\documentclass[12pt]{article}
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\title{Congruent Numbers and Cubic Curves}
\author{William A. Stein\\{\sf was@math.harvard.edu}\\{\sf http://modular.fas.harvard.edu/$\sim$was}}
\date{3 March 2001 at MIT}
\begin{document}
\maketitle
\section{Congruent Numbers}

\begin{definition}[Congruent number]
An integer~$n$ is {\em congruent} if
it is the area of a right triangle with rational side lengths.
\end{definition}
Why ``congruent''?: One can show that $n$ is congruent if and only if there
is an~$x$ such that $x-n$, $x$ and $x+n$ are all squares of rational
numbers ($n$ is the ``congruence'').

\begin{example}\mbox{}
\begin{itemize}
\item $6$ is congruent; it is the area of the triangle
with sides $3$, $4$, and $5$.  [[Display on laptop.]]
\item The numbers $1$, $2$, $3$, and $4$ are not congruent,
but $5$, $6$, and $7$ are.
\end{itemize}
\end{example}


\begin{question}
Is there a simple criterion to determine whether
or not a number is a congruent number?
\end{question}
This tantalizing problem has remained open for over a millenium, though perhaps you will 
live to see (or find?) a complete solution!

The following theorem is not very difficult to prove.
\begin{theorem}
An integer $n$ is a congruent number if and only if the equation
$$y^2 = x^3-n^2x$$
has more than the three obvious solutions.
\end{theorem}



\begin{theorem}[Tunnel]
If $n$ is a congruent number then the number
of triples $(x,y,z)$ such that $2x^2+y^2+8z^2=n$
is twice the number of triples $(x,y,z)$ such that
$2x^2+y^2+32z^2=n$. 
If the Birch and Swinnerton-Dyer conjecture (see below)
for $y^2=x^3-n^2x$ is true, then the converse is true.
\end{theorem}

This is almost a complete solution to the ancient congruent number
problem.  Unfortunately, nobody knows how to prove the Birch and
Swinnerton-Dyer conjecture yet.


\section{The Birch and Swinnerton-Dyer Conjecture}
The Birch and Swinnerton-Dyer conjecture is a conjecture
about whether or not $y^2=x^3-n^2x$ has (infinitely many) 
non-obvious solutions. [[Display pictures of Birch and Swinnerton-Dyer on my laptop.]]

\begin{conjecture}
The equation $y^2=x^3-n^2x$ has infinitely many
solutions {\em only} in the following two cases:
\begin{itemize}
\item $n$ is odd and $-2$ is a square modulo $n$, or $n$ is even and $-2$ is a square modulo $n/2$
\item the sum 
$$\sum_{n=1}^{\infty} \frac{a_n}{n} e^{-\pi n/M}$$
is~$0$.
Here $M = 2n\sqrt{2}$ if $n$ is odd, or $2n$ if $n$ is even,
and $a_p$ is $p$ minus the number of solutions to 
$y^2=x^3-n^2x$ modulo~$p$. The $a_n$ are
got from the $a_p$ by a simple recipe.
\end{itemize}
\end{conjecture}

In fact, Birch and Swinnerton-Dyer made a more general conjecture,
which applies to any curve of the form $y^2 = x^3 + ax + b$.  If you
prove it, then you will be famous and the Clay Math Institute will
award you one million dollars.


\section{Fun examples (using my computer)}
Have somebody from the audience choose an~$n$, and determine whether or not
the conjecture asserts that it is congruent.  If it is 
congruent, give the rational right triangle.



\end{document}