Time of birth
A woman was born in a year that was a square number, lived a square
number of years and died in a year that was also a square number.
When was she born?
Age
11 to 14
Challenge level
Problem
A woman was born in a year that was a square number, lived a square
number of years and died in a year that was also a square number.
When was she born?
Student Solutions
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.