{S}$, there does not exist any Diophantine quadruple in
$\mathbb{Z}[\sqrt{d}]$ with the property $D(n)$.

There are many results concerning the integers which can be represented as
$n=\alpha^{2}-\beta^{2}$ in $\mathbb{Z}[\sqrt{d}]$. These integers can be used
in the construction of quadruples in $\mathbb{Z}[\sqrt{d}]$. In this context,
we deduce the following lemma from [10, Propositions 5 and 6].

######  Lemma 2.2.

For any square-free $d\in\mathbb{Z}$, all the numbers of the form
$n=(4m+2,4n)\in\mathbb{Z}[\sqrt{d}]$ can be written as the difference of two
squares if and only if $x^{2}-dy^{2}=\pm 2$ is solvable in integers. Further,
all the numbers of the form $n=(4m+2,4n+2)\in\mathbb{Z}[\sqrt{d}]$ can be
written as the difference of two squares if and only if $d\equiv 2\pmod{4}$
and $x^{2}-dy^{2}=\pm 2$ is solvable in integers.

The next lemma follows from the definition.

######  Lemma 2.3.

For a square-free $d\in\mathbb{Z}$, let $\\{a_{1},a_{2},a_{3},a_{4}\\}$ be a
quadruple with the property $D(n)$. Then for any non-zero
$w\in\mathbb{Z}[\sqrt{d}]$, $\\{wa_{1},wa_{2},wa_{3},wa_{4}\\}$ is a quadruple
in $\mathbb{Z}[\sqrt{d}]$ with the property $D(w^{2}n)$.

Let us consider the set $\\{a,b,a+b+2r,a+4b+4r\\}$ having non-zero distinct
elements of $\mathbb{Z}[\sqrt{d}]$. The condition ‘non-zero distinctness’ in
the above set can be verified using the following lemma.

######  Lemma 2.4.

Let $a_{1},b_{1},c_{1},d_{1},e_{1},a_{2},b_{2},c_{2},d_{2}\in\mathbb{Z}$ with
$a_{1},b_{1},a_{2}\neq 0$. Then the following system of equations

$\displaystyle\begin{cases}a_{1}x^{2}+b_{1}y^{2}+c_{1}x+d_{1}y+e_{1}=0,\\\
a_{2}xy+b_{2}x+c_{2}y+d_{2}=0,\end{cases}$ (2.1)

has only finitely many integer solutions.

###### Proof.

Consider $a_{2}xy+b_{2}x+c_{2}y+d_{2}=0$. This implies that

$x=-\dfrac{c_{2}y+d_{2}}{a_{2}y+b_{2}},$ (2.2)

where $y\neq-\dfrac{b_{2}}{a_{2}}$. If $y\neq-\dfrac{b_{2}}{a_{2}}$, we put
this value of $x$ in the first equation of (2.1). This gives a fourth degree
equation in $y$ with the leading coefficient $a_{2}^{2}b_{1}\neq 0$. Thus, we
get at most four values of $y$ and therefore by (2.2), one gets at most four
values of $x$. On the other hand, for $y=-\dfrac{b_{2}}{a_{2}}$, the first
equation of (2.1) gives a two degree equation in $x$ with leading coefficient
$a_{2}^{2}a_{1}$ which provides at most two values for $x$. This completes the
proof. ∎

We now assume that $n\in\mathbb{Z}[\sqrt{d}]$ such that $ab+n=r^{2}$. Then
$a(a+b+2r)+n=(a+r)^{2}$, $b(a+b+2r)+n=(b+r)^{2}$ and
$b(a+4b+4r)+n=(2b+r)^{2}$. Now if $a(a+4b+4r)+n=\alpha^{2}$ for some
$\alpha\in\mathbb{Z}[\sqrt{d}]$, then
$a^{2}+4ar+4r^{2}=(a+2r)^{2}=\alpha^{2}+3n$. Thus, we have the following
lemma.

######  Lemma 2.5.

For any $n\in\mathbb{Z}[\sqrt{d}]$, the set $\\{a,b,a+b+2r,a+4b+4r\\}$ is a
quadruple in $\mathbb{Z}[\sqrt{d}]$ with the property $D(n)$ when $ab+n=r^{2}$
and $3n=\alpha_{1}\alpha_{2}$, where $\alpha_{1}=a+2r+\alpha$ and
$\alpha_{2}=a+2r-\alpha$ for some $a,b,r,\alpha\in\mathbb{Z}[\sqrt{d}]$ such
that all the element of this set are non-zero and distinct.

##  3\. Pell type equations $x^{2}-dy^{2}=\pm 6$ and $x^{2}-dy^{2}=\pm 1$

We now consider only quadratic ring of integers $\mathbb{Z}[\sqrt{d}]$, where
the square-free integer $d$ satisfies $d\equiv 2\pmod{4}$, and the equations
$x^{2}-dy^{2}=-1$, $x^{2}-dy^{2}=6$ are solvable in integers. This implies
that the equation $x^{2}-dy^{2}=-6$ will be also solvable in integers. Also it
is well known that the ring of integers of real quadratic field has infinitely
many elements of norm 1. Now, we discuss about the elements of norm $\pm 1$
and $\pm 6$ in $\mathbb{Z}[\sqrt{d}]$. First, let $x^{2}-dy^{2}=6$ be solvable
in integers, then $x^{2}-dy^{2}\equiv 0\pmod{3}.$ Clearly, $x$ and $y$ can not
be $0$ modulo $3$, since $9$ can not divide $6$. If $x\equiv 1\pmod{3}$ and
$y\equiv 0\pmod{3}$, then $1\equiv 0\pmod{3}$, which is not possible, and
similarly, $x\equiv 2\pmod{3}$ and $y\equiv 0\pmod{3}$ is not possible. Now,
assume that $x\equiv 0\pmod{3}$ and $y\equiv\pm 1\pmod{3}$. Then $3$ divides
$d$ and thus by the assumption $d\equiv 2\pmod{4}$ one gets $d\equiv
6\pmod{12}.$ Since the equation $x^{2}-dy^{2}=-1$ is solvable in integers, so
that reading it modulo $12$ we see that $x\equiv 0\pmod{3}$ and $y\equiv\pm
1\pmod{3}$ are not possible. Therefore $x,y\equiv\pm 1\pmod{3}$ and, after
reducing the equation $x^{2}-dy^{2}=6$ to modulo 4, we see that $x$ is even
and $y$ is odd. This implies that $(x,y)=(4,1),(4,5),(2,1),(2,5)\pmod{(6,6)}$.
Taking $x,y\equiv\pm 1\pmod{3}$, one gets $d\equiv 1\pmod{3}$. Therefore
$d\equiv 10\pmod{12}$ since by hypothesis $d\equiv 2\pmod{4}$.

Consider the Pell equation $x^{2}-dy^{2}=1$. After using $d\equiv 10\pmod{12}$
and reducing this equation to modulo $3$ and modulo $4$, we get $x\equiv\pm
1\pmod{6}$ and $y\equiv 0\pmod{6}$. Similarly for $x^{2}-dy^{2}=-1$, we get
$x\equiv 3\pmod{6}$ and $y\equiv\pm 1\pmod{6}$. We also know that there exist
infinitely many solutions of these two equations.

Let $\alpha=\beta\gamma$, where $\beta=(6a+4,6b+1)$ and
$\gamma=(6a_{1}+3,6b_{1}+1)$ for some $a,b,a_{1},b_{1}\in\mathbb{Z}$, with
norm of $\beta$ and $\gamma$ are $\pm 6$ and -1, respectively. If $a$ is even
then $\beta=(12a+4,6b+1)$, for some $a,b\in\mathbb{Z}$ with norm $\pm 6$, and
$\alpha=(12a_{1}-2,6b_{1}+1)$ with norm $\mp 6$, for some
$a_{1},b_{1}\in\mathbb{Z}$. In the similar way, if $a$ is odd, then
$\alpha=(12M+4,6N+1)$ for some $M,N\in\mathbb{Z}$ with norm $\mp 6$.
Multiplying this $\alpha$ by $\gamma$ one gets an element of the form
$(12M+2,6N+1)$ with norm $\pm 6$ for some $M,N\in\mathbb{Z}$ because if
$x^{2}-dy^{2}=\pm 6$ has a solution $(X,Y)$ then $(\pm X,\pm Y)$ are also
solutions. Now, we prove that there does not exist any norm $-6$ element of
the form $(12M\pm 4,6N\pm 1)$ for some $M,N\in\mathbb{Z}$. Suppose it exists
then the equation $(12M\pm 4)^{2}-d(6N\pm 1)^{2}=-6$ gives $d\equiv
22\pmod{24}$. In the above paragraph, we have proved that all elements whose
norm are $-1$ have the form $(6a\pm 3,6b\pm 1)$ with $a,b\in\mathbb{Z}$. So
that $(6a\pm 3)^{2}-d(6b\pm 1)^{2}=-1$. After rearranging this equation we get
$(36(a)(a+1)+9)-d(24b^{2}+12b(b+1)+1)=-1$. Using $d\equiv 22\pmod{24}$, we get
a contradiction. To conclude, in these quadratic ring of integers, norm $6$
and norm $-6$ element will be of form $(12M\pm 4,6N\pm 1)$ and $(12M\pm
2,6N\pm 1)$ respectively for some $M,N\in\mathbb{Z}$, i.e. $(12M\pm
4)^{2}-d(6N\pm 1)^{2}=6$. And if we reduce this equation modulo $48$, we get
$d\equiv 10\pmod{48}$, since $d\equiv 2\pmod{4}$.

Now we prove that there are infinitely many elements of norm $\pm 6$. Let
$\alpha=\beta\gamma$, where $\beta=(12a+2,6b+1)$ and
$\gamma=(6a_{1}+1,6b_{1})$ for some $a,b,a_{1},b_{1}\in\mathbb{Z}$, with norm
of $\beta$ and $\gamma$ are $-6$ and $1$ respectively. Then $\alpha$ will be
of the form $(12M+2,6N+1)$, for some $M,N\in\mathbb{Z}$ with norm $-6$. Since
we have infinitely many choices of $\gamma$, there exist infinitely many
elements of norm $-6$ and of the form $(12M\pm 2,6N\pm 1)$, for
$M,N\in\mathbb{Z}$ because if $x^{2}-dy^{2}=-6$ has a solution $(X,Y)$, then
$(\pm X,\pm Y)$ are also solutions. Similarly, there exist infinitely many
elements of norm $6$ and of the form $(12M\pm 4,6N\pm 1)$ for some
$M,N\in\mathbb{Z}$. We sum up the above discussion in the following lemma.

######  Lemma 3.1.

Assume that $d\equiv 2\pmod{4}$ is a square-free integer and the equations
(1.1) and (1.2) are solvable in integers. Then for the ring of integers
$\mathbb{Z}[\sqrt{d}]$, the following statements hold:

  * (i)

all elements whose norm is $1$ have the form $(6a_{1}\pm 1,6b_{1})$ and there
are infinitely many such elements;

  * (ii)

all elements whose norm is $-1$ have the form $(6a\pm 3,6b\pm 1)$ and there
are infinitely many such elements;

  * (iii)

$d\equiv 10\pmod{48}$;

  * (iv)

all elements whose norm is $6$ have the form $(12M\pm 4,6N\pm 1)$ and there
are infinitely many such elements;

  * (v)

all elements whose norm is $-6$ have the form $(12M\pm 2,6N\pm 1)$ and there
are infinitely many such elements, where $a,b,a_{1},b_{1},M,N\in\mathbb{Z}$.

##  4\. Proof of Theorem 1.1

The proof is divided in various parts depending on the values of $n$ and it
relies upon Lemmas 3.1 and 2.5. Lemma 3.1 helps us for choosing elements with
specific norms whereas Lemma 2.5 helps us to construct quadruples.

###  Case 1: $n=(4m+1,4k)$

In this case $3n=(-1)(-3)(4m+1,4k)$ and thus by (i) of Lemma 3.1, we have

$\displaystyle 3n$
$\displaystyle=(-6\alpha-3,6\beta+3)(2\alpha+1,2\beta+1)(4m+1,4k),\text{ for
some }\alpha,\beta\in\mathbb{Z}$
$\displaystyle=(-6\alpha-3,6\beta+3)(8m\alpha+2\alpha+4m+1+d(8\beta
k+4k),8\alpha k+4k+8\beta m+2\beta+4m+1).$

Let us set:

$\begin{cases}\alpha_{1}=(-6\alpha-3,6\beta+3)\\\
\alpha_{2}=(8m\alpha+2\alpha+4m+1+d(8\beta k+4k),8\alpha k+4k+8\beta
m+2\beta+4m+1).\end{cases}$

Then by Lemma 2.5, we have

$\alpha_{1}+\alpha_{2}=2a+4r,$

which implies that

$a+2r=(4m\alpha-2\alpha+2m-1+2d(2\beta k+k),4\alpha k+2k+4\beta
m+4\beta+2m+2).$ (4.1)

Now by (i) of Lemma 3.1, we have infinitely many $a\in\mathbb{Z}[\sqrt{d}]$
such that $a=(6a_{1}+1,6b_{1})$ with $Nm(a)=1$. Note that out of these
infinitely many choices of $a$, by Lemma 2.4, only finitely many provide a
zero and distinct element in the set $\\{a,b,a+b+2r,a+4b+4r\\}$ as in Lemma
2.5. Thus (4.1) gives

$r=(2m\alpha-\alpha+m-1-3a_{1}+d(2\beta k+k),2\alpha k+k+2\beta
m+2\beta+m+1-3b_{1}),$

which is

$r=(m(2\alpha+1)+kd(2\beta+1)-\alpha-1-3a_{1},m(2\beta+1)+k(2\alpha+1)+2\beta+1-3b_{1}).$

Again from Lemma 2.5, we use $ab+n=r^{2}$ to calculate $b$ which turned out to
be

$\displaystyle b=$
$\displaystyle\Big{(}(m(2\alpha+1)+kd(2\beta+1)-\alpha-1-3a_{1})^{2}+d(m(2\beta+1)+k(2\alpha+1)+2\beta+1-$
$\displaystyle
3b_{1})^{2}-4m-1,2(m(2\alpha+1)+kd(2\beta+1)-\alpha-1-3a_{1})(m(2\beta+1)+k(2\alpha+1)+$
$\displaystyle 2\beta+1-3b_{1})-4k\Big{)}\Big{(}6a_{1}+1,-6b_{1}\Big{)}.$

Thus by Lemma 2.5, the elements $a,b$ and $r$ give the quadruple
$\\{a,b,a+b+2r,a+4b+4r\\}$ in $\mathbb{Z}[\sqrt{d}]$ with the property
$D(4m+1,4k)$. Since there are infinitely many choices for $a$, we conclude
that there are infinitely many quadruples of the form
$\\{a,b,a+b+2r,a+4b+4r\\}$ in $\mathbb{Z}[\sqrt{d}]$ with the property
$D(4m+1,4k)$.

### Case 2: $n=(4m+1,4k+2)$

Here, we will use the same $\alpha_{1}$ and $\alpha_{2}$ as in the previous
case. Analogous to (4.1), we get

$a+2r=\Big{(}4m\alpha-2\alpha+2m-1+d(4\beta k+2\beta+2k+1),4\alpha
k+2\alpha+2k+3+4m\beta+4\beta+2m\Big{)}.$

Now we choose $a=(6a_{1}+3,6b_{1}+1)$ with $Nm(a)=-1$ and the above equation
gives

$r=\Big{(}2m\alpha-\alpha+m-2-3a_{1}+(d/2)(4\beta k+2\beta+2k+1),2\alpha
k+\alpha+k+1-3b_{1}+2m\beta+2\beta+m\Big{)}.$

We complete this case by computing $b$ using $ab+n=r^{2}$, which is given by

$\displaystyle b=$
$\displaystyle\Big{(}(2m\alpha-\alpha+m-2-3a_{1}+(d/2)(4\beta
k+2\beta+2k+1)^{2}+d(2\alpha k+\alpha+k+1-3b_{1}+$ $\displaystyle
2m\beta+2\beta+m)^{2}-4m-1,2(2m\alpha-\alpha+m-2-3a_{1}+(d/2)(4\beta
k+2\beta+2k+1)$ $\displaystyle(2\alpha
k+\alpha+k+1-3b_{1}+2m\beta+2\beta+m)-4k-2\Big{)}\times\Big{(}-6a_{1}-3,6b_{1}+1\Big{)}.$

### Case 3: $n=(4m+3,4k)$

In this case, $3n=3(4m+3,4k)$ and we consider $\alpha_{1}=3$ and
$\alpha_{2}=(4m+3,4k)$. Using Lemma 2.5, we get

$a+2r=(2m+3,2k).$

Employing Lemma 3.1, we can find infinitely many elements in
$\mathbb{Z}[\sqrt{d}]$ of the form

$a=(6a_{1}+1,6b_{1})$

with $Nm(a)=1$, and thus the last equation gives

$r=(m+1-3a_{1},k-3b_{1}).$

Finally $ab+n=r^{2}$ gives

$b=\Big{\\{}(m+1-3a_{1})^{2}+d(k-3b_{1})^{2}-4m-3,2(m+1-3a_{1})(k-3b_{1})-4k\Big{\\}}\Big{(}6a_{1}+1,-6b_{1}\Big{)}.$

### Case 4: $n=(4m+3,4k+2)$

Let $\alpha_{1}$ and $\alpha_{2}$ be as in Case 3. Then by Lemma 2.5, we have
$a+2r=(2m+3,~{}2k+1).$ We choose

$a=(6a_{1}+3,~{}6b_{1}+1)$

in $\mathbb{Z}[\sqrt{d}]$ with $Nm(a)=-1$. Then

$r=(m-3a_{1},k-3b_{1}).$

Again the equation $ab+n=r^{2}$ gives

$b=\Big{\\{}(m-3a_{1})^{2}+d(k-3b_{1})^{2}-4m-3,~{}2(m-3a_{1})(k-3b_{1})-4k-2\Big{\\}}\Big{(}-6a_{1}-3,6b_{1}+1\Big{)}.$

### Case 5: $n=(4m+2,4k+2)$

Here, we have

$\displaystyle 3n$ $\displaystyle=3(4m+2,4k+2)$ $\displaystyle=6(2m+1,2k+1)$
$\displaystyle=(12M+4,-6N-1)(12M+4,6N+1)(2m+1,2k+1)\text{ for some
}M,N\in\mathbb{Z}.$ (4.2)

Now we choose $\alpha_{1}$, $\alpha_{2}$ as follows:

$\displaystyle\alpha_{1}$ $\displaystyle=(12M+4,-6N-1),$
$\displaystyle\alpha_{2}$ $\displaystyle=(12M+4,6N+1)(2m+1,2k+1)$
$\displaystyle=(24Mm+12M+8m+4+d(12Nk+6N+2k+1),24Mk+12M+8k+5$
$\displaystyle+12Nm+6N+2m).$

As before, utilizing Lemma 2.5, we obtain

$a+2r=(12Mm+12M+4m+4+\frac{d}{2}(12Nk+6N+2k+1),12Mk+6M+4k+2+6Nm+m).$ (4.3)

Again by Lemma 3.1, we can find infinitely many $a=(6a_{1}+1,6b_{1})$ in
$\mathbb{Z}[\sqrt{d}]$ with $Nm(a)=1$, and thus (4.3) gives

$\displaystyle r=$
$\displaystyle(6Mm+6M+2m+2+\frac{d}{2}(6Nk+3N+k)+\frac{1}{2}(\frac{d}{2}-1)-3a_{1},6Mk+3M+$
$\displaystyle 2k+1+3Nm+\frac{m}{2}-3b_{1}).$

For $r\in\mathbb{Z}[\sqrt{d}]$, $m$ should be even, and thus the equation
$ab+n=r^{2}$ entails

$\displaystyle b=$
$\displaystyle\Big{(}(6Mm+6M+2m+2+(d/2)(6Nk+3N+k)+(1/2)(d/2-1)-3a_{1})^{2}+d(6Mk+$
$\displaystyle 3M+2k+1+3Nm+(m/2)-3b_{1})^{2}-4m-2,2(6Mm+6M+2m+2+(d/2)(6Nk$
$\displaystyle+3N+k)+(1/2)(d/2-1)-3a_{1})(6Mk+3M+2k+1+3Nm+(m/2)-3b_{1})-4k$
$\displaystyle-2\Big{)}\times\Big{(}6a_{1}+1,-6b_{1}\Big{)}.$

On the other hand for odd $m$, we choose $a=(6a_{1}+3,6b_{1}+1)$ with
$Nm(a)=-1$. In this case (4.3) gives

$\displaystyle r=$
$\displaystyle\Big{(}6Mm+6M+2m+2+(d/2)(6Nk+3N+k)+(1/2)(d/2-3)-3a_{1},6Mk+3M$
$\displaystyle+2k+1+3Nm+(m/2)-3b_{1}-(1/2)\Big{)},$

and hence

$\displaystyle b=$
$\displaystyle\Big{(}(6Mm+6M+2m+2+(d/2)(6Nk+3N+k)+(1/2)(d/2-3)-3a_{1})^{2}+d(6Mk+$
$\displaystyle 3M+2k+1+3Nm+(m/2)-3b_{1}-(1/2))^{2}-4m-2,2(6Mm+6M+2m+2+(d/2)$
$\displaystyle(6Nk+3N+k)+(1/2)(d/2-3)-3a_{1})(6Mk+3M+2k+1+3Nm+(m/2)-3b_{1}-$
$\displaystyle(1/2))-4k-2\Big{)}\times\Big{(}-6a_{1}-3,6b_{1}+1\Big{)}.$

This completes the proof of Theorem 1.1.

##  5\. Counter-examples of Conjecture 1.1

We have $d\equiv 10\pmod{48}$ by Lemma 3.1. Thus utilizing Lemma 2.2 we see
that all the numbers of the form $(4m+2)+(4k+2)\sqrt{d}$ can be expressed as
the difference of two square in $\mathbb{Z}[\sqrt{d}]$ if and only if
$x^{2}-dy^{2}=\pm 2$ is solvable in integers. We now consider the equation
$x^{2}-dy^{2}=\pm 2$. Reducing this equation modulo $48$, we see that no
integers $x$ and $y$ satisfy it. To conclude, we state the following lemma.

######  Lemma 5.1.

Let $d\equiv 2\pmod{4}$ be square-free integer. Assume that the ring
$\mathbb{Z}[\sqrt{d}]$ contains elements of norm $-1$ and norm $6$. Then
$\mathbb{Z}[\sqrt{d}]$ does not contain any element of norm $\pm 2$.

Sum up these, one sees that for $d\equiv 10\pmod{48}$, there are integers $m$
and $k$ such that $(4m+2)+(4k+2)\sqrt{d}$ can not be expressed as the
difference of two square in $\mathbb{Z}[\sqrt{d}]$. Further we will prove in
fact that there exist infinitely many $n=(4m+2)+(4k+2)\sqrt{d}$ which are not
representable as a difference of two squares in $\mathbb{Z}[\sqrt{d}]$. Then
by employing Theorem 1.1 on these $n$’s, we get that a Diophantine quadruple
in $\mathbb{Z}[\sqrt{d}]$ with the property $D(n)$ exists but $n$ is not
expressible as the difference of two squares in $\mathbb{Z}[\sqrt{d}]$.

Assume that $n=(4m+2)+(4k+2)\sqrt{d}$. Dirichlet announced that among the
primes represented by the quadratic form $ax^{2}+2bxy+cy^{2}$, where
$\gcd(a,2b,c)=1$, infinitely many of them are representable by any given
linear form $Mx+N$, with $\gcd(M,N)=1$, provided $a,b,c,M,N$ are such that the
linear and quadratic forms can represent the same number. Meyer [24] gave a
complete proof of this result, whereas Mertens [23] gave an elementary proof
of the same. We conclude by this result that there exist infinitely many
primes $p$ of the form $p\equiv 3\pmod{4}$ such that $x^{2}-dy^{2}=p.$ As
$d\equiv 2\pmod{4}$, so that both $x$ and $y$ are odd. Thus, there exist
infinitely many primes $p$ of the form $p\equiv 3\pmod{4}$ such that
$(2m+1)^{2}-d(2k+1)^{2}=p.$ Therefore due to the existence of infinitely many
such primes, we get infinitely many pairs of integers $m$ and $k$. We claim
that for these $m$ and $k$, $n=(4m+2)+(4k+2)\sqrt{d}$ can not be written as
difference of two squares. To prove this claim, we follow [10, Proposition 6,
(ii)], which gives that $u^{2}-dv^{2}=\pm 2$ must be solvable in integers
$u,v$. But from Lemma 5.1, it is not possible. Hence there exist infinitely
many $n$ which can not be written as a difference of two squares in
$\mathbb{Z}[\sqrt{d}]$, however Theorem 1.1 confirms the existence of a
Diophantine quadruple in $\mathbb{Z}[\sqrt{d}]$ with the property $D(n)$.

Constructing a counter-example for Conjecture 1.1, we can take square-free
positive integer $d\equiv 2\pmod{4}$ such that $x^{2}-dy^{2}=6$ and
$x^{2}-dy^{2}=-1$ are solvable in integers $x$ and $y$. Let $d=48l+10$ with
$l\in\mathbb{N}\cup\\{0\\}$ (see Lemma 3.1). In the following table, we record
some positive integer $l$ for which (1.1) and (1.2) are solvable:

$l$ |  $d$ |  $(x,y)$ for (1.1) |  $(x,y)$ for (1.2)
---|---|---|---
0 |  10 |  (3,1) |  (4,1)
1 |  58 |  (99,13) |  (8,1)
2 |  106 |  (4005,389) |  (1184,115)
4 |  202 |  (3141,221) |  (668,47)
6 |  298 |  (409557,23725) |  (328,19)

We now determine $l$ for which $x^{2}-dy^{2}=6$ and $x^{2}-dy^{2}=-1$ are
solvable in integers $x$ and $y$. First we consider $x^{2}-(48l+10)y^{2}=6$.
Putting $y=1$ and $l=3l^{\prime 2}\pm 2l^{\prime}$ in this equation, we get
$x=(12l^{\prime}\pm 4)$, where $l^{\prime}\in\mathbb{Z}$. Hence for
$d=48(3l^{\prime 2}\pm 2l^{\prime})+10=2(24l^{\prime}(3l^{\prime}\pm 2)+5)$,
the equation $x^{2}-dy^{2}=6$ will be solvable. If $d=2p$ , where $p$ is prime
and $p\equiv 5\pmod{8}$, then $x^{2}-2py^{2}=-1$ is solvable in integers (see
[21]). Let $p(l^{\prime})=24l^{\prime}(3l^{\prime}\pm 2)+5$. Clearly,
$p(l^{\prime})\equiv 5\pmod{8}$. Since coefficient of $l^{\prime 2}$ is
positive, $p(l^{\prime})$ is irreducible over $\mathbb{Z}$ and gcd of $p(1)$
and $p(2)$ is one. To proceed further, we state a famous conjecture of
Bunyakovsky [5].

######  Conjecture 5.1 (Bunyakovsky conjecture).

Let f(x) be a polynomial of one variable with integer coefficient having the
following property.

  * (i)

The leading coefficient must be positive;

  * (ii)

Polynomial must be irreducible over the integers;

  * (iii)

The values $f(1),f(2),f(3),\cdots$ must have no common factor.

Then there are infinitely many primes in the sequence $f(1),f(2),f(3),\cdots.$

Applying this conjecture, we conclude that there exist infinitely many primes
of form $24l^{\prime}(3l^{\prime}\pm 2)+5$. Therefore, there are infinitely
many $d$’s for which Conjecture 1.1 fails.

###  5.1. Toy examples

Here we will discuss a couple of explicit counter-examples to Conjecture 1.1.

### Example 1

Assume that $d=10$ and $n=26+6\sqrt{10}$. Since $(13)^{2}-10(3)^{2}=79$ is a
prime number and $79\equiv 3\pmod{4}$, so that as discussed earlier, $n$ can
not be represented as a difference of two squares in $\mathbb{Z}[\sqrt{10}]$.
However, by Theorem 1.1 we can generate infinitely many quadruples in
$\mathbb{Z}[\sqrt{10}]$ with the property $D(n)$. One of them is
$\\{19+6\sqrt{10},-8+6\sqrt{10},35+18\sqrt{10},35+42\sqrt{10}\\}$ since

$(19+6\sqrt{10})(-8+6\sqrt{10})+n=(12+3\sqrt{10})^{2},$
$(19+6\sqrt{10})(35+18\sqrt{10})+n=(31+9\sqrt{10})^{2},$
$(19+6\sqrt{10})(35+42\sqrt{10})+n=(39+13\sqrt{10})^{2},$
$(-8+6\sqrt{10})(35+18\sqrt{10})+n=(4+9\sqrt{10})^{2},$
$(-8+6\sqrt{10})(35+42\sqrt{10})+n=(-4+15\sqrt{10})^{2},$
$(35+18\sqrt{10})(35+42\sqrt{10})+n=(39+27\sqrt{10})^{2}.$

### Example 2

Let $d=58$ and $n=18+2\sqrt{58}$. As $(9)^{2}-58(1)^{2}=23$ is a prime number
and $23\equiv 3\pmod{4}$, so that $n$ can not be represented as a difference
of two squares in $\mathbb{Z}[\sqrt{58}]$. On the other hand, utilizing the
case 5 of the proof of Theorem 1.1 we find a quadruple in
$\mathbb{Z}[\sqrt{58}]$ with the property $D(n)$, viz.
$\\{19603+2574\sqrt{58},543627-70094\sqrt{58},543616-70094\sqrt{58},2154883-282950\sqrt{58}\\}$
since

$(19603+2574\sqrt{58})(543627-70094\sqrt{58})+n=(9807+1287\sqrt{58})^{2},$
$(19603+2574\sqrt{58})(543616-70094\sqrt{58})+n=(9796+1287\sqrt{58})^{2},$
$(19603+2574\sqrt{58})(2154883-282950\sqrt{58})+n=(-3+\sqrt{58})^{2},$
$(543627-70094\sqrt{58})(543616-70094\sqrt{58})+n=(-533820+71381\sqrt{58})^{2},$
$(543627-70094\sqrt{58})(2154883-282950\sqrt{58})+n=(1077447-141475\sqrt{58})^{2},$
$(543616-70094\sqrt{58})(2154883-282950\sqrt{58})+n=(-1077436+141475\sqrt{58})^{2}.$

## Acknowledgements

The authors thank the anonymous referee for his/her valuable
remarks/suggestions that immensely improved the results as well as the
presentation of the paper. The third author would like to appreciate the
hospitality provided by Kerala School of Mathematics, Kozhikode, Kerala, where
the a part of the work was done. The third author acknowledges SERB MATRICS
grant (No. MTR/2021/00762), Govt. of India.

### Declaration

The authors declare that there are no conflict of interests.

### Data Availability Statement

This manuscript has no associate data.

## References

  * [1] F. S. Abu Muriefah and A. Al Rashed, Some Diophantine quadruples in the ring $\mathbb{Z}[\sqrt{-2}]$, Math. Commun. 9 (2004), 1–8.
  * [2] A. Baker and H. Davenport, The equations $3x^{2}-2=y^{2}$ and $8x^{2}-7=z^{2}$, Quart. J. Math. Oxford Ser. (2) 20 (1969), 129–137.
  * [3] N. C. Bonciocat, M. Cipu and M. Mignotte, There is no Diophantine $D(-1)$-quadruple, J. London Math. Soc. 105 (2022), 63-69.
  * [4] E. Brown, Sets in which $xy+k$ is always a square, Math. Comp. 45 (1985), 613–620.
  * [5] V. Buniakovsky, Sur les diviseurs numeriques invariables des fonctions rationnelles entieres, Mem Acad. Sci. St Petersburg 6 (1857), 305–329.
  * [6] K. Chakraborty, S. Gupta and A. Hoque, Diophantine triples with the property $D(n)$ for distinct $n$’s, Mediterr. J. Math. (to appear).
  * [7] A. Dujella, Generalization of a problem of Diophantus, Acta Arith. 65 (1993), 15–27.
  * [8] A. Dujella, The problem of Diophantus and Davenport for Gaussian integers, Glas. Mat. Ser. III 32 (1997), 1–10.
  * [9] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math. 566 (2004), 183–214.
  * [10] A. Dujella and Franušić, On differences of two squares in some quadratic fields, Rocky Mountain J. Math. 37 (2007), 429–453.
  * [11] A. Dujella, Number Theory, Školska knjiga, Zagreb, 2021.
  * [12] C. Elsholz, A. Filipin and Y. Fujita, On Diophantine quintuples and $D(-1)$-quadruples, Monats. Math. 175 (2014), 227–239.
  * [13] Z. Franušić, Diophantine quadruples in the ring $\mathbb{Z}[\sqrt{2}]$, Math. Commun. 9 (2004), 141–148.
  * [14] Z. Franušić, Diophantine quadruples in $\mathbb{Z}[\sqrt{4k+3}]$, Ramanujan J. 17 (2008), 77–88.
  * [15] Z. Franušić, A Diophantine problem in $\mathbb{Z}[\sqrt{(1+d)/2}]$, Studia Sci. Math. Hungar. 46 (2009), 103–112.
  * [16] Z. Franušić, Diophantine quadruples in the ring of integers of the pure cubic field $\mathbb{Q}(\sqrt[3]{2})$, Miskolc Math. Notes 14 (2013), 893–903.
  * [17] Z. Franušić and B. Jadrijević, $D(n)$-quadruples in the ring of integers of $\mathbb{Q}(\sqrt{2},\sqrt{3})$, Math. Slovaca 69 (2019), 1263–1278.
  * [18] Z. Franušić and I. Soldo, The problem of Diophantus for integers of $\mathbb{Q}(\sqrt{-3})$, Rad Hrvat. Akad. Znan. Umjet. Mat. Znan. 18 (2014), 15–25.
  * [19] B. He and A. Togbé, On the $D(-1)$-triple $\\{1,k^{2}+1,k^{2}+2k+2\\}$ and its unique $D(1)$-extension, J. Number Theory 131 (2011), 120–137.
  * [20] B. He, A. Togbé and V. Ziegler, There is no Diophantine quintuple, Trans. Amer. Math. Soc. 371 (2019), 6665–6709.
  * [21] F. Lemmermeyer, Higher descent on Pell conics I. From Legendre to Selmer, preprint. math.NT/0311309.
  * [22] Lj. Jukić Matić, Non-existence of certain Diophantine quadruples in rings of integers of pure cubic fields, Proc. Japan Acad. Ser. A Math. Sci. 88 (2012), no. 10, 163–167.
  * [23] F. Mertens, Üeber Dirichletsche Reihen, Ak. Wiss. Wien. (Math.) 104 (1895), 1093–1153.
  * [24] A. Meyer, Üeber einen Satz von Dirichlet, J. Reine Angew. Math. 103 (1888), 98–117.
  * [25] S. P. Mohanty and M. S. Ramamsamy, On $P_{r,k}$ sequences, Fibonacci Quart. 23 (1985), 36–44.
  * [26] I. Soldo, On the existence of Diophantine quadruples in $\mathbb{Z}[\sqrt{-2}]$, Miskolc Math. Notes 14 (2013), 265–277.