2460.81 – An Unlikely Coincidence

This could be solved with a lot of guessing and checking, but let’s see if we can use some algebra. Let $k$ be Keisha’s age. Then $k+17$ is Gil’s age. To put Keisha’s age before Gil’s means to consider the number $100k + k+17$. This is supposed to be a perfect square, say, $s^2$,

$s^2 = 101 k + 17.$

In 13 years, it will happen again. Let $m^2$ be the new perfect square so that

$m^2 = 100 (k+13) + (k + 17 + 13) = 101 k + 1330.$

Hmmm. Observe that we now have

$101 k + 17 = s^2$

$101 k + 1330 = m^2.$

So,

$1313 = m^2 – s^2 = (m + s) (m – s).$

Now 1313 does factor into 101 times 13 (both primes!). This suggests that $m + s$ must be = 101 and $m – s$ is 13. Solving gives m = 57 and s = 44. Are these correct? Look at the squares. First,

$s^2 = 44^2 = 1936,$

suggesting that Keisha is 19, and Gil is $19 + 17 = 36$. Then in 13 years, Keisha will be 32 and Gil will be 49. Well look:

$3249 = 57^2 = m^2.$

Ta Daa! Keisha is 19.