We are looking for solutions to the equation $x^2 = y^2 + z^2$
where the woman was born in the year $y$ (between $0$ A.D. and
$1997$), she lived $z$ years (at least $1$ year and not more than
$120$ years) and she died in the year $x$ where $x < 1997$ .
There are twenty possible solutions. If she was born in $0$ A.D.
there are ten possible solutions. The remaining solutions
are:
Born |
Lived |
Died |
|
|
|
$y^2$ |
$z^2$ |
$x^2$ |
|
|
|
$9$ |
$16$ |
$25$ |
$16$ |
$9$ |
$25$ |
$36$ |
$64$ |
$100$ |
$64$ |
$36$ |
$100$ |
$144$ |
$25$ |
$169$ |
$144$ |
$81$ |
$225$ |
$225$ |
$64$ |
$289$ |
$576$ |
$49$ |
$625$ |
$576$ |
$100$ |
$676$ |
$1600$ |
$81$ |
$1681$ |
This can never happen in the future (taking $x^2 > 1997$). Even
if the woman had a longer life there are still no solutions in the
past for a lifespan of $121$ years, but there are four solutions
for a lifespan of $144$ years. In the next $2000$ years (assuming a
lifespan of no more than $400$ years) the only solutions are:
Born |
Lived |
Died |
|
|
|
$y^2$ |
$z^2$ |
$x^2$ |
|
|
|
$2304$ |
$196$ |
$2500$ |
$2304$ |
$400$ |
$2704$ |
$3600$ |
$121$ |
$3721$ |
$3969$ |
$256$ |
$4225$ |
The sets of numbers $x$, $y$ and $z$ are Pythagorean triples.