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Age 11 to 14 Challenge Level:

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.