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\begin{center}
\vskip 1cm{\LARGE\bf Congruent Number Elliptic Curves
Related \\
\vskip .1in
to Integral Solutions of $m^2=n^2+nl+l^2$}
\vskip 1cm \large
Lorenz Halbeisen and Norbert Hungerb\"uhler\\
Department of Mathematics\\ 
ETH Zentrum\\
8092 Z\"urich\\ 
Switzerland\\ 
\href{lorenz.halbeisen@math.ethz.ch}{\tt lorenz.halbeisen@math.ethz.ch}\\
\href{norbert.hungerbuehler@math.ethz.ch}{\tt norbert.hungerbuehler@math.ethz.ch}
\end{center}

\vskip .2 in

\begin{abstract} 
We construct an infinite family of congruent number elliptic curves, 
each with rank at least two, which are related to integral solutions
of $m^2=n^2+nl+l^2$.
\end{abstract}

\section{Introduction}
Elliptic curves and their geometric and algebraic structure have been a flourishing
field of research in the past. They find prominent applications in cryptography
and played a key role in the proof of Fermat's Last Theorem. A salient feature of the algebraic
structure of an elliptic curve is its rank. Among general elliptic curves, 
congruent number curves of high rank are of particular interest 
(see, for example, \cite{HighRank}). 
More difficult than finding an individual congruent number curve
of high rank is to find infinite families of such curves. Johnstone and Spearman~\cite{JS}
constructed such a family with rank at least three, which is related
to rational points on the biquadratic curve $w^2 = t^4 + 14t^2 + 4$. 
In the present paper,
we show an elementary construction for an infinite family of 
congruent number curves of rank at least two that are related
to the quadratic diophantine equation $m^2=n^2+nl+l^2$, and which
have three integral points with positive $y$-coordinate 
on a straight line. Incidentally, 
some members of the family exhibit surprisingly high individual rank,
namely rank five (whereas the members of the family given by
Johnstone and Spearman~\cite{JS} all have rank three).
We start by recalling some basic results on congruent numbers.

A positive integer $A$ is called a {\it congruent number} if 
$A$ is the area of a right-angled triangle with three rational 
sides. So $A$ is congruent if and only if there exists a
rational Pythagorean triple $(a,b,c)$ (\ie, $a,b,c\in\Q$,
$a^2+b^2=c^2$, and $ab\neq 0$) such that $\frac{ab}2=A$.
The sequence of integer congruent numbers starts with
$$
5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34, 37,\ldots
$$
(see, for example, the On-Line Encyclopedia of Integer Sequences~\cite[\seqnum{A003273}]{oeisA003273}).

It is well-known that $A$ is a congruent number if
and only if the cubic curve $$C_A:\ y^2=x^3-A^2 x$$
has a rational point $(x_0,y_0)$ with $y_0\neq 0$.
The cubic curve $C_A$ is called a {\it congruent number elliptic curve}
or just {\it congruent number curve}.
With respect to some congruent number $A$, the correspondence between rational 
points $(x,y)$ with $y\neq 0$ on the congruent number curve $C_A$ on the one hand, and 
rational Pythagorean triples $(a,b,c)$ with $ab=2A$ on the other hand, is given by 
\begin{equation}\label{psi}
\bigl(a,b,c\bigr)\ \mapsto\ \Biggl(\frac{b(b+c)}{2}\,,\,
\frac{b^2(b+c)}{2}\Biggr),
\end{equation}
and
\begin{equation}\label{psi-1}
        \bigl(x,y\bigr)\ \mapsto\ \Biggl(\frac{2x A}{y}\,,\;
        \frac{x^2-A^2}{y}\,,\;\frac{x^2+A^2}{y}\Biggr).
\end{equation}

For a positive integer $A$, a triple $(a,b,c)$ of 
non-zero rational (integral) numbers such that
$a^2+b^2=c^2$ and $A=\big{|}\frac{ab}{2}\big{|}$
is called a {\it rational (integral) Pythagorean $A$-triple}.
Notice that if $(a,b,c)$ is a {\rPAt}, then $A$ 
is a congruent number and $|a|,|b|,|c|$ are the 
lengths of the sides of a right-angled triangle 
with area $A$. Notice also that we allow $a,b,c$
to be negative. 
In particular, for any positive integers $m$ and $n$ with $m>n$, the triple
$$\bigl(\,\underset{\text{\small $a$}}{\underbrace{\,\mathstrut 2mn\,}}\,,\; 
\underset{\text{\small $b$}}{\underbrace{\mathstrut m^2-n^2}}\,,\; 
\underset{\text{\small $c$}}{\underbrace{\mathstrut m^2+n^2}}\,\bigr)$$ 
is an {\iPAt}. In this case, we obtain $A=mn(m^2-n^2)$ and by~(\ref{psi})
\begin{equation}\label{eq-psi}
(a,b,c)\mapsto\bigl(
\underset{\text{\small $x$}}{\underbrace{m^2(m^2-n^2)}}\,,\;
\underset{\text{\small $y$}}{\underbrace{m^2(m^2-n^2)^2}}\bigr)\,.
\end{equation}
Notice that in this case, the point $(x,y)$ on $C_A$ that corresponds to the 
{\iPAt} $(a,b,c)$ is an integral point. 

Concerning the equations
$$m=n^2+nl+l^2\notag\eqno{\eqm}$$
and 
$$m^2=n^2+nl+l^2\notag\eqno{\eqmm}$$
we first prove the following result
(for a geometric representation of integral solutions of $x^2+xy+y^2=m^2$,
see Halbeisen and Hungerb\"uhler~\cite{HHAnning}):

\begin{proposition}\label{prp:m2}
 Let $p_1<p_2<\cdots <p_j$ be primes, such that
 $p_i\equimod{1}{6}$ for $1\le i\le j$, and let $$m=\prod_{i=1}^j p_i.$$
 
\begin{enumerate}[label=(\alph*)]
 \item\label{itm:a} The number of positive, integral solutions $(n,l)$ of
 $\eqm$ with $l<n$ is $2^j$. 
 
 \item\label{itm:b} For each integral solution of $\eqm$, 
 $n$ and $l$ are relatively prime and neither $n$ nor $l$ 
 is a multiple of $p_i$ (for $1\le i\le j$).
 
 \item\label{itm:c} The number of positive, integral solutions $(n,l)$ of
 $\eqmm$ with $l<n$ is $\frac{3^j-1}2$.
 
 \item\label{itm:d} Among the $\frac{3^j-1}2$ integral solutions 
 solutions $(n,l)$ of
 $\eqmm$ with $l<n$ we find $2^{j-1}$ solutions $(n,l)$ such that 
 $n$ and $l$ are relatively prime. 
 In particular, if $j=1$ and $p\equimod 16$, then
 the solution in positive integers $l<n$ of 
 $$p^2=n^2+nl+l^2,$$ such that $n$ and $l$ are relatively prime, is unique.
\end{enumerate}
\end{proposition}

\begin{proof}
\leavevmode

\noindent \ref{itm:a} By Dickson~\cite[Exercises\,XXII.2, p.\,80]{Dickson}, the number
of integral solutions of $\eqm$ is $6E(m)$, where $E(m)$ is the excess
of the number of divisors $3h+1$ of $m$ over the number of divisors of the form
$3h+2$. By definition of $m$, $E(m)=2^j$. Now, with each
positive, integral solution $(n,l)$ of $\eqm$ with $0<l<n$ we obtain the
following~$12$ pairwise different integral solutions:
 
$$(n,l),\;(l,n),\;(-n,-l),\;(-l,-n),$$
$$(-n,n+l),\;(n+l,-n),\;(n,-n-l),\;(-n-l,n),$$
$$(-l,n+l),\;(n+l,-l),\;(l,-n-l),\;(-n-l,l).$$

So, if $e(m)$ denotes the number of
positive, integral solutions $(n,l)$ of $\eqm$ with $l<n$, then
$6E(m)\ge 12e(m)$. On the other hand, every integral solution of $m=n^2+nl+l^2$
corresponds to a unique positive, integral solution $(n,l)$ with $l<n$,
which implies that $6E(m)=12e(m)$ and consequently we obtain $e(m)=2^{j-1}$. 

\bigskip
\noindent\ref{itm:b} This follows immediately from the definition of $m$. 

\bigskip
\noindent\ref{itm:c} Again by Dickson~\cite[Exercises\,XXII.2, p.\,80]{Dickson}, the number
of integral solutions of $\eqmm$ is $6E(m^2)$, where $E(m^2)=3^j$.
Let $e(m^2)$ denote the number of
positive, integral solutions $(n,l)$ of $m^2=n^2+nl+l^2$ with $l<n$. In addition
to the $12e(m^2)$ integral solutions of $\eqmm$, we have the~$6$ solutions
$$(m,0),\;(0,m),\;(-m,0),\;(0,-m),\;(-m,m),\;(m,-m).$$
So, $6E(m)=12e(m)+6$ and consequently we obtain $e(m)=\frac{3^j-1}2$. 

\bigskip
\noindent\ref{itm:d} For the sake of simplicity, let us call a positive,
integral solution $(n,l)$ of $\eqmm$ with $l<n$ a {\it normal solution}.
Among the normal solutions $(n,l)$ of $\eqmm$, we distinguish between
the ones with $n$ and $l$ relatively prime, which we call {\it primitive
solutions}, and the other ones, which we call {\it composite}.
For a given $m$, there is a one-to-one correspondence between composite solutions
of $\eqmm$ with $(n,l) = d > 1$ and primitive solutions of $\eq{(m/d)^2}$ via
the correspondence
$$m^2 = n^2 + nl + l^2 \;\iff\; (m/d)^2 = (n/d)^2 + (n/d)(l/d) + (l/d)^2\,.$$

The proof is now by induction on $j$. For $j=1$, the 
statement follows from the definition of $m$. 
Suppose the statement holds when $m$ has $j-1$ or less prime factors. 
Then, let $m$ have $j$ prime factors. We observe by the above correspondence 
that the number of composite solutions to $\eqmm$ is
$$\sum_{\substack{r\mid m\\ 0<r<m}}P(r^2)$$
where $P(r^2)$ is the number of primitive solutions to $r^2 = n^2 + nl + l^2$. 
By the inductive step, if $r$ is comprised of $i$ prime factors of $m$ then 
$P(r^2) = 2^{i-1}$. Furthermore, the number of distinct factors $r$ of $m$ with 
$i$ prime factors is $\binom ji$, and therefore,
by the binomial theorem, the number of composite solutions to $\eqmm$ is
$$\sum_{\substack{r\mid m\\ 0<r<m}}P(r^2)=
\sum_{i=1}^{j-1}\binom ji 2^{i-1}=\frac 12\bigl((1+2)^j-2^j-1\bigr)=
\frac{3^j-1}{2}-2^{j-1}\,.$$
Since the total number of solutions to $\eqmm$ is $\frac{3^j-1}{2}$,
by subtracting the number of composite solutions we finally 
obtain $P(m^2) = 2^{j-1}$, which completes the proof.
\end{proof}

Let us now consider the relationship between positive, integral solutions
of $m^2=n^2+nl+l^2$ and integral Pythagorean triples.

If $m,n,l$ are positive integers such that $m^2=n^2+nl+l^2$, then, for $k:=n+l$,
each of the following three triples
$$\bigl(\,\underset{\text{\small $a_1$}}{\underbrace{\,\mathstrut 2mn\,}}\,,\; 
\underset{\text{\small $b_1$}}{\underbrace{\mathstrut m^2-n^2}}\,,\; 
\underset{\text{\small $c_1$}}{\underbrace{\mathstrut m^2+n^2}}\,\bigr)\,,$$
$$\bigl(\,\underset{\text{\small $a_2$}}{\underbrace{\,\mathstrut 2ml\,}}\,,\; 
\underset{\text{\small $b_2$}}{\underbrace{\mathstrut m^2-l^2}}\,,\; 
\underset{\text{\small $c_2$}}{\underbrace{\mathstrut m^2+l^2}}\,\bigr)\,,$$
$$\bigl(\,\underset{\text{\small $a_3$}}{\underbrace{\,\mathstrut 2mk\,}}\,,\; 
\underset{\text{\small $b_3$}}{\underbrace{\mathstrut k^2-m^2}}\,,\; 
\underset{\text{\small $c_3$}}{\underbrace{\mathstrut k^2+m^2}}\,\bigr)\,,$$ 
is an {\iPAt} for $$A=mn(m^2-n^2)=ml(m^2-l^2)=km(k^2-m^2)=klmn$$ (see Hungerb\"uhler~\cite{Noebi}).
In particular, with $m,n,l$ and~(\ref{eq-psi}) we obtain three distinct integral points on $C_A$.

As a matter of fact we would like to mention that the 
three integral points on $C_A$ that correspond to an integral solution 
of $m^2=n^2+nl+l^2$ lie on a straight line. 

Let us now turn back to the curve $C_A$, where $A$ is a 
congruent number. One can readily check that  
the three points $(0,0)$ and $(\pm A,0)$ are the 
only points on $C_A$ of order~$2$ 
with respect to the group law of elliptic curves.
Moreover, one can show that these three points, together with
the point at infinity, are the only points of finite order
(for an elementary proof
of this result, which is based on a theorem of Fermat; see
Halbeisen and Hungerb\"uhler~\cite{Fermat}).
This implies that if
$A$ is a congruent number and $(x_0,y_0)$ is a rational
point on $C_A$ with $y_0\neq 0$, then the order of $(x_0,y_0)$
is infinite. So, Mordell's theorem (which states that
the group of rational points on $C_A$ is
finitely generated) and the
fundamental theorem of finitely generated abelian groups
imply that the group of rational points on a congruent number curve 
$C_A$ is isomorphic to
$$\underset{\text{\scriptsize torsion group}}
{\underbrace{\Z/2\Z\times\Z/2\Z}}\times\Z^r,$$
where $r> 0$ is the aforementioned {\it rank\/} of $C_A$. 

\section{Rank at least two}

Based on integral solutions of $m^2=n^2+nl+l^2$, we will show that there
are infinitely many congruent number curves $C_A$ with rank at 
least two\,---\,where many of the curves $C_A$ have rank three or four
and several curves have even rank five (see Section~\ref{sec-three}).

The following result, which can be found in Silverman and 
Tate\;\cite[Chapter\,III.6.]{SilvermanTate}, allows us to compute the 
rank\,---\,or at least a lower bound of the rank\,---\,of 
certain elliptic curves. For simplicity, we state the result just 
for congruent number curves.

\begin{proposition}\label{prp:SilvermanTate}
Let $b$ be a non-zero integer and let $\bar b:=-4b$. Furthermore,
let $$B:=\bigl{\{}b_1\in\Z\,:\,b_1\mid b,\hspace{1.3ex}\text{and\hspace{1.3ex}$b_1$ is square-free}\bigr{\}}$$
and $$\bar B:=\bigl{\{}\bar{b}_1\in\Z\,:\,\bar{b}_1\mid\bar{b},\hspace{1.3ex}
\text{and\hspace{1.3ex}$\bar{b}_1$ is square-free}\bigr{\}}.$$
Finally, let $\beta_b$ and $\beta_{\bar b}$ be the number of integers $b_1\in B$ 
and $\bar{b}_1\in\bar{B}$, respectively, such that the corresponding
equation
\begin{eqnarray}
 b_1 M^4+\tfrac b{b_1} e^4&=&N^2\label{eqn:A}\\[1.2ex]
 \bar{b}_1\bar{M}^4+\tfrac{\bar{b}}{\bar{b}_1}\bar{e}^4&=&\bar{N}^2 \label{eqn:barA}
\end{eqnarray}
has integral solutions, where $e\neq 0$, $\bar{e}\neq 0$, and
$(M,e)=(\bar{M},\bar{e})=1$.

Then the rank $r$ of the curve $y^2=x^3+bx$ satisfies the equation
$$2^r=\frac{\beta_b\cdot\beta_{\bar b}}{4}.$$

Moreover,  if $(x,y)$ is a rational point on 
the curve $y^2=x^3+bx$ with $y\neq 0$, then one can write that point in the form
\begin{equation}\label{eq-x_b1}
x=\frac{b_1 M^2}{e^2},\qquad y=\frac{b_1 M N}{e^3},
\end{equation}
where $M,e,N$ is an integral solution of~(\ref{eqn:A}) with $N>0$ and $(M,e)=1$,
and vice versa. The analogous statement holds for rational points on the curve 
$y^2=x^3+\bar{b}x$ with respect to equations of the form~(\ref{eqn:barA}).
\end{proposition}

Now we are ready to prove

\begin{theorem}\label{thm:main}
Let $m,n,l$ be pairwise relatively prime positive integers, where
$m=\prod_{i=1}^j p_i$ is a product of pairwise distinct primes
$p_i\equimod{1}{6}$ and $m^2=n^2+nl+l^2$. Furthermore, let $k:=n+l$
and let $A:=klmn$. Then the rank of 
the congruent number curve $$C_A: y^2=x^3-A^2x$$ is at least two.
\end{theorem}

\begin{proof} Since we have at least one rational point $(x,y)$ on $C_A$
with $y\neq 0$, we know that the rank $r$ of $C_A$ is positive. So,
to show that the rank of the curve $C_A$ is at least two, it
would be enough to show that $\beta_{-A^2}\ge 9$. For this, we have
to show that there are integral solutions for~(\ref{eqn:A}) for at least~9 
distinct square-free integers $b_1$ dividing $-A^2$, or equivalently, 
we have to find at least~9 rational points on $C_A$, such that the~9 
corresponding integers $b_1$ are pairwise distinct. Even though it would
be enough to find integral solutions for~(\ref{eqn:A}) for at least~9 
distinct square-free integers $b_1$, we shall give~$16$ solutions,
such that a single additional solution for~(\ref{eqn:A}) would give
us a rank of at least three (see Proposition~\ref{prp:rank3}).

Notice that because of~(\ref{eq-x_b1}), to compute $b_1$ 
from a rational point $P=(x,y)$ on $C_A$
with $x\neq 0$, it is enough to know the $x$-coordinate of $P$ and then
compute $x$~mod~${\Q^*}^2$ (\ie, we compute $x$ modulo squares of rationals).
The $x$-coordinates of the three integral points we get by~(\ref{psi}) from the three
{\iPAts} $(a_1,b_1,c_1)$, $(a_2,b_2,c_2)$, $(a_3,b_3,c_3)$ generated by $m,n,l,k$, are
$$x_1=m^2(m^2-n^2)=m^2kl\,,\quad x_2=m^2(m^2-l^2)=m^2kn\,,\quad x_3=k^2(k^2-m^2)=k^2nl\,,$$ 
and modulo squares, this gives us three values for $b_1$ modulo squares,
namely $$b_{1,1}\equiv kl\,,\quad b_{1,2}\equiv kn\,,\quad b_{1,3}\equiv nl\,.$$ 
Now, exchanging in each of the three {\iPAts} the two catheti $a_i$ and $b_i$ (for $i=1,2,3$),
we obtain again three distinct integral points on $C_A$, whose $x$-coordinates gives 
us again three values for $b_1$ modulo squares, namely
$$b_{1,4}\equiv mn\,,\quad b_{1,5}\equiv ml\,,\quad b_{1,6}\equiv mk\,.$$
Finally, if we replace each hypotenuse $c_j$ of these six {\iPAts} with $-c_j$, we 
obtain again six distinct integral points on $C_A$, whose $x$-coordinates 
give us six values for $b_1$ modulo squares, namely
$$
\begin{array}{rclrclrcl}
  b_{1,7}&\hspace{-1.5ex}\equiv\hspace{-1.5ex}&-kl\,,
  & b_{1,8}&\hspace{-1.5ex}\equiv\hspace{-1.5ex}&-kn\,,
  & b_{1,9}&\hspace{-1.5ex}\equiv\hspace{-1.5ex}&-nl\,\\[1.6ex]
  b_{1,10}&\hspace{-1.5ex}\equiv\hspace{-1.5ex}&-mn\,,
  & b_{1,11}&\hspace{-1.5ex}\equiv\hspace{-1.5ex}&-ml\,,& 
  b_{1,12}&\hspace{-1.5ex}\equiv\hspace{-1.5ex}&-mk\,.
\end{array}
$$
In addition to these~$12$ integral points on $C_A$ (and the 
corresponding $b_1$'s), we have the two integral points $(\pm A,0)$ on $C_A$,
which give us two more values for $b_1$ modulo squares, namely
$$b_{1,13}\equiv klmn\qquad\text{and}\qquad b_{1,14}\equiv -klmn.$$

Now, it remains to show that the square-free parts of the $b_{1,j}$'s 
are pairwise distinct. By assumption, $m$ is square-free and 
$k,l,n$ are pairwise relatively prime.
Therefore, if for some $i,j$ with $1\le i<j\le 14$, $b_{1,i}\equimod{b_{1,j}}{{\Q^*}^2}$,
at least two of the integers $k,l,n$ are squares, say $n=u^2$, and $l=v^2$ or $k=v^2$. 
Then $$m^2=u^4+u^2v^2+v^4\quad\text{(in the case when $l=v^2$),}$$
or $$m^2=u^4-u^2v^2+v^4\quad\text{(in the case when $k=v^2$).}$$
If $l=v^2$, this implies that $u^2=1$ and $v=0$, or $u=0$ and $v^2=1$, and
if $k=v^2$, this implies that $u^2=1$ and $v=0$, $u=0$ and $v^2=1$, or $u^2=v^2=1$
(see, for example, Mordell~\cite[p.\,19{\sl f\/}]{MordellBook} or 
Euler~\cite[p.\,16]{Euler1780}). So, at most one of the integers $k,l,n$ is a square,
which shows that at least~14 equations of the form~(\ref{eqn:A})\,---\,for different 
square-free integers $b_1$ dividing $A^2$\,---\,have integral solutions.
Notice that so far, because the corresponding points on $C_A$ are integral, 
we always had $|e|=1$.

Now, we show that there are also solutions for~(\ref{eqn:A})
with $b_{1,15}=1$ and $b_{1,16}=-1$.
Assume first that there is a solution for~(\ref{eqn:A})
with $b_{1,15}=1$ and $e=1$, \ie, there
are positive integers $M$ and $N$ such that
$$M^4-A^2=N^2.$$
Then we also have $$A^2 M^4-A^4=(AN)^2,$$ which shows that 
$\tilde{M}:=A$, $\tilde{e}:=M$, and $\tilde{N}:=AN$, satisfy
$$-\tilde{M}^4+A^2\tilde{e}^4=\tilde{N}^2,$$ and hence,
there is a solution for~(\ref{eqn:A}) with $b_{1,16}=-1$. 
Notice that since $|\tilde{e}|\neq 1$, the corresponding point
on $C_A$ is not an integral point.

It remains to find a solution for~(\ref{eqn:A}) with $b_{1,15}=1$
and $e=1$. Since $m,n,l,k$ are pairwise relatively prime positive integers
and $k=n+l$, exactly one of $n,l,k$ is even, \ie, at least one of
$n$ and $l$ is odd. Without loss of generality, assume
that $n$ is odd. Furthermore, by definition 
of $m$, $m$ is odd. Let $p:=\frac{m+n}2$ and $q:=\frac{m-n}2$. Then $p$ and
$q$ are positive integers. Now, $m=p+q$, $n=p-q$, and $m^2-n^2=4pq$, and since
$m^2-n^2=kl$, we have $$A=klmn=4pq(p+q)(p-q)\,.$$
Notice that since $pq(p+q)$ is even, we have $A\equimod 08$.
An easy calculation shows that $M:=p^2+q^2$ and $N:=(p^2 - q^2)^2 - (2 p q)^2$
satisfy $$M^4-A^2=N^2.$$ So, there is a solution for~(\ref{eqn:A}) with $b_{1,15}=1$
and $e=1$, which gives us again an integral point on $C_A$

This shows that $\beta_{-A^2}\ge 16$ and completes the proof.
\end{proof}

As an immediate consequence we get the following

\begin{corollary} Let $m,n,l$ be as in Theorem~\ref{thm:main} and let
$q$ be a non-zero integer. Then the rank of 
the curve $C_{Aq^4}$ is at least two.
\end{corollary}

\begin{proof}
Notice that if $m,n,l$ are such that $m^2=n^2+nl+l^2$, then, 
for $mq,nq,lq$, we have $(mq)^2=(nq)^2+nq\cdot lq+(lq)^2$, 
which implies that for $\tilde A=kq\cdot lq\cdot mq\cdot nq=Aq^4$,
the rank of the curve $C_{\tilde A}$ is at least two.
\end{proof}


\section{Rank at least three}\label{sec-three}

\begin{proposition}\label{prp:rank3}
For $A=341\,880$, the rank of the curve $C_A$ is at least three.
\end{proposition}

\begin{proof} For $k=40$, $l=7$, $m=37$, $n=33$, we have $A=klmn$,
$m^2=n^2+nl+l^2$, and $k=n+l$. Thus, by Theorem~\ref{thm:main},
the rank of the curve $C_A$ is a least two. Now, for $b_{1,17}=-30$,
which is distinct from the square-free values of $b_{1,1},\ldots,b_{1,16}$,
we get that $M=98$, $e=1$, $N=33\,600$ is an integral solution of~(\ref{eqn:A}),
which implies that the rank of $C_A$ is at least three.
In fact, with the help of {\scriptsize\sagelogo} one can show that 
the rank of $C_A$ is equal to three.
\end{proof}

As a final remark concerning the rank of congruent number curves,
we would like to mention that with the help
of {\scriptsize \sagelogo} we found that many of the curves
which correspond to an integral solution of $m^2=n^2+nl+l^2$ have
rank~3 or higher. In fact, we found plenty of curves of rank~$3$ or~$4$,
as well as the following curves of rank~$5$:

\begin{center}
\begin{tabular}{rcrrr}
$A=klmn$\hspace{1ex} &  \hspace{3ex}$m=\prod p_{i\mathstrut}$\hspace{3ex} & 
\hspace{5ex}$l$\hspace{1ex} & \hspace{9ex}$n$\hspace{1ex} & \hspace{3ex}$k=n+l$\\
\hline\\
$237\,195\,512\,400$ & 
$7\cdot 127$ & $464$ & $561$ & $1\,025$\\
$8\,813\,542\,297\,560$ & 
$7\cdot 13\cdot 37$ & $232$ & $3\,245$ & $3\,477$\\
$10\,280\,171\,942\,040$ & 
$37\cdot 67$ & ${741}$ & $2\,024$ & $2\,765$\\
$81\,096\,660\,783\,600$ & 
$37\cdot 103$ & $2\,139$ & $2\,261$ & $4\,400$\\
$225\,722\,120\,463\,840$ & 
$13\cdot 19\cdot 31$ & $505$ & $7\,392$ & $7\,897$\\
$457\,485\,316\,904\,280$ & 
$7\cdot 31\cdot 37$ & $895$ & $7\,544$ & $8\,439$\\
$5\,117\,352\,889\,729\,080$ & 
$67\cdot 223$ & $1\,551$ & $14\,105$ & $15\,656$\\
$281\,692\,457\,452\,791\,000$ & 
$79\cdot 409$ & $9\,064$ & $26\,811$ & $35\,875$\\
$24\,666\,188\,870\,481\,576\,600$ & 
$13\cdot 31\cdot 223$ & $46\,169$ & $57\,400$ & $103\,569$
\end{tabular}
\end{center}

Of special interest might be values of $A$ which are related
to an $m$ with few factors, especially to primes $m$.
Recall that if $m$ is prime, then the integers $n,l,n+l$ such that
\hbox{$m^2=n^2+nl+l^2$} are unique, and therefore, $A(m):=(n+l)lmn$ is 
determined by $m$. Among the $666$ prime numbers $m\le 11\,113$ 
with $m\equimod 16$, we found the following $30$~values of $m$ 
such that $C_{A(m)}$ has rank~$4$:
\begin{align*}
&127, 139, 181, 277, 337, 709, 769, 823, 829, 883, 1051, 1087, 1213, \\
&1747, 1777, 1873, 2137, 2287, 2377, 2467, 2521, 3529, 3877, 3931, \\
&4129, 4999, 5521, 7573, 9601, 10711.
\end{align*}
However, we did not find any prime $m\equimod 16$ 
such that $C_{A(m)}$ has rank~$5$.

The preceding observations might indicate that the congruent number curves $C_A$ constructed in Theorem~\ref{thm:main}
are candidates for high rank congruent number elliptic curves (for another approach, see Dujella, Janfada, Salami~\cite{HighRank}).

\section{Acknowledgment}
We would like to thank the referee for his or her valuable remarks 
which greatly helped to improve this article.

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\end{thebibliography}

\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11G05; Secondary 11D09.

\noindent \emph{Keywords: } 
congruent number elliptic curves, Pythagorean triples.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A003273}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 29 2018; revised versions received  
January 25 2019; February 19 2019; May 15 2019; May 16 2019.
Published in {\it Journal of Integer Sequences}, May 16 2019.

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\noindent
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