# Equal values of certain partition functions via Diophantine equations

Szabolcs Tengely  Szabolcs Tengely
Mathematical Institute
University of Debrecen
P.O.Box 12
4010 Debrecen
Hungary<EMAIL_ADDRESS>and  Maciej Ulas  Maciej Ulas
Jagiellonian University
Faculty of Mathematics and Computer Science
Institute of Mathematics
Łojasiewicza 6
30-348 Kraków
Poland<EMAIL_ADDRESS>
###### Abstract.

Let $A\subset\mathbb{N}_{+}$ and by $P_{A}(n)$ denotes the number of
partitions of an integer $n$ into parts from the set $A$. The aim of this
paper is to prove several result concerning the existence of integer solutions
of Diophantine equations of the form $P_{A}(x)=P_{B}(y)$, where $A,B$ are
certain finite sets.

###### Key words and phrases:

partitions, Diophantine equation, polynomial, identities

###### 2000 Mathematics Subject Classification:

11D41, 11P83

The research of the first author was supported in part by the NKFIH grants
115479, 128088 and130909 and by the project EFOP-3.6.1-16-2016-00022, co-
financed by the European Union and the European Social Fund. Research of the
second author was supported by a grant of the National Science Centre (NCN),
Poland, no. UMO-2019/34/E/ST1/00094

##  1\. Introduction

Let $\mathbb{N}$ be the set of non-negative integers, $\mathbb{N}_{+}$ the set
of positive integers and for $k\in\mathbb{N}_{+}$ we write $\mathbb{N}_{\geq
k}$ for the set of integers $\geq k$.

Let $A\subset\mathbb{N}_{+}$ be given and take $n\in\mathbb{N}$. By an
$A$-partition $\lambda=(\lambda_{1},\ldots,\lambda_{k})$, of a non-negative
integer $n$ with parts in $A$, we mean representation of $n$ in the form

$n=\lambda_{1}+\ldots+\lambda_{k},$

where $\lambda_{i}\in A$. The representations of $n$ differing only in order
of the terms are counted as a one. We also put

$\operatorname{Part}_{A}(n)=\\{\lambda:\;\lambda\;\mbox{is}\;A\mbox{-partition
of}\;n\\},$

and consider the corresponding partition function

$P_{A}(n):=\\#\operatorname{Part}_{A}(n).$

It is well know that the ordinary generating function of the sequence
$(P_{A}(n))_{n\in\mathbb{N}}$ takes the form

$\prod_{a\in A}\frac{1}{1-x^{a}}=\sum_{n=0}^{\infty}P_{A}(n)x^{n}.$

In particular, if $A=\mathbb{N}_{+}$, then $P_{A}(n)$, simply denoted as
$p(n)$, is the famous partition function introduced by L. Euler and
extensively studied by S. Ramanujan. The function $p(n)$ counts the number of
partitions with parts in $\mathbb{N}_{+}$, i.e., unrestricted partitions of
$n$. Many questions concerning arithmetic properties of $p(n)$ is still
unsolved. Probably, the most famous one, is the question whether the sequence
$\\{p(n)\mod{m}\\}_{n\in\mathbb{N}}$ is equidistributed modulo $m$, i.e.,
whether, for any given $m\in\mathbb{N}_{\geq 2}$ and each
$r\in\\{0,\ldots,m-1\\}$, there is an equality

$\limsup_{N\rightarrow+\infty}\frac{\\#\\{n\leq N:\;p(n)\equiv r\mkern
4.0mu({\operator@font mod}\mkern 6.0mum)\\}}{N}=\frac{1}{m}.$

For discussion on this topic see [1].

The literature concerning arithmetic properties of functions counting various
partitions is enormous. One of the central problems in partition theory is to
obtain partition identities [3]. More precisely, if $\mathcal{W}$ is a certain
property which can be applied to the parts of a given $A$-partition $\lambda$
of a positive integer $n$, then by $P_{A}(\mathcal{W},n)$ we denote the number
of $A$-partitions of $n$ which have the property $\mathcal{W}$. Thus, by a
partition identity we mean an identity of the form

$P_{A_{1}}(\mathcal{W}_{1},n)=P_{A_{2}}(\mathcal{W}_{2},n),$

where $A_{1},A_{2}\subset\mathbb{N}_{+}$ and $\mathcal{W}_{1},\mathcal{W}_{2}$
are given properties. The basic identity of the kind is Euler’s famous
identity

$\prod_{n=1}^{\infty}(1+x^{n})=\prod_{n=1}^{\infty}\frac{1}{1-x^{2n-1}},$

which gives the equality between the number of partitions into distinct parts
and the number of partitions into odd parts. Equivalently

$P_{\mathbb{N}_{+}}(\mbox{distinct parts},n)=P_{\mathbb{N}_{+}}(\mbox{odd
parts},n).$

Euler’s identity can be proved by a simple manipulation of infinite products.
Much deeper partition identities are two (of many known) Rogers-Ramanujan type
identities. These identities can be deduced from the equalities of the type:
infinite product $=$ infinite series, i.e.,

$\displaystyle\prod_{n=1}^{\infty}\frac{1}{(1-x^{5n-1})(1-x^{5n-4})}$
$\displaystyle=\sum_{n=1}^{\infty}\frac{x^{n^{2}}}{(1-x)(1-x^{2})\cdot\ldots\cdot(1-x^{n})},$
$\displaystyle\prod_{n=1}^{\infty}\frac{1}{(1-x^{5n-2})(1-x^{5n-3})}$
$\displaystyle=\sum_{n=1}^{\infty}\frac{x^{n^{2}+n}}{(1-x)(1-x^{2})\cdot\ldots\cdot(1-x^{n})}.$

The first identity implies that the number of partitions of $n$ such that the
adjacent parts differ by at least 2 is the same as the number of partitions of
$n$ such that each part is congruent to either 1 or 4 modulo 5, i.e.,

$P_{\mathbb{N}_{+}}(\mbox{adjacent parts differ by at least
2},n)=P_{\mathbb{N}_{+}}(\mbox{parts}\equiv 1,4\mkern 4.0mu({\operator@font
mod}\mkern 6.0mu5),n).$

The second one implies that the number of partitions of $n$ such that the
adjacent parts differ by at least 2 and such that the smallest part is at
least 2 is the same as the number of partitions of $n$ such that each part is
congruent to either 2 or 3 modulo 5. Mentioned identities are only special
cases of more general one which can be found in the literature. However,
according to our best knowledge nothing is known about identities of the form

(1) $P_{A_{1}}(\mathcal{W}_{1},n)=P_{A_{2}}(\mathcal{W}_{2},m)\neq 0,$

where $n,m$ are different positive integers.

Of course, in the above, we are interested in non-trivial identities. More
precisely, if the set of values of $P_{A_{i}}(\mathcal{W}_{i},n)$ contains all
positive integers for some $i\in\\{1,2\\}$ then the problem is trivial. For
example, if $A_{1}=\\{2^{j}:\;j\in\mathbb{N}\\}$ and the property
$\mathcal{W}_{1}$ says that at least two parts in $A_{1}$-partition of $n$ are
equal, then, for $s(n):=P_{A_{1}}(\mathcal{W}_{1},n)$ we clearly have

$\sum_{n=0}^{\infty}s(n)x^{n}=\prod_{n=0}^{\infty}(1+x^{2^{n}}+x^{2^{n+1}}),$

and the sequence $\\{s(n)\\}_{n\in\mathbb{N}}$ is a famous Stern sequence
satisfying the recurrence relations

$s(0)=1,\quad s(2n)=s(n)+s(n-1),\quad s(2n+1)=s(n).$

One can easily prove that $s(2^{n})=n+1$ and thus the sequence
$(s(n))_{n\in\mathbb{N}}$ contains all positive integers. Thus the question
concerning the existence of positive integer solutions of related equation is
trivial.

Of great interests would be the proof of non-existence of
$A_{i},\mathcal{W}_{i},i=1,2$ such that the corresponding partition functions
$P_{A_{i}}(\mathcal{W}_{i},n),i=1,2,$ have exponential growth and equation (1)
has infinitely many solutions in positive integers. Note that if
$P_{A_{i}}(\mathcal{W}_{i},n)$ has an exponential growth then the set $A_{i}$
is necessarily infinite.

A related question is to whether given partition function takes values in a
given infinite set. Especially, in the set of values of a given polynomial. In
other words we are interested in the solvability in non-negative integers of
the equation

(2) $f(m)=P_{A}(\mathcal{W},n)\neq 0,$

where $f\in\mathbb{Q}[x]$ is of positive degree and positive leading
coefficient. Again we are interested in non-trivial situations only. Here if
$f$ is linear, then we enter in the realm of partition congruences. The
Ramanujan congruences

$p(5n+4)\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mu5),\quad
p(7n+5)\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mu7),\quad
p(11n+7)\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mu11)$

and its various generalizations give some non-trivial examples when equation
(2) has infinitely many solutions in positive integers. Indeed, each equation

$p(n)=5m,\quad p(n)=7m,\quad p(n)=11m$

has infinitely many solutions in positive integers. In fact, one can prove
that for each prime $q\neq 3$ the Diophantine equation $p(n)=qm$ has
infinitely many solutions in positive integers. Indeed, Nicolas, Ruzsa and
Sárközy [13] proved this for $q=2$, and Ono proved that such a statement is
true for each $q\geq 5$ [18].

In case of equation (2) more can be said provided we know some arithmetic
properties of the partition function $P_{A}(\mathcal{W},n)$. For example, if
$A=\\{2^{i}:\;i\in\mathbb{N}\\}$ then so called binary partition function
$b(n):=P_{A}(n)$ counting the partitions with all parts being powers of $2$,
satisfies the recurrence relation

$b(0)=1,\quad b(2n)=b(2n-1)+b(n),\quad b(2n+1)=b(2n).$

A classical result of Churchhouse states that $\nu_{2}(b(n))\in\\{1,2\\}$ for
$n\in\mathbb{N}_{\geq 2}$ [7]. Thus, if

$\\{1,2\\}\cap\\{\nu_{2}(f(m)):\;m\in\mathbb{N}_{+}\\}=\emptyset$

then equation (2) has at least $\operatorname{deg}f$ solutions possibly coming
from the integer solutions of the equation $f(m)=1$.

Our discussion above shows that the problem of solvability of equation (1) or
equation (2) is interesting, difficult and worth of further investigations. It
is clear that our questions are on the intersection of discrete mathematics,
combinatorics and Diophantine equations. This suggest to start investigations
with the case of $A$ finite. This is reasonable due to the fact that
$p_{A}(n)$ is a quasi polynomial (see for example [14]). More precisely, if
$A=(a_{1},\ldots,a_{k})$ and $L_{A}:=\operatorname{lcm}(a_{1},\ldots,a_{k})$,
then

$p_{A}(L_{A}n+i)\in\mathbb{Q}[n]\quad\mbox{for}\quad i=0,1,\ldots,L_{A}-1.$

Thus, if $A_{1},A_{2}$ are finite, then solvability in positive integers of
equation (1) or a more general equation (2), is equivalent with the
solvability in non-negative integers, of at least one equation of the type

$p_{A_{1}}(L_{A_{1}}m+i)=p_{A_{2}}(L_{A_{2}}n+j),$

where $i\in\\{0,1,\ldots,L_{A_{1}}-1\\},j\in\\{0,1,\ldots,L_{A_{2}}-1\\}$, or

$f(m)=p_{A_{1}}(L_{A_{1}}n+i),\quad i\in\\{0,1,\ldots,L_{A_{1}}-1\\},$

respectively. Thus, we enter into realm of polynomial Diophantine equations
with separable variables where great deal of methods are for our disposal.

In the rest of the paper, in case of the set $A=\\{1,\ldots,m\\}$ instead of
writing $P_{A}(x)$ and $\operatorname{Part}_{A}(n)$, we will simply write
$P_{m}(x)$ and $\operatorname{Part}_{m}(n)$, respectively. Let us describe the
content of the paper in some details.

In Section 2 we prove that for each $f\in\mathbb{Z}[x]$ with positive degree
and positive leading coefficient and
$A=\\{a_{1},a_{2}\\}\subset\mathbb{N}_{+}$ with $\gcd(a_{1},a_{2})=1$, the
Diophantine equation $P_{A}(x)=f(y)$ has infinitely many solutions in positive
integers (Theorem 2.1).

Section 3 is mainly devoted to the study of the equation $P_{3}(x)=P_{n}(y)$
for $n=4,5$. In particular, we describe all positive integer solutions in both
cases and present some related results (Theorem 3.1 and Theorem 3.2).

In Section 4 we study some equation involving $P_{A}(x)$, where
$A=\\{1,2,a\\},a\in\mathbb{N}_{\geq 3}$. In particular, we obtain a general
result concerning the existence of infinitely many positive integer solutions
of the equation $P_{A}(x)=P_{4}(y)$ (Theorem 4.3). We also obtain, under weak
assumptions on $a,b\in\mathbb{N}_{\geq 3},a\neq b$, that for
$A=\\{1,2,a\\},B=\\{1,2,b\\}$, the Diophantine equation $P_{A}(x)=P_{B}(y)$
has infinitely many solutions in positive integers (Theorem 4.5).

In Section 5 we obtain several results concerning the square values of
$P_{A}(x)$. In particular we describe the set of positive integer solutions of
the equation $y^{2}=P_{n}(x)$ for $n=3,4,5$. We also discuss results of some
computations. Finally, in the last section we collect some general questions
and conjectures concerning various aspects of the Diophantine equations of the
form $P_{A}(x)=P_{B}(y)$ and report results of various computations.

##  2\. The case of $A=\\{a_{1},a_{2}\\}$

In this short section we prove a general result concerning the existence of
positive integer solutions of the Diophantine equation

$P_{A}(x)=f(y),$

where $A=\\{a_{1},a_{2}\\},a_{1}<a_{2}$ and $f\in\mathbb{Z}[y]$ is a non-
constant polynomial with positive leading coefficient. More precisely, based
on the formula obtained by Sertöz in [15] we easily prove the following.

######  Theorem 2.1.

Let $A=\\{a_{1},a_{2}\\}\subset\mathbb{N}_{+}$. Then, for each
$f\in\mathbb{Z}[x]$ with positive leading coefficient, the Diophantine
equation $P_{A}(x)=f(y)$ has infinitely many solutions in positive integers.

###### Proof.

Without loss of generality we can assume that $\gcd(a_{1},a_{2})=1$. Then, by
Sertöz result [15], we know that the following formula holds

$P_{A}(n)=\frac{n+a_{1}\cdot a_{1}^{\prime}+a_{2}\cdot
a_{2}^{\prime}}{a_{1}a_{2}}-1,$

where $a_{i}\cdot a_{i}^{\prime}\equiv-n\mkern 4.0mu({\operator@font
mod}\mkern 6.0mua_{i+1})$ and $1\leq a_{i}^{\prime}(n)\leq a_{i+1}$ for
$i=1,2$ and if $i=2$, then $i+1$ is taken modulo 2. In particular, for each
$n\in\mathbb{N}_{+}$

$P_{A}(n)=\left\lfloor\frac{n}{a_{1}a_{2}}\right\rfloor\quad\mbox{or}\quad
P_{A}(n)=\left\lfloor\frac{n}{a_{1}a_{2}}\right\rfloor+1.$

Thus, let us take $n=a_{1}a_{2}(f(m)-1)$, with $m\in\mathbb{N}_{+}$ chosen in
such a way that $f(m)>1$. We thus consider the congruence

$a_{i}\cdot a_{i}^{\prime}\equiv-n\equiv-a_{1}a_{2}(f(m)-1)\equiv 0\mkern
4.0mu({\operator@font mod}\mkern 6.0mua_{i+1})$

and due to co-primality condition and the bound for
$a_{1}^{\prime},a_{2}^{\prime}$, we get that $a_{i}^{\prime}=a_{i+1}$, i.e.,
we have that $a_{1}^{\prime}=a_{2},a_{2}^{\prime}=a_{1}$. In consequence

$P_{A}(a_{1}a_{2}(f(m)-1))=\frac{a_{1}a_{2}(f(m)-1)+a_{1}a_{2}+a_{2}a_{1}}{a_{1}a_{2}}-1=f(m)$

and our theorem is proved. ∎

##  3\. The equation $P_{3}(x)=P_{n}(y)$ for $n=4,5$

In this section we are interesting in the characterization of positive integer
solutions of the Diophantine equation

$P_{3}(x)=P_{n}(y)$

for $n=4,5$.

######  Theorem 3.1.

The Diophantine equation $P_{3}(x)=P_{4}(y)$ has infinitely many solutions in
integers.

###### Proof.

First of all we recall that

$\displaystyle P_{3}(n)$
$\displaystyle=\left\lfloor\frac{(n+3)^{2}}{12}\right\rceil,$ $\displaystyle
P_{4}(n)$
$\displaystyle=\left\lfloor\frac{(n+1)(n^{2}+23n+85)}{144}-\frac{n+4}{8}\left\lfloor\frac{n+1}{2}\right\rfloor\right\rceil,$

where $\lfloor x\rceil$ denotes the nearest integer to $x$. For concise proofs
of these two identities see [2, p. 57-60]. An alternative recent proof of
these and related equalities were obtained by Castillo et al. [5].

From the form of $P_{3}(n)$ we see that for $i\in\\{0,\ldots,5\\}$ the
expression $P_{3}(6n+i)$ is a polynomial in $n$. More precisely, we define
$P_{i,6,3}(n)=P_{3}(6n+i)$ and observe that

$\begin{array}[]{ll}P_{0,6,3}(n)=3n^{2}+3n+1,&P_{1,6,3}(n)=(n+1)(3n+1),\\\
P_{2,6,3}(n)=(n+1)(3n+2),&P_{3,6,3}(n)=3(n+1)^{2},\\\
P_{4,6,3}(n)=(n+1)(3n+4),&P_{5,6,3}(n)=(n+1)(3n+5).\\\ \end{array}$

We can similar treat the case of $P_{4}(n)$ and by defining
$P_{2i+1,6,4}(n)=P_{4}(6n+2i+1)$ for $i=0,1,2$ and
$P_{2i,12,4}(n)=P_{4}(12n+2i)$ for $i=0,1,\ldots,5$, we get

$\begin{array}[]{ll}P_{1,6,4}(n)=\frac{1}{2}(n+1)\left(3n^{2}+6n+2\right),&P_{3,6,4}(n)=\frac{3}{2}(n+1)^{2}(n+2),\\\
P_{5,6,4}(n)=\frac{3}{2}(n+1)(n+2)^{2},&\\\
P_{0,12,4}(n)=12n^{3}+15n^{2}+6n+1,&P_{2,12,4}(n)=12n^{3}+21n^{2}+12n+2,\\\
P_{4,12,4}(n)=(n+1)(12n^{2}+15n+5),&P_{6,12,4}(n)=3(n+1)^{2}(4n+3),\\\
P_{8,12,4}(n)=3(n+1)^{2}(4n+5),&P_{10,12,4}(n)=(n+1)(12n^{2}+33n+23).\end{array}$

In order to characterize all integers solutions of the Diophantine equation
$P_{3}(x)=P_{4}(y)$ we need to perform case by case analysis. More precisely,
we consider all possible combinations of the equations

${\rm(I)}\quad
P_{i,6,3}(x)=P_{2j+1,6,4}(y)\quad\quad\mbox{or}\quad\quad{\rm(II)}\quad
P_{i,6,3}(x)=P_{2j,12,4}(y),$

i.e., we deal with 54 equations. In each case we deal in the same way. Because
the degree of $P_{i,6,3}$ is 2, each equation of interests can be reduced to
the equation of the type $Y^{2}=f(X)$ for some $f\in\mathbb{Z}[X]$ and the
degree of $f$ is 3. If $f$ has no multiple roots, by classical Siegel result,
we know that the curve defined by the equation $Y^{2}=f(X)$ has only finitely
many integral points. On the other hand, if $f$ has multiple roots then there
is a chance that our equation has infinitely many integral solutions which can
be parameterized via polynomials.

In order to see what is going on, let us consider the equation

(3)
$P_{1,6,3}(x)=P_{3,6,4}(y),\;\mbox{i.e.},\;(x+1)(3x+1)=\frac{3}{2}(y+1)^{2}(y+2)$

or equivalently $Y^{2}=X^{3}-108X+1728$, where we put $X=6(3y+4),Y=36(3x+2)$.
Our equation represents an elliptic curve, say $E$, in the plane $(X,Y)$ and
standard methods allow to find that the curve $E$ has trivial torsion and the
rank of $E$ is equal to 2, with the generators $(X,Y)=(6,-36),(-2,44)$. Using
internal Magma [8] procedures

E:=EllipticCurve([-108,1728]);

IntegralPoints(E);
(the background for this latter routine is found in [11] and [17]; see also
[19]) we find that the point $(X,Y)\in E(\mathbb{Q})$ has integer coordinates,
if and only if

$\displaystyle(X,Y)\in\\{$ $\displaystyle(-12,\pm 36),(-3,\pm 45),(-2,\pm
44),(6,\pm 36),(16,\pm 64),$ $\displaystyle(22,\pm 100),(78,\pm 684),(96,\pm
936),(7926,\pm 705636)\\}.$

Direct check shows that only the points $P_{1}=(96,-936)$ and
$P_{2}=(7926,-705636)$ correspond to the solutions of our equation. We thus
get that the only integer solutions of equation (3) are
$(x,y)=(8,4),(6533,439)$. We thus get the equalities

$\displaystyle P_{1,6,3}(8)$ $\displaystyle=P_{3,6,4}(4)=225,$ $\displaystyle
P_{1,6,3}(6533)$ $\displaystyle=P_{3,6,4}(439)=128066400.$

Case by case analysis reveals that the solutions exist only in the following
cases:

$\displaystyle(i,j)$ $\displaystyle=(0,0),(1,0),(1,1),(1,2),(5,1)\quad\mbox{in
the case (I)}\quad\mbox{and}$ $\displaystyle(i,j)$
$\displaystyle=(0,0),(1,0),(2,1),(3,4),(5,2)\quad\mbox{in the case (II)}.$

In the table below we give all integral solutions in these cases.

$\begin{array}[]{|l|l|}\hline\cr\hline\cr(i,j)&\mbox{integral
solutions}\;(x,y)\;\mbox{of}\;P_{i,6,3}(x)=P_{2j+1,6,4}(y)\\\
\hline\cr\hline\cr(0,0)&(0,0)\\\ (0,1)&(0,0)\\\ (1,1)&(8,4),(6533,439)\\\
(1,2)&(293,54)\\\ (3,1)&((t-1)(2t^{2}+2t+1),2(t-1)(t+1)),t\in\mathbb{N}_{+}\\\
(3,2)&(2t^{3}+t-1,2t^{2}-1),t\in\mathbb{N}_{+}\\\ (5,1)&(5,3)\\\
\hline\cr\hline\cr(i,j)&\mbox{integral
solutions}\;(x,y)\;\mbox{of}\;P_{i,6,3}(x)=P_{2j,12,4}(y)\\\
\hline\cr\hline\cr(0,0)&((t-1)(2t^{2}-t+1),2(t-1)t),t\in\mathbb{N}_{+}\\\
(1,0)&(0,0)\\\ (2,1)&(0,0)\\\
(3,4)&(2t^{3}+3t^{2}+t-1,t^{2}+t-1),t\in\mathbb{N}_{+}\\\ (5,2)&(0,0)\\\
\hline\cr\end{array}$

Table. Values of $(i,j)$ such that the corresponding equations of types (I),
(II) have solutions in non-negative integers.

Having the form of solutions presented in the table above, we can easily back
to our original Diophantine equation $P_{3}(X)=P_{4}(Y)$ and found that
solutions take the form $(X,Y)=(6x+i,6y+2j)$ in the first type equation, and
$(X,Y)=(6x+i,12y+2j+1)$ in the second type equation. ∎

In our next theorem we characterize the set of positive integer solutions of
the Diophantine equation $P_{3}(x)=P_{5}(y)$.

######  Theorem 3.2.

The equation $P_{3}(x)=P_{5}(y)$ has only finitely many solutions in positive
integers. More precisely, the pair $(x,y)$ is a solution if and only if
$(x,y)\in\mathcal{A}$, where

$\displaystyle\mathcal{A}=\\{$
$\displaystyle(1,1),(2,2),(3,3),(5,4),(6,5),(8,6),(16,10),(18,11),(26,14),$
$\displaystyle(45,20),(174,45),(217,51),(457,77),(468,78),(701,97),(10093,388)\\}.$

###### Proof.

Direct check reveals that $P_{5}(60n+i),i\in\\{0,\ldots,59\\}$ is a polynomial
in variable $n$. Thus, one can perform the same analysis as in the case of the
equation $P_{3}(x)=P_{4}(y)$. However, here the situation is a bit more
complicated because, after necessary simplifications, we need to work with the
equations of the type $Y^{2}=f(X)$, where $f$ is a polynomial of degree 4. We
need to consider $6\cdot 60=360$ equations in order to get the result. Here we
may apply the Magma procedure IntegralQuarticPoints() based on the paper [19].
It worked well in all except the 8 cases, where the Magma function failed to
determine the complete set of integral solutions. These 8 problematic
equations are of the form

$P_{3}(6y+i)=P_{5}(60x+j)$

for

$(i,j)\in\mathcal{A}=\\{(3,9),(3,12),(3,21),(3,24),(3,33),(3,36),(3,48),(3,57)\\}.$

The equations corresponding to $(i,j)=(3,48),(3,57)$ are of the following form

$\displaystyle Y^{2}$ $\displaystyle=$ $\displaystyle
u(54000u^{3}-16200u^{2}+1410u-18),$ $\displaystyle Y^{2}$ $\displaystyle=$
$\displaystyle u(54000u^{3}+16200u^{2}+1410u+18),$

respectively, where $u=x+1.$ In both cases we obtain that $u$ is a square
multiplied by a divisor of 18. Therefore we need to handle the equations

$(2\delta^{2}v)^{2}=(60\delta u)^{3}-18\delta(60\delta
u)^{2}+94\delta^{2}(60\delta u)-72\delta^{3},$

where $\delta\in\\{\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18\\}.$ One more time we
use the MAGMA procedure IntegralPoints() to determine the integral points on
these elliptic curves. We only need to consider points having first coordinate
divisible by $60\delta.$ It turns out that $u=0$ is the only solution, that is
$(x,Y)=(-1,0).$

In the remaining 6 cases, we observed that the discriminant of
$P_{3}(6y+i)=P_{5}(60x+j)$ with respect to $y$ is equal to

$F(u)=432u^{4}+648u^{3}+282u^{2}+18u,$

for suitable substitution of the form $u=ax+b$ (depending on values of $i,j$).
The expression for $u$ are given below

$\begin{array}[]{|l|l|l|l|l|l|l|}\hline\cr(i,j)&(3,9)&(3,12)&(3,21)&(3,24)&(3,33)&(3,36)\\\
\hline\cr u&5x+1&-5x-2&5x+2&-5x-3&5x+3&-5x-4\\\ \hline\cr\end{array}$

Therefore we only need to determine integral points on the curve

$72u^{4}+108u^{3}+47u^{2}+3u=30v^{2}.$

We obtain that 3 divides $u,$ so $u=3u_{1}$ for some integer $u_{1}.$ We have
that

$u_{1}(648u_{1}^{3}+324u_{1}^{2}+47u_{1}+1)=30v_{1}^{2},\mbox{ where
}v=3v_{1}.$

The factorization yields the following elliptic curves

$X^{3}+324\delta X^{2}+30456\delta^{2}X+419904\delta^{3}=Y^{2},\mbox{ where
}\delta\in\\{1,2,3,5,6,10,15,30\\}.$

We determined the integral points on these curves and checked if $X$ is
divisible by $648\delta,$ the only such solution corresponds to $X=0.$ Hence
we do not obtain integral solution in case of these six curves. ∎

In the light of the result one can ask for which sequences $A$ of the form
$A=\\{1,2,3,a\\},a\geq 4$, the Diophantine equation $P_{3}(x)=P_{A}(y)$ has
infinitely many solutions in positive integers. It is not difficult to find
many values of $a$ with this property. Indeed, for a fixed $a$ the equation
has the form that a quadratic polynomial is equal to a cubic polynomial, hence
we may expect a genus 1 curve. However, we for certain values of $a$ we may
obtain infinitely many integral solutions. The strategy we follow is simple,
we determine polynomials $P_{A}(6an+k)$ in $n$ that are not square-free and
then deal with the equation $P_{3}(6m+3)=3(m+1)^{2}=P_{A}(6an+k)$. This works
for $a\in\\{4,6,12,14,20\\}$. In these cases the equation $P_{3}(x)=P_{A}(y)$
has a polynomial solution and hence infinitely many solutions in positive
integers.

We close this section with the following

######  Conjecture 3.3.

There are infinitely many values of $a\in\mathbb{N}_{\geq 4}$ such that for
$A=\\{1,2,3,a\\}$, the Diophantine equation $P_{3}(x)=P_{A}(y)$ has infinitely
many solutions in positive integers.

##  4\. Some properties of $P_{A}(x)$ for $A=\\{1,2,a\\},a\geq 3$, and related
equations

In this section we obtain explicit expression for $P_{A}(n)$ in case of
$A=\\{1,2,a\\}$. As an application we deduce several results concerning
Diophantine properties of $P_{A}(n)$.

######  Theorem 4.1.

Let $a\in\mathbb{N}_{\geq 3}$ and put $A=\\{1,2,a\\}$. If $a=2c$ for some
$c\in\mathbb{N}_{\geq 2}$ then

$P_{A}(4cn+i)=2cn^{2}+\left(c+2\left\lfloor\frac{i}{2}\right\rfloor+2\right)n+\begin{cases}\begin{array}[]{ll}\left\lfloor\frac{i+2}{2}\right\rfloor,&i\in\\{0,\ldots,2c-1\\}\\\
2\left\lfloor\frac{i}{2}\right\rfloor+2-a,&i\in\\{2c,\ldots,4c-1\\}\end{array}\end{cases}.$

If $a=2c+1$ for some $c\in\mathbb{N}_{+}$ then

$\displaystyle P_{A}(2(2c+1)n$ $\displaystyle+i)=(2c+1)n^{2}+(c+i+2)n$
$\displaystyle+\begin{cases}\begin{array}[]{ll}\left\lfloor\frac{i+2}{2}\right\rfloor,&i\in\\{0,\ldots,2c\\}\\\
i+1-a,&i\in\\{2c+1,\ldots,4c+1\\}\end{array}\end{cases}.$

###### Proof.

Let $A=\\{1,2,a\\}$ and recall that

$\frac{1}{(1-x)(1-x^{2})(1-x^{a})}=\sum_{n=0}^{\infty}P_{A}(n)x^{n},\quad\frac{1}{(1-x)(1-x^{2})}=\sum_{n=0}^{\infty}\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)x^{n}.$

Thus, using the identity $(1-x^{a})F_{A}(x)=\frac{1}{(1-x)(1-x^{2})}$ by
comparison of coefficients of like powers on both sides of we get that
$P_{A}(n)$ satisfies the following recurrence relation:

$P_{A}(n)=\left\lfloor\frac{n}{2}\right\rfloor+1,n\leq a-1\quad\mbox{and}\quad
P_{A}(n)=P_{A}(n-a)+\left\lfloor\frac{n}{2}\right\rfloor+1\;\mbox{for}\;n\geq
a.$

Knowing that $P_{A}(n)$ satisfies recurrence relation of the presented form
and using the (conjectural) form of the solution it is easy to perform the
rest of the proof by induction. We omit the tiresome details. ∎

There are many papers devoted to the explicit computation of the function
$P_{A}(n)$ for given $A=\\{a_{1},\ldots,a_{k}\\}$ under various conditions on
$a_{1},\ldots,a_{k}$. Although the result above can also be deduced from known
results (see for example [9, 16]), the explicit form with exact values of
coefficients is very useful in what follows.

Having the explicit form of the
$P_{A}(2an+i),i\in\\{0,\ldots,2a-1\\},A=\\{1,2,a\\}$ one can obtain certain
results concerning polynomial values taken by the partition function
$P_{A}(n),n\in\mathbb{N}_{+}$. We start with the following simple

######  Corollary 4.2.

Let $a\in\mathbb{N}_{\geq 3}$ and put $A=\\{1,2,a\\}$.

  1. (1)

If $a\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mu2)$, then
$P_{A}(2n)=P_{A}(2n+1)$ for each $n\in\mathbb{N}$.

  2. (2)

If $a\equiv 1\mkern 4.0mu({\operator@font mod}\mkern 6.0mu2)$, then

$P_{A}(n)=P_{A}(n+1)\;\Longleftrightarrow\;n=2j,j\in\left\\{1,\ldots,\frac{a-3}{2}\right\\}.$

###### Proof.

(1) Let us put $a=2c,c\in\mathbb{N}_{\geq 2}$. The statement is an immediate
consequence of the first formula from Theorem 4.1. Indeed, for each
$i\in\\{0,\ldots,2c-1\\}$ and $m\in\mathbb{N}$ we have the equality

$P_{A}(4cm+2i)=P_{A}(4cm+2i+1).$

Because of the equality
$\\{4cn+2i:\;n\in\mathbb{N},\;i\in\\{0,\ldots,2c-1\\}\\}=2\mathbb{N}$ (the set
of even non-negative integers) we get the result.

(2) Let us put $a=2c+1,c\in\mathbb{N}_{+}$. Because
$\\{2(2c+1)n+i:\;n\in\mathbb{N},\;i\in\\{0,\ldots,4c+1\\}\\}=\mathbb{N}$ to
get the solutions of $P_{A}(n)=P_{A}(n+1)$ it is enough to consider the
solutions (in $m\in\mathbb{N}$) of equations
$P_{A}(2(2c+1)m+2i)=P_{A}(2(2c+1)m+2i+1)$ or
$P_{A}(2(2c+1)m+2i+1)=P_{A}(2(2c+1)m+2i+2)$. We consider the former equation
first. If $i<c$ then we deal with the equation

$(2c+1)m^{2}+(c+2i+2)m+i+1=(2c+1)m^{2}+(c+2i+3)m+i+1,$

i.e., $m=0$ and $n=2i$ for $i=0,\ldots,c-1=\frac{a-3}{2}$. If $i=c$ then we
work with the equation

$(2c+1)m^{2}+(c+2i+2)m+i+1=(2c+1)m^{2}+(c+2i+3)m+2i+2-c,$

i.e., $m=-1$ and we do not get any new solution.

The same analysis can be applied to the equation
$P_{A}(2(2c+1)m+2i+1)=P_{A}(2(2c+1)m+2i+2)$ and we easily get that it has no
solutions in $\mathbb{N}$. We omit the simple details. ∎

######  Theorem 4.3.

Let $a\in\mathbb{N}_{\geq 3}$ and put $A=\\{1,2,a\\}$.

  1. (1)

If $a\not\equiv 2\mkern 4.0mu({\operator@font mod}\mkern 6.0mu4)$ then the
Diophantine equation $P_{A}(m)=P_{4}(n)$ has infinitely many solutions in
positive integers.

  2. (2)

If $a\equiv 2\mkern 4.0mu({\operator@font mod}\mkern 6.0mu4)$ then the
Diophantine equation $P_{A}(m)=P_{4}(n)$ has only finitely many solutions in
integers.

###### Proof.

The general strategy of the proof is the following. The set of integer
solutions of the equation $P_{A}(m)=P_{4}(n)$ is the sum of sets
$U_{i,j},i\in\\{0,\ldots,2a-1\\},j\in\\{0,\ldots,11\\}$, where

$U_{i,j}=\\{(2am+i,12n+j):\;P_{A}(2am+i)=P_{4}(12n+j),\;m,n\in\mathbb{N}\\}.$

Thus, in order to show that the Diophantine equation $P_{A}(x)=P_{4}(y)$ has
infinitely many solutions it is enough to prove that for some
$i\in\\{0,\ldots,2a-1\\},j\in\\{0,\ldots,11\\}$ the set $U_{i,j}$ is infinite.
However, because $\operatorname{deg}P_{A}(2am+i)$ has degree 2 and
$\operatorname{deg}P_{4}(12n+j)=3$, the set $U_{i,j}$ is infinite if and only
if the discriminant (with respect to the variable $m$) of the polynomial

$F_{i,j,a}(m,n)=P_{A}(2am+i)-P_{4}(12n+j)$

is a square. Here, we treat $i,j$ and $n\in\mathbb{N}$ as variables. Moreover,
because $P_{4}(12n+j)$ is o degree three, then the discriminant

$G_{i,j,a}(n)=\operatorname{Disc}_{m}(F_{i,j,a}(m,n))$

is a polynomial of degree three in the variable $n$. Thus, the value of
$G_{i,j,a}(n)$ is square for infinitely many values of $n\in\mathbb{N}$ if and
only if $G_{i,j,a}$, treated as a polynomial in the variable $n$, has double
root. This, in turn, is equivalent with the vanishing of the discriminant of
$G_{i,j,a}$. Summing up we get the following implication

$U_{i,j,a}\;\mbox{is
infinite}\;\Longrightarrow\;H_{i,j,a}:=\operatorname{Disc}_{n}(\operatorname{Disc}_{m}(F_{i,j,a}(m,n)))=0.$

From our discussion it follows that we need to investigate the vanishing of
$H_{i,j}$. However, before we will go one, let us note that a priori
$U_{i,j,a}$ can be finite and the condition $H_{i,j,a}=0$ can be still
satisfied (in other words we can not expect to have equivalence between the
conditions above). This is clear. Due to the form of the polynomial
$F_{i,j,a}$ to get an element of $U_{i,j,a}$ some additional congruence
conditions need to be satisfied (which are not seen in discriminant
computations). Indeed, let us take $a=9,i=0,j=3$, i.e.,

$F_{0,3,9}(m,n)=9m^{2}+6m-(12n^{3}-24n^{2}-15n-2).$

Then $G_{0,3,9}(n)=108(n+1)(2n+1)^{2}$. Thus $G_{0,3,9}(n)$ is a square if and
only if $n=3u^{2}-1$. However,
$F_{0,3,9}(m,3u^{2}-1)=(3m-18u^{3}+3u+1)(3m+18u^{3}-3u+1)$ and it is clear
that our equation has no solutions.

After this discussion let us back to the proof of the statement.

To get the first part of our theorem we perform case by case analysis. If
$a\equiv 0\mkern 4.0mu({\operator@font mod}\mkern 6.0mu4)$, say $a=4s$ then
$H_{2s-2,3,4s}=0$ and we have that

$G_{2s-2,3,4s}(n)=48(n+1)(2n+1)^{2}s.$

Thus, in order to make the above expression a square, we need to take
$n=3su^{2}-1$. Then

$F_{2s-2,3,4s}(m,3su^{2}-1)=s(-2m+18su^{3}-3u-1)(2m+18su^{3}-3u+1)$

and $m=\frac{1}{2}\left(18su^{3}-3u-1\right)$. Summing up: if $u$ is odd
positive integer then the numbers

$\displaystyle x$ $\displaystyle=8sm+2s-2=2\left(36s^{2}u^{3}-6su-s-1\right),$
$\displaystyle y$ $\displaystyle=12n+3=9\left(4su^{2}-1\right)$

solve the equation $P_{A}(x)=P_{4}(y)$.

Because in the case $a\equiv 1,3\mkern 4.0mu({\operator@font mod}\mkern
6.0mu4)$ the reasoning goes in exactly the same way we present only the
appropriate values of $i,j$ and the corresponding solutions $x,y$.

If $a\equiv 1\mkern 4.0mu({\operator@font mod}\mkern 6.0mu4)$, i.e., $a=4s+1$
then we take $i=2s-1,j=2$ and $u$ positive odd number and get

$\displaystyle x$
$\displaystyle=\frac{1}{2}(9(4s+1)^{2}u^{3}-3(4s+1)u-4(s+1)),$ $\displaystyle
y$ $\displaystyle=(9s+4)u^{2}-7,$

positive integers solving the equation $P_{A}(x)=P_{4}(y)$.

If $a\equiv 1\mkern 4.0mu({\operator@font mod}\mkern 6.0mu4)$, i.e., $a=4s+3$
then we take $i=s+1,j=0$ and $u$ positive odd number and get

$\displaystyle x$
$\displaystyle=\frac{1}{2}(9(4s+3)^{2}u^{3}+(4s+3)u-2(2s+3)),$ $\displaystyle
y$ $\displaystyle=3((4s+3)u^{2}-1),$

positive integers solving the equation $P_{A}(x)=P_{4}(y)$.

To get the second part of our theorem we need to investigate the vanishing of
$H{i,j,4s+2}$. Because we have exact expression for $P_{A}(4(2s+1)m+2i)$
(which is equal to $P_{A}(4(2s+1)m+2i)$) it is enough to consider
$H_{i,j,4s+2}$ for $j=0,\ldots,11$ as a polynomial in two variables: $s$ and
$i$. Because we need to consider two cases $i\in\\{0,\ldots,2s\\}$ and
$i\in\\{2s+1,\ldots,4s-1\\}$ we work with 24 polynomials $H_{i,j,4s+2}$. It is
easy compute these polynomials. Each has the form

$C(i,j)(2s+1)^{2}Q_{j}(i,s)R_{j}(i,s),$

where $C(i,j)\in\mathbb{Z}$ and $Q_{j},R_{j}$ are quadratic inhomogeneous
polynomials. In each case the quadratic forms $Q_{j},R_{j}$ has no integer
zeros. Because in each case the reasoning is the same we present only one
typical example. So let us suppose that $i\in\\{0,\ldots,2s\\}$ and take
$j=0$. Then

$H_{i,0,4s+2}=-27648(2s+1)^{2}(4(i^{2}+s^{2})-4i(2s-1)+3)(36(i^{2}+s^{2})-36i(2s-1)-4s+25)$

and quick computation reveals that each factor is non-zero for
$s\in\mathbb{Z}$ and $i\in\mathbb{N}$. Performing the same analysis for the
rest of polynomials we get the statement of our theorem. ∎

The first part of the above result can be further generalized. In order to get
the generalization we will need the following simple

######  Theorem 4.4.

Let $a\in\mathbb{N}_{\geq 3}$ and put $A=\\{1,2,a\\}$. The Diophantine
equation $y^{2}=P_{A}(x)$ has infinitely many solutions in positive integers.

###### Proof.

First we consider the case $a$ is not a square.

Let $a$ be even, i.e., $a=2c$ for some $c$. Take $i=0$ in the first formula in
Theorem 4.1, i.e., we work with the Diophantine equation

$P_{A}(4cn)=2cn^{2}+(c+2)n+1=y^{2}.$

In order to show that the equation $P_{A}(4cn)=y^{2}$ has infinitely many
integral solutions, we will follow the standard argument to parameterize
(rational) solutions since we know that $(n,y)=(0,1)$ solves the equation. The
lines through $(0,1)$ can be written as $y=mn+1.$ Therefore we get that
$2cn^{2}+(c+2)n+1=\left(mn+1\right)^{2},$ that is $n=0$ or

$n=\frac{c+2-2m}{m^{2}-2c}.$

Here $m=u/v$ is a rational parameter, so we have that

$n=\frac{(c+2)v^{2}-2uv}{u^{2}-2cv^{2}}.$

For our assumption, $a=2c$ is not a square, then we consider the Pell-equation
$u^{2}-2cv^{2}=1$ and denote the sequence of positive integer solutions by
$(u_{k},v_{k}).$ In this case it follows that $n=(c+2)v_{k}^{2}-2u_{k}v_{k}$
and $y=(c+2)u_{k}v_{k}-2u_{k}^{2}.$

Let us now consider the case with $a=2c+1$ odd. Applying the second formula in
Theorem 4.1 with $i=0$, we work with the Diophantine equation

$P_{A}(2(2c+1)n)=(2c+1)n^{2}+(c+2)n+1=y^{2}.$

Again, in order to show that the equation $P_{A}(4cn)=y^{2}$ has infinitely
many integral solutions, we follow the standard argument. The pair
$(n,y)=(0,1)$ solves the equation. The lines through $(0,1)$ can be written as
$y=mn+1.$ Therefore, we get that $(2c+1)n^{2}+(c+2)n+1=\left(mn+1\right)^{2},$
that is $n=0$ or

$n=\frac{c+2-2m}{m^{2}-2c-1}.$

Here $m=u/v$ is a rational parameter, so we have that

$n=\frac{(c+2)v^{2}-2uv}{u^{2}-(2c+1)v^{2}}.$

For our assumption, $a=2c+1$ is not a square, then we consider the Pell-
equation $u^{2}-(2c+1)v^{2}=1$ and denote the sequence of positive integer
solutions by $(u_{k},v_{k}).$ In this case it follows that
$n=(c+2)v_{k}^{2}-2u_{k}v_{k}$ and $y=(c+2)u_{k}v_{k}-2u_{k}^{2}$.

Summing up: we proved that if $a$ is not a square then the Diophantine
equation $y^{2}=P_{A}(x)$ has infinitely many solutions in positive integers.

It remains to deal with the case when $a$ is a square (even or odd). We follow
similar lines, so we only provide details in case of $a=4t^{2}$, that is when
$a$ is an even square. Again, using the first formula from Theorem 4.1 with
$c=2t^{2},i=2t^{2}-2<c-1$ we get that

$P_{A}(8t^{2}n+2t^{2}-2)=4t^{2}n^{2}+4t^{2}n+t^{2}=t^{2}(2n+1)^{2},$

and for each $n\in\mathbb{N}_{+}$ the number $P_{A}(8t^{2}n+2t^{2}-2)$ is a
square and the Diophantine equation $y^{2}=P_{A}(x)$ has infinitely many
solutions in positive integers.

If $a=(2t+1)^{2}$ is an odd square, then using the second formula from Theorem
4.1 with $c=2t(2t+1)$ and $i=2t^{2}-2$ we find that

$P_{A}(2(2t+1)^{2}n+2t^{2}-2)=((2t+1)n+t)^{2}.$

For each $n\in\mathbb{N}$ the number $P_{A}(2(2t+1)^{2}n+2t^{2}-2)$ is a
square and our theorem is proved. ∎

######  Theorem 4.5.

Let $a,b\in\mathbb{N}_{\geq 3},a<b$ such that $a,b$ are divisible by 4 and
either $a/2$ or $b/2$ is not a square. Put $A=\\{1,2,a\\},B=\\{1,2,b\\}$. The
Diophantine equation $P_{A}(x)=P_{B}(y)$ has infinitely many solutions in
positive integers.

###### Proof.

Let $a=2s$ and $b=2t.$ It follows from Theorem 4.1 that

$\displaystyle P_{A}(4sn)$ $\displaystyle=$ $\displaystyle 2sn^{2}+(s+2)n+1,$
$\displaystyle P_{B}(4tm)$ $\displaystyle=$ $\displaystyle 2tm^{2}+(t+2)m+1.$

Suppose that $a/2=s$ is not a square. We have the solution $(0,0)$ of the
equation $P_{A}(4sn)=P_{B}(4tm)$ so we write $n=km$ for some rational number
$k=u/v.$ Solving the equation for $m$ provides that either $m=0$ or

$m=\frac{(t+2)v^{2}-suv-2uv}{2su^{2}-2v^{2}}.$

Since $s$ and $t$ are even integers the numerator is divisible by 2 and we
obtain the expression

$m=\frac{(s+2)/2uv-(t+2)/2v^{2}}{v^{2}-su^{2}}.$

The integer $s$ is not a square, hence we consider the sequence of positive
solutions $(u_{k},v_{k})$ of the Pell-equation $v^{2}-su^{2}=1.$ For these
solutions we have $m=(s+2)/2u_{k}v_{k}-(t+2)/2v_{k}^{2}$ and
$n=(s+2)/2u_{k}^{2}-(t+2)/2u_{k}v_{k}.$ ∎

In view of theorem above we formulate the following conjecture.

######  Conjecture 4.6.

Let $a,b\in\mathbb{N}_{\geq 3},a<b$ and put $A=\\{1,2,a\\},B=\\{1,2,b\\}$. The
Diophantine equation $P_{A}(x)=P_{B}(y)$ has infinitely many solutions in
positive integers.

Let us note that to prove the above conjecture it is enough to find a pair
$(i,j)$ of integers such that
$i\in\\{0,\ldots,2a-1\\},j\in\\{0,\ldots,2b-1\\}$ and the Diophantine equation
$P_{A}(2ax+i)=P_{B}(2by+j)$ has infinitely many solutions in integers.
Although for any fixed values of $a,b$ it is easy to find suitable $i,j$ we
were unable to get the general result.

##  5\. Remarks on the Diophantine equation $y^{2}=P_{A}(x)$

A difficult and still unsolved question is whether the number $p(n)$ can be a
perfect power. Let us recall that $p(n)$ counts the number of all partitions
of $n$, i.e.,

$\prod_{n=1}^{\infty}\frac{1}{1-x^{n}}=\sum_{n=0}^{\infty}p(n)x^{n}.$

In other words, we do not know any example of $n\geq 2$ such that $y^{k}=p(n)$
for some $k\in\mathbb{N}_{\geq 2}$. In fact Zhi-Wei Sun conjectured that the
equation $y^{k}=p(n)$ has no solutions in positive integers $n,y,k$ with
$k\geq 2$. Let us also note that Alekseyev checked that there are no solutions
with $n\leq 10^{8}$ [20].

A question arises whether some results concerning the equation
$y^{2}=P_{k}(x)$ can be proved for some values of $k\in\mathbb{N}_{+}$. We
know that the Diophantine equation $y^{2}=P_{k}(x)$ has infinitely many
solutions in positive integers for $k\leq 4$. Indeed, to get the result for
$k=3$ it is enough to back to the explicit form of $P_{3}(6n+i)$ presented in
the proof of Theorem 3.1. It is easy to see that for $i\in\\{0,1,4,5\\}$ the
Diophantine equation

$y^{2}=P_{3}(6n+i)$

has infinitely many solutions in positive integer. For example, if $i=4$ we
deal with the equation $y^{2}=(n+1)(3n+4)$. Thus, if $(u_{k},v_{k})$ is a
solution of the Pell equation $v^{2}-3u^{2}=1$ then the pair
$(n,y)=(u_{k}^{2}-1,u_{k}v_{k})$ solves our equation.

If $k=4$ then again we back to the explicit form of $P_{4}(6n+2i+1)$ for
$i=0,1,2$ and $P_{4}(12n+2i)$ for $i=0,1,\ldots,5$. A quick inspection reveals
that the equation $y^{2}=P_{4}(6n+2i+1)$ has infinitely many solutions for
$i=1,2$. For example, if $i=1$ then it is enough to take $n=6u^{2}-2$ and
$y=3u(6u^{2}-1)$. Similarly, it is easy to see that equation
$y^{2}=P_{4}(12n+2i)$ has infinitely many solutions for $i=3,4$.

The first non-trivial problem is characterization of the positive integer
solutions of the Diophantine equation $y^{2}=P_{5}(x)$. We prove the following

######  Theorem 5.1.

The equation $y^{2}=P_{5}(x)$ has only finitely many solutions in positive
integers. More precisely, the pair $(x,y)$ is a solution if and only if
$(x,y)=(1,1),(2027,77129)$.

###### Proof.

We have 60 curves of the form $y^{2}=P_{5}(60n+i),i\in\\{0,\ldots,59\\}$. If
$i\in\\{5,20,25,40\\}$ the corresponding quartic has no
$\mathbb{Q}_{5}$-rational points, and thus has no rational points at all.

Similarly as in the proof of Theorem 3.1 and Theorem 3.2, we apply the
procedure IntegralQuarticPoints() to determine all integral solutions in the
remaining 56 cases. In particular, the solution $(1,1)$ comes from the
equation $y^{2}=P_{5}(60n+1)$ with $n=0$. The solution $(2027,77129)$ comes
from the solution $(n,y)=(33,77129)$ of the equation $y^{2}=P_{5}(60n+47)$.
The procedure works well, except in 6 special cases. Here we do not get any
error message like in the special cases appearing in the proof of Theorem 3.2,
but warnings about time-consuming final enumerations. The 6 problematic
polynomials correspond to $i\in\\{21,24,48,51,54,57\\}$. The equations (up to
multiplication by $16$) corresponding to $i=21,24$ give the following
equations

$\displaystyle 5y^{2}$ $\displaystyle=$ $\displaystyle
u(36u^{3}+108u^{2}+34u+12),\quad u=5(2n+1),$ $\displaystyle 5y^{2}$
$\displaystyle=$ $\displaystyle u(36u^{3}-108u^{2}+34u-12),\quad u=2(5n+3),$

respectively. Hence we need to resolve the following elliptic equations

$Y^{2}=X^{3}+108\delta X^{2}+3384\delta^{2}X+15552\delta^{3},$

where $\delta$ divides 60. We only get the trivial solution given by $u=0$.

For $i\in\\{48,51,54,57\\}$ after the substitution $u=2(n+1)$ we get the
following quartic equations

$\displaystyle y^{2}$ $\displaystyle=$ $\displaystyle
u(4500u^{3}-2700u^{2}+470u-12),$ $\displaystyle y^{2}$ $\displaystyle=$
$\displaystyle u(4500u^{3}-900u^{2}-70u+4),$ $\displaystyle y^{2}$
$\displaystyle=$ $\displaystyle u(4500u^{3}+900u^{2}-70u-4),$ $\displaystyle
y^{2}$ $\displaystyle=$ $\displaystyle u(4500u^{3}+2700u^{2}+470u+12).$

We obtain elliptic equations in a similar way as before, so we omit details.
It turns out that we get only the trivial solution with $u=0$ from these
cases. ∎

The case of the equation $y^{2}=P_{6}(x)$ is far more difficult. To get the
solutions we need to consider 60 genus 2 curves

$C_{i}:\;y^{2}=P_{6}(60n+i),\quad i=0,\ldots,59.$

Let $J_{i}=\operatorname{Jac}(C_{i})$ be the Jacobian of the curve $C_{i}$ and
by $r_{i}$ denote the rank of $J_{i}$. We checked that $r_{i}\leq 5$ for
$0\leq i\leq 59$.

$\begin{array}[]{|l||l|}\hline\cr r&\mbox{values of}\;i\;\mbox{such
that}\;r_{i}\leq r\\\ \hline\cr 0&3,14,34,47,50,51,55,59\\\
1&18,22,27,32,35,38,41,43,44,45,46,54\\\
2&0,7,8,9,15,23,24,25,26,28,29,30,33,36,37,39,40,42,49,52,53,57,58\\\
3&2,5,6,11,17,31,48\\\ 4&4,10,13,16,19,20,21,56\\\ 5&1,12\\\
\hline\cr\end{array}$

Table. Upper bounds for the $\mathbb{Q}$-rank of the Jacobian $J_{i}$ of the
curve $C_{i}:\;y^{2}=P_{6}(60x+i)$ for $i=0,\ldots,59$.

It is curious that the polynomial $P_{6}(60n+i)$ is reducible (in the ring
$\mathbb{Q}[n]$) for $i\in\\{40,\ldots,59\\}$ and thus, instead of working
with genus two curve we need to play with certain curves of the type
$y^{2}=Q_{i}(x)$, where $Q_{i}$ is a quartic polynomial.

If the rank of the Mordell-Weil group is less than the genus of the curve,
that is 2 in these cases, then classical Chabauty’s method [6] may be applied
to determine all rational points on the hyperelliptic curves. If the rank is
greater than or equal to 2, then there are two different approaches to compute
the set of integral points on the curves (see [4, 10]). The difficulty is that
one needs a Mordell-Weil basis. Among the above curves there are some for
which we were not able to obtain such bases, these are as follows $C_{i}$ with

$i\in\\{15,16,23,24,27,28,29,31,32,33,35,36,38,39\\}.$

The most interesting one may be the hyperelliptic curve given by

$y^{2}=12x^{5}+1125x^{4}+41960x^{3}+778050x^{2}+7171020x+26276400,$

which, as computed with the help of Magma, is the minimal model of the curve
$C_{27}:~{}~{}y^{2}=P_{6}(60n+27).$ In this case the rank is 1, however we
were unable to found a generator of the Mordell-Weil group.

We finish with the following

######  Conjecture 5.2.

Let $n\in\mathbb{N}_{\geq 6}$. The only positive integer solution of the
Diophantine equation $y^{2}=P_{n}(x)$ is $x=y=1$.

Motivated by the results above one can ask a more general

######  Question 5.3.

Let $A\subset\mathbb{N}_{+}$ and suppose that the Diophantine equation
$y^{2}=P_{A}(x)$. How large the number $\\#A$ can be?

In case of $\\#A=5$ there is a large number of sets such that
$P_{A}(L_{A}n+i)$ is a square of a polynomial in $n$. More precisely, with the
constraint $\operatorname{max}(A)\leq 15$, there are exactly 119 different
pairs $(A,i)$ such that $P_{A}(L_{A}n+i)$ is a square of a polynomial with
integer coefficients. For example, if $A=\\{1,2,8,10,15\\}$, then $L_{A}=120$
and for $i=1,11,41,43,73,83,91,113$ we have $P_{A}(L_{A}n+i)$ is a square of a
polynomial. In particular,

$P_{A}(120n+1)=(4n+1)^{2}(15n+1)^{2}.$

In the table below we collect data concerning our search.

$\begin{array}[]{|l|l|l|}\hline\cr A&L_{A}&i\\\
\hline\cr\\{1,2,8,10,15\\}&120&1,11,41,43,73,83,91,113\\\
\\{1,4,5,10,12\\}&60&12,16,36,52\\\
\\{1,4,8,9,12\\}&72&1,13,19,25,37,43,49,61,67\\\
\\{1,5,6,8,10\\}&120&2,8,13,17,32,37,53,58,73,77,82,88,97,98,112,113\\\
\\{2,3,7,8,14\\}&168&32,102,144,158\\\
\\{2,4,5,6,10\\}&60&12,16,17,21,36,41,52,57\\\
\\{3,4,6,9,12\\}&36&3,7,11,27,31,35\\\
\\{3,5,6,9,15\\}&90&18,23,24,28,29,34,54,59,64,78,83,88\\\
\\{4,5,6,12,15\\}&60&27,51\\\ \\{4,7,9,12,14\\}&252&58,64,142,148,226,232\\\
\\{5,6,8,9,10\\}&360&8,29,53,74,89,98,104,113,128,149,173,194,209,\\\
&&218,224,233,248,269,293,314,329,338,344,353\\\
\\{5,7,9,14,15\\}&630&47,113,173,197,257,323,383,407,467,533,593,617\\\
\\{7,8,10,14,15\\}&840&182,212,364,422,574,604,812,814\\\
\hline\cr\end{array}$

Table. The sets $A$ such that $\\#A=5,\operatorname{max}(A)\leq 15$ and there
is an $i\in\\{0,\ldots,L_{A}-1\\}$ such that $P_{A}(L_{A}n+i)$ is a square of
a polynomial in $\mathbb{Z}[n]$,

In case of $\\#A=6$ there is a large number of sets such that
$P_{A}(L_{A}n+i)$ is a square of a polynomial (with rational coefficients) in
$n$ times a linear factor (note that this is only possibility to get
infinitely many square values). However, in each case the values of a
corresponding linear factor nor the value of $P_{A}(L_{A}n+i)$ can be a square
of an integer.

We were able to find only the one set $A$ with 7 elements,
$\operatorname{max}(A)\leq 10$ and such that $y^{2}=P_{A}(x)$ has infinitely
many solutions in positive integers. More precisely, if
$A=\\{1,2,4,5,8,9,10\\}$ then

$\displaystyle P_{A}(360n+95)$
$\displaystyle=25(3n+1)^{2}(18n+5)^{2}(36n+13)(40n+13),$ $\displaystyle
P_{A}(360n+226)$ $\displaystyle=25(3n+2)^{2}(18n+13)^{2}(36n+23)(40n+27).$

One can easily check that the factor $(36n+13)(40n+13)$ is a square infinitely
often. The smallest values of $n$ which makes this factor a square, are
$n=0,494,712842,\ldots$. However, the factor $(36n+23)(40n+27)$ takes square
values for infinitely many values negative values of $n$ and thus is not of
interests for us.

##  6\. Problems, questions and conjectures

Besides the conjectures stated in previous sections, we formulate now several
question and conjectures which hopefully will stimulate further research.

######  Question 6.1.

Let $k\in\mathbb{N}_{\geq 2}$ and $f\in\mathbb{Z}[x]$ be given. Does there
exist an ascending sequence of sets
$A_{2}=\\{a_{1},a_{2}\\}\subset\ldots\subset
A_{k}=\\{a_{1},\ldots,a_{k}\\}\ldots\subset\mathbb{N}_{+}$ such that the
Diophantine equation $P_{A_{k}}(x)=f(y)$ has at least $C_{k,f}$ solutions in
positive integers and $C_{k,f}\rightarrow+\infty$ as $k\rightarrow+\infty$?

Let us observe that without the condition $C_{k,f}\rightarrow+\infty$ the
question is not difficult. Indeed, let us take
$A_{2}=\\{a_{1},a_{2}\\}\subset\mathbb{N}_{+}$ and suppose that
$\gcd(a_{1},a_{2})=1$. As we already proved in Theorem 2.1 the Diophantine
equation $P_{A_{2}}(x)=f(y)$ has infinitely many solutions in positive
integers. If $C\in\mathbb{N}$ is fixed let us take an increasing sequence
$\\{a_{3},a_{4},\ldots,a_{k}\\}$ of positive integers such that $a_{3}$ is
grater then the smallest integer $N$ such that there is at least $C$ values of
$x$ for which there is an integer $y$ satisfying $P_{A_{2}}(x)=f(y)$. Then,
for $A_{k}=\\{a_{1},a_{2},a_{3},\ldots,a_{k}\\}$ the Diophantine equation

$P_{A_{k}}(x)=f(y)$

has at least $C$ solutions in positive integers. Indeed, this is simple
consequence of the recurrence relation satisfied by the sequence
$\\{P_{A_{k}}(n)\\}_{n\in\mathbb{N}}$. Indeed, because
$P_{A_{k}}(n)=P_{A_{k-1}}(n)$ for $n<a_{k}$, then $P_{A_{k}}(n)=P_{A_{2}}(n)$
for $n<\min\\{a_{3},\ldots,a_{k}\\}=a_{3}$ and hence the result.

We proved that the equation $P_{3}(x)=P_{5}(x)$ has only finitely many
solutions in positive integers and it is quite natural to ask whether there
are $A,B$ satisfying $\\#A=3,\\#B=5$, such that the equation
$P_{A}(x)=P_{B}(y)$ has infinitely many solutions in positive integers. To get
the result in this direction we will need the following.

######  Lemma 6.2.

Let $b\in\mathbb{N}_{\geq 4}$ and put $B=\\{1,2,3,4,b\\}$.

  1. (1)

If $b=4(6k+1),j=3(8k-1)$ for some $k\in\mathbb{N}_{+}$, then
$P_{B}(3bn+j)=(3n+2)((6k+1)n+2k)Q_{1}(k,n)$, where

$Q_{1}(k,n)=3(6k+1)^{2}n^{2}+2(9k+1)(6k+1)n+6k(4k+1).$

  2. (2)

If $b=4(6k+5),j=24k+13$ for some $k\in\mathbb{N}_{+}$, then
$P_{B}(3bn+j)=(3n+1)((6k+5)n+4k+3)Q_{2}(k,n)$, where

$Q_{2}(k,n)=3(6k+5)^{2}n^{2}+2(6k+5)(9k+7)n+24k^{2}+36k+1).$

  3. (3)

If $b=4(12k+2),j=48k+1$ for some $k\in\mathbb{N}_{+}$, then
$P_{B}(3bn+j)=(3n+1)(2(6k+1)n+8k+1)Q_{3}(k,n)$, where

$Q_{3}(k,n)=12(6k+1)^{2}n^{2}+2(6k+1)(36k+5)n+96k^{2}+24k+1.$

  4. (4)

If $b=4(12k+10),j=48k+1$ for some $k\in\mathbb{N}_{+}$, then
$P_{B}(3bn+j)=(3n+2)(2(6k+5)n+4k+3)Q_{4}(k,n)$, where

$Q_{4}(k,n)=12(6k+5)^{2}n^{2}+2(6k+5)(36k+29)n+3(4k+3)(8k+7).$

###### Proof.

Let us note that the sequence $\\{P_{B}(n)\\}_{n\in\mathbb{N}}$ satisfies the
following recurrence relation

$P_{B}(n)=\begin{cases}\begin{array}[]{ll}P_{A}(n),&n<b,\\\
P_{B}(n-b)+P_{A}(n),&b\leq n,\end{array}\end{cases}$

where $A=\\{1,2,3,4\\}$. We know the polynomial expressions for
$P_{A}(12n+i),i\in\\{0,\ldots,11\\}$ and that
$P_{B}(L_{b}n+j),j\in\\{0,\ldots,L_{b}-1\\}$, where
$L_{b}=\operatorname{LCM}(1,2,3,4,b)$, is a polynomial of degree 4 with
rational coefficients. Using induction one can obtain the expression for the
polynomials of interests. We omit tiresome details. ∎

######  Theorem 6.3.

Let $a\in\mathbb{N}_{\geq 3},b\in\mathbb{N}_{\geq 4}$ and put
$A=\\{1,2,a\\},B=\\{1,2,3,4,b\\}$. If $a\equiv 1,2,5,7,11,10\mkern
4.0mu({\operator@font mod}\mkern 6.0mu12)$ and $b=4a$, then the Diophantine
equation $P_{A}(x)=P_{B}(y)$ has infinitely many solutions in positive
integers.

###### Proof.

Note that if $a\equiv 1,2,5,7,11,10\mkern 4.0mu({\operator@font mod}\mkern
6.0mu12)$, then $a$ can be written in one of the following form:
$a=6k+1,a=6k+5,a=12k+2$ $a=12k+10$. Thus, in each case, the value of $b=4a$ is
exactly the value of $b$ considered in Lemma 6.2. Following the idea of proof
of Theorem 4.3 we present the values of $i,j$ such that the polynomial
$P_{A}(2am+i)-P_{B}(3bn+j)$ is reducible and the coefficient in of the linear
factor (in $m$) near $m$ is equal to 1.

Let $a=6k+1,b=4a,i=11k,j=3(8k-1)$. Then
$P_{A}(2am+i)-P_{B}(3bn+j)=R_{1}(m,n)R_{2}(m,n)$, where

$\displaystyle R_{1}(m,n)$ $\displaystyle=m-3(6k+5)n^{2}-2(9k+7)n-4k+1,$
$\displaystyle R_{2}(m,n)$
$\displaystyle=(6k+1)m+3(6k+1)^{2}n^{2}+2(6k+1)(9k+1)n+24k^{2}+12k+1.$

Thus, if $m=3(6k+5)n^{2}+2(9k+7)n+4k-1$ then $P_{A}(2am+i)=P_{B}(3bn+j)$ and
our equation has infinitely many solutions.

Because in each case we proceed in the same way we present only the values of
$a,i,b,j$ and the corresponding solution for $m$.

If $a=6k+5,i=7k+4,b=4(6k+4),j=24k+13$, then

$m=3(6k+5)n^{2}+2(9k+7)n+2(2k+1).$

If $a=2(6k+1),i=14k,b=8(6k+1),j=48k+1$, then

$m=6(6k+1)n^{2}+(36k+5)n+8k.$

If $a=2(6k+5),i=2(11k+8),b=8(6k+5),j=3(16k+11)$, then

$m=6(6k+5)n^{2}+(36k+29)n+8k+5.$

∎

We proved that for many choices of sequences $A,B$, the corresponding
Diophantine equation $P_{A}(x)=P_{B}(y)$ has infinitely many solutions in
positive integers. However, in each case under consideration we had
$\operatorname{min}\\{\\#A,\\#B\\}\leq 3$. This observation lead us to the
following.

######  Question 6.4.

Let $A,B\subset\mathbb{N}_{+}$. Let us suppose that the Diophantine equation
$P_{A}(x)=P_{B}(y)$ has infinitely many (non-trivial) solutions in positive
integers. How large the number $\operatorname{min}\\{\\#A,\\#B\\}$ can be?

Let us explain what a trivial solution means. More precisely, if for example
$A=\\{1,pa_{2},...,pa_{k}\\}$ then if $P_{A}(pn)$ is a non-zero, then in each
representation

$1\cdot x_{1}+\sum_{i=2}^{k}pa_{i}x_{i}=pn$

we need to have $p|x_{1}$ and thus we get a representation

$1\cdot y_{1}+\sum_{i=2}^{k}a_{i}x_{i}=n.$

It is clear that this mapping can be reversed. Thus, by taking
$B=\\{1,a_{2},...,a_{k}\\}$ we have the boring identity $P_{A}(pn)=P_{B}(n)$.

Thus, in regards to question above, we considered equations of the form
$P_{A}(x)=P_{B}(y),$ where $A,B$ are sets having 5 elements from
$\\{1,2,\ldots,10\\}$ and one of the elements is 1. We searched for reducible
polynomials $P_{A}(x)-P_{B}(y)$ having a linear or quadratic factor. We
implemented a parallel algorithm and used SageMath on a machine having 16
cores. It took about 10 hours to determine the appropriate polynomials. There
are 44982 such cases. Among these polynomials we looked for examples providing
infinitely many integral solutions. To reduce the time of computation a
timeout was set to be 60 seconds. There are 392 cases for which the 60 seconds
were not sufficient to compute the result. There are 2338 quadratic equations
that yield infinitely many integral solutions and 2100 linear equations that
provide parametric solutions. However, even in the case of reducibility we
sometimes get factors without positive integer solutions. We present several
examples.

Let $A=\\{1,2,4,5,6\\}$ and $B=\\{1,4,6,9,10\\}.$ Here we obtain that
$P_{A}(60m+22)-P_{B}(180n+111)$ is, up to a constant factor, equal to
$f_{1}(m,n)f_{2}(m,n)$, where

$\displaystyle f_{1}(m,n)$ $\displaystyle=$ $\displaystyle
150m^{2}+450n^{2}+155m+630n+259,$ $\displaystyle f_{2}(m,n)$ $\displaystyle=$
$\displaystyle 30m^{2}-90n^{2}+31m-126n-36.$

The equation $f_{1}(m,n)=0$ has no solution modulo 5. The equation
$f_{2}(m,n)=0$ has infinitely many integral solutions. However, all are
negative and are not of interest for us.

As a second example consider $A=\\{1,2,4,6,10\\}$ and $B=\\{1,2,5,6,8\\}.$ We
get that $P_{A}(60m+17)-P_{B}(120n+17)$ is, up to a constant factor, equal to
$g_{1}(m,n)g_{2}(m,n)g_{3}(m,n)$, where

$\displaystyle g_{1}(m,n)$ $\displaystyle=$ $\displaystyle m-2n$
$\displaystyle g_{2}(m,n)$ $\displaystyle=$ $\displaystyle 15m+30n+14,$
$\displaystyle g_{3}(m,n)$ $\displaystyle=$ $\displaystyle
75m^{2}+300n^{2}+70m+140n+31.$

We obtain infinitely many integral solutions from the equation $g_{1}(m,n)=0$
(however, these are trivial solutions). The other two equations have no
solutions modulo 5.

As a third example let $A=\\{1,2,3,4,6\\},B=\\{1,2,4,5,10\\}.$ It follows that
$P_{A}(12m+1)-P_{B}(20n+1)=1/6h_{1}(m,n)h_{2}(m,n)$, where

$\displaystyle h_{1}(m,n)$ $\displaystyle=$ $\displaystyle
6m^{2}+10n^{2}+9m+12n+5$ $\displaystyle h_{2}(m,n)$ $\displaystyle=$
$\displaystyle 6m^{2}-10n^{2}+9m-12n.$

The equation $h_{1}(m,n)=0$ can be written as

$15(36m+27)^{2}+(180n+108)^{2}=6399,$

and it follows that the only integral solution is given by $(m,n)=(-1,-1).$
The equation $h_{2}(m,n)=0$ has infinitely many positive integral solutions,
the two smallest being $(m,n)=(2928,2268),(11252256,8715960)$.

For given $A\in\mathbb{N}_{+}$ the function $P_{A}(n)$ has a dual nature: from
one side it is a quasi-polynomial and hance an algebraic object. On the other
side $P_{A}(n)$ is counting function and thus live in a realm of
combinatorics. In this paper we mainly operated on the former side. Thus, it
is natural to state the following general question.

######  Problem 6.5.

Let $A,B\subset\mathbb{N}_{+}$. Does there exist combinatorial conditions on
$A$ and $B$ which guarantees non-existence (or finiteness) of integral
solutions of the Diophantine equation $P_{A}(x)=P_{B}(y)$?

It is clear that the above problem can be stated in a grater generality. More
precisely, we can ask whether there are some combinatorial conditions which
guarantee that for not necessarily finite sets $A_{1},A_{2}$, and
corresponding properties $\mathcal{W}_{1},\mathcal{W}_{2}$, the equation
$p_{A_{1}}(\mathcal{W}_{1},x)=p_{A_{2}}(\mathcal{W}_{2},y)$ has only finitely
many solutions in positive integers.

As we mentioned above, the partition functions count combinatorial objects.
Thus, equality between different partition functions at certain integer
arguments is equivalent with the statement that certain finite sets have the
same number of elements. This suggest the following

######  Problem 6.6.

Let $A,B\subset\mathbb{N}_{+}$ and suppose that the Diophantine equation
$p_{A}(x)=p_{B}(y)$ has infinitely many solutions in integers. Moreover, let
$x=\varphi(n),y=\psi(n)$ be parametrization of one (of possibly many) infinite
part of the solution set, i.e., $p_{A}(\varphi(n))=p_{B}(\psi(n))$ for each
$n\in\mathbb{N}_{+}$. Describe the bijection (in combinatorial or other way)
between the sets
$\operatorname{Part}(\varphi(n))=\operatorname{Part}(\psi(n))$.

Motivated by our findings presented in Theorem 3.2 and related results we
formulate the following

######  Conjecture 6.7.

Let $m,n\in\mathbb{N}_{+}$. If $(m,n)\neq(3,4)$ and $3\leq m<n$, then the
Diophantine equation $P_{m}(x)=P_{n}(y)$ has only finitely many solutions in
non-negative integers.

######  Remark 6.8.

Let us note that from the recurrence relation satisfied by the sequence
$\\{P_{m}(k)\\}_{k\in\mathbb{N}}$, i.e.,

$P_{m}(k)=P_{m-1}(k),k<m,\quad P_{m}(k)=P_{m-1}(k)+P_{m}(k-m),\;k\geq m,$

we know that the equation $P_{m}(x)=P_{n}(y)$ has trivial solutions
$x=y=i,i\leq m$. So, it is reasonable to consider the set

$C_{m,n}:=\\{(x,y)\in\mathbb{Z}\times\mathbb{Z}:\;P_{m}(x)=P_{n}(y)\wedge
y\geq n\\}.$

We believe that much stronger property is true, i.e.,

$\bigcup_{\min\\{m,n\\}\geq 3,m<n,(m,n)\neq(3,4)}C_{m,n}(\mathbb{N})<+\infty.$

######  Acknowledgement.

The authors are grateful to Nikolaos Tzanakis for his ideas to complete the
proof of Theorem 3.2. We are also grateful for an anonymous referee for
remarks which led to improving the presentation.

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