  It is easy to find a right-angled triangle with integer sides whose area is
6. There is no such triangle with area 5, but there is one with rational sides
(a `\emph{Pythagorean triangle}').
  For historical reasons, integers such as 6 or 5 that are (the squarefree part
of) the area of some Pythagorean triangle are called `\emph{congruent
numbers}'. These numbers actually are interesting for the following reason:
Notice the sequence $\frac14$, $6\frac14$, $12\frac14$. It is an arithmetic
progression with common difference 6, consisting of squares $(\frac12)^2$,
$(\frac52)^2$, $(\frac72)^2$ of rational numbers. Indeed the common difference
of three rational squares in AP is a congruent number and every congruent
number is the common difference of three rational squares in arithmetic
progression.
  The triangle given by $9^{2}+40^{2}=41^{2}$ has area $180=5\cdot6^{2}$ and
the numbers $x-5$, $x$ and $x+5$ all are rational squares if $x=11{97/144}$.
Recall one obtains all Pythagorean triangles with relatively prime integer
sides by taking $x=4uv$, $y=\pm(4u^{2}-v^{2})$, $z=4u^{2}+v^{2}$ where $u$ and
$v$ are integers with $2u$ and $v$ relatively prime.
  Fermat proved that there is no AP of more than three squares of rationals.
