#  Generalized Fruit Diophantine equation and Hyperelliptic curves

Om Prakash and Kalyan Chakraborty  Kerala School of Mathematics, Kozhikode -
673571, Kerala, India<EMAIL_ADDRESS>Kerala School of Mathematics,
Kozhikode - 673571, Kerala, India<EMAIL_ADDRESS>
###### Abstract.

We show the insolvability of the Diophantine equation
$ax^{d}-y^{2}-z^{2}+xyz-b=0$ in $\mathbb{Z}$ for fixed $a$ and $b$ such that
$a\equiv 1\pmod{12}$ and $b=2^{d}a-3$, where $d$ is an odd integer and is a
multiple of $3$. Further, we investigate the more general family with
$b=2^{d}a-3^{r}$, where $r$ is a positive odd integer. As a consequence, we
found an infinite family of hyperelliptic curves with trivial torsion over
$\mathbb{Q}$. We conclude by providing some numerical evidence corroborating
the main results.

###### Key words and phrases:

Diophantine equation, Quadratic residue, Elliptic curves, Hyperelliptic
curves.

###### 2010 Mathematics Subject Classification:

Primary: 11D41, 11D72. Secondary: 11G30

##  1\. Introduction

One of the earliest topics in number theory is the study of Diophantine
equations. In the third century, Greek mathematician Diophantus of
‘Alexandria’ began this study. A polynomial equation of the form

$P(x_{1},x_{2},\cdots,x_{n})=0$

is known as a Diophantine equation. Finding all of its integer solutions, or
all of the $n-$tuples $\left(x_{1},x_{2},\cdots,x_{n}\right)\in\mathbb{Z}$
that satisfy the above equation, is of prime interest. The main task is to
investigate whether solutions exist for a given Diophantine equation. If they
do, it would be the aim to know how many are there and how to find all. There
are certain Diophantine equations which has no non zero integer solutions, for
example, Fermat’s equation $x^{n}+y^{n}=z^{n}$ for $n\geq 3$. The tenth of
Hilbert’s 23 problems, which he presented in 1900, dealt with Diophantine
equations. Hilbert asked, is there an algorithm to determine weather a given
Diophantine equation has a solution or not? and Matiyasevich in 1970 answered
it negatively.

We investigate a class of Diophantine equations of the form
$ax^{d}-y^{2}-z^{2}+xyz-b=0$ for fixed $a$ and $b$. Due to its emergence when
attempting to solve an equation involving fruits, this type of Diophantine
equations were given the name “Fruit Diophantine equation” by B. Sury and D.
Majumdar [5] and they proved the following:

######  Theorem 1.1.

[5] The equation

$y^{2}-xyz+z^{2}=x^{3}-5$

has no integer solution in $x$, $y$ and $z$.

Similar type of equations were previously studied by F. Luca and A. Togbé. In
particular, Luca and Togbé [4] studied the solution of the Diophantine
equation $x^{3}+by+1-xyz=0$ and later, Togbé [7] independently studied the
equation $x^{3}+by+4-xyz=0$.

As a consequence of Theorem 1.1 Majumdar and Sury proved the following:

######  Theorem 1.2.

[5] For any integer $m$, the elliptic curve

$E_{m}:y^{2}-mxy=x^{3}+m^{2}+5$

has no integral point.

L. Vaishya and R. Sharma expanded on Majumdar and Sury’s work in [8]. A class
of fruit Diophantine equations without an integer solution was found by them.
In particular Vaishya and Sharma showed,

######  Theorem 1.3.

[8] For fixed integers $a$ and $b$ with $a\equiv 1\pmod{12}$ and $b=8a-3$. The
Diophantine equation

$ax^{3}-y^{2}-z^{2}+xyz-b=0$

has no integer solution.

Using Nagell-Lutz theorem [6] and Theorem 1.3 they got hold of an infinite
family of elliptic curves with torsion-free Mordell-Weil group over
$\mathbb{Q}$.

######  Theorem 1.4.

[8] Let $a$ and $b$ be as in Theorem 1.3.

  * •

For any even integer $m$ the elliptic curve

$E_{m,a,b}^{e}:y^{2}=x^{3}+\frac{1}{4}m^{2}x^{2}-a^{2}\left(m^{2}+b\right)$

has torsion-free Mordell-Weil group.

  * •

For any odd integer $m$ the elliptic curve

$E_{m,a,b}^{o}:y^{2}=x^{3}+m^{2}x^{2}-64a^{2}\left(m^{2}+b\right)$

has torsion-free Mordell-Weil group.

We extend Vaishya and Sharma’s results [8] for higher exponents. We obtain a
family of hyperelliptic curves, by carrying out some appropriate
transformations. In 2013, D. Grant gave an analogue of Nagell-Lutz theorem for
hyperelliptic curves [3], using which we conclude that the Mordell-Weil group
of each member of the corresponding family of hyperelliptic curves is torsion-
free.

##  2\. Insolvability

Here we state and prove the main theorem and derive a couple of interesting
corollaries. We end this section by looking into a couple of examples.

######  Theorem 2.1.

The equation

$ax^{d}-y^{2}-z^{2}+xyz-b=0$

has no integer solutions for fixed $a$ and $b$ such that $a\equiv 1\pmod{12}$
and $b=2^{d}a-3$, where $d$ is an odd integer and divisible by $3$.

###### Proof.

Consider

$\displaystyle ax^{d}-y^{2}-z^{2}+xyz-b=0.$ (2.1)

If possible, let $(x,y,z)$ be an integer solution of (2.1). Let us fix
$x=\alpha$. Then (2.1) can be re-written as,

$\displaystyle y^{2}+z^{2}+b=a{\alpha}^{d}+\alpha yz.$ (2.2)

We consider the cases of $\alpha$ being even or odd separately.

######  Case 1.

If $\alpha$ is even. Then, we write (2.2) as:

$\displaystyle\left(y-\frac{\alpha
z}{2}\right)^{2}-\left(\frac{{\alpha}^{2}}{4}-1\right)z^{2}=a{\alpha}^{d}-b$
(2.3)

and set $Y=y-\frac{\alpha z}{2}$, $\beta=\frac{\alpha}{2}$ and $z=Z$. Thus
(2.3) becomes,

$\displaystyle
Y^{2}-\left(\beta^{2}-1\right)Z^{2}=a{\alpha}^{d}-b=2^{d}\beta^{d}a-b.$ (2.4)

  * •

If $\beta$ is even, say $\beta=2n$ for some integer $n$, then reducing (2.4)
modulo $4$ gives,

$\displaystyle Y^{2}+Z^{2}\equiv 3\pmod{4},$ (2.5)

which is not possible in $\mathbb{Z}/4\mathbb{Z}$.

  * •

If $\beta$ is odd, then $\beta=2n+1$ for some integer $n$. Reduction of (2.4)
modulo $4$ entails,

$\displaystyle Y^{2}\equiv 3\pmod{4}$ (2.6)

which is impossible.

######  Case 2.

If $\alpha$ is odd, say, $\alpha=2n+1$ for some integer $n$. Then,

$\displaystyle y^{2}+z^{2}+b$ $\displaystyle=$ $\displaystyle
a\alpha^{d}+\alpha yz$ $\displaystyle y^{2}+z^{2}+a2^{d}-3$ $\displaystyle=$
$\displaystyle a\left(2n+1\right)^{d}+\alpha yz$ $\displaystyle
y^{2}+z^{2}-\left(2n+1\right)yz$ $\displaystyle=$ $\displaystyle
a\left(2n+1\right)^{d}-a2^{d}+3.$

Now

$\displaystyle y^{2}+z^{2}+yz$ $\displaystyle\equiv$ $\displaystyle
a+3\pmod{2},$ $\displaystyle\Rightarrow y^{2}+z^{2}+yz$ $\displaystyle\equiv$
$\displaystyle 0\pmod{2}.$

Note that $y^{2}+z^{2}+yz\equiv a+3\pmod{2}$ has only solution $y\equiv
0\equiv z$ in $\mathbb{Z}/2\mathbb{Z}$, that is, $y$ and $z$ are even. Thus
(2.3) becomes

$a\alpha^{d}-b\equiv 0\pmod{4}.$

If we write $a=12l+1$ for some integer $l$, then,

$\displaystyle\alpha^{d}-\left(a2^{d}-3\right)$ $\displaystyle\equiv$
$\displaystyle 0\pmod{4},$ $\displaystyle\Rightarrow\alpha^{d}+3$
$\displaystyle\equiv$ $\displaystyle 0\pmod{4},$
$\displaystyle\Rightarrow\alpha^{d}$ $\displaystyle\equiv$ $\displaystyle
1\pmod{4},$ $\displaystyle\Rightarrow\alpha$ $\displaystyle\equiv$
$\displaystyle 1\pmod{4}.$

Let us consider

$\displaystyle\left(y-\frac{\alpha
z}{2}\right)^{2}-\left(\frac{{\alpha}^{2}}{4}-1\right)z^{2}$ $\displaystyle=$
$\displaystyle a{\alpha}^{d}-b,$
$\displaystyle\mbox{i.e.}~{}~{}\left(y-\frac{\alpha
z}{2}\right)^{2}-\left(\alpha^{2}-4\right)\left(\frac{z}{2}\right)^{2}$
$\displaystyle=$ $\displaystyle a{\alpha}^{d}-b.$

Further, we set $Y=y-\frac{\alpha z}{2}$ and $Z=\frac{z}{2}$. Then,

$\displaystyle Y^{2}-\left(\alpha^{2}-4\right)Z^{2}=a{\alpha}^{d}-b$ (2.7)

where $\alpha\equiv 1\pmod{4}$, $a\equiv 1\pmod{12}$ and $b=a2^{d}-3$. Three
sub cases need to be considered.

######  Sub-case 1.

If $\alpha\equiv 1\pmod{12}$, write $\alpha=12l+1$ for some integer $l$. Then,

$\displaystyle\alpha\equiv 1\pmod{3}$ $\displaystyle\Rightarrow\alpha+2\equiv
0\pmod{3}.$

Substituting $\alpha=12l+1$ in 2.7, we get

$\displaystyle Y^{2}-\left(\left(12l+1\right)^{2}-4\right)Z^{2}$
$\displaystyle=$ $\displaystyle a{\alpha}^{d}-b,$ $\displaystyle\Rightarrow
Y^{2}\equiv a\alpha^{d}-b\pmod{3},$ $\displaystyle\Rightarrow Y^{2}\equiv
a\left(12l+1\right)^{d}-a2^{d}+3\pmod{3},$ $\displaystyle\Rightarrow Y\equiv
1-2^{d}\pmod{3},$ $\displaystyle\Rightarrow Y^{2}\equiv 2\pmod{3}.$

A contradiction as $2$ is not square modulo $3$.

######  Sub-case 2.

If $\alpha\equiv 9\pmod{12}$. Then, there is a prime factor $p\equiv 5$ or
$7\pmod{12}$ of $(\alpha-2)$. Let $p\equiv 5$ or $7$ $\pmod{12}$ be a prime
factor of $(\alpha-2)$. Thus,

$\displaystyle Y^{2}\equiv a\alpha^{d}-b\pmod{p}.$

Let $\alpha=pl+2$ for some integer $l$. Then,

$\displaystyle Y^{2}$ $\displaystyle\equiv$ $\displaystyle
a\left(pl+2\right)^{d}-b\pmod{p},$ $\displaystyle\Rightarrow Y^{2}$
$\displaystyle\equiv$ $\displaystyle 3\pmod{p}.$

This leads to a contradiction as $3$ is not a quadratic residue modulo $p$.

######  Sub-case 3.

When $\alpha\equiv 5\pmod{12}$, we substitute $\alpha=3k+2$ for some integer
$k$ and get,

$\displaystyle Y^{2}-\left(\left(3l+2\right)^{2}-4\right)Z^{2}$
$\displaystyle=$
$\displaystyle\left(12l+1\right)\left(3k+2\right)-2^{d}\left(12l+1\right)+3,$
$\displaystyle\Rightarrow Y^{2}$ $\displaystyle\equiv$ $\displaystyle
2-2^{d}\equiv 0\pmod{3},$ $\displaystyle\Rightarrow Y$ $\displaystyle\equiv$
$\displaystyle 0\pmod{3}.$

Further, we substitute $Y=3m$ and $\alpha=12n+5$ for some integers $n$ and $m$
in 2.7 and arrive onto,

$\displaystyle 9m^{2}-\left(12n+3\right)\left(12n+7\right)Z^{2}$
$\displaystyle=$ $\displaystyle a\left(12n+5\right)^{d}-b,$
$\displaystyle\Rightarrow-\left(n+1\right)Z^{2}$ $\displaystyle\equiv$
$\displaystyle\sum_{i=0}^{d-1}\left(12n+5\right)^{d-1-i}2^{i}\pmod{3},$
$\displaystyle\Rightarrow-\left(n+1\right)Z^{2}$ $\displaystyle\equiv$
$\displaystyle 1\pmod{3},$ $\displaystyle\Rightarrow n$ $\displaystyle\equiv$
$\displaystyle 1\pmod{3}.$

Hence, $\alpha\equiv 17\pmod{36}$.

Note that $3$ divides $(\alpha-2)$. Thus there is a prime factor $p\equiv 5$
or $7\pmod{12}$ of $\frac{(\alpha-2)}{3}$, otherwise it would mean that
$\frac{\alpha-2}{3}$ is congruent to $\pm 1$, which is not the case.
Therefore,

$\displaystyle\alpha-2\equiv 0\pmod{p}.$

Thus,

$\displaystyle Y^{2}\equiv a\alpha^{d}-b\pmod{p}.$

Substituting $\alpha=pl+2$ for some integer $l$, we have

$\displaystyle Y^{2}\equiv 3\pmod{p},$

which contradicts the fact that $3$ is quadratic residue modulo $p$ if
$p\equiv\pm 1\pmod{12}$.

∎

######  Remark 1.

The result of Sury and Majumdar [5] follows by substituting $a=1$ and $d=3$ in
Theorem 2.1. The particular case $d=3$ in the same theorem deduces the results
of Vaishya and Sharma [8].

By increasing the exponents in the expression for $b$ to 3, we will now
examine the Diophantine equation with a little more generality. The potential
of a solution in this scenario is described by the following two corollaries,
along with a few examples.

######  Corollary 2.1.

The equation

$ax^{d}-y^{2}-z^{2}+xyz-b=0$

has no integer solution $(x,y,z)$ with $x$ even for fixed integers $a$ and $b$
such that $a\equiv 1\pmod{12}$ and $b=2^{d}a-3^{r}$ with positive odd integers
$r$ and $d$ as in Theorem 2.1.

###### Proof.

We follow exactly the same steps as in Case 1 of Theorem 2.1. Suppose there is
a solution with $x=\alpha$ even, then we write (2.2) as:

$\displaystyle\left(y-\frac{\alpha
z}{2}\right)^{2}-\left(\frac{{\alpha}^{2}}{4}-1\right)z^{2}=a{\alpha}^{d}-b.$
(2.8)

Let $Y=y-\frac{\alpha z}{2},\beta=\frac{\alpha}{2}$ and $z=Z$. Then (2.8) can
be written as,

$\displaystyle
Y^{2}-\left(\beta^{2}-1\right)Z^{2}=a{\alpha}^{d}-b=2^{d}\beta^{d}a-b.$ (2.9)

  * •

If $\beta$ is even, say $\beta=2n$ for some integer $n$, then the reduction
modulo $4$ of (2.9) will give,

$\displaystyle Y^{2}+Z^{2}\equiv 3^{r}\equiv 3\pmod{4},$ (2.10)

which is not feasible in $\mathbb{Z}/4\mathbb{Z}$.

  * •

If $\beta$ is odd, say $\beta=2n+1$ for some integer $n$. Then, the reduction
modulo $4$ of (2.9) provides,

$\displaystyle Y^{2}\equiv 3^{r}\equiv 3\pmod{4},$ (2.11)

which again is not possible.

∎

The following corollary deals with solutions having $x$, an odd integer:

######  Corollary 2.2.

The equation

$ax^{d}-y^{2}-z^{2}+xyz-b=0$

has no integer solution in $x$, $y$ and $z$ with $x\equiv 1$ or $9\pmod{12}$,
for fixed integers $a,b$ such that $a\equiv 1\pmod{12}$ and $b=2^{d}a-3^{r}$,
for $r$ and $d$ as in Corollary 2.1.

###### Proof.

Analogous steps as in Sub-case 2 and 3 of Theorem 2.1 will give the proof. ∎

######  Remark 2.

Corollary 2.2 says that, if there is a solution of
$ax^{d}-y^{2}-z^{2}+xyz-b=0$ with $a$ and $b$ as described in the Corollary
2.2, then $x$ must be $5$ modulo $12$.

We will see some examples.

######  Example 1.

For $a=25$, $d=3$ and $r=3$. The equation

$\displaystyle 25x^{3}-y^{2}-z^{2}+xyz-173=0$ (2.12)

has no integer solution.

Example 2 shows that the equation may not have solution even with $x\equiv
5\pmod{12}$. However, the next examples tell us the other possibility as well.

######  Example 2.

If $a=13$, $d=3$ and $r=3$, then

$\displaystyle 13x^{3}-y^{2}-z^{2}+xyz-77=0$ (2.13)

has an integer solution $(5,=18,-102)$.

######  Remark 3.

The condition that $r$ should be odd is rigid.

######  Example 3.

For $a=13$, $d=3$ and $r=2$, the equation

$\displaystyle 13x^{3}-y^{2}-z^{2}+xyz-95=0$ (2.14)

has an integer solution $(2,-10,-7)$.

##  3\. Hyperelliptic curves

A hyperelliptic curve $H$ over $\mathbb{Q}$ is a smooth projective curve
associated to an affine plane curve given by the equation
$y^{2}=f\left(x\right)$, where $f$ is a square-free polynomial of degree at
least $5$. If the degree of $f$ is $2g+1$ or $2g+2$, then the curve has genus
$g$. We write $H\left(\mathbb{Q}\right)$ for the set of $\mathbb{Q}$-points on
$H$. Determining rational points on hyperelliptic curve is one of the major
problems in mathematics. The following is the general result regarding the
size of $H\left(\mathbb{Q}\right)$, which was conjectured by Mordell and was
proved by Faltings:

######  Theorem 3.1.

[2] If $C$ is a smooth, projective and absolutely irreducible curve over
$\mathbb{Q}$ of genus at least $2$, then $C\left(\mathbb{Q}\right)$ is finite.

We may thus, at least theoretically, write down the finite set
$C\left(\mathbb{Q}\right)$. It is still a significant unresolved problem to
perform this practically for a given curve.

Given a hyperelliptic curve $H$, we can define the height (classical) function
to be the maximum of absolute values of the coefficients. The Northcott
property tells us that there are finitely many equations with bounded height.
Thus, one may talk about the density and averages. In this regard, Bhargava
[1] has proved that most of the hyperelliptic curve over $\mathbb{Q}$ has no
rational point. So, most of the times calculating $H\left(\mathbb{Q}\right)$
means proving $H\left(\mathbb{Q}\right)=\phi$.

In this section, we construct hyperelliptic curves corresponding to the
equation $ax^{d}-y^{2}-z^{2}+xyz-b=0$ with $a$ and $b$ as mentioned in Theorem
2.1. Then, we prove that $H\left(\mathbb{Q}\right)=\phi$ (corroborating
Bhargava [1]). The main ingredient to prove this is the following Nagell-Lutz
type theorem (Theorem 3, [3]) proved by D. Grant.

######  Theorem 3.2.

[3] Let $C$ be a nonsingular projective curve of genus $g\geq 1$ given by
$y^{2}=x^{2g+1}+b_{1}x^{2g}+\cdots+b_{2g}x+b_{2g+1}$, where
$b_{i}\in\mathbb{Z}$. Suppose

$\psi:C\left(\mathbb{Q}\right)\rightarrow J\left(\mathbb{Q}\right)$

be the Abel-Jacobi map, defined by $\psi\left(p\right)=\left[p-\infty\right]$,
where $J\left(\mathbb{Q}\right)$ is the Jacobian variety. If
$p=\left(x,y\right)\in C\left(\mathbb{Q}\right)\setminus\\{\infty\\}$ and
$\psi\left(p\right)\in J\left(\mathbb{Q}\right)_{\text{tors}}$, then,
$x,y\in\mathbb{Z}$ and either $y=0$ or $y^{2}$ divides discriminant of the
polynomial $x^{2g+1}+b_{1}x^{2g}+\cdots+b_{2g}x+b_{2g+1}$.

For fixed $m$ we define hyperelliptic curves,

$\displaystyle H_{m,a,b}:y^{2}-mxy=ax^{d}-m^{2}-b.$

  * •

Suppose $m$ is even. Then write (2.1) as:

$\displaystyle\left(y-\frac{mx}{2}\right)^{2}-\frac{m^{2}x^{2}}{4}=ax^{d}-m^{2}-b.$
(3.1)

Multiplying (3.1) by $a^{d-1}$ throughout, and using the fact that $d$ is odd
and divisible by $3$, we have,

$\displaystyle\left(\left(y-\frac{mx}{2}\right)a^{\frac{d-1}{2}}\right)^{2}-a^{d-1}\frac{m^{2}x^{2}}{4}=\left(ax\right)^{d}-m^{2}a^{d-1}-ba^{d-1}.$
(3.2)

We get the following hyperelliptic curve by substituting
$\left(\left(y-\frac{mx}{2}\right)a^{\frac{d-1}{2}}\right)=Y$ and $ax=X$,

$\displaystyle
H_{m,a,b}^{e}:Y^{2}-a^{d-3}\frac{m^{2}X^{2}}{4}=X^{d}-m^{2}a^{d-1}-ba^{d-1}.$
(3.3)

  * •

Now if $m$ is odd, multiply (3.2) by $4^{d}$ throughout to get

$\displaystyle\left(\left(y-\frac{mx}{2}\right)a^{\frac{d-1}{2}}2^{d}\right)^{2}-\left(4a\right)^{d-1}{m^{2}x^{2}}=\left(4ax\right)^{d}-m^{2}a^{d-1}4^{d}-ba^{d-1}4^{d}.$

Finally substitute
$\left(\left(y-\frac{mx}{2}\right)a^{\frac{d-1}{2}}2^{d}\right)=Y$ and
$4ax=X$, to get

$\displaystyle
H_{m,a,b}^{o}:Y^{2}-\left(4a\right)^{d-3}{m^{2}X^{2}}=X^{d}-m^{2}a^{d-1}4^{d}-ba^{d-1}4^{d}.$
(3.4)

Let,

$\displaystyle H_{m,a,b}=\begin{cases}H_{m,a,b}^{e}&{\rm~{}if~{}}m\hskip
2.84544pt\text{is even}\\\ H_{m,a,b}^{o}&{\rm~{}if~{}}m\hskip
2.84544pt\text{is odd},\end{cases}$ (3.5)

be the hyperelliptic curves.

######  Theorem 3.3.

Let $a$ and $b$ be as defined in Theorem 2.1. For any $m\in\mathbb{N}$, the
hyperelliptic curve $H_{m,a,b}$ has torsion-free Mordell-Weil group over
$\mathbb{Q}$.

###### Proof.

Let $a$ and $b$ be fixed positive integers with $a\equiv 1\pmod{12}$ and
$b=2^{d}a-3$.

  * •

For any even integer $m$, consider the hyperelliptic curve

$H_{m,a,b}^{e}:Y^{2}-a^{d-3}\frac{m^{2}X^{2}}{4}=X^{d}-m^{2}a^{d-1}-ba^{d-1}.$
(3.6)

By Theorem $3$ of [3], if (3.6) has an integer solution
$\left(X_{0},Y_{0}\right)$,
then
$\left(aX_{0},\left(\left(Y_{0}-\frac{mX_{0}}{2}\right)a^{\frac{d-1}{2}}\right),m\right)$
is a solution of (2.1). However, in Theorem 2.1 we have proved that it has no
integer solutions.

  * •

For an odd integer $m$, consider the hyperelliptic curve

$H_{m,a,b}^{o}:Y^{2}-\left(4a\right)^{d-3}{m^{2}X^{2}}=X^{d}-m^{2}a^{d-1}4^{d}-ba^{d-1}4^{d}.$
(3.7)

Suppose (3.7) has a solution $\left(X_{0},Y_{0}\right)$, then
$\left(4aX_{0},\left(\left(Y_{0}-\frac{mX_{0}}{2}\right)a^{\frac{d-1}{2}}2^{d}\right),m\right)$
is a solution of (2.1), which is a contradiction.

∎

##  4\. Numerical examples

In this section we give some numerical examples corroborating our results in
Corollary 2.2 and Remark 2.

$a$ | $d$ | $r$ | Equation | Solution
---|---|---|---|---
1 | 3 | 3 | $x^{3}-y^{2}-z^{2}+xyz+19=0$ | $\left(5,0,-12\right)$
1 | 3 | 5 | $x^{3}-y^{2}-z^{2}+xyz+235=0$ | $\left(29,12,-60\right)$
1 | 3 | 7 | $x^{3}-y^{2}-z^{2}+xyz+2179=0$ | $\left(5,0,-48\right)$
1 | 3 | 9 | $x^{3}-y^{2}-z^{2}+xyz+19675=0$ | $\left(-31,12,-30\right)$
13 | 3 | 3 | $13x^{3}-y^{2}-z^{2}+xyz-77=0$ | $\left(5,-18,-102\right)$
13 | 3 | 5 | $13x^{3}-y^{2}-z^{2}+xyz+139=0$ | $\left(5,0,-42\right)$
13 | 3 | 7 | $13x^{3}-y^{2}-z^{2}+xyz+2083=0$ | ?
25 | 3 | 3 | $25x^{3}-y^{2}-z^{2}+xyz-173=0$ | $\left(5,0,-42\right)$

## Acknowledgement

This work is done during the first author’s visit to Institute of Mathematical
Sciences (IMSc), Chennai, and he is grateful to the Institute for the
hospitality and the wonderful working ambience. Both the authors are grateful
to Kerala School of Mathematics(KSoM), Kozhikode, for it’s support and
wonderful ambience.

## References

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