# A question of Erdös on $3$-powerful numbers and an elliptic curve analogue
of the Ankeny-Artin-Chowla conjecture.

P.G. Walsh  Department of Mathematics
University of Ottawa<EMAIL_ADDRESS>
###### Abstract.

We describe how the Mordell-Weil group of rational points on a certain family
of elliptic curves give rise to solutions to a conjecture of Erdös on
$3$-powerful numbers, and state a related conjecture which can be viewed as an
elliptic curve analogue of the famous Ankeny-Artin-Chowla conjecture.

###### Key words and phrases:

powerful number, diophantine equation, elliptic curve

###### 2020 Mathematics Subject Classification:

11D25,11G05

##  1\. Introduction

In a series of papers [4],[5],[6], Erdös posed a number of problems concerning
powerful numbers. Quite recently, the arithmetic of elliptic curves has been
used by Bajpai, Bennett and Chan [2] to prove the existence of infinitely many
quadruples of four pairwise coprime powerful numbers in arithmetic
progression, thereby answering one of Erdös’ problems. In this article we
consider a different problem of Erdös. In particular, he asked for solutions
to $a+b=c$ in coprime $3$-powerful numbers. This was solved by Nitaj [9], and
later in a different way by Cohn [3]. We will describe how one can produce
solutions to Erdös’ problem by using the group of rational points on an
elliptic curve. It is worth noting here that despite the abundance of coprime
solutions to $a+b=c$ in $3$-powerful numbers, the abc conjecture implies that
there are only finitely many triples $(a,b,c)$ consisting of pairwise coprime
$4$-powerful numbers which satisfy $a+b=c$, and that no such example is even
known to exist (see [7] for details).

Toward this end, it is fairly straightforward to see that a solution to the
problem is equivalent to solving the equation

$ax^{3}+by^{3}=cz^{3}$

in integers for which $rad(a)|x$, $rad(b)|y$, $rad(c)|z$, and $\gcd(ax,by)=1$.
We will make things as simple as possible for ourselves by restricting our
attention to the case $a=b=1$ and $rad(c)=p$, an odd prime, which results in
the problem of solving one of

$x^{3}+y^{3}=p^{\mu}z^{3}\;\;\;\;(\mu\in{1,2})$

with $(x,y)=1$ and $p|z$. We will focus primarily on the case $\mu=1$, as the
other case is quite similar. The following result provides a solution to
Erdös’ problem, however it seems to be merely a first step in this direction.

######  Theorem 1.1.

Let $p$ denote an odd prime for which the curve

$E:Y^{2}=X^{3}-432p^{2}$

has positive rank. Then there are infinitely many pairwise coprime integer
solutions $(x,y,z)$ to $x^{3}+y^{3}=p^{4}z^{3}$. Furthermore, if $P$ denotes a
generator of infinite order on $E$, then pairwise coprime integer solutions to
$x^{3}+y^{3}=p^{4}z^{3}$ can be derived from $(3pk)P$ for every integer $k\geq
1$.

To be more explicit we elucidate the above statement. Assume that $u,v,d$ are
integers for which $(uv,d)=1$ and $(X=u/d^{2},Y=v/d^{3})$ is a point on the
curve. Assume further that $p$ divides $d$. We leave it for the reader to
verify that the triple $(x,y,z)$ given by
$x=num((36pd^{3}+v)/6ud)$,
$y=num((36pd^{3}-v)/6ud)$,
$z=denom((36pd^{3}+v)/6ud)$
(all in lowest terms) gives solutions to Erdös’ problem.

It is worth remarking at this point that in order to ensure $(x,y)=1$ and $p$
divides $z$, a sufficient condition is for $p$ to divide the integer $d$
defined just above. However, this condition is often not actually necessary.
This corresponds to the extra factor of $3$ in front of $P$ in the statement
of the theorem, which seems to be necessary only when $p\equiv 1\;(\bmod\;3)$.

Adam Logan has pointed out that the $3$-Selmer group of $E$ has order $1$ for
$p\equiv 4,7,8\;(\bmod\;9)$. Consequently, one would expect the rank of $E$ to
be $1$ for all of these cases.

##  2\. Proof of Theorem 1.1

We begin by connecting the $p$-divisibility of $d$ to that of $z$. The
definitions of $x$, $y$ and $z$ from the previous section imply the existence
of a non-zero integer $k$ for which $36pd^{3}+v=kx,36pd^{3}-v=ky,6ud=kz.$
Assuming that $p|d$, then because of the fact that $u$ and $v$ are coprime to
$d$, it follows that $p$ does not divide $kx$, and therefore cannot be a
factor of $k$. The equation $6ud=kz$ shows that $p$ must be a divisor of $z$.

We therefore need to pin down the set of points $(X,Y)=(u/d^{2},v/d^{3})$ on
$E:Y^{2}=X^{3}-432p^{2}$ which have the property that $p$ divides $d$. Let $P$
denote a point of infinite order on $E$, then regarded as a point on
$E(\mathbb{Q}_{p})$, the multiple of $P$ having denominator divisible by $p$
is equivalent to the order of $P$ in
$E(\mathbb{Q}_{p})/E_{0}(\mathbb{Q}_{p})$, where $E_{0}(\mathbb{Q}_{p})$
consists of those points whose reduction is non-singular. Since $E$ has
additive reduction, this quotient consists of the additive group
$\mathbb{Z}/p\mathbb{Z}^{+}$ of order $p$ times the order of the group of
components of the Neron model (see p.359 of [11]), which in this case is $3$.
It follows that the point $(3p)P$ has coordinates $(u/d^{2},v/d^{3})$
satisfying the property that $d$ is divisible by $p$, which translates into
the corresponding value for $z$ in the solution to $x^{3}+y^{3}=pz^{3}$ being
divisible by $p$, and thus giving integer solutions to Erdös’ problem on
$3$-powerful numbers.

It is worth noting that if $m$ is squarefree, the additive reduction at each
prime can be pieced together so that the group structure coming from each
prime dividing $m$ can also be glued together to get pairwise coprime integer
solutions to $x^{3}+y^{3}=m^{4}z^{3}$ by multiplying a generator by $3m$. A
simple example of this phenomenon is given by $m=35$, which we leave to the
reader for verification.

##  3\. An Elliptic Curve Analogue of the Ankeny-Artin-Chowla Conjecture

In 1952, Ankeny, Artin and Chowla [1] published a congruence involving various
quantities related to a quadratic field and a Bernoulli number, which resulted
in a statement which is now referred to as the AAC conjecture. Specifically,
the AAC conjecture states that if $p\equiv 1\;(\bmod\;4)$ is prime, and if
$\epsilon_{p}=\frac{t+u\sqrt{p}}{2}$ is the fundamental unit of the quadratic
field $K=\mathbb{Q}(\sqrt{p})$, then $p\nmid u$. Mordell [8] later conjectured
this to be true for $p\equiv 3\;(\bmod\;4)$.

An equivalent formulation of this can be stated as follows. If
$u=\frac{a+b\sqrt{p}}{2}>1$ is a unit in the ring of integers of $K$ with
$b\equiv 0\;(\bmod\;p)$, then $u=\epsilon^{kp}$ for some integer $k$.

This last statement appears to have similarities with the required properties
that were required to solve $x^{3}+y^{3}=p^{4}z^{3}$ in pairwise coprime
integers.

Let $p$ denote any odd prime, and let $E_{p,2}:Y^{2}=X^{3}-432p^{2}$ and
$E_{p,4}:Y^{2}=X^{3}-432p^{4}$. Let $P=(X,Y)$ denote a generator of one of the
curves $E_{p,2}$ or $E_{p,4}$, and $Q=kP=(u/d^{2},v/d^{3})$ for integers
$k,u,v,d$. We have seen that if the denominator $d$ is divisible by $p$, then
$Q$ gives rise to a solution in pairwise coprime integers $(x,y,z)$ of
$x^{3}+y^{3}=p^{4}z^{3}$ and $x^{3}+y^{3}=p^{5}z^{3}$ respectively. Although
Theorem 1.1 states that the integer $k$ should be divisible by $3p$, our
computations have shown that the desired condition holds very often when $k$
is divisible by $p$, and thus we state the following as an analogue of AAC for
these two families of elliptic curves.

######  Conjecture 3.1.

EC-AAC
If $P$ is a generator of $E_{p,2}$ (resp. $E_{p,4}$), and $k$ is a positive
integer for which $Q=kP=(u/d^{2},v/d^{3})$ with $p|d$, then $p$ divides $k$.

A few cautionary historical remarks are in order. At the time that this author
learned of the AAC conjecture, the obvious question of whether a similar
result holds for composite discriminants led to the immediate discovery that
the fundamental units $\frac{t+u\sqrt{d}}{2}$ in $\mathbb{Q}(\sqrt{d})$, with
$d=46$, $430$ and $1817$, actually have the property that $d|u$. This led to
an investigation by Stephens and Williams [12], who produced a longer list of
composite values with this property, and very recently, new examples have been
found by Reinhart [10], with one of the examples being a prime of the form
$4k+3$, and hence a counterexample to Mordell’s extension of AAC. Therefore,
one should be extremely wary about the truth of the AAC conjecture. Similarly,
the composite value $m=1349$ has the property that $E_{m,4}$ is generated by a
point $P$ having $1349|d$, and presumably a larger search would produce other
such curves. We have yet to find a composite integer $m$ for which $m$ divides
the integer $d$ arising from a generator of the curve $E_{m,2}$, nor have we
found a counterexample to Conjecture 3.1.

Acknowledgements The author would like to express gratitude to Noam elkies,
Adam Logan and Joseph Silverman for their extremely helpful discussions.

## References

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