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\begin{document}

\title{Congruent Numbers}
\maketitle

The Congruent Number problem is: {\em Let $n$ be a positive integer.
Is there a right triangle, with the lengths of all three of its sides
being rational numbers, that has area~$n$?}

Fermat showed that $n=1$ is not a congruent number.
Euler showed that $n=7$ is a congruent number.
Eventually, it became known that $n=1,2,3,4$ are not congruent,
but $n=5,6,7$ are.  But an efficient procedure to answer the question
for general~$n$ seemed hopelessly out of reach.

A terribly inefficient procedure does exist that will eventually answer
``Yes'' if $n$ is congruent and run forever if not.
But if we ask the procedure whether some value of~$n$ (say, $n=157$)
is a congruent number, and the procedure then runs for a century without
giving us an answer, then we don't know if this is because $n=157$ congruent
but the procedure hasn't figured this out yet, or because $n=157$ is not
congruent.

\medskip

In fact, $n=157$ {\em is} congruent.  Let
$$
a=\frac{411340519227716149383203}{21666555693714761309610}
$$
and
$$
b=\frac{6803298487826435051217540}{411340519227716149383203}
$$
and
$$
c=\frac{224403517704336969924557513090674863160948472041}
{8912332268928859588025535178967163570016480830}\,.
$$
Then
$$
a^2+b^2=c^2,
$$
and
$$
{\textstyle\frac12}ab=157.
$$
But an example this gigantic must be using ideas beyond brute-force search.

\vfil\pagebreak

In fact, the following result holds:

\medskip

\noindent{\em Let $n$ be an odd squarefree natural number.
Consider the following two conditions:
\begin{itemize}
\item The number $n$ is congruent.
\item The number of integer triples $(x,y,z)$ satisfying the equation
$$
2x^2+y^2+8z^2=n
$$
is equal to twice the number of integer triples satisfying the equation
$$
2x^2+y^2+32z^2=n.
$$
\end{itemize}
Then the first condition implies the second; and if a weak form of the
so-called Birch--Swinnerton-Dyer conjecture is true then also the second
condition implies the first.}

\medskip

Note that checking the second condition for a given~$n$ terminates in a
finite number of steps, and in fact in a number of steps that we can
bound explicitly.

And a similar result holds for even~$n$.
\end{document}
