# Diophantine problems related to cyclic cubic and quartic fields

Szabolcs Tangely and Maciej Ulas

###### Abstract.

We are interested in solving the congruences $f^{3}+g^{3}+1\equiv 0\pmod{fg}$
and $f^{4}-4g^{2}+4\equiv 0\pmod{fg}$ in polynomials $f,g$ with rational
coefficients. Moreover, we present results of computations of all integer
points on certain one parametric curves of genus 1 and 3, related to cubic and
quartic fields, respectively.

The research of the first author was supported in part by the NKFIH grants
115479, 128088 and 130909 and by the project EFOP-3.6.1-16-2016-00022, co-
financed by the European Union and the European Social Fund. The research of
the second author is supported by the grant of the National Science Centre
(NCN), Poland, no. UMO-2019/34/E/ST1/00094

##  1\. Introduction

A family of cubic fields studied by Cohn [4], Ennola [6] and Shanks [17] is
related to the parametric polynomial

$X^{3}+(n+3)X^{2}+nX-1=h_{n}(X).$

The family $(h_{n}(X)){n\in\mathbb{N}}$ has been further investigated by many
authors e.g. by Lecacheux [10], Lettl [11] and Washington [20]. Kishi [8] and
Washington [21] provided other one-parameter families with similar properties.
Later, Balady [1, 3] generalized Washington’s procedure [21] and described a
method for generating other one-parameter families. In this construction
polynomials $f$ and $g$ having integral coefficients such that the fraction

$\frac{f(t)^{3}+g(t)^{3}+1}{f(t)g(t)}=:\lambda(t)$

is also a polynomial with integral coefficients play an important role. More
precisely, if $\lambda\in\mathbb{Z}[t]$, then the pair of polynomials $(f,g)$
determines one-parameter family of polynomials

$P_{f,g}(X)=X^{3}+a(n)X^{2}+\lambda(n)X-1,$

where $a(n)=3(f(n)^{2}+g(n)^{2}-f(n)g(n))-\lambda(n)(f(n)+g(n))$. For all but
finitely many values of $n$, the polynomial $P_{f,g}$ has three real roots
which are units and more importantly the field generates by these roots is
cyclic cubic field.

These families turned out to be interesting from Diophantine number theory
point of view as well. Indeed, for any given $n$, the equation
$x^{3}+y^{3}+\lambda(n)xy+1=0$, where $n$ is an integer parameter, defines a
curve in cubic Hessian form and defines genus 1 curve (and thus can be seen as
a one possible model of an appropriate elliptic curve). Moreover, it satisfies
Runge type condition and thus, for any given $n$, one can offer an efficient
algorithm to finding all integral points without necessity of computing
generators of the Mordell-Weil group of an associate elliptic curve.

On the other side, the equation $Y^{3}h_{n}(X/Y)=\pm 1$ associated with the
simplest cubic fields of Shanks, defines a (parametric) Thue type equation and
it is natural to ask about characterization of its integral solutions. In this
direction one may consult the results [12, 13, 14, 19]. For any given $n$, one
can also consider the equation $h_{n}(X)=Y^{2}$ as the equation defining an
elliptic curve and mention [5], where questions concerning the existence of
integer/rational points, characterization of torsion groups and ranks are
investigated. It is clear that analogous question can be asked to the more
general family $(P_{f(n),g(n)})_{n\in\mathbb{N}}$.

Similarly, families of cyclic quartic fields with explicit units have been
studied by many authors (see e.g. [7, 9, 10]). Based on ideas given in the
paper [1] Balady and Washington [2] studied cyclic quartic fields. They
considered the polynomial

$X^{4}+(2s^{3}+Ls^{2}-4ps+2Lp)X^{3}-(3s^{2}+3Ls-6p)X^{2}+2LX+1,$

where

$L=-\frac{s^{4}-4p^{2}+4}{4sp}.$

If $L$ or $2L\in\mathbb{Z}$ then the roots of the above polynomial are units.
Under suitable specialization we get a quartic number field with cyclic Galois
group of order 4 and the roots generate the group of units or a subgroup of
index 5. We thus see that the necessary condition to get something interesting
is the investigation of the divisibility condition $4fg|(f^{4}-4g^{2}+4)$
which can be seen as a quartic analogue of the condition $fg|(f^{3}+g^{3}+1)$
considered in a cubic case.

The findings by Shanks, Balady and Balady and Washington motivated us to look
for solutions of the congruences

(1) $f^{3}+g^{3}+1\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mufg)$

and

(2) $f^{4}-4g^{2}+4\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mufg)$

where $f,g$ are polynomials with rational coefficients. Note that if $f,g$ are
treated as integers, then from the recent results of Schinzel [16] (which
extend earlier results of Mordell [15]) we know that each congruence above has
infinitely many solutions. Thus, the question concerning the existence of
polynomial solutions is quite natural in this direction too. It is clear that
finding polynomial solutions is more difficult question. Moreover, before we
start let us observe that if $f,g\in\mathbb{Z}[t]$ is a solution, then for
each $h\in\mathbb{Z}[t]$ the polynomials $f\circ h,g\circ h$ is also a
solution. Thus, our first attempt to remedy this situations is to looking
solutions of the following form

(3) $f(t)=\sum_{i=0}^{m-2}a_{i}t^{i}+a_{m}t^{m},\quad
g(t)=\sum_{i=0}^{n}b_{i}t^{i}.$

In other words, instead of asking for $f,g$ with integer coefficients we are
asking for $f,g$ with rational coefficients. Here, as usual,
$m=\operatorname{deg}f,n=\operatorname{deg}g$, i.e., $a_{m}b_{n}\neq 0$. In
order to deal with the case of substitutions $h$ of the form
$h(t)=pt,p\in\mathbb{Q}\setminus\\{0\\}$ or $h\in\mathbb{Q}[t]$ is of degree
$\geq 2$, we will be interested only in the non-trivial solutions. More
precisely, we say that a pair of polynomials $f,g\in\mathbb{Q}[t]$ of the form
given above, is a non-trivial solution of the congruence (1) (or the
congruence (2)) if and only if:

  1. (a)

there are no polynomials $f_{1},g_{1}\in\mathbb{Q}[t]$ and the polynomial
$h\in\mathbb{Q}[t]$ of degree $\geq 2$ such that $f=f_{1}\circ h,g=g_{1}\circ
h$;

  2. (b)

the leading coefficient $a_{m}$ of the polynomial $f$ is free of $m$-th power
of a rational integer. In other words, for each prime $p$, we have
$|\nu_{p}(a_{m})|<m$, where $\nu_{p}(q_{1}/q_{2})$ is the $p$-adic valuation
of the rational number $q_{1}/q_{2}$.

Let us describe the content of the paper in some details. In Section 2 and
Section 3 we investigate congruences (1), (2), respectively. In particular, we
characterize all polynomial solutions $(f,g)$ satisfying $\deg f\leq 2$. In
case of the congruence (2) we also characterize solutions with $\deg f=3,\deg
g=2$.

In Section 4 we investigate the existence and characterizations of all integer
points on the curves $H_{d}:\;x^{3}+y^{3}+dxy+1=0$ form computational point on
view. In fact, for any $d\in[-2\cdot 10^{6},2\cdot 10^{6}]$ we found all
integer points on $H_{d}$. A similar investigations are presented in the case
of genus three curves $X_{T}:\;(x+y)^{4}-4x^{2}y^{2}+Txy(x+y)+4=0$, where $T$
is even and $2\leq T\leq 15\cdot 10^{6}$ and $Q_{T}:\;x^{4}-4y^{2}+4Txy+4=0$
for $T\leq 5\cdot 10^{6}$.

##  2\. Cubic case

It is clear that the study of the existence of solutions of the congruence (1)
is equivalent with studying the existence of solutions of the system of
congruences

(4) $f^{3}+1\equiv 0\pmod{g},\quad g^{3}+1\equiv\pmod{f}.$

Moreover, due to symmetric nature of (1) without loss of generality we can
assume that $m\leq n$. As an immediate consequence of the equivalent
formulation we get that $n\leq 3m$. In other words, if we fix the degree of a
polynomial $f$ then the degree $n$ of $g$ is bounded by $3m$.

Let

$\bar{A}=(a_{0},a_{1},\ldots,a_{m-2},a_{m}),\quad\bar{B}=(b_{0},b_{1},\ldots,b_{n-1},b_{n})$

be the vectors of variables. We define the
$F_{i}=F_{i}(\bar{A},\bar{B}),i=0,\ldots n-1$ and
$G_{i}(\bar{A},\bar{B}),i=0,\ldots,m-1$ as the numerators of the coefficients
in the remainders of divisibility of $f^{3}+1\pmod{g}$ and $g^{3}+1\pmod{f}$
respectively. More precisely,

$\displaystyle(f(t)^{3}+1)\pmod{g(t)}$
$\displaystyle=\sum_{i=0}^{n-1}F_{i}(\bar{A},\bar{B})t^{i},$
$\displaystyle(g(t)^{3}+1)\pmod{f(t)}$
$\displaystyle=\sum_{i=0}^{m-1}G_{i}(\bar{A},\bar{B})t^{i},$

i.e., the above congruences are unique up to multiplication by appropriate
power of $b_{n}$ and $a_{n}$ respectively.

Our first aim is to attack the congruence (1) in a systematic way. More
precisely, we are interested in finding all non-trivial solutions $f,g$ of (1)
in the case when $\operatorname{deg}f\leq 2$. It is clear that in order to
find the pairs of polynomials satisfying the system (4) we need to study
rational solutions of the system

$S(m,n):\;\begin{cases}\begin{array}[]{cc}F_{i}(\bar{A},\bar{B})=0,&\quad
i=0,\ldots,n-1,\\\ G_{j}(\bar{A},\bar{B})=0,&\quad
j=0,\ldots,m-1.\end{array}\end{cases}$

Let us note that for given $m,n$, the system $S(m,n)$ is defined by $m+n$
equations in $m+n+1$ variables. We could expect, that in general $S(m,n)$
defines a curve. This is not the case due to the problem with trivial
solutions. More precisely, let $d$ be a divisor of $\gcd(m,n)$ and observe
that with any given solution $f,g$ of the system $S(m/d,n/d)$ we get the
solution $f\circ h,g\circ h$ of system $S(m,n)$, where $h$ is of degree $d$.
Thus in this case, in some sense we will get $d$ dimensional set of trivial
solutions. In particular, because of $d=1$ we will always have a one
dimensional set of trivial solutions. The problem is that the system “doesn’t
seem” the difference between trivial and non-trivial solutions. Thus, we
expect that essentially we have $m+n$ true variables (which is the case of
$\gcd(m,n)=1$). In other words, the set of non-trivial solutions in case of
fixed value of $(m,n)$ is finite. Thus, the strategy, at least in theory, is
clear. Compute the Gröbner basis of the ideal generated by the polynomials
defined the system $S(m,n)$ and extract for it all non-trivial solutions. It
is clear that the necessary and sufficient condition for solvability of
$S(m,n)$ is the vanishing of each polynomial in the Gröbner basis.
Unfortunately, the corresponding ideals are quite big and the computations
start to be difficult very fast. This is the reason, why we will deal with
small values of $m=\operatorname{deg}f$, i.e., with $m\leq 2$. One can think
that our aim is modest, but as we will see even in such relatively small cases
of $\deg f$, the corresponding systems of equations are quite big and
difficult to solve. Note that the same discussion applies also to the
congruence (2).

All Gröbner basis computations were performed with the Mathematica [22]. The
program was running on a laptop with 32 GB of RAM memory and i7 type
processor. We used standard lexicographic order in our computations, i.e.,

$b_{n}<b_{n-1}<\ldots b_{1}<b_{0}<a_{m}<a_{m-2}<\ldots<a_{1}<a_{0}.$

All the systems considered in this section are presented in the full form in
Appendix A.

###  2.1. The case $\deg f=1$

Without loss of generality we can write $f(t)=t$. Thus $g(t)$ is a divisor of

$f(t)^{3}+1=(t+1)(t^{2}-t+1).$

Thus, the polynomial $g$ takes one of the following form: $g(t)=b_{1}(t+1)$ or
$g(t)=b_{2}(t^{2}-t+1)$ or $g(t)=b_{3}(t^{3}+1)$. A quick inspection reveals
that $b_{1}=-1$ or $b_{2}=-1$ or $b_{3}=-1$. Thus, the only non-trivial
solutions of the congruence (1) in case of $\deg f=1$ are given by

(5) $\begin{array}[]{lll}f(t)=t,&g(t)=-t-1,&\lambda(t)=3,\\\
f(t)=t,&g(t)=-t^{2}+t-1,&\lambda(t)=t^{3}-2t^{2}+3t-3\\\
f(t)=t,&g(t)=-t^{3}-1,&\lambda(t)=t^{2}\left(t^{3}+2\right).\end{array}$

In particular, in case of the system $S(m,im)$ with $i=1,2,3$ we will always
have trivial solutions of the form $f\circ h,g\circ h$, where $h$ is a general
polynomial of degree $m$ and the pair $(f,g)$ is the one from (5).

In the following subsection we offer systematic analysis of the rational
solutions of the system $S(2,n)$ for $n\in\\{2,\ldots,6\\}$.

###  2.2. The case $\deg f=2,\deg g=2$

Here we investigate rational solutions of the polynomial system $S(2,2)$.

First we consider the case $b_{1}=0$. If $b_{1}=0$ then the polynomials
$F_{1},G_{1}$ vanish and we deal with the system

$S^{\prime}(2,2):\;P_{1}P_{2}=0,\quad Q_{1}Q_{2}=0,$

where

$\begin{array}[]{ll}P_{1}=a_{2}b_{0}-a_{0}b_{2}-b_{2},&P_{2}=(b_{0}^{2}+b_{2}^{2})a_{2}^{2}-2a_{0}a_{2}b_{2}b_{0}+a_{2}b_{2}b_{0}-a_{0}b_{2}^{2}+b_{2}^{2},\\\
Q_{1}=a_{2}b_{0}-a_{0}b_{2}+a_{2},&Q_{2}=(b_{0}^{2}+b_{2}^{2})a_{2}^{2}-a_{2}^{2}b_{0}+a_{0}a_{2}b_{2}-2a_{0}a_{2}b_{0}b_{2}+a_{2}^{2}.\end{array}$

In order to solve the system $S^{\prime}(2,2)$ it is enough to solve the four
systems $P_{i}=Q_{j}=0$ for $i,j\in\\{1,2\\}$.

If $P_{1}=Q_{1}=0$, then $a_{0}=-b_{0}-1,b_{2}=-a_{2}$. This leads to the
trivial solution which is re-parametrization of the first solution in (5) with
$t$ replaced by $a_{2}t^{2}-b_{0}-1$.

If $P_{1}=0,Q_{2}=0$, then from the first equation we get
$a_{0}=(a_{2}b_{0}-b_{2})/b_{2}$. Putting this into the second one we get the
equation $a_{2}^{2}-a_{2}b_{2}+b_{2}^{2}=0$ without non-zero solutions in
rationals. In exactly the same way we deal with the system $P_{2}=0,Q_{1}=0$.

Finally, if $P_{2}=0,Q_{2}=0$, then we observe that
$P_{2}-Q_{2}=(a_{2}+b_{2})(a_{2}b_{0}-a_{0}b_{2}-a_{2}+b_{2})$. If
$a_{2}=-b_{2}$ then we get the equation $(a_{0}+b_{0})^{2}-(a_{0}+b_{0})+1=0$
without solutions in rationals. If $a_{2}+b_{2}\neq 0$, then we need to have
$b_{0}=(a_{0}b_{2}+a_{2}-b_{2})/a_{2}$ and our system reduces to the equation
$a_{2}^{2}-a_{2}b_{2}+b_{2}^{2}=0$ without non-zero solutions in rationals.

Now let $b_{1}\neq 0$. In order to get the solutions of the system $S(2,2)$ we
compute the Gröbner basis, say $G(2,2)$, of the ideal generated by the set of
polynomials $\\{F_{0},F_{1}/b_{1},G_{0},G_{1}/b_{1}\\}$. The set $G(2,2)$
contains polynomials $H_{i},i=1,\ldots,25$. We have that

$\begin{array}[]{ll}H_{1}=(8a_{0}-3)(64a_{0}^{2}+24a_{0}+9)a_{2}^{5},&H_{2}=a_{2}^{5}(8b_{0}-7)(64b_{0}^{2}+56b_{0}+49),\\\
H_{3}=-a_{2}^{4}(8192a_{0}^{2}a_{2}b_{0}^{2}-441b_{1}^{2}),&H_{8}=-a_{2}^{2}(2a_{0}b_{1}^{4}-3a_{2}b_{2}).\end{array}$

Let us recall that $a_{2}\neq 0$. We thus solve the triangular (with respect
to the variables $a_{0},b_{0},a_{2},b_{2}$) system defined by the vanishing of
the polynomials from the set $H$ and get

$a_{0}=\frac{3}{8},\quad b_{0}=\frac{7}{8},\quad
a_{2}=\frac{b_{1}^{2}}{2},\quad b_{2}=\frac{b_{1}^{2}}{2}.$

In fact, the expressions for $a_{0},b_{0},a_{2},b_{2}$ given above solve also
the whole system $S(2,2)$. We thus get the solution
$f(t)=(4b_{1}^{2}t^{2}+3)/8,g(t)=(4b_{1}^{2}t^{2}+8b_{1}t+7)/8$, which after
the re-parametrization $t\rightarrow(2t-1)/2b_{1}$, takes the form

(6) $f(t)=\frac{1}{2}(t^{2}-t+1),\quad g(t)=\frac{1}{2}(t^{2}+t+1)=f(-t).$

We also have $\lambda(t)=t^{2}+5$.

We thus proved

######  Theorem 2.1.

The only non-trivial solution of the system (4) with
$\operatorname{deg}f=\operatorname{deg}g=2$ is the one given by (6).

Let us also note

######  Corollary 2.2.

The only (non-trivial) solution of the congruence $f^{3}+g^{3}+8\equiv
0\pmod{fg}$ with $\operatorname{deg}f=\operatorname{deg}g=2$ is of the form
$f(t)=t^{2}-t+1,g(t)=f(-t)$.

In the above the notion of a non-trivial solution of the congruence
$f^{3}+g^{3}+8\equiv 0\pmod{fg}$ is understood in exactly the same way as in
the case of the congruence $f^{3}+g^{3}+1\equiv 0\pmod{fg}$.

###  2.3. The case $\deg f=2,\deg g=3$

We are interested in solving the system (4) in polynomials
$f,g\in\mathbb{Q}[t]$, with $\operatorname{deg}f=2,\operatorname{deg}g=3$.
Equivalently, we are interested in the rational solutions of the system
$S(2,3)$.

We computed the Gröbner basis, say $G(2,3)$, of the set of polynomials
defining the system $S(2,3)$. The set $G(2,3)$ contains the polynomials
$H_{i},i=1,\ldots,451$. We have

$\displaystyle H_{1}=$ $\displaystyle
b_{2}b_{3}^{4}(8b_{2}^{3}+b_{3}^{2})(64b_{2}^{6}-280b_{3}^{2}b_{2}^{3}-243b_{3}^{4})(64b_{2}^{6}-8b_{3}^{2}b_{2}^{3}+b_{3}^{4})\times$
$\displaystyle(4096b_{2}^{12}+17920b_{3}^{2}b_{2}^{9}+93952b_{3}^{4}b_{2}^{6}-68040b_{3}^{6}b_{2}^{3}+59049b_{3}^{8}),$
$\displaystyle H_{2}=$ $\displaystyle
b_{3}^{4}(8b_{2}^{3}+b_{3}^{2})(64b_{2}^{6}-8b_{3}^{2}b_{2}^{3}+b_{3}^{4})(512b_{2}^{11}-96218b_{3}^{6}b_{2}^{2}+42939b_{1}b_{3}^{7}),$
$\displaystyle H_{5}=$ $\displaystyle
b_{3}^{4}(-23396352b_{2}^{13}+35946496b_{1}b_{3}b_{2}^{11}+43985895b_{3}^{6}b_{2}^{4}+$
$\displaystyle\quad-11830222b_{1}b_{3}^{7}b_{2}^{2}+1190043b_{1}^{2}b_{3}^{8}).$

We consider two cases: $b_{2}\neq 0$ or $b_{2}=0$.

Let $b_{2}\neq 0$. We are interested in rational solutions of
$H_{1}=H_{2}=H_{5}=0$ (together with the condition $b_{3}\neq 0$), from the
first two equations we get that $8b_{2}^{3}+b_{3}^{2}=0$. This is equivalent
with $b_{2}=-2u^{2},\quad b_{3}=8u^{3}$ for some $u\in\mathbb{Q}$. However,
substituting the computed values in the equation $H_{5}=0$, we left with the
equation $C_{0}u^{36}\left(-20b_{1}u+4b_{1}^{2}+37u^{2}\right)=0$, where
$C_{0}$ is a non-zero rationals. This equation has a rational solution only
for $u=0$. We get a contradiction because $b_{3}\neq 0$.

If $b_{2}=0$, from the equation $H_{2}=0$ we get that $b_{1}=0$. Next, using
the computed values we get that

$H_{53}=C_{1}(b_{0}+1)(b_{0}^{2}-b_{0}+1)b_{3}^{8},\quad
H_{120}=C_{2}b_{3}^{5}(b_{0}b_{3}^{2}-a_{2}^{3}),\quad
H_{261}=C_{3}a_{0}a_{2}b_{3}^{6},$

where $C_{1},C_{2},C_{3}$ are non-zero rationals. We get that

$b_{0}=-1,\quad b_{3}=u^{3},\quad a_{2}=-u^{2},\quad a_{0}=0,$

for some $u\in\mathbb{Q}$. The solution obtained in this way is
$f(t)=-t^{2}u^{2},g(t)=-1+t^{3}u^{3}$. After the reparametrization
$t\rightarrow t/u$, it takes the form

(7) $f(t)=-t^{2},\quad g(t)=t^{3}-1.$

We also have $\lambda(t)=-t(t^{3}-3)$.

We thus proved

######  Theorem 2.3.

The only non-trivial solution of the system (4) with
$\operatorname{deg}f=2,\operatorname{deg}g=3$ is the one given by (7).

###  2.4. The case $\deg f=2,\deg g=4$

In this section we look for polynomial solutions of the system $S(2,4)$. Let
us note that the set of rational solutions of the system $S(2,4)$ is the set
theoretic sum of the set of rational solutions of the systems $S_{1}(2,4)$ and
$S_{2}(2,4)$, where

$\displaystyle S_{1}(2,4):\;F_{0}=F_{1}=F_{2}=G_{0}=G_{1,1}=0,$ $\displaystyle
S_{2}(2,4):\;F_{0}=F_{1}=F_{2}=G_{0}=G_{1,2}=0,$

where $G_{1,1}=a_{2}b_{1}-a_{0}b_{3}$ and $G_{1,2}=G_{1}/G_{1,1}$.

Let $G_{i}(2,4)$ be the Gröbner basis of the set of polynomials defining the
systems $S_{i}(2,4)$ for $i=1,2$.

We consider the case $i=1$. Then $G_{1}(2,4)$ contains polynomials
$H_{i},i=1,\ldots,110$. We have $H_{1}=b_{3}b_{4}^{3},H_{2}=b_{1}b_{4}^{3}$
and

$\displaystyle H_{4}=$ $\displaystyle
b_{4}^{4}(b_{2}^{2}-4b_{0}b_{4}-3b_{4})(b_{2}^{4}-8b_{0}b_{4}b_{2}^{2}+3b_{4}b_{2}^{2}+16b_{0}^{2}b_{4}^{2}-12b_{0}b_{4}^{2}+9b_{4}^{2}),$
$\displaystyle H_{63}=$ $\displaystyle
b_{4}^{2}(a_{2}^{2}+b_{4})(-a_{2}^{2}b_{4}+a_{2}^{4}+b_{4}^{2}),$
$\displaystyle H_{72}=$ $\displaystyle
9(2a_{0}b_{4}-a_{2}b_{2})b_{4}^{4}-(b_{4}b_{2}^{4}-b_{3}^{2}b_{2}^{3}-8b_{0}b_{4}^{2}b_{2}^{2}+16b_{0}^{2}b_{4}^{3})a_{2}^{4}.$

The unique rational solution of the system $H_{1}=H_{2}=H_{4}=H_{63}=H_{72}$
with the respect to the variables $b_{0},b_{1},b_{2},b_{3},b_{4}$ is

$b_{0}=-a_{0}^{2}+a_{0}-1,\quad b_{1}=0,\quad b_{2}=(1-2a_{0})a_{2},\quad
b_{3}=0,\quad b_{4}=-a_{2}^{2}.$

However, with $b_{i},i=0,\ldots,4$ chosen in this way we get that
$g(t)=-(a_{2}t^{2}+a_{0})^{2}+(a_{2}t^{2}+a_{0})-1$, i.e., trivial solution
coming from the second solution in (5) with $t$ replaced by
$a_{2}t^{2}+a_{0}$.

We consider the case $i=2$ now. Then $G_{2}(2,4)$ contains polynomials
$H_{i},i=1,\ldots,479$. We have

$\displaystyle H_{1}=$ $\displaystyle
b_{4}^{3}\left(b_{3}^{4}-4b_{4}^{3}\right)\left(b_{3}^{8}+4b_{4}^{3}b_{3}^{4}+16b_{4}^{6}\right)$
, $\displaystyle H_{2}=$ $\displaystyle-
b_{4}^{4}\left(9b_{3}^{2}-8b_{2}b_{4}\right),$ $\displaystyle H_{11}=$
$\displaystyle-b_{4}^{3}\left(11b_{3}^{3}-16b_{1}b_{4}^{2}\right),$
$\displaystyle H_{28}=$ $\displaystyle-
b_{4}^{3}\left(77b_{3}^{4}-256b_{0}b_{4}^{3}\right),$ $\displaystyle H_{163}=$
$\displaystyle-b_{4}^{3}\left(8b_{2}^{2}b_{3}^{2}-81a_{2}^{3}b_{4}\right).$

From the equation $H_{1}=0$ we get that $b_{3}=4u^{3},b_{4}=4u^{4}$ for some
$u\in\mathbb{Q}\setminus\\{0\\}$. Then, one can easily solve the system
$H_{2}=H_{11}=H_{28}=H_{163}=0$ and we get

$b_{0}=\frac{77}{64},\quad b_{1}=\frac{11u}{4},\quad
b_{2}=\frac{9u^{2}}{2},\quad a_{2}=2u^{2}.$

With $b_{i}$ and $a_{2}$ chosen in this way the equation $H_{254}=0$ takes the
form $Cu^{20}(-3+8a_{0})=0$, where $C\in\mathbb{Q}\setminus\\{0\\}$, and thus
$a_{0}=3/8$. Finally, if we replace $t$ by $(2t-1)/4u$ we get a non-trivial
solution

(8) $f(t)=\frac{1}{2}(t^{2}-t+1),\quad
g(t)=\frac{1}{4}(t^{2}+t+1)(t^{2}-t+3).$

We also have
$\lambda(t)=\frac{1}{8}\left(t^{6}+t^{5}+6t^{4}+9t^{3}+18t^{2}+21t+33\right)$.

We thus proved the following

######  Theorem 2.4.

The only non-trivial solution of the system (4) with
$\operatorname{deg}f=2,\operatorname{deg}g=4$ is the one given by (8).

###  2.5. The case $\deg f=2,\deg g=5$

We deal with the system $S(2,5)$. First we consider the case $b_{4}=0$. Then
the equations $F_{0}=F_{1}=F_{3}=0$ implies that $a_{0}=-1,b_{0}=b_{2}=0$.
However, then the equation $G_{0}=0$ reduces to $a_{2}^{7}=0$ \- a
contradiction.

Let us suppose that $b_{4}\neq 0$. It is easy to see that the (sub)system
consisting the equations $F_{i}=0$ for $i=1,2,3,4$ is triangular with respect
to the variables $b_{0},b_{1},b_{2},b_{3}$. The required solution is given by

$\displaystyle b_{0}=$
$\displaystyle-\frac{(a_{0}^{3}+1)b_{5}^{2}}{a_{2}^{3}b_{4}},$ $\displaystyle
b_{1}=$ $\displaystyle-\frac{(a_{0}^{3}+1)b_{5}^{3}}{a_{2}^{3}b_{4}^{2}},$
$\displaystyle b_{2}=$
$\displaystyle-\frac{b_{5}^{2}\left(a_{0}^{3}b_{5}^{2}+3a_{2}a_{0}^{2}b_{4}^{2}+b_{5}^{2}\right)}{a_{2}^{3}b_{4}^{3}},$
$\displaystyle b_{3}=$
$\displaystyle-\frac{b_{5}^{3}\left(a_{0}^{3}b_{5}^{2}+3a_{2}a_{0}^{2}b_{4}^{2}+b_{5}^{2}\right)}{a_{2}^{3}b_{4}^{4}}$

With $b_{i},i=0,1,2,3$, we consider the remaining part of the system $S(2,5)$,
i.e., $F_{4}=G_{0}=G_{1}=G_{2}=0$. It is easy to compute the Gröbner basis,
say $G(2,5)$ of the corresponding ideal. The basis contains polynomials
$H_{i},i=1,\ldots,41$ and we have

$\displaystyle H_{1}=$
$\displaystyle(a_{0}^{3}+1)^{6}(4a_{0}+3)(16a_{0}^{2}-12a_{0}+9)a_{2}^{13}b_{5}^{16},$
$\displaystyle H_{5}=$
$\displaystyle(a_{0}^{3}+1)^{6}b_{5}^{16}(a_{2}^{5}+b_{5}^{2})(-a_{2}^{5}b_{5}^{2}+a_{2}^{10}+b_{5}^{4}),$
$\displaystyle H_{9}=$
$\displaystyle(a_{0}^{3}+1)^{6}a_{2}^{7}b_{5}^{16}(8a_{0}^{2}a_{2}^{7}+9b_{4}b_{5}^{2}).$

If $a_{0}=-1$ then the equation $H_{13}=0$ reduces to
$Ca_{2}^{16}b_{4}^{2}b_{5}^{12}=0$, where $C\in\mathbb{Q}\setminus\\{0\\}$.
Thus, $b_{4}=0$ and we back to the case we already considered on the
beginning.

If $a_{0}\neq-1$, then the unique rational solution of the system
$H_{1}=H_{5}=H_{9}=0$ is

$a_{0}=-\frac{3}{4},\quad a_{2}=u^{5},\quad b_{4}=-\frac{u^{31}}{2},\quad
b_{5}=u^{2}.$

Finally, to get the solution of our initial system $S(2,5)$ we need to take
$u=-1$. Tracing back our reasoning we get the obtained solution after the
change of variables $t\rightarrow(2t-1)/2$, leads to a non-trivial solution of
the system (4)

(9) $f(t)=-t^{2}+t-1,\quad g(t)=t(t^{4}-2t^{3}+4t^{2}-3t+3).$

We also have
$\lambda(t)=-t^{8}+3t^{7}-8t^{6}+11t^{5}-15t^{4}+10t^{3}-8t^{2}-1$.

We thus proved

######  Theorem 2.5.

The only non-trivial solution of the system (4) with
$\operatorname{deg}f=2,\operatorname{deg}g=5$ is the one given by (9).

###  2.6. The case $\deg f=2,\deg g=6$

This case is easy. Indeed, if $g\in\mathbb{Q}[t]$ is of degree 6, then the
condition $f(t)^{3}+1\equiv 0\mkern 4.0mu({\operator@font mod}\mkern
6.0mug(t))$ implies the equality $f(t)^{3}+1=cg(t)$ for some
$c\in\mathbb{Q}\setminus\\{0\\}$. Because $f(t)=a_{2}t^{2}+a_{0}$, thus
$b_{1}=b_{3}=b_{5}=0$ and comparing coefficients on both sides of the equality
$f(t)^{3}+1=cg(t)$ we get that

$b_{0}=\frac{a_{0}^{3}+1}{c},\quad b_{2}=\frac{3a_{0}^{2}a_{2}}{c},\quad
b_{4}=\frac{3a_{0}a_{2}^{2}}{c},\quad b_{6}=\frac{a_{2}^{3}}{c}.$

The condition $g(t)^{3}+1\equiv 0\mkern 4.0mu({\operator@font mod}\mkern
6.0muf(t))$ implies that $c=-1$. A quick computation reveals that
$g(t)=-f(t)^{3}-1$. However, this is trivial solution coming form the third
solution in (5) with $t$ replaced by $f(t)$.

###  2.7. Remark concerning co-prime solutions

Let us also note that each non-trivial solution $f,g$ of (1) can be extended
to an infinite family. Indeed, if the pair $(f,g)$ solves (1) and we define

$f_{1}=f(t),g_{1}=g(t),\quad f_{n}=g_{n-1},\quad
g_{n}=\frac{g_{n-1}^{3}+1}{f_{n-1}},$

then the pair of polynomials $(f_{n},g_{n})$ also solves (1). Moreover, it is
also clear that in some senses the most interesting solutions of the
congruence (1) are those satisfying the condition $(\deg f,\deg g)=1$. Indeed,
if additionally $3\nmid\deg f$ then the corresponding sequence of polynomials
generated by the algorithm above produces solutions with co-prime degrees. In
this direction one can ask the following

######  Question 2.6.

Does there exists infinitely many non-trivial solutions $(F,G)$ of (1) such
that there is no solution $(f,g)$ of (1) and that for each $n\geq 2$ we have
$(F,G)\neq(f_{n},g_{n})$? In other words, does the congruence (1) has
infinitely many disjoint orbits of non-trivial solutions?

##  3\. Quartic case

In this section we consider the system of congruences

(10) $f^{4}+4\equiv 0\pmod{g},\quad 4-4g^{2}\equiv 0\pmod{f}.$

Let $m=\deg f$ and $n=\deg g.$ We have that $m\leq 2n$ and $n\leq 4m.$ Let

$\bar{A}=(a_{0},a_{1},\ldots,a_{m-2},a_{m}),\quad\bar{B}=(b_{0},b_{1},\ldots,b_{n-1},b_{n})$

be the vectors of variables. We define the
$I_{i}=I_{i}(\bar{A},\bar{B}),i=0,\ldots n-1$ and
$J_{i}(\bar{A},\bar{B}),i=0,\ldots,m-1$ as the numerators of the coefficients
in the remainders of divisibility of $f^{4}+4\pmod{g}$ and $4-4g^{2}\pmod{f}$
respectively. More precisely,

$\displaystyle(f(t)^{4}+4)$ $\displaystyle\mkern 4.0mu({\operator@font
mod}\mkern 6.0mug(t))=\sum_{i=0}^{n-1}F_{i}(\bar{A},\bar{B})t^{i},$
$\displaystyle(4-4g(t)^{2})$ $\displaystyle\mkern 4.0mu({\operator@font
mod}\mkern 6.0muf(t))=\sum_{i=0}^{m-1}G_{i}(\bar{A},\bar{B})t^{i}.$

Let us also note that in the sequel we can assume that $a_{m}>0,b_{n}>0$,
indeed, if $(f,g)$ is a solution of (10) then $(\pm f,\pm g)$ is also a
solution.

To determine the pairs of polynomials satisfying the system (10) we need to
study rational solutions of the system

$R(m,n):\;\begin{cases}\begin{array}[]{cc}F_{i}(\bar{A},\bar{B})=0,&\quad
i=0,\ldots,n-1,\\\ F_{j}(\bar{A},\bar{B})=0,&\quad
j=0,\ldots,m-1.\end{array}\end{cases}$

As in the case of cubic case we perform case by case analysis for $\deg f\leq
2$. Moreover, at the end of the section we characterize solutions of (10)
under assumption $\deg f=3,\deg g=2$.

All remarks concerning symbolic computations mentioned in cubic case are in
order here too.

All the systems considered in this section are presented in the full form in
Appendix B.

###  3.1. The case $\deg f=1$

Without loss o generality we can assume that $f(t)=t$. As a consequence we see
that $g(t)$ is a divisor of

$t^{4}+4=(t^{2}-2t+2)(t^{2}+2t+2).$

Thus, the polynomial $g$ takes the form: $g(t)=b_{2}(t^{2}-2t+2)$ or
$g(t)=b_{2}(t^{2}+2t+2)$ or $g(t)=b_{4}(t^{4}+4)$. A quick inspection reveals
that $b_{2}=\pm\frac{1}{2}$ and $b_{4}=\pm\frac{1}{4}$. Thus, the only non-
trivial solutions are

(11) $\begin{array}[]{ll}f(t)=t,&g(t)=\frac{1}{2}(t^{2}+2t+2),\\\
f(t)=t,&g(t)=\frac{1}{4}(t^{2}+2t+2)(t^{2}-2t+2).\end{array}$

The corresponding values of $L$ are: $L(t)=-2$ and $L(t)=-t^{3}/4$. Note that
if we replace $t$ by $2t$ then the corresponding polynomials have integer
coefficients.

###  3.2. The case $\deg f=2,\deg g=1$

Because $\deg(4-4g(t)^{2})=2$ we do not need to consider the whole system. It
is enough to note that the part corresponding to the condition
$4-4g(t)^{2}\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0muf(t))$ takes
the form

$G_{0}=4\left(a_{2}b_{0}^{2}-a_{0}b_{1}^{2}-a_{2}\right),G_{1}=-8b_{0}b_{1}.$

Thus $b_{0}=0$ and the vanishing of $f(t)^{4}+4$ at $t=0$ implies
$a_{0}^{4}+4=0$ \- a contradiction.

###  3.3. The case $\deg f=2,\deg g=2$

We consider the system $R(2,2)$. Let $G(2,2)$ be the Gröbner basis of the
ideal generated by the polynomials defining the system $R(2,2)$. The set
$G(2,2)$ contains polynomials $H_{i},i=1,\ldots,33$. We have

$H_{1}=b_{1}b_{2}^{7}\left(2b_{1}^{4}+5b_{2}^{2}\right),\quad
H_{12}=b_{2}^{4}\left(125a_{2}^{4}+320b_{2}^{4}+512b_{0}b_{1}^{2}b_{2}^{3}\right).$

Vanishing of $H_{1}$ implies that $b_{1}=0$. Then the equation $H_{12}=0$ is
equivalent with $5b_{2}^{4}\left(25a_{2}^{4}+64b_{2}^{4}\right)=0$ and thus
$b_{2}=0$ \- a contradiction.

Summing up: there is no solutions of the system (2) with $\deg f=2,\deg g=2$.

###  3.4. The case $\deg f=2,\deg g=3$

We consider the system $R(2,3)$. Let $G(2,3)$ be the Gröbner basis of the
ideal generated by the polynomials defining the system $R(2,3)$. The set
$G(2,3)$ contains polynomials $H_{i},i=1,\ldots,630$. We have

$\displaystyle H_{1}=$ $\displaystyle
b_{2}b_{3}^{6}\left(256b_{2}^{6}-5b_{3}^{4}\right)\left(256b_{2}^{18}+15775b_{3}^{4}b_{2}^{12}+42400b_{3}^{8}b_{2}^{6}+32000b_{3}^{12}\right),$
$\displaystyle H_{23}=$ $\displaystyle
b_{3}^{6}\left(1376b_{2}b_{3}b_{1}^{4}-1312b_{2}^{3}b_{1}^{3}+9500b_{2}b_{3}^{2}b_{1}+3820b_{0}b_{3}^{3}-10885b_{2}^{3}b_{3}\right)$

The unique rational solution of the system $H_{1}=H_{23}=0$ with respect to
$b_{2},b_{0}$ is $b_{0}=b_{2}=0$. Then, with $b_{0},b_{2}$ chosen in this way
we get that the equation $H_{22}=0$ reduces to

$1343488b_{3}^{7}\left(b_{1}^{3}+10b_{3}\right)\left(2b_{1}^{6}+10b_{3}b_{1}^{3}+25b_{3}^{2}\right)=0$

and thus $b_{3}=-b_{1}^{3}/10$. Consequently, the equation $H_{201}=0$ reduces
to

$Cb_{1}^{21}\left(-10a_{2}b_{1}^{2}+25a_{2}^{2}+2b_{1}^{4}\right)\left(10a_{2}b_{1}^{2}+25a_{2}^{2}+2b_{1}^{4}\right)=0,$

where $C\in\mathbb{Q}\setminus\\{0\\}$. Thus $b_{1}=0$. However, under all
these equalities we get that the equation $H_{628}=0$ reduces to
$5a_{2}^{3}=0$ \- a contradiction.

Summing up: there is no solutions of the system (2) with $\deg f=2,\deg g=3$.

###  3.5. The case $\deg f=2,\deg g=4$

In this section we look for polynomial solutions of the system $R(2,4)$. Let
us note that the set of rational solutions of the system $R(2,4)$ is the set
theoretic sum of the set of rational solutions of the systems $R_{1}(2,4)$ and
$R_{2}(2,4)$, where

$\displaystyle S_{1}(2,4):\;F_{0}=F_{1}=F_{2}=G_{0}=G_{1,1}=0,$ $\displaystyle
S_{2}(2,4):\;F_{0}=F_{1}=F_{2}=G_{0}=G_{1,2}=0,$

where $G_{1,1}=a_{2}b_{1}-a_{0}b_{3}$ and $G_{1,2}=G_{1}/G_{1,1}$.

Let $G_{i}(2,4)$ be the Gröbner basis of the set of polynomials defining the
systems $S_{i}(2,4)$ for $i=1,2$.

We consider the case $i=1$. Then $G_{1}(2,4)$ contains polynomials
$H_{i},i=1,\ldots,588$. We have
$H_{1}=b_{3}b_{4}^{6},H_{12}=b_{1}^{7}b_{4}^{5}$ and

$\displaystyle H_{15}=$ $\displaystyle
b_{4}^{6}(b_{2}^{4}-8b_{0}b_{4}b_{2}^{2}+16b_{0}^{2}b_{4}^{2}-20b_{4}^{2})(b_{2}^{4}-8b_{0}b_{4}b_{2}^{2}+16b_{0}^{2}b_{4}^{2}-5b_{4}^{2}),$
$\displaystyle H_{395}=$ $\displaystyle-
b_{4}^{2}\left(25a_{2}^{8}b_{3}^{2}-25a_{2}^{8}b_{2}b_{4}+256b_{2}b_{4}^{5}\right).$

The unique rational solution of the system $H_{1}=H_{12}=H_{395}=0$ is
$b_{1}=b_{2}=b_{3}=0$. However, with $b_{1},b_{2},b_{3}$ chosen in this way,
the equation $H_{15}=0$ reduces to
$4(4b_{0}^{2}-5)(16b_{0}^{2}-5)b_{4}^{10}=0$ and thus $b_{4}=0$ \- a
contradiction.

Let us consider the case $i=2$. The set $G_{2}(2,4)$ contains polynomials
$H_{i},i=1,\ldots,1092$. We have

$H_{1}=b_{4}^{5}\left(b_{3}^{8}+540b_{4}^{6}\right)\left(b_{3}^{16}+25b_{4}^{12}\right)$

and hence $b_{4}=0$ \- a contradiction.

Hence, we do not get a new non-trivial solution of the system (2) with $\deg
f=2,\deg g=4$.

###  3.6. The case $\deg f=2,\deg g=5$

The condition $G_{1}=0$ implies that

(12)
$b_{0}=\frac{a_{0}a_{2}b_{2}-a_{0}^{2}b_{4}}{a_{2}^{2}}\quad\mbox{or}\quad
b_{1}=\frac{a_{0}a_{2}b_{3}-a_{0}^{2}b_{5}}{a_{2}^{2}}.$

After the substitution of the expression for $b_{0}$ and necessary
simplifications we get the system $R_{1}(2,5)$. Let $G_{1}(2,5)$ be the
Gröbner basis of the set of polynomials defining $R_{1}(2,5)$ computed with
respect to the standard order $b_{5}<b_{4}<b_{3}<b_{2}<b_{1}<a_{2}<a_{0}$. The
set $G_{1}(2,5)$ consists of polynomials $H_{i},i=1,\ldots,1011$. In
particular, we have

$\displaystyle H_{1}=$ $\displaystyle
b_{4}b_{5}^{4}\left(4096b_{4}^{30}+3130855b_{5}^{8}b_{4}^{20}-8716800b_{5}^{16}b_{4}^{10}+8192000b_{5}^{24}\right),$
$\displaystyle H_{34}=$ $\displaystyle
b_{5}^{4}\left(-22b_{2}b_{4}^{3}+33b_{3}^{2}b_{4}^{2}-64b_{2}b_{3}b_{5}b_{4}+50b_{2}^{2}b_{5}^{2}\right)$

The equation $H_{1}=0$ implies $b_{4}=0$ and then from $H_{34}=0$ we get that
$b_{2}=0$. With $b_{2},b_{4}$ chosen in this way we get the equation
$H_{23}=0$ reduces to

$6656b_{5}^{5}\left(b_{3}^{5}+40b_{5}^{3}\right)\left(b_{3}^{10}+30b_{5}^{3}b_{3}^{5}+156250b_{5}^{6}\right)=0,$

We thus get that $b_{5}=40^{3}u^{5}$ and $b_{3}=-40^{2}u^{3}$ for some
$u\in\mathbb{Q}$. However, substituting the computed values of
$b_{2},b_{3},b_{4},b_{5}$ into the equation $F_{0}=0$ we obtain that it
reduces to the equation $2^{36}\cdot
5^{12}\left(a_{0}^{2}-2a_{0}+2\right)\left(a_{0}^{2}+2a_{0}+2\right)u^{20}=0$.
Thus $u=0$ and this leads to $b_{5}=0$ \- a contradiction.

Next, we turn to the second possibility in (12). After the substitution of the
expression for $b_{1}$ from (12 and necessary simplifications we get the
system $R_{2}(2,5)$. Let $G_{2}(2,5)$ be the Gröbner basis of the set of
polynomials defining $R_{1}(2,5)$ computed with respect to the standard order
$b_{6}<b_{5}<b_{4}<b_{3}<b_{2}<b_{0}<a_{2}<a_{0}$. The set $G_{2}(2,5)$
consists of polynomials $H_{i},i=1,\ldots,705$. In particular, we have
$H_{1}=b_{5}^{4}(4096b_{4}^{10}-5b_{5}^{8})$. Hence the vanishing of $H_{1}$
implies that $b_{5}=0$ \- a contradiction.

Summing up: there is no solutions of the system (2) with $\deg f=2,\deg g=5$.

###  3.7. The case $\deg f=2,\deg g=6$

We consider the system $R(2,6)$ and are looking for its rational solutions.
The condition $G_{1}=0$ implies that

(13)
$b_{0}=\frac{a_{0}^{3}b_{6}-a_{2}a_{0}^{2}b_{4}+a_{2}^{2}a_{0}b_{2}}{a_{2}^{3}}\quad\mbox{or}\quad
b_{1}=\frac{a_{0}a_{2}b_{3}-a_{0}^{2}b_{5}}{a_{2}^{2}}.$

After the substitution of the expression for $b_{0}$ and necessary
simplifications we get the system $R_{1}(2,5)$. Let $G_{1}(2,6)$ be the
Gröbner basis of the set of polynomials defining $R_{1}(2,6)$ computed with
respect to the standard order
$b_{6}<b_{5}<b_{4}<b_{3}<b_{2}<b_{1}<a_{2}<a_{0}$. The set $G_{1}(2,6)$
consists of polynomials $H_{i},i=1,\ldots,628$. In particular, we have
$H_{1}=b_{6}^{3}(b_{5}^{12}+10b_{6}^{10})$. Thus $H_{1}=0$ implies that
$b_{6}=0$ \- a contradiction.

Next we turn to the second possibility in (13). After the substitution of the
expression for $b_{1}$ from (13 and necessary simplifications we get the
system $R_{2}(2,6)$. Let $G_{2}(2,6)$ be the Gröbner basis of the set of
polynomials defining $R_{2}(2,6)$ computed with respect to the standard order
$b_{6}<b_{5}<b_{4}<b_{3}<b_{2}<b_{0}<a_{2}<a_{0}$. The set $G_{2}(2,6)$
consists of polynomials $H_{i},i=1,\ldots,337$. In particular, we have

$\displaystyle H_{1}=$ $\displaystyle b_{5}b_{6}^{4},$ $\displaystyle H_{4}=$
$\displaystyle-b_{6}^{3}(b_{4}b_{5}-b_{3}b_{6}),$ $\displaystyle H_{9}=$
$\displaystyle-
b_{6}^{3}\left(b_{4}^{6}-9b_{2}b_{6}b_{4}^{4}+27b_{2}^{2}b_{6}^{2}b_{4}^{2}+10b_{6}^{4}-27b_{2}^{3}b_{6}^{3}\right),$
$\displaystyle H_{96}=$ $\displaystyle
b_{6}^{4}\left(5a_{2}^{4}-8b_{4}^{2}+24b_{2}b_{6}\right)$

The system $H_{1}=H_{4}=H_{96}=0$ implies that Solving the system
$H_{1}=H_{4}=H_{96}=0$ we get that

$b_{3}=b_{5}=0,\quad b_{2}=\frac{8b_{4}^{2}-5a_{2}^{4}}{24b_{6}}.$

However, with $b_{3},b_{5},b_{2}$ chosen in this way, the equation $H_{96}=0$
reduces to the form

$-\frac{5}{512}b_{6}^{3}\left(25a_{2}^{12}+1024b_{6}^{4}\right)=0.$

Hence $b_{6}=0$ \- a contradiction.

Summing up: there is no solutions of the system (2) with $\deg f=2,\deg g=6$.

###  3.8. The case $\deg f=2,\deg g=7$

This case is easy. Indeed, if $g\in\mathbb{Q}[t]$ is of degree 7, then the
condition $f(t)^{4}+4\equiv 0\mkern 4.0mu({\operator@font mod}\mkern
6.0mug(t))$ implies the equality $f(t)^{4}+4=g(t)L(t)$, where $\deg L=1$.
Thus, if $r\in\mathbb{Q}$ is a root of $L(t)=0$, then $f(r)$ is a rational
root of the polynomial $t^{4}+4$ \- a contradiction.

###  3.9. The case $\deg f=2,\deg g=8$

This case is equally easy. Indeed, if $g\in\mathbb{Q}[t]$ is of degree 8, then
the condition $f(t)^{4}+4\equiv 0\mkern 4.0mu({\operator@font mod}\mkern
6.0mug(t))$ implies the equality $f(t)^{4}+4=cg(t)$ for some
$c\in\mathbb{Q}\setminus\\{0\\}$. Because $f(t)=a_{2}^{2}+a_{0}$, thus
$b_{1}=b_{3}=b_{5}=b_{7}=0$ and comparing coefficients on both sides of the
equality $f(t)^{4}+4=cg(t)$ we get that

$b_{0}=\frac{a_{0}^{4}+4}{c},\quad b_{2}=\frac{4a_{0}^{3}a_{2}}{c},\quad
b_{4}=\frac{6a_{0}^{2}a_{2}^{2}}{c},\quad
b_{6}=\frac{4a_{0}a_{2}^{3}}{c},\quad b_{8}=\frac{a_{2}^{4}}{c}.$

The condition $4-4g(t)^{2}\equiv 0\mkern 4.0mu({\operator@font mod}\mkern
6.0muf(t))$ implies that $c=\pm 4$. Thus, we get that
$g(t)=\pm\frac{1}{2}(f(t)^{2}+f(t)+2)$. However, this is trivial solution
coming form the first solution in (11) with $t$ replaced by $f(t)$.

###  3.10. Some remarks concerning the case of $\deg f=3$

From our computations in previous subsections we find that there are no non-
trivial solutions of (2) in the case of $\deg f=2$. A question arises whether
there are any nontrivial solution with $\deg f>2$. Although the resulting
systems of polynomials are more and more complicated, this natural question
motivated us to look for new solutions. It is clear that the most natural
candidate is the next simplest one, i.e., $\deg f=3$. Unfortunately we were
unable to find all solutions of the congruence (2) in this case but we were
able to cover the case $\deg f=3,\deg g=2$. More precisely, we prove the
following

######  Theorem 3.1.

The only solution $f,g\in\mathbb{Q}[t]$ of the congruence (2) satisfying $\deg
f=3,\deg g=2$ is

$f(t)=4t(t^{2}-t+1),\quad g(t)=2t^{2}-2t+1.$

The corresponding value of $L$ is
$L(t)=8t^{7}-16t^{6}+28t^{5}-20t^{4}+14t^{3}+t+1$.

###### Proof.

The subsystem of the system $R(3,2)$ corresponding to the condition
$4-4g(t)^{2}\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0muf(t))$ has
the form

$\displaystyle G_{0}=$
$\displaystyle(a_{3}b_{0}^{2}-2a_{0}b_{1}b_{2}-a_{3}=0,$ $\displaystyle
G_{1}=$ $\displaystyle-a_{0}b_{2}^{2}-2a_{1}b_{1}b_{2}+2a_{3}b_{0}b_{1}=0,$
$\displaystyle G_{2}=$ $\displaystyle
a_{3}b_{1}^{2}-a_{1}b_{2}^{2}+2a_{3}b_{0}b_{2}=0.$

It can be easily solved with respect to $b_{0},a_{0},a_{1}$. The solutions are

(14) $b_{0}=\frac{-2b_{1}^{2}-b_{2}}{b_{2}},\quad
a_{0}=\frac{2a_{3}b_{1}\left(b_{1}^{2}+b_{2}\right)}{b_{2}^{3}},\quad
a_{1}=\frac{-3a_{3}b_{1}^{2}-2a_{3}b_{2}}{b_{2}^{2}}$

or

(15) $b_{0}=\frac{b_{2}-2b_{1}^{2}}{b_{2}},\quad
a_{0}=-\frac{2a_{3}b_{1}\left(b_{2}-b_{1}^{2}\right)}{b_{2}^{3}},\quad
a_{1}=\frac{2a_{3}b_{2}-3a_{3}b_{1}^{2}}{b_{2}^{2}}.$

Let us consider first (14). Substituting computed values of
$b_{0},a_{0},a_{1}$ into the equations coming from the condition
$f(t)^{4}+4\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mug(t))$ we get
the system

$F^{\prime}_{0}=a_{3}^{4}b_{2}^{2}+15a_{3}^{4}b_{1}^{2}b_{2}+27a_{3}^{4}b_{1}^{4}+4b_{2}^{8}=0,\quad
F^{\prime}_{1}=-3a_{3}^{4}b_{1}b_{2}^{4}(9b_{1}^{2}+2b_{2})=0.$

The only solutions of interest is

$b_{2}=-\frac{9b_{1}^{2}}{2},\quad a_{3}=\pm\frac{27b_{1}^{3}}{2}.$

Up to the sign and replacement of $t$ by $(6t-2)/9b_{1}$ we get exactly the
solution presented in the statement of our theorem.

Exactly the same solution is obtained in the case (15) and thus we omit the
simple details. ∎

We finish with the following

######  Question 3.2.

Does the congruence (2) has infinitely many non-trivial solutions in
polynomials $f,g\in\mathbb{Q}[t]$?

We expect that the answer is yes.

##  4\. Diophantine problems related to cyclic cubic and quartic fields

###  4.1. A different look on the congruence $f^{3}+g^{3}+1\equiv 0\mkern
4.0mu({\operator@font mod}\mkern 6.0mufg)$

In case of cubic case we deal with the congruence $f^{3}+g^{3}+1\equiv 0\mkern
4.0mu({\operator@font mod}\mkern 6.0mufg)$. As we mentioned in the
introduction, there are infinitely many solutions of this congruence (both in
integers and in polynomials with integer coefficients). However, our approach
was to find solutions without worrying about the corresponding value of
$(f^{3}+g^{3}+1)/fg$. This motivated us to concentrate on the value of $d$ and
asks about integer points on the curve

$H_{d}:\quad x^{3}+y^{3}+dxy+1=0.$

Let us note that for $d\neq-3$, the curve $H_{d}$ is a genus one curve and
thus has only finitely many integer solutions. Thus, for any given $d$, the
problem concerning characterization of the set $H_{d}(\mathbb{Z})$ is
interesting and non-trivial. Moreover, the curve $H_{d}$ is the so-called
Hessian form of an elliptic curve. One can prove that over some extension of
$\mathbb{Q}$, an elliptic curve with torsion point of order 3, is birationally
equivalent with $H_{d}$ for some (not necessarily rational) value of $d$.

######  Theorem 4.1.

If $(x,y)\in\mathbb{Z}^{2}$ is a solution of equation $H_{d}$ for some
$d\neq-3$ and $d_{1}$ is an integer dividing $d^{3}+27,$ then we have

$\displaystyle x$
$\displaystyle=\frac{3dd_{1}+3\,d_{1}^{2}\pm\sqrt{-27d^{2}d_{1}^{2}-18\,dd_{1}^{3}-3\,d_{1}^{4}-12\left(d^{3}+27\right)d_{1}}}{18d_{1}},$
$\displaystyle y$ $\displaystyle=\frac{d+d_{1}-3x}{3}.$

###### Proof.

We have that

$27(x^{3}+y^{3}+dxy+1)-(d^{2}+3\,dx+9\,x^{2}+3\,dy-9\,xy+9\,y^{2})(3x+3y-d)=d^{3}+27.$

Let $d_{1}$ be a divisor of $d^{3}+27.$ We obtain that

$\displaystyle 3x+3y-d$ $\displaystyle=$ $\displaystyle d_{1},$ $\displaystyle
d^{2}+3\,dx+9\,x^{2}+3\,dy-9\,xy+9\,y^{2}$ $\displaystyle=$
$\displaystyle\frac{-d^{3}-27}{d_{1}}.$

The first equation gives the formula for $y,$ that is
$y=\frac{1}{3}\,d+\frac{1}{3}\,d_{1}-x.$ Using this expression we eliminate
$y$ from the second equation and the solutions for $x$ are coming from a
simple quadratic equation. ∎

The proof of the theorem provides an algorithm to determine all the integral
solutions, we implemented the method in SageMath [18] and applied it for all
values in the range $[-2\cdot 10^{6},2\cdot 10^{6}]$ except $d=-3.$ The
complete list of integral solutions can be downloaded from
https://tengely.github.io/Hd.pdf. We note that for any $d$ we have the trivial
solutions $(-1,0),(0,-1).$ Also, if $d$ is a square, let say $d=s^{2}$ for
some integer $s,$ then $(\pm s,-1),(-1,\pm s)$ are solutions. Hence there are
many equations having at least 6 solutions (in the given range 333). There are
2755 cases in the range $[-2\cdot 10^{6},2\cdot 10^{6}]$ having non-trivial
solutions (solutions other than the above mentioned $(-1,0),(0,-1)$). There
are 1317 equations with 4 solutions, 1 equation with 5 solutions, 1413
equations with 6 solutions, 13 equations with 8 solutions, 7 equations with 10
solutions, 3 equations with 12 solutions and 1 equation with 14 solutions. The
table containing those $d$ in the considered range such that $X_{d}$ has at
least 10 solutions is presented below. The parallel SageMath code (using 3
cores of a Core I5 desktop machine) was running for 3 days and 8 hours.

$\begin{array}[]{|l|l|}\hline\cr\hline\cr d&solutions(x,y)\\\
\hline\cr\hline\cr-1056&\left(1063,-1406\right),\left(-1406,1063\right),\left(38,-201\right),\left(-201,38\right),\left(7,-86\right),\\\
&\left(-86,7\right),\left(4,-65\right),\left(-65,4\right),\left(0,-1\right),\left(-1,0\right)\\\
\hline\cr
25&\left(27,-19\right),\left(-19,27\right),\left(5,-1\right),\left(-1,5\right),\left(0,-1\right),\left(-1,0\right),\left(-1,-5\right),\\\
&\left(-5,-1\right),\left(-2,-7\right),\left(-7,-2\right),\left(-9,-13\right),\left(-13,-9\right)\\\
\hline\cr
49&\left(19,-7\right),\left(-7,19\right),\left(7,-1\right),\left(-1,7\right),\left(0,-1\right),\left(-1,0\right),\left(-1,-7\right),\\\
&\left(-7,-1\right),\left(-7,-18\right),\left(-18,-7\right)\\\ \hline\cr
225&\left(161,-93\right),\left(-93,161\right),\left(26,-3\right),\left(-3,26\right),\left(15,-1\right),\left(-1,15\right),\\\
&\left(0,-1\right),\left(-1,0\right),\left(-1,-15\right),\left(-15,-1\right),\left(-63,-109\right),\left(-109,-63\right)\\\
\hline\cr
1369&\left(1159,-733\right),\left(-733,1159\right),\left(37,-1\right),\left(-1,37\right),\left(0,-1\right),\left(-1,0\right),\\\
&\left(-1,-37\right),\left(-37,-1\right),\left(-373,-657\right),\left(-657,-373\right),\left(-437,-691\right),\\\
&\left(-691,-437\right)\\\ \hline\cr
1609&\left(1267,-772\right),\left(-772,1267\right),\left(468,-133\right),\left(-133,468\right),\left(0,-1\right),\left(-1,0\right),\\\
&\left(-11,-133\right),\left(-133,-11\right),\left(-133,-457\right),\left(-457,-133\right),\left(-331,-693\right),\\\
&\left(-693,-331\right),\left(-365,-721\right),\left(-721,-365\right)\\\
\hline\cr
9801&\left(7791,-4771\right),\left(-4771,7791\right),\left(99,-1\right),\left(-1,99\right),\left(0,-1\right),\left(-1,0\right),\\\
&\left(-1,-99\right),\left(-99,-1\right),\left(-2989,-4881\right),\left(-4881,-2989\right)\\\
\hline\cr
43089&\left(172333,-158006\right),\left(-158006,172333\right),\left(6697,-1037\right),\left(-1037,6697\right),\\\
&\left(6135,-871\right),\left(-871,6135\right),\left(0,-1\right),\left(-1,0\right),\left(-16226,-22583\right),\\\
&\left(-22583,-16226\right)\\\ \hline\cr
66049&\left(53793,-33361\right),\left(-33361,53793\right),\left(257,-1\right),\left(-1,257\right),\left(0,-1\right),\\\
&\left(-1,0\right),\left(-1,-257\right),\left(-257,-1\right),\left(-20511,-33073\right),\left(-33073,-20511\right)\\\
\hline\cr
212521&\left(27729,-3610\right),\left(-3610,27729\right),\left(461,-1\right),\left(-1,461\right),\left(0,-1\right),\\\
&\left(-1,0\right),\left(-1,-461\right),\left(-461,-1\right),\left(-59,-3541\right),\left(-3541,-59\right)\\\
\hline\cr
455625&\left(367667,-226929\right),\left(-226929,367667\right),\left(675,-1\right),\left(-1,675\right),\left(0,-1\right),\\\
&\left(-1,0\right),\left(-1,-675\right),\left(-675,-1\right),\left(-140529,-227683\right),\\\
&\left(-227683,-140529\right)\\\ \hline\cr\end{array}$

Table 1. Values of $d\in[-2\cdot 10^{6},2\cdot 10^{6}]$ such that $H_{d}$ has
at least 10 solutions together with the complete set of integer solutions.

###  4.2. A genus three curve of Balady and Washington

Let $T\in\mathbb{Z}$ and consider the curve

$X_{T}:\quad(x+y)^{4}-4x^{2}y^{2}+Txy(x+y)+4=0.$

Balady and Washington showed a clear connection between integer points on
$X_{4T}$ with $2T\in\mathbb{Z}$ and cyclic quartic fields generated by the
polynomial

(16) $X^{4}+(2s^{3}+Ls^{2}-4ps+2Lp)X^{3}-(3s^{2}+3Ls-6p)X^{2}+2LX+1.$

In fact, each integer point on $X_{4T}$ leads to suitable polynomial of the
above form with square discriminant. For most values of $T$, the curve $X_{T}$
represents genus 3 curve. Thus, famous Faltings theorem implies that the set
of rational (and thus integer) points on $X_{T}$ is finite. Balady and
Washington observed that $X_{-8}$ is a singular curve and were able to prove
the existence of infinitely many quartic number fields with cyclic Galois
group of order 4 and such that the roots of the defining polynomial (16)
generate the group of units or a subgroup of index 5. They also noted some
sporadic values of $T$ such that $X_{T}$ has non-trivial integer points, i.e.,
different then $(\pm 1,\mp 1)$. It is an interesting question whether there
are infinitely many values of $T\in\mathbb{Z}$ such that on $X_{T}$ we have
non-trivial integer points. Similarly, as in the case of the family $H_{d}$ we
investigated this question from a computational point of view. More precisely,
we have the following result.

######  Theorem 4.2.

Let $T$ be an even integer such that $T\neq\pm 8.$ If $(x,y)\in\mathbb{Z}^{2}$
is a solution of $X_{T}$ and $d_{1}$ is a divisor of

$\frac{(T^{2}+64)(T-8)(T+8)}{16},$

then we have that

$(4x+4y)^{2}=d_{1}+\frac{(T^{2}+64)(T-8)(T+8)}{16d_{1}}-\frac{T^{2}}{2}.$

###### Proof.

We have that

$\displaystyle\left(8x^{2}+32xy+8y^{2}-2Tx-2Ty+\frac{T^{2}}{4}\right)\left(8x^{2}+8y^{2}+2Tx+2Ty+\frac{T^{2}}{4}\right)=$
$\displaystyle=\frac{(T^{2}+64)(T-8)(T+8)}{16}.$

Since $T$ is even we obtain that

$\left(8x^{2}+32xy+8y^{2}-2Tx-2Ty+\frac{T^{2}}{4}\right)=d_{1}$

and

$\left(8x^{2}+8y^{2}+2Tx+2Ty+\frac{T^{2}}{4}\right)=\frac{(T^{2}+64)(T-8)(T+8)}{16d_{1}},$

where $d_{1}$ divides $\frac{(T^{2}+64)(T-8)(T+8)}{16}.$ The statement follows
by taking the sum of these equations. ∎

Based on the above theorem we have two strategies to determine the complete
set of integral solutions of $X_{T}.$ If we can factorize the integer
$\frac{(T^{2}+64)(T-8)(T+8)}{16},$ then we collect together the divisors
$d_{1}$ for which the number

$d_{1}+\frac{(T^{2}+64)(T-8)(T+8)}{16d_{1}}-\frac{T^{2}}{2}$

is a square divisible by 16. After that the problem can be reduced to
determine integral roots of univariate polynomials. If the factorization of
the integer $\frac{(T^{2}+64)(T-8)(T+8)}{16}$ is too expensive, then we may
compute the resultant of the polynomials

$(x+y)^{4}-4x^{2}y^{2}+Txy(x+y)+4$

and

$(4x+4y)^{2}-d_{1}-\frac{(T^{2}+64)(T-8)(T+8)}{16d_{1}}+\frac{T^{2}}{2}$

with respect to $y.$ Let us denote it by $R_{d_{1},T}(x).$ This is a degree 8
reducible polynomial that is a product of two degree 4 polynomials. Therefore
we can provide bound for $|x|$ as a function of $d_{1}$ and $T$ and finally we
note that we can bound $d_{1}$ as well.

######  Corollary 4.3.

Let $T\neq 8$ be an even integer for which $2\leq T\leq 15\cdot 10^{6}.$ The
Diophantine equation $X_{T}$ has at least 3 integral solutions in the
following cases

$T$ |  solutions $(x,y)$
---|---
40 | $(\pm 1,\mp 1),(-1,-5),(-5,-1)$
154 | $(\pm 1,\mp 1),(-5,37),(37,-5)$
77236 | $(\pm 1,\mp 1),(-5,629),(629,-5)$

.

In the study of cyclic quartic fields one needs to consider the Diophantine
equation $X_{4L},$ where $L$ or $2L$ is an integer. Balady and Washington [2]
considered the special cases $L=-2$ and $L=2$ and they also made a computer
search and provided a table containing 5 examples (with $L=\pm 2$ and
$L=-77/2$). The solutions in Corollary 4.3 yield the following polynomials

$x$ | $y$ | $L$ | Polynomial
---|---|---|---
$-1$ | $-5$ | 10 | $t^{4}+148t^{3}+102t^{2}+20t+1$
$-5$ | 37 | $77/2$ | $t^{4}+114395t^{3}-7878t^{2}+77t+1$
$-5$ | 629 | 19309 | $t^{4}+7890798742t^{3}-37333446t^{2}+38618t+1$

We note that the example with $L=77/2$ corresponds to the one with $L=-77/2$
in the table of Balady and Washington by $t\mapsto-t.$

Based on the polynomial solutions of the congruences, see the formulas at (11)
and in Theorem 3.1, we determined integral points on $X_{2t^{3}}$ such that
$1\leq t\leq 80000$ and on $X_{8t^{7}-16t^{6}+28t^{5}-20t^{4}+14t^{3}+t+1}$
with $1\leq t\leq 1001.$ There are only trivial solutions.

Our computations clearly show that the problem of finding values of $T\neq 8$
such that $X_{T}$ contains non-trivial integer points, is difficult. We thus
formulate the following explicit problem.

######  Problem 4.4.

Find more integers $T$ such that $X_{T}$ contains non-trivial integer points.

######  Remark 4.5.

Let us note that each integer point $(x,y)$ on $X_{T}$ for some $T$, leads to
the simultaneous congruences

(17) $x^{4}+4\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0muy),\quad
y^{4}+4\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mux),\quad
4-4x^{2}y^{2}\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mux+y).$

However, there are integer solutions of the system (17) which do not lead to
the solutions of the equation defining $X_{T}$ for any integer $T$, i.e., the
value

$T=T(x,y)=\frac{(x+y)^{4}-4x^{2}y^{2}+4}{xy(x+y)}$

is not an integer. We performed small numerical search for solutions of (17)
satisfying $x\leq y$ with $|x|\leq 10^{6}$ and which do not leading to integer
value of $T(x,y)$. Let us note that if $(x,y)$ is a solution and $x,y$ are
both positive/negative then $(y,x)$ is also a solution. Moreover, if $x<0,y>0$
and $(x,y)$ is a solution of (17) with corresponding value of $T$, then
$(-y,-x)$ is also a solution with corresponding value $-T$. Thus in our
computations we deal only with $y>|x|$. The results of our computations are
presented in the table below.

$\begin{array}[]{|l|l|l|l|}\hline\cr x&y&x+y&T(x,y)\\\
\hline\cr-20&26&6&\frac{1385}{4}\\\ -20&34&14&\frac{761}{4}\\\
-4&10&6&\frac{85}{4}\\\ -2&4&2&\frac{59}{4}\\\ 2&2&4&\frac{49}{4}\\\
2&4&6&\frac{87}{4}\\\ 2&10&12&\frac{319}{4}\\\ 4&26&30&\frac{983}{4}\\\
10&122&132&\frac{7393}{4}\\\ 148&4322&4470&\frac{556227}{4}\\\
\hline\cr\end{array}$

Table 2. Solutions $(x,y)$ of (17) satisfying the conditions:
$|x|<10^{7},|x|<y$ and $T(x,y)\not\in\mathbb{Z}$.

In the light of our computations we state the following

######  Question 4.6.

Does the system (17) has infinitely many solutions in integers $x,y$?

Let us observe that there is natural map (of degree 2)

$\varphi:\;X_{T}\ni(x,y)\mapsto(x+y,xy)=(X,Y)\in Q_{T},$

where $Q_{T}:\;X^{4}-4Y^{2}+TXY+4=0$. Thus, it is also natural to ask about
values of $T$ which lead to curves $Q_{T}$ with non-trivial integer points.
Similarly, as in the case of the curve $X_{T}$ one can propose an efficient
method which for a given value of $T$ allows to find all integer points on
$Q_{T}$. In fact, we can work with slightly more general form

$Q_{a,b}:\quad x^{4}-4y^{2}+axy+b=0.$

Note that $Q_{T,4}=Q_{T}$ and recall that $Q_{4a,4}$ were studied in [2].

In case of the equation $Q_{a,b}$ we have the following result.

######  Theorem 4.7.

Let $a,b$ be integers such that $a^{4}\neq 1024b.$ Let $g=\gcd(32,8a,a^{2}).$
If $(x,y)\in\mathbb{Z}^{2}$ is a solution of $Q_{a,b}$ and $d_{1}$ is a
divisor of

$\frac{a^{4}-1024b}{g^{2}},$

then we have that

$64x^{2}=d_{1}g+\frac{a^{4}-1024b}{d_{1}g}-2a^{2}.$

###### Proof.

We have that

$\frac{1}{g^{2}}\left(32x^{2}-8ax+64y+a^{2}\right)\left(32x^{2}+8ax-64y+a^{2}\right)=\frac{a^{4}-1024b}{g^{2}}.$

We obtain that

$\frac{1}{g}\left(32x^{2}-8ax+64y+a^{2}\right)=d_{1}$

and

$\frac{1}{g}\left(32x^{2}+8ax-64y+a^{2}\right)=\frac{a^{4}-1024b}{d_{1}g^{2}},$

where $d_{1}$ divides $\frac{a^{4}-1024b}{g^{2}}.$ The statement follows by
taking the sum of these equations.

∎

We determined all solutions of $Q_{4a,4}$ for $3\leq a\leq 5\cdot 10^{6}$ by
using SageMath, the complete list of integral solutions can be downloaded from
https://tengely.github.io/Qab.pdf. It is interesting to note that among the
552 cases that yield non-trivial (solutions other than $(0,\pm 1)$) there are
544 with 6 solutions and 8 with 10 solutions. In what follows we list the
latter 8 cases. We also considered the equation $Q_{2a,4}$ for odd values of
$a$ such that $3\leq a\leq 5\cdot 10^{6},$ since in case of the equation
$Q_{4a,4}$ those values are also interesting for which $2a$ is an integer. In
the given range there are 111 equations having non-trivial solutions. In all
these cases there are 6 integral solutions. The computation was performed on a
Core I5 desktop computer and we used 3 cores parallel in SageMath, it took
23.3 hours to complete. We provide the list of equations that have 10 integral
solutions in the table below.

$\begin{array}[]{|l|l|}\hline\cr U&\mbox{integral solutions
of}\;x^{4}-4y^{2}+4axy+4=0\\\ \hline\cr
250&(0,1),\left(156,42485\right),\left(-10,1\right),\left(-156,3485\right),\left(10,-1\right),\left(-10,-2501\right),\\\
&\left(0,-1\right),\left(10,2501\right),\left(156,-3485\right),\left(-156,-42485\right)\\\
\hline\cr
691&\left(0,1\right),\left(112,77897\right),\left(-112,505\right),\left(-24,-16589\right),\left(24,16589\right),\\\
&\left(112,-505\right),\left(-112,-77897\right),\left(-24,5\right),\left(0,-1\right),\left(24,-5\right)\\\
\hline\cr
6750&\left(-30,-202501\right),\left(0,1\right),\left(-2142,355265\right),\left(30,-1\right),\left(2142,14813765\right),\\\
&\left(-30,1\right),\left(-2142,-14813765\right),\left(30,202501\right),\left(2142,-355265\right),\left(0,-1\right)]\\\
\hline\cr
13718&\left(0,1\right),\left(-38,1\right),\left(38,-1\right),\left(7964,-8538133\right),\left(38,521285\right),\\\
&\left(7964,117788285\right),\left(-38,-521285\right),\left(0,-1\right),\left(-7964,-117788285\right),\\\
&\left(-7964,8538133\right)\\\ \hline\cr
19924&\left(0,1\right),\left(-4216,-84929585\right),\left(-904,9265\right),\\\
&\left(-4216,930001\right),\left(-904,-18020561\right),\left(904,18020561\right),\left(4216,-930001\right),\\\
&\left(904,-9265\right),\left(4216,84929585\right),\left(0,-1\right)\\\
\hline\cr
50510&\left(0,1\right),\left(-4556,-230590685\right),\left(4556,-467125\right),\left(-258,85\right),\\\
&\left(-4556,467125\right),\left(258,13031665\right),\left(4556,230590685\right),\\\
&\left(-258,-13031665\right),\left(258,-85\right),\left(0,-1\right)\\\
\hline\cr
236788&\left(0,1\right),\left(23212,5509495865\right),\left(-23212,-5509495865\right),\\\
&\left(-23212,13172809\right),\left(659512,-152988152785\right),\\\
&\left(-659512,152988152785\right),\left(23212,-13172809\right),\\\
&\left(659512,309152680241\right),\left(0,-1\right),\left(-659512,-309152680241\right)\\\
\hline\cr 715822&(-413444,-318861253093),(413444,-22908942125),(142,-1),\\\
&(-142,-101646725),(-413444,22908942125),(0,-1),(0,1),\\\
&(413444,318861253093),(-142,1),(142,101646725)\\\ \hline\cr\end{array}$

Table 3. Values of $a\in[0,5\cdot 10^{6}]$ such that $Q_{4a,a}$ has 10
integral solutions

######  Acknowledgement.

The authors are very grateful to Attila Pethő for bringing the problem to
their attention.

Appendix A

The polynomials defining the system $S(m,n)$ for $m=2,n\in\\{1,\ldots,6\\}$.

The case $(m,n)=(2,2)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{0}^{3}b_{2}^{5}-3a_{0}^{2}a_{2}b_{0}b_{2}^{4}+3a_{0}a_{2}^{2}b_{0}^{2}b_{2}^{3}-a_{2}^{3}b_{0}^{3}b_{2}^{2}-3a_{0}a_{2}^{2}b_{0}b_{1}^{2}b_{2}^{2}+3a_{2}^{3}b_{0}^{2}b_{1}^{2}b_{2}+$
$\displaystyle-a_{2}^{3}b_{0}b_{1}^{4}+b_{2}^{5},$ $\displaystyle F_{1}=$
$\displaystyle
a_{2}b_{1}(a_{2}^{2}b_{1}^{4}+3a_{0}a_{2}b_{2}^{2}b_{1}^{2}-4a_{2}^{2}b_{0}b_{2}b_{1}^{2}+3a_{0}^{2}b_{2}^{4}-6a_{0}a_{2}b_{0}b_{2}^{3}+3a_{2}^{2}b_{0}^{2}b_{2}^{2}),$
$\displaystyle G_{0}=$ $\displaystyle
a_{2}^{3}b_{0}^{3}-3a_{0}a_{2}^{2}b_{0}b_{1}^{2}-3a_{0}a_{2}^{2}b_{0}^{2}b_{2}+3a_{0}^{2}a_{2}b_{0}b_{2}^{2}+3a_{0}^{2}a_{2}b_{1}^{2}b_{2}-a_{0}^{3}b_{2}^{3}+a_{2}^{3},$
$\displaystyle G_{1}=$ $\displaystyle
b_{1}(3a_{2}^{2}b_{0}^{2}-6a_{0}a_{2}b_{2}b_{0}-a_{0}a_{2}b_{1}^{2}+3a_{0}^{2}b_{2}^{2}).$

The case $(m,n)=(2,3)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{0}^{3}b_{3}^{4}+a_{2}^{3}b_{0}^{2}b_{3}^{2}+3a_{0}a_{2}^{2}b_{0}b_{2}b_{3}^{2}-2a_{2}^{3}b_{0}b_{1}b_{2}b_{3}+a_{2}^{3}b_{0}b_{2}^{3}+b_{3}^{4},$
$\displaystyle F_{1}=$ $\displaystyle-
a_{2}^{2}(b_{1}b_{2}-b_{0}b_{3})(-a_{2}b_{2}^{2}-3a_{0}b_{3}^{2}+2a_{2}b_{1}b_{3}),$
$\displaystyle F_{2}=$ $\displaystyle
a_{2}^{2}b_{2}^{4}+3a_{0}a_{2}b_{3}^{2}b_{2}^{2}-3a_{2}^{2}b_{1}b_{3}b_{2}^{2}+2a_{2}^{2}b_{0}b_{3}^{2}b_{2}+3a_{0}^{2}b_{3}^{4}-3a_{0}a_{2}b_{1}b_{3}^{3}+a_{2}^{2}b_{1}^{2}b_{3}^{2},$
$\displaystyle G_{0}=$ $\displaystyle
3a_{0}^{4}b_{2}b_{3}^{2}-a_{2}a_{0}^{3}b_{2}^{3}-3a_{2}a_{0}^{3}b_{0}b_{3}^{2}-6a_{2}a_{0}^{3}b_{1}b_{2}b_{3}+3a_{2}^{2}a_{0}^{2}b_{0}b_{2}^{2}+3a_{2}^{2}a_{0}^{2}b_{1}^{2}b_{2}+$
$\displaystyle
6a_{2}^{2}a_{0}^{2}b_{0}b_{1}b_{3}-3a_{2}^{3}a_{0}b_{0}b_{1}^{2}-3a_{2}^{3}a_{0}b_{0}^{2}b_{2}+a_{2}^{4}b_{0}^{3}+a_{2}^{4},$
$\displaystyle G_{1}=$ $\displaystyle(a_{2}b_{1}-a_{0}b_{3})\times$
$\displaystyle(-a_{0}^{3}b_{3}^{2}+3a_{2}a_{0}^{2}b_{2}^{2}+2a_{2}a_{0}^{2}b_{1}b_{3}-a_{2}^{2}a_{0}b_{1}^{2}-6a_{2}^{2}a_{0}b_{0}b_{2}+3a_{2}^{3}b_{0}^{2}).$

The case $(m,n)=(2,4)$.

$\displaystyle F_{0}=$ $\displaystyle-
a_{2}^{3}b_{0}b_{3}^{2}+a_{2}^{3}b_{0}b_{2}b_{4}-3a_{0}a_{2}^{2}b_{0}b_{4}^{2}+a_{0}^{3}b_{4}^{3}+b_{4}^{3},$
$\displaystyle F_{1}=$ $\displaystyle
a_{2}b_{1}b_{3}^{2}-a_{2}b_{0}b_{4}b_{3}+3a_{0}b_{1}b_{4}^{2}-a_{2}b_{1}b_{2}b_{4},$
$\displaystyle F_{2}=$ $\displaystyle
3a_{0}^{2}b_{4}^{3}-a_{2}^{2}b_{0}b_{4}^{2}-3a_{0}a_{2}b_{2}b_{4}^{2}+a_{2}^{2}b_{2}^{2}b_{4}+a_{2}^{2}b_{1}b_{3}b_{4}-a_{2}^{2}b_{2}b_{3}^{2},$
$\displaystyle F_{3}=$ $\displaystyle
a_{2}b_{3}^{3}+3a_{0}b_{4}^{2}b_{3}-2a_{2}b_{2}b_{4}b_{3}+a_{2}b_{1}b_{4}^{2},$
$\displaystyle G_{0}=$ $\displaystyle
a_{0}^{6}b_{4}^{3}-3a_{2}a_{0}^{5}b_{2}b_{4}^{2}-3a_{2}a_{0}^{5}b_{3}^{2}b_{4}+3a_{2}^{2}a_{0}^{4}b_{2}b_{3}^{2}+3a_{2}^{2}a_{0}^{4}b_{0}b_{4}^{2}$
$\displaystyle+3a_{2}^{2}a_{0}^{4}b_{2}^{2}b_{4}+6a_{2}^{2}a_{0}^{4}b_{1}b_{3}b_{4}-a_{2}^{3}a_{0}^{3}b_{2}^{3}-3a_{2}^{3}a_{0}^{3}b_{0}b_{3}^{2}-6a_{2}^{3}a_{0}^{3}b_{1}b_{2}b_{3}+$
$\displaystyle-3a_{2}^{3}a_{0}^{3}b_{1}^{2}b_{4}-6a_{2}^{3}a_{0}^{3}b_{0}b_{2}b_{4}+3a_{2}^{4}a_{0}^{2}b_{0}b_{2}^{2}+3a_{2}^{4}a_{0}^{2}b_{1}^{2}b_{2}+6a_{2}^{4}a_{0}^{2}b_{0}b_{1}b_{3}$
$\displaystyle+3a_{2}^{4}a_{0}^{2}b_{0}^{2}b_{4}-3a_{2}^{5}a_{0}b_{0}b_{1}^{2}-3a_{2}^{5}a_{0}b_{0}^{2}b_{2}+a_{2}^{6}b_{0}^{3}+a_{2}^{6},$
$\displaystyle G_{1}=$
$\displaystyle(a_{2}b_{1}-a_{0}b_{3})(3a_{0}^{4}b_{4}^{2}-a_{2}a_{0}^{3}b_{3}^{2}-6a_{2}a_{0}^{3}b_{2}b_{4}+3a_{2}^{2}a_{0}^{2}b_{2}^{2}+2a_{2}^{2}a_{0}^{2}b_{1}b_{3}+$
$\displaystyle
6a_{2}^{2}a_{0}^{2}b_{0}b_{4}-a_{2}^{3}a_{0}b_{1}^{2}-6a_{2}^{3}a_{0}b_{0}b_{2}+3a_{2}^{4}b_{0}^{2}).$

The case $(m,n)=(2,5)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{0}^{3}b_{5}^{2}+a_{2}^{3}b_{0}b_{4}+b_{5}^{2},\quad
F_{1}=b_{1}b_{4}-b_{0}b_{5},\quad
F_{2}=a_{2}^{2}b_{2}b_{4}-a_{2}^{2}b_{1}b_{5}+3a_{0}^{2}b_{5}^{2},$
$\displaystyle F_{3}=$ $\displaystyle b_{3}b_{4}-b_{2}b_{5},\quad
F_{4}=a_{2}b_{4}^{2}+3a_{0}b_{5}^{2}-a_{2}b_{3}b_{5},$ $\displaystyle G_{0}=$
$\displaystyle-3a_{0}^{7}b_{4}b_{5}^{2}+a_{2}a_{0}^{6}b_{4}^{3}+3a_{2}a_{0}^{6}b_{2}b_{5}^{2}+6a_{2}a_{0}^{6}b_{3}b_{4}b_{5}-3a_{2}^{2}a_{0}^{5}b_{2}b_{4}^{2}+$
$\displaystyle-3a_{2}^{2}a_{0}^{5}b_{0}b_{5}^{2}-3a_{2}^{2}a_{0}^{5}b_{3}^{2}b_{4}-6a_{2}^{2}a_{0}^{5}b_{2}b_{3}b_{5}-6a_{2}^{2}a_{0}^{5}b_{1}b_{4}b_{5}+3a_{2}^{3}a_{0}^{4}b_{2}b_{3}^{2}$
$\displaystyle+3a_{2}^{3}a_{0}^{4}b_{0}b_{4}^{2}+3a_{2}^{3}a_{0}^{4}b_{2}^{2}b_{4}+6a_{2}^{3}a_{0}^{4}b_{1}b_{3}b_{4}+6a_{2}^{3}a_{0}^{4}b_{1}b_{2}b_{5}+6a_{2}^{3}a_{0}^{4}b_{0}b_{3}b_{5}+$
$\displaystyle-
a_{2}^{4}a_{0}^{3}b_{2}^{3}-3a_{2}^{4}a_{0}^{3}b_{0}b_{3}^{2}-6a_{2}^{4}a_{0}^{3}b_{1}b_{2}b_{3}-3a_{2}^{4}a_{0}^{3}b_{1}^{2}b_{4}-6a_{2}^{4}a_{0}^{3}b_{0}b_{2}b_{4}+$
$\displaystyle-6a_{2}^{4}a_{0}^{3}b_{0}b_{1}b_{5}+3a_{2}^{5}a_{0}^{2}b_{0}b_{2}^{2}+3a_{2}^{5}a_{0}^{2}b_{1}^{2}b_{2}+6a_{2}^{5}a_{0}^{2}b_{0}b_{1}b_{3}+3a_{2}^{5}a_{0}^{2}b_{0}^{2}b_{4}+$
$\displaystyle-3a_{2}^{6}a_{0}b_{0}b_{1}^{2}-3a_{2}^{6}a_{0}b_{0}^{2}b_{2}+a_{2}^{7}b_{0}^{3}+a_{2}^{7},$
$\displaystyle G_{1}=$
$\displaystyle(a_{0}^{2}b_{5}-a_{2}a_{0}b_{3}+a_{2}^{2}b_{1})(-a_{0}^{5}b_{5}^{2}+3a_{2}a_{0}^{4}b_{4}^{2}+2a_{2}a_{0}^{4}b_{3}b_{5}-a_{2}^{2}a_{0}^{3}b_{3}^{2}+$
$\displaystyle\quad-6a_{2}^{2}a_{0}^{3}b_{2}b_{4}-2a_{2}^{2}a_{0}^{3}b_{1}b_{5}+3a_{2}^{3}a_{0}^{2}b_{2}^{2}+2a_{2}^{3}a_{0}^{2}b_{1}b_{3}+6a_{2}^{3}a_{0}^{2}b_{0}b_{4}+$
$\displaystyle\quad-
a_{2}^{4}a_{0}b_{1}^{2}-6a_{2}^{4}a_{0}b_{0}b_{2}+3a_{2}^{5}b_{0}^{2}).$

Appendix B

The polynomials defining the system $R(m,n)$ for $m=2,n\in\\{2,\ldots,9\\}$
and $(m,n)=(3,2)$.

The case $(m,n)=(2,2)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{0}^{4}b_{2}^{7}-4a_{0}^{3}a_{2}b_{0}b_{2}^{6}+6a_{0}^{2}a_{2}^{2}b_{0}^{2}b_{2}^{5}-4a_{0}a_{2}^{3}b_{0}^{3}b_{2}^{4}-6a_{0}^{2}a_{2}^{2}b_{0}b_{1}^{2}b_{2}^{4}+a_{2}^{4}b_{0}^{4}b_{2}^{3}$
$\displaystyle+12a_{0}a_{2}^{3}b_{0}^{2}b_{1}^{2}b_{2}^{3}-4a_{0}a_{2}^{3}b_{0}b_{1}^{4}b_{2}^{2}-6a_{2}^{4}b_{0}^{3}b_{1}^{2}b_{2}^{2}+5a_{2}^{4}b_{0}^{2}b_{1}^{4}b_{2}-a_{2}^{4}b_{0}b_{1}^{6}+4b_{2}^{7},$
$\displaystyle F_{1}=$ $\displaystyle
b_{1}(a_{2}b_{1}^{2}+2a_{0}b_{2}^{2}-2a_{2}b_{0}b_{2})\times$
$\displaystyle(a_{2}^{2}b_{1}^{4}+2a_{0}a_{2}b_{2}^{2}b_{1}^{2}-4a_{2}^{2}b_{0}b_{2}b_{1}^{2}+2a_{0}^{2}b_{2}^{4}-4a_{0}a_{2}b_{0}b_{2}^{3}+2a_{2}^{2}b_{0}^{2}b_{2}^{2}),$
$\displaystyle G_{0}=$
$\displaystyle-4a_{2}^{2}b_{0}^{2}+4a_{0}a_{2}b_{1}^{2}+8a_{0}a_{2}b_{0}b_{2}-4a_{0}^{2}b_{2}^{2}+5a_{2}^{2},$
$\displaystyle G_{1}=$ $\displaystyle b_{1}\left(a_{2}b_{0}-a_{0}b_{2}\right)$

The case $(m,n)=(2,3)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{0}^{4}b_{3}^{6}+4a_{0}a_{2}^{3}b_{0}^{2}b_{3}^{4}+6a_{0}^{2}a_{2}^{2}b_{0}b_{2}b_{3}^{4}-2a_{2}^{4}b_{0}^{2}b_{1}b_{3}^{3}-8a_{0}a_{2}^{3}b_{0}b_{1}b_{2}b_{3}^{3}$
$\displaystyle+4a_{0}a_{2}^{3}b_{0}b_{2}^{3}b_{3}^{2}+3a_{2}^{4}b_{0}^{2}b_{2}^{2}b_{3}^{2}+3a_{2}^{4}b_{0}b_{1}^{2}b_{2}b_{3}^{2}-4a_{2}^{4}b_{0}b_{1}b_{2}^{3}b_{3}+a_{2}^{4}b_{0}b_{2}^{5}+4b_{3}^{6},$
$\displaystyle F_{1}=$ $\displaystyle(b_{1}b_{2}-b_{0}b_{3})\times$
$\displaystyle\;(a_{2}^{2}b_{2}^{4}+4a_{0}a_{2}b_{3}^{2}b_{2}^{2}-4a_{2}^{2}b_{1}b_{3}b_{2}^{2}+2a_{2}^{2}b_{0}b_{3}^{2}b_{2}+6a_{0}^{2}b_{3}^{4}-8a_{0}a_{2}b_{1}b_{3}^{3}+3a_{2}^{2}b_{1}^{2}b_{3}^{2}),$
$\displaystyle F_{2}=$ $\displaystyle
a_{2}^{3}b_{2}^{6}+4a_{0}a_{2}^{2}b_{3}^{2}b_{2}^{4}-5a_{2}^{3}b_{1}b_{3}b_{2}^{4}+4a_{2}^{3}b_{0}b_{3}^{2}b_{2}^{3}+6a_{0}^{2}a_{2}b_{3}^{4}b_{2}^{2}-12a_{0}a_{2}^{2}b_{1}b_{3}^{3}b_{2}^{2}$
$\displaystyle+6a_{2}^{3}b_{1}^{2}b_{3}^{2}b_{2}^{2}+8a_{0}a_{2}^{2}b_{0}b_{3}^{4}b_{2}-6a_{2}^{3}b_{0}b_{1}b_{3}^{3}b_{2}+4a_{0}^{3}b_{3}^{6}-6a_{0}^{2}a_{2}b_{1}b_{3}^{5}+a_{2}^{3}b_{0}^{2}b_{3}^{4}$
$\displaystyle+4a_{0}a_{2}^{2}b_{1}^{2}b_{3}^{4}-a_{2}^{3}b_{1}^{3}b_{3}^{3},$
$\displaystyle G_{0}=$ $\displaystyle
4a_{0}^{3}b_{3}^{2}-4a_{2}a_{0}^{2}b_{2}^{2}-8a_{2}a_{0}^{2}b_{1}b_{3}+4a_{2}^{2}a_{0}b_{1}^{2}+8a_{2}^{2}a_{0}b_{0}b_{2}-4a_{2}^{3}b_{0}^{2}+5a_{2}^{3},$
$\displaystyle G_{1}=$
$\displaystyle(a_{2}b_{0}-a_{0}b_{2})(a_{2}b_{1}-a_{0}b_{3}).$

The case $(m,n)=(2,4)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{0}^{4}b_{4}^{5}-6a_{0}^{2}a_{2}^{2}b_{0}b_{4}^{4}+a_{2}^{4}b_{0}^{2}b_{4}^{3}+4a_{0}a_{2}^{3}b_{0}b_{2}b_{4}^{3}-a_{2}^{4}b_{0}b_{2}^{2}b_{4}^{2}-4a_{0}a_{2}^{3}b_{0}b_{3}^{2}b_{4}^{2}+$
$\displaystyle-2a_{2}^{4}b_{0}b_{1}b_{3}b_{4}^{2}+3a_{2}^{4}b_{0}b_{2}b_{3}^{2}b_{4}-a_{2}^{4}b_{0}b_{3}^{4}+4b_{4}^{5},$
$\displaystyle F_{1}=$ $\displaystyle
a_{2}^{2}b_{1}b_{3}^{4}-a_{2}^{2}b_{0}b_{4}b_{3}^{3}+4a_{0}a_{2}b_{1}b_{4}^{2}b_{3}^{2}-3a_{2}^{2}b_{1}b_{2}b_{4}b_{3}^{2}-4a_{0}a_{2}b_{0}b_{4}^{3}b_{3}$
$\displaystyle+2a_{2}^{2}b_{1}^{2}b_{4}^{2}b_{3}+2a_{2}^{2}b_{0}b_{2}b_{4}^{2}b_{3}+6a_{0}^{2}b_{1}b_{4}^{4}-2a_{2}^{2}b_{0}b_{1}b_{4}^{3}-4a_{0}a_{2}b_{1}b_{2}b_{4}^{3}+a_{2}^{2}b_{1}b_{2}^{2}b_{4}^{2},$
$\displaystyle F_{2}=$ $\displaystyle
4a_{0}^{3}b_{4}^{5}-4a_{0}a_{2}^{2}b_{0}b_{4}^{4}-6a_{0}^{2}a_{2}b_{2}b_{4}^{4}+a_{2}^{3}b_{1}^{2}b_{4}^{3}+4a_{0}a_{2}^{2}b_{2}^{2}b_{4}^{3}+2a_{2}^{3}b_{0}b_{2}b_{4}^{3}$
$\displaystyle+4a_{0}a_{2}^{2}b_{1}b_{3}b_{4}^{3}-a_{2}^{3}b_{2}^{3}b_{4}^{2}-a_{2}^{3}b_{0}b_{3}^{2}b_{4}^{2}-4a_{0}a_{2}^{2}b_{2}b_{3}^{2}b_{4}^{2}-4a_{2}^{3}b_{1}b_{2}b_{3}b_{4}^{2}+a_{2}^{3}b_{1}b_{3}^{3}b_{4}$
$\displaystyle+3a_{2}^{3}b_{2}^{2}b_{3}^{2}b_{4}-a_{2}^{3}b_{2}b_{3}^{4},$
$\displaystyle F_{3}=$ $\displaystyle
a_{2}^{2}b_{3}^{5}+4a_{0}a_{2}b_{4}^{2}b_{3}^{3}-4a_{2}^{2}b_{2}b_{4}b_{3}^{3}+3a_{2}^{2}b_{1}b_{4}^{2}b_{3}^{2}+6a_{0}^{2}b_{4}^{4}b_{3}-2a_{2}^{2}b_{0}b_{4}^{3}b_{3}+$
$\displaystyle-8a_{0}a_{2}b_{2}b_{4}^{3}b_{3}+3a_{2}^{2}b_{2}^{2}b_{4}^{2}b_{3}+4a_{0}a_{2}b_{1}b_{4}^{4}-2a_{2}^{2}b_{1}b_{2}b_{4}^{3},$
$\displaystyle G_{0}=$ $\displaystyle
a_{0}^{4}b_{4}^{2}-a_{2}a_{0}^{3}b_{3}^{2}-2a_{2}a_{0}^{3}b_{2}b_{4}+a_{2}^{2}a_{0}^{2}b_{2}^{2}+2a_{2}^{2}a_{0}^{2}b_{1}b_{3}+2a_{2}^{2}a_{0}^{2}b_{0}b_{4}-a_{2}^{3}a_{0}b_{1}^{2}+$
$\displaystyle-2a_{2}^{3}a_{0}b_{0}b_{2}+a_{2}^{4}b_{0}^{2}-a_{2}^{4},$
$\displaystyle G_{1}=$
$\displaystyle(a_{2}b_{1}-a_{0}b_{3})(a_{0}^{2}b_{4}-a_{2}a_{0}b_{2}+a_{2}^{2}b_{0}).$

The case $(m,n)=(2,5)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{2}^{4}b_{0}b_{4}^{3}+a_{2}^{4}b_{0}b_{2}b_{5}^{2}-2a_{2}^{4}b_{0}b_{3}b_{4}b_{5}+4a_{0}a_{2}^{3}b_{0}b_{4}b_{5}^{2}+a_{0}^{4}b_{5}^{4}+4b_{5}^{4},$
$\displaystyle F_{1}=$ $\displaystyle
a_{2}b_{1}b_{4}^{3}-a_{2}b_{0}b_{5}b_{4}^{2}+4a_{0}b_{1}b_{5}^{2}b_{4}-2a_{2}b_{1}b_{3}b_{5}b_{4}-4a_{0}b_{0}b_{5}^{3}$
$\displaystyle+a_{2}b_{1}b_{2}b_{5}^{2}+a_{2}b_{0}b_{3}b_{5}^{2},$
$\displaystyle F_{2}=$ $\displaystyle
4a_{0}^{3}b_{5}^{4}-4a_{0}a_{2}^{2}b_{1}b_{5}^{3}+a_{2}^{3}b_{2}^{2}b_{5}^{2}+a_{2}^{3}b_{1}b_{3}b_{5}^{2}+a_{2}^{3}b_{0}b_{4}b_{5}^{2}+4a_{0}a_{2}^{2}b_{2}b_{4}b_{5}^{2}+$
$\displaystyle-
a_{2}^{3}b_{1}b_{4}^{2}b_{5}-2a_{2}^{3}b_{2}b_{3}b_{4}b_{5}+a_{2}^{3}b_{2}b_{4}^{3},$
$\displaystyle F_{3}=$ $\displaystyle-
a_{2}b_{3}b_{4}^{3}+a_{2}b_{2}b_{5}b_{4}^{2}-a_{2}b_{1}b_{5}^{2}b_{4}-4a_{0}b_{3}b_{5}^{2}b_{4}+2a_{2}b_{3}^{2}b_{5}b_{4}+a_{2}b_{0}b_{5}^{3}$
$\displaystyle+4a_{0}b_{2}b_{5}^{3}-2a_{2}b_{2}b_{3}b_{5}^{2},$ $\displaystyle
F_{4}=$ $\displaystyle
a_{2}^{2}b_{4}^{4}+4a_{0}a_{2}b_{5}^{2}b_{4}^{2}-3a_{2}^{2}b_{3}b_{5}b_{4}^{2}+2a_{2}^{2}b_{2}b_{5}^{2}b_{4}+6a_{0}^{2}b_{5}^{4}-a_{2}^{2}b_{1}b_{5}^{3}+$
$\displaystyle-4a_{0}a_{2}b_{3}b_{5}^{3}+a_{2}^{2}b_{3}^{2}b_{5}^{2},$
$\displaystyle G_{0}=$ $\displaystyle
4a_{0}^{5}b_{5}^{2}-4a_{2}a_{0}^{4}b_{4}^{2}-8a_{2}a_{0}^{4}b_{3}b_{5}+4a_{2}^{2}a_{0}^{3}b_{3}^{2}+8a_{2}^{2}a_{0}^{3}b_{2}b_{4}+8a_{2}^{2}a_{0}^{3}b_{1}b_{5}+$
$\displaystyle-4a_{2}^{3}a_{0}^{2}b_{2}^{2}-8a_{2}^{3}a_{0}^{2}b_{1}b_{3}-8a_{2}^{3}a_{0}^{2}b_{0}b_{4}+4a_{2}^{4}a_{0}b_{1}^{2}+8a_{2}^{4}a_{0}b_{0}b_{2}-4a_{2}^{5}b_{0}^{2}+5a_{2}^{5},$
$\displaystyle G_{1}=$
$\displaystyle(a_{0}^{2}b_{4}-a_{2}a_{0}b_{2}+a_{2}^{2}b_{0})(a_{0}^{2}b_{5}-a_{2}a_{0}b_{3}+a_{2}^{2}b_{1}).$

The case $(m,n)=(2,6)$.

$\displaystyle F_{0}=$ $\displaystyle
a_{0}^{4}b_{6}^{3}-4a_{2}^{3}a_{0}b_{0}b_{6}^{2}-a_{2}^{4}b_{0}b_{5}^{2}+a_{2}^{4}b_{0}b_{4}b_{6}+4b_{6}^{3},$
$\displaystyle F_{1}=$ $\displaystyle
a_{2}b_{1}b_{5}^{2}-a_{2}b_{0}b_{6}b_{5}+4a_{0}b_{1}b_{6}^{2}-a_{2}b_{1}b_{4}b_{6},$
$\displaystyle F_{2}=$ $\displaystyle-
a_{2}^{3}b_{2}b_{5}^{2}-a_{2}^{3}b_{0}b_{6}^{2}+a_{2}^{3}b_{2}b_{4}b_{6}+a_{2}^{3}b_{1}b_{5}b_{6}-4a_{0}a_{2}^{2}b_{2}b_{6}^{2}+4a_{0}^{3}b_{6}^{3},$
$\displaystyle F_{3}=$ $\displaystyle
a_{2}b_{3}b_{5}^{2}-a_{2}b_{2}b_{6}b_{5}+a_{2}b_{1}b_{6}^{2}+4a_{0}b_{3}b_{6}^{2}-a_{2}b_{3}b_{4}b_{6},$
$\displaystyle F_{4}=$ $\displaystyle
6a_{0}^{2}b_{6}^{3}-a_{2}^{2}b_{2}b_{6}^{2}-4a_{0}a_{2}b_{4}b_{6}^{2}+a_{2}^{2}b_{4}^{2}b_{6}+a_{2}^{2}b_{3}b_{5}b_{6}-a_{2}^{2}b_{4}b_{5}^{2},$
$\displaystyle F_{5}=$ $\displaystyle
a_{2}b_{5}^{3}+4a_{0}b_{6}^{2}b_{5}-2a_{2}b_{4}b_{6}b_{5}+a_{2}b_{3}b_{6}^{2},$
$\displaystyle G_{0}=$ $\displaystyle
a_{0}^{6}b_{6}^{2}-a_{2}a_{0}^{5}b_{5}^{2}-2a_{2}a_{0}^{5}b_{4}b_{6}+a_{2}^{2}a_{0}^{4}b_{4}^{2}+2a_{2}^{2}a_{0}^{4}b_{3}b_{5}+2a_{2}^{2}a_{0}^{4}b_{2}b_{6}-a_{2}^{3}a_{0}^{3}b_{3}^{2}+$
$\displaystyle-2a_{2}^{3}a_{0}^{3}b_{2}b_{4}-2a_{2}^{3}a_{0}^{3}b_{1}b_{5}-2a_{2}^{3}a_{0}^{3}b_{0}b_{6}+a_{2}^{4}a_{0}^{2}b_{2}^{2}+2a_{2}^{4}a_{0}^{2}b_{1}b_{3}$
$\displaystyle+2a_{2}^{4}a_{0}^{2}b_{0}b_{4}-a_{2}^{5}a_{0}b_{1}^{2}-2a_{2}^{5}a_{0}b_{0}b_{2}+a_{2}^{6}b_{0}^{2}-a_{2}^{6},$
$\displaystyle G_{1}=$
$\displaystyle(a_{0}^{2}b_{5}-a_{2}a_{0}b_{3}+a_{2}^{2}b_{1})(-a_{0}^{3}b_{6}+a_{2}a_{0}^{2}b_{4}-a_{2}^{2}a_{0}b_{2}+a_{2}^{3}b_{0}).$

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Szabolcs Tengely, Institute of Mathematics, University of Debrecen, P.O.Box
400, 4010 Debrecen, Hungary. email<EMAIL_ADDRESS>

Maciej Ulas, Jagiellonian University, Faculty of Mathematics and Computer
Science, Institute of Mathematics, Łojasiewicza 6, 30-348 Kraków, Poland;
email<EMAIL_ADDRESS>