*Sorry, this is an answer to an other question. (I did not read the question carefully.)* 

**Question:** *For which $k$, $k$ squares can tile the surface of cube.* 

**Answer:** $k=6\cdot(n^2+m^2)$.

Here is a tiling with $k=30$, $n=1$ and $m=2$.

![$k=30$.][1]

It is obvious if the tiling is vertex-to-vertex.


If the tiling is not vertex-to-vertex, you get a closed geodesic formed by overlaping sides.
Then you can shift squares on one side of the geodesic to make the tiling "more vertex-to-vertex".
Repeating this operation you can make the tiling to be vertex-to-vertex.


  [1]: http://www.math.psu.edu/petrunin/wiki/2-1.png