How to prove that the minimum square partition of a 3X2 rectangle has 3 squares
The case $w = 3$ is somewhat boring. Here is how to show that you need at least three squares. There is some square $S_1$ touching the bottom-left corner. There is some square $S_2$ touching the bottom-right corner. These squares must be different. At least one of them doesn't go all the way to the top, and so one of the other two corners remains uncovered, requiring another square.

I'm not sure how to extend this proof beyond $w &gt; 3$. Let me just mention that a classical result of Dehn shows that if an $h \times w$ rectangle can be tiled by squares then $h/w$ is rational, and moreover there is some integer $c$ such that if you scale everything by $c$ then all squares have integer sides.