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\title{An elliptic Diophantine equation from the study of partitions}

\author{Dorin Andrica}
\address{``Babe\c{s}-Bolyai'' University, \\ Faculty of Mathematics and Computer Sciences\\
1, Kog\u{a}lniceanu Street,\\
400084 Cluj-Napoca,\\
Romania}
\email{dandrica@math.ubbcluj.ro}
%
\author{George C. \c Turca\c s}
\address{University of Warwick, \\ Mathematics Institute\\
CV4 7AL, Coventry\\
United Kingdom}
\email{g.c.turcas@warwick.ac.uk}
%

\subjclass{Primary 14G05, Secondary 11P57, 11Y50.}
\keywords{Elliptic curves, partitions of a set, Mordell equations, Lutz-Nagell theorem}
\thanks{The second author is supported by
	an EPSRC PhD studentship.}
\begin{abstract}
We present the elliptic equation $X^3+2=Y^2$ as the first in a sequence of Diophantine equations arising from some new results in the theory of partitions of multisets with equal sums. Two proofs for Theorem 2.3, showing that the only integer solutions to this equation are $(-1,1)$ and $(-1,-1)$, are given.
\end{abstract}
\maketitle

\section{Introduction and motivation}
For a positive integer $k \geq 2$ and an arbitrary positive integer $n$, in the papers \cite{AndBag2} and \cite{AndBag1} the authors introduced the sequence $\left( Q_k(n) \right)_{n \geq 1}$,

\begin{equation} \label{integral}
Q_k(n) = \frac{1}{2 \pi} \int_{0}^{2 \pi} \prod_{s=1}^n \left( k-2+ 2 \cos{st} \right) \dif t.
\end{equation}

An enumerative formula for $Q_k(n)$ is given by the number of ordered partitions of $[n]= \{1, \dots, n \}$ into $k$ disjoint sets $A_1, \dots, A_k$ with the property that $\sigma(A_1)= \sigma(A_k)$, where $\sigma(A)$ denotes the sum of all elements in $A$.

Clearly, $Q_k(n)$ is a monic polynomial of degree $n$ in $k-2$. Moreover, in the paper \cite{AndBag2} is proved that
\begin{equation}
Q_k(n) = \sum_{d=0}^n N(d,n) (k-2)^{n-d},
\end{equation}
where for each $d=0, \dots, n$, the coefficient $N(d,n)$ is the number of ordered partitions of $[n]$ into $3$ subsets $A,B,C$ such that $|B|=d$ and $\sigma(A)= \sigma(C)$, where $|B|$ is the cardinality of $B$.

Therefore, $Q_k(n)$ has non-negative integer coefficients, and each coefficient has a combinatorial meaning in terms of partitions of the set $[n]$. A simple direct computation of the integral \eqref{integral} shows that for $n=3,4,5,6$ and $k \geq 2$, we have
\[\def\arraystretch{1.5} 
\begin{array}{l} Q_k(3)=(k-2)^3 +2 ; \\
Q_k(4)=(k-2)^4+4(k-2)+2 ;\\
Q_k(5) = (k-2)^5 + 8 (k-2)^2+6(k-2); \\
Q_k(6) = (k-2)^6 + 12(k-2)^3+16(k-2)^2+6(k-2).  \end{array}\]

The sequence $Q_k(3)$ is indexed as A084380 in OEIS \cite{OEIS}, where it is mentioned that it does not contain any perfect squares, i.e. the elliptic equation $X^3+2=Y^2$ has no solutions in positive integers. This is linked to a Catalan-type conjecture related to Pillai's equation $X^{U}-Y^{V}=m$, with $X,Y,U,V \geq 2$ integers. The conjecture states that for any given integer $m$, there are finitely many perfect powers whose difference is $m$ (see \cite{Wald}, Conjecture 1.6). For $m=2$, it was computationally checked that the only solution involving perfect powers smaller than $10^{18}$ is $2=3^3-5^2$. The number of such solutions is linked to A076427 in OEIS.

Motivated by the property that the sequence $Q_k(3)$ does not contain any perfect squares, in the papers \cite{AndBag2} and \cite{AndBag1}, the authors suggested the following problems: study if the sequence $Q_k(n)$ contains any $n-1$ powers, where $n=4,5$ or $6$. These are equivalent to the study of the following Diophantine equations:
\[\def\arraystretch{1.5} 
\begin{array}{l}
X^4+4X+2=Y^3; \\
X^5+8X^2+6X=Y^4; \\
X^6+12X^3+16X^2+6X=Y^5.
\end{array} \]
Using effective methods for identifying integral points on curves, we will discuss these equations and variations of them in a following series of papers.

In Theorem 2.3 of the present paper we prove that the equation $X^3+2=Y^2$ has only integer solutions $(-1,1)$ and $(-1,-1)$. We give two proofs for this statement. In the first we use the fact that $\mathbb Q(\sqrt{2})$ has trivial class group, property that allows us to pass from factorisations of ideals to nice factorisations in the ring $\mathbb Z[\sqrt{2}]$. The second proof uses the geometry of the elliptic curve defining the equation.

\section{The equation $Y^2=X^3+2$}



Although the family of Mordell equations $Y^2=X^3+D$, where $D \in \Z \setminus \{ 0\}$ (see \cite{Conr}) was extensively studied, we were unable to find in the literature an explicit solution for the case $D=2$. In this section, we give two different solutions to the problem of finding all integral $x,y$ satisfying the aforementioned equation. In the first one we combine factorisations in the ring of integers of $\Q(\sqrt{2})$ with an algorithm for solving cubic Thue equations \cite{BiluHan}. Our second solution relies on the geometric structure of the elliptic curve defined by the given affine equation.

Given a polynomial Diophantine equation, by what are by now commonly called ``local methods", there exist algorithms to  decide whether or not the equation has solutions in $\mathbb R$ or in the $p$-adic numbers $\mathbb Q_p$, where $p$ is any prime. Let us call any of these fields ``local".  The rational numbers can be seen as a subfield of the local fields, therefore if our equation has no solutions in any of the local fields then one can conclude that the equation does not have rational solutions. 

Unfortunately, in our situation the reader can immediately see that trying to deploy such a strategy is doomed to failure, as the equation under attention has $(x,y)=(-1,1)$ and $(x,y)=(-1,-1)$ as solutions, hence it is solvable in all of the local fields. Before going further, let us make a few remarks about the finiteness of the set of integral points on various curves.

For any bivariate polynomial $f \in \Z[X,Y]$, let $C_f := \{ (x,y) \in \overline{\Q}^2 : f(x,y) = 0 \}$ be an affine algebraic curve. The points of $C_f$ with coordinates in $\Q$ are called rational and, in general, for any $S \subseteq \overline{\Q}$, we denote by $C_f(S)=C_f \cap S^2$. Curves can be classified by their genus, a non-negative integer associated to their projectivization. The genus is a geometric invariant. A classical result in number theory is the following theorem

\begin{theorem}[Siegel, 1929] If $f \in \mathbb Z[X,Y]$ defines an irreducible curve $C_f$ of genus $g(C_f)>0$, then $C_f(\mathbb Z)$ is finite.
\end{theorem}

If additionally $g_f(C_f) \geq 2$, this result is superseded by the notorious Falting's theorem, which says that $C_f(\Q)$ is also finite. Although both Siegels' and Faltings' theorems are milestones in number theory, they are ``ineffective" results, meaning that their proof does not even allow one to control the size of the sets known to be finite. Therefore, they cannot be used to explicitly determine $C_f(\mathbb Z)$ or $C_f(\Q)$.

Effectively finding rational points on curves is an incredible difficult task and a very active topic of research. The toolbox for determining $C_f(\Z)$ became a lot richer starting with the monumental work of Baker on linear forms in logarithms. As one of the first applications to his theory, Baker proved the following result.

\begin{theorem}[Baker, 1969] \label{baker} Suppose $f(X,Y)=Y^2-a_nX^n- a_{n-1}X^{n-1}- \dots - a_0 \in \Z[X,Y]$,  the polynomial $a_nX^n+ \dots + a_0$ is irreducible in $\Z[X]$, $a_n \neq 0$ and $n \geq 5$. Let $H = \max\{ |a_0|, \dots, |a_n| \}$. Then, any integral point $(x,y) \in C_f(\Z) $ satisfies $\max(|x|,|y|) \leq \exp \exp \exp \{(n^{10n}H)^{n^2} \}.$
\end{theorem}

Bounds on such solutions have been improved by many authors, but they remain astronomical and often involve inexplicit constants. Let us proceed to the resolution of our Diophantine equations.



To settle the conjecture posed by Andrica and Bagdasar in \cite{AndBag2} and \cite{AndBag1} which inferred that $X^3+2$ does not contain perfect squares when $X$ runs through the set of positive integers, we prove the following theorem.

\begin{theorem} The only solutions of $X^3+2=Y^2$ in the set of integer numbers are $(-1,1)$ and $(-1,-1)$. 
\end{theorem}

A few remarks are in order before giving the proof of this theorem. Since the genus of (the projectivization of) the curve determined by this equation is $1$, we can use Siegel's theorem to deduce that there are finitely many points with integer coordinates. By Theorem \ref{baker}, we know that if $(x,y) \in \Z^2$ is a point lying on this curve, then $$\max(|x|, |y|) \leq \exp \exp \exp((3^{30} \cdot 2)^{3^2}).$$ Although theoretically one could now run a for loop through all possible values of $x$ and check for which $x^3+2$ is a perfect square, the triple exponential bound presented above is astronomical and way out of the current computational limitations. In practice, one could check values of $x$ up to $10^{18}$, but could not hope to even get close to the aforementioned triple exponential. We proceed with the first proof of for our theorem.

\section{Proof to Theorem 2.3}

Let $K= \Q(\sqrt{2})$ and denote by $\mathcal O_K = \Z[\sqrt{2}]$ its ring of integers. The later is a Dedekind domain, i.e. it is Noetherian, integrally closed in its field of fractions $\Frac(\mathcal O_K)=K$ and all its non-zero prime ideals are maximal. For any element $o \in \mathcal O_K$, we are going to denote by $(o) \subseteq O_K$ the principal ideal $o$ generates.

Suppose that $x,y \in \Z \setminus\{0\}$ are such that $y^2=x^3+2$. Therefore, in $\mathcal O_K$ we have the factorization $(y- \sqrt{2}) \cdot (y + \sqrt{2}) = x^3$ and the same holds for the ideals generated by these factors. It is known that ideals of $\mathcal O_K$ factor uniquely into prime ideals. Suppose the prime ideal $\mathfrak p \subset \mathcal O_K$ divides both of the non-zero ideals $(y-\sqrt{2})$ and $(y+\sqrt{2})$. Then, $\mathfrak p$ must divide the ideal generated by the difference $ y+\sqrt{2}- y + \sqrt{2}= 2\sqrt{2}= \sqrt{2^3}$. As $(\sqrt{2}) \subset \mathcal O_K$ is the only prime ideal of $\mathcal O_K$ that lies above $2$, we must have $\mathfrak p= (\sqrt{2})$. Hence, the ideals $(y-\sqrt{2})$ and $(y+\sqrt{2})$ are coprime outside of $(\sqrt{2})$. From the previous factorization, we deduce that for every prime ideal $\mathfrak p \neq (\sqrt{2})$, if $\mathfrak p$ divides $(y-\sqrt{2})$, then $\mathfrak p^3$ divides the same ideal.

It is an easy exercise, using for example the Minkowski bound, to prove that the class group of $K$ is trivial. In particular, this means that every ideal of $\mathcal O_K$ is  principal. Considering the remarks above, we have
$$(y - \sqrt{2}) = (\sqrt{2})^{i} \cdot (x_0)^3=(\sqrt{2}^i \cdot x_0^3), \text{ as ideals, where } i \in \{0,1,2\} \text{  and } x_0 \in \mathcal O_K.$$

We deduce that $y-\sqrt{2}$ and $\sqrt{2}^i x_0^3$ are the same up to a unit in the ring $\mathcal O_K$, that is there exists a unit $u \in U(\mathcal O_K)$ such that $y-\sqrt{2}= u \cdot \sqrt{2}^i \cdot x_0^3$.

By Dirichlet unit's theorem we know that $U(\mathcal O_K)$ is isomorphic to $T \cdot \Z$, where $T$ is the finite group formed by the roots of unity that lie in $K$. It is an easy exercise to verify that $U(\mathcal O_K)= \langle -1 \rangle \cdot \langle 1-\sqrt{2} \rangle,$ so  $1-\sqrt{2}$ is the fundamental unit of $\mathcal O_K$. Observing that every element $u \in U(U_K)$ can be written as $u=(1-\sqrt{2})^{k} \cdot (u_0)^3$ where $j \in \{0,1,2\}$ and $u_0 \in U(\mathcal O_K) \subseteq \mathcal O_K$, we derive that
$$y- \sqrt{2} = (1-\sqrt{2})^j \cdot \sqrt{2}^i \cdot x_1^3,$$
for some $ i,j \in \{0,1,2\}$ and $x_1 \in \mathcal O_K$. The element $x_1$ is of the form $a+b\sqrt{2}$ for $a,b \in \mathbb Z$. For each of the nine choices of a pair $(i,j) \in \{0,1,2 \}^2$, by equating the coefficients of $\sqrt{2}$ in the left and right hand side of the above equation, we obtain an equality of the form
\begin{equation} \label{Thuez}
f(a,b)=-1
\end{equation}
where $f \in \Z[x,y]$ is a homogeneous cubic polynomial. When $f$ is reducible \eqref{Thuez} can be easily solved using factorization in $\Z$. If this is not the case and $f$ is irreducible, the equation \eqref{Thuez} is a cubic Thue equation. It is known that the latter have finitely many integral solutions and routines for determining them have been implemented in various computer algebra packages. We use \textbf{Magma}'s \cite{magma} implementation of the algorithm given in \cite{BiluHan} to solve them.

Purely for space reasons we present here only five of the nine cases considered. These give a complete picture of the possible phenomena and all the remaining cases can be solved analogously.

\noindent\textbf{Case 1.} $i=j=0 \Rightarrow y-\sqrt{2}=(a+b\sqrt{2})^3.$

\noindent Expanding the right hand side, we see that
$$y - \sqrt{2}= a^3+6ab^2+\sqrt{2}\left( 3a^2b+2b^3 \right)$$
and since $1,\sqrt{2}$ are linearly independent over $\Q$ we must have
$$-1 = b \cdot (3a^2+2b^2).$$ Trying $b= \pm 1$, we see that $3a^2+2= \mp 1$ is not solvable. Hence this case does not give us any solutions.

\noindent\textbf{Case 2.} $i=0, j=1 \Rightarrow y-\sqrt{2}=\left(1-\sqrt{2} \right) \cdot (a+b\sqrt{2})^3$.

\noindent This gives us
$$y - \sqrt{2}= a^3+6a^2b+6ab^2+4b^3  + \sqrt{2} \left(a^3+3a^2b+6ab^2+2b^3 \right),$$
which implies that $$-1=a^3+3a^2b+6ab^2+2b^3.$$ \textbf{Magma}'s built-in Thue solver outputs that the only solution to the last equation is $a=-1$ and $b=0$, which gives $y=-1$. Substituting in the original equation $y^2=x^3+2$, we obtain $x=-1$.

\noindent\textbf{Case 3.} $i=0,j=2 \Rightarrow y-\sqrt{2}=(1-\sqrt{2})^2 \cdot (a+b \sqrt{2})^3$. Equating the coefficients of $\sqrt{2}$ on both sides, we get $$-1=-2a^3+9a^2b-12ab^2+6b^3.$$ Using again the Thue solver implemented in \textbf{Magma}, we find that the only solution in integers to the last equation is $a=b=-1$. Substituting, we see that this corresponds to $y=-1$, which implies that $x=-1$ as above.

\noindent\textbf{Case 4.} $i=1, j= 0 \Rightarrow y - \sqrt{2}=\sqrt{2} \cdot (a+b\sqrt{2})^3.$ Expanding and equating the coefficients of $\sqrt{2}$, we see that $$a\cdot(a^2+6b^2)=-1,$$ which has the unique solution $a=-1$, $b=0$. This implies that $y=0$, hence returning to the original equation we find $x^3+2=0$. The latter is insoluble in the integers.

\noindent\textbf{Case 5.} $i=j=1 \Rightarrow y - \sqrt{2}=\sqrt{2} \cdot (1- \sqrt{2})\cdot(a+b\sqrt{2})^3.$ As before, one obtains the cubic Thue equation
$$-1=a^3-6a^2b+6ab^2-4b^3.$$
Using \textbf{Magma}, we compute that the only solutions in integers to the equations above are $(a,b)=(-1,0)$ and $(a,b)=(-5,-1)$. The first pair gives $y=2$ and hence $x^3+2=4$, which is unsolvable in the integers. The second pair gives $y=156$. Returning to the initial equation, we see that $x^3+2=156^2 \Leftrightarrow x^3=24334$, which does not have integral solutions.

Similarly we analysed all the other four cases and obtained exactly one more solution $(x,y)=(-1,1)$, proving our theorem.

In the proof above we made explicit use of the fact that $\Q(\sqrt{2})$ has trivial class group, information that allowed us to pass from factorisations of ideals to nice factorisations of elements in the ring $\Z[\sqrt{2}]$. In general, for $D \in \Z$ the ideal class group of $\Q(\sqrt{D})$ can be arbitrary large so our first strategy will not work for more general Mordell equations. The second proof of our theorem can be adapted to find all the integral solutions of $Y^2=X^3+D$ for any fixed $D \in \Z$.

The given problem is one of explicitly determining the integral points on the affine curve given by $Y^2=X^3+2$. These can be found by exploiting its rich geometric structure, as presented below.

\section{Alternate proof to Theorem 2.3}

The geometry of the curve is better captured by its projectivization
\begin{equation} \label{elcurve}
E := Y^2Z =X^3+2Z^3 \in \mathbb P^2(\C),
\end{equation}
a non-singular projective curve of genus $1$, which contains the point $\mathcal O = [0:1:0] \in \mathbb P^2(\Q)$, commonly called ``the point at infinity". The point at infinity is the only one on the projective curve that does not naturally project on our chosen affine model. The set of complex points on $E$ can be given an abelian group structure for which the distinguished point $\mathcal O$ acts as the identity element. The group law is given by chord-tangent formulas and therefore it is easy to see that $E(\mathbb Q)$ is a subgroup of $E(\mathbb C)$. By a famous theorem of Mordell, we know that $E(\mathbb Q) \cong T \times \mathbb Z^r$ (as abstract abelian groups) where $T$ is a finite group, commonly called \textit{the torsion subgroup} and $r$ is a positive integer called the \textit{rank}.

Using the Lutz-Nagell theorem (see Corollary 7.2 in \cite{Silv}), it is easy to deduce that $T$ is included in $\{\mathcal O, P, -P\}$, where $P=[-1:1:1]$ and $-P=[-1:-1:1]$ are inverses of each other under the group law. Using the formulae for addition on the elliptic curve, we compute all the values of $2\cdot P, \dots, 12 \cdot P$ and observe that none of them is equal to the origin $\mathcal O$.
For example, $5 \cdot P =
[108305279/48846121 : 1226178094681/341385539669 : 1] \neq \mathcal O,$
and the larger multiples of $P$ involve denominators that are too big to fit in one line. In his seminal article \cite{Maz}, Mazur gave a classification of all the possible isomorphisms types for the torsion group of an elliptic curve defined over $\mathbb Q$. From there, we see that the order of any torsion point is at most $12$ and therefore we can conclude that $P$ has infinite order. 

The non-torsion part of $E(\mathbb Q)$ is in general extremely difficult to compute. Even computing the rank of a given elliptic curve is, in general, a notorious problem. The latter quantity features in the famous Birch and Swinnerton-Dyer conjecture, one of the Millenium Problems. There are implementations of algorithms that succeed most of the times in computing the rank and finding generators. By running one such, namely  John Cremona's \textbf{mwrank} algorithm implemented in \textbf{Sage} \cite{sage}, we prove unconditionally that $r=1$ and $P$ is the generator of $E(\mathbb Q)$. Just to sum up,
$$E(\mathbb Q) = \langle P \rangle \cong \Z,$$
so all the points with rational coordinates on the projective curve are of the form $k \cdot P$, for $k \in \mathbb Z$.  By computing with the group law, one can observe that $2 \cdot P = [17/4 : -71/8 : 1]$ and $-2 \cdot P= [17/4 : 71/8 : 1]$. As $|k| \geq 2$, the experiments suggest that the coordinates of $k \cdot P$ have denominators that grow extremely fast. We should remark that we always set the last coordinate $Z=1$, as we are interested in the image of these points on the affine curve. Suspecting that $P$ and $-P$ are the only points with integral affine coordinates, we are now going to proceed to prove it.

Using a deep generalization of Baker's theorem due to David and Hirata-K\" ohno \cite{DavidLin}, we derive an explicit very large exponential bound for $C$ such that if $|k| > C$ then $k \cdot P$ does not have integral coordinates on our affine model. By a clever application of the \textbf{LLL} latice reduction algorithm, one can significantly reduce the previous bound and run a computer check to see which multiples $k$ multiples of $P$ are integral. This is a routine well known to experts by now, described in Section 8.7 of \cite{Coh} and is implemented in \textbf{Sage}. After running this routine, we conclude that the only integral points are indeed $P=[-1:1:1]$ and $-P=[-1:-1:1]$ which proves our theorem.

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\end{thebibliography}

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