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

$$\begin{align*}101 k + 17 &= s^2 \\101 k + 1330 &= m^2\end{align*}$$

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 $m – s = 13$. 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.