jmc

Let $s$ be the length of the side of the square. The circles have radii of $1$ and $2$. We can then draw the triangle shown in the figure above and write expressions for the sides of the triangle in terms of $s$. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can use the Pythagorean Theorem: $$\begin{array}{rl}{\left(\frac{s}{2}\right)}^2+(s+1{)}^2& =2^2\\ \frac{1}{4}{s}^2+s^2+2s+1& =4\\ \frac{5}{4}{s}^2+2s-3& =0\\ 5s^2+8s-12& =0.\end{array}$$Finally, we can use the quadratic formula to solve for $s$: $$s=\frac{-8+\sqrt{8^2-4(5)(-12)}}{10}=\frac{-8+\sqrt{304}}{10}=\frac{-8+4\sqrt{19}}{10}=\frac{2\sqrt{19}-4}{5}.$$Thus, our answer is $2+19+4+5=\boxed{30}$.