Pythagorean Triangle in Square

I thought this was a really nice question.

We have $\angle CEF = 90^{\circ} - \angle DEA - \angle AED$ , and $\angle ADE = \angle ECF = 90^{\circ}$ . Thus, $\Delta ADE \sim \Delta ECF$ . Comparing proportional sides, we have $\frac{AD}{AE} = \frac{EC}{EF}$ , so $AD = \frac{4}{3} EC$ , so $AD = 4DE$ . We also have $AD^2 + DE^2 = AE^2$ , so $17DE^2 = 16$ . Thus, $DE = \frac{4}{\sqrt{17}}$ , so $AD = \frac{16}{\sqrt{17}}$ . Thus, the area of the square is $AD^2 = \frac{256}{17}$ . Thus, the answer is $256+17 = \boxed{273}$ .