jmc

We start by drawing a diagram. When the paper is folded, sides $AD$ and $CD$ coincide on the longer dashed line, and points $A$ and $C$ meet at $G$, as you can see below. Now, we assign variables. We are looking for the length of $AE$, so let $AE=x$. Then, $BE=1-x$. Because of the symmetry of the square and the fold, everything to the left of line $BD$ is a mirror image of everything to the right of $BD$. Thus, $△BEF$ is an isosceles right triangle (45-45-90), so $EF=\sqrt{2}EB=\sqrt{2}(1-x)$. Also, $△EGB$ and $△FGB$ are congruent 45-45-90 triangles, so $GB=\frac{EB}{\sqrt{2}}=\frac{(1-x)}{\sqrt{2}}$.

Also, notice that because the way the paper is folded (its original position versus its final position), we have more congruent triangles, $△AED\cong △GED$. This means that $AD=GD=1$.

Lastly, notice that since $G$ is on $BD$, we have $BD=BG+GD$. $BD$ is a diagonal of the square, so it has side length $\sqrt{2}$, $GD=1$, and $GB=\frac{(1-x)}{\sqrt{2}}$. Thus, our equation becomes $$\sqrt{2}=1+\frac{(1-x)}{\sqrt{2}}.$$ Multiplying both sides by $\sqrt{2}$ yields $2=\sqrt{2}+1-x$; solving for $x$ yields $x=\sqrt{2}-1$. Thus, $AE=\sqrt{2}-1=\sqrt{k}-m$, and we see that $k+m=2+1=\boxed{3}$.