A square can't be divided into 2, 3 or 5 smaller squares I have just read that a square can't be divided into 2, 3 or 5 smaller squares (not necessarily distinct). I am unable to think of a way to prove it. Can anyone help me by providing a hint?
If $n&lt;4$, at least one of the squares must cover two vertices of the original square. These vertices must also be vertices of the smaller square and it must not protrude; we conclude that it has side length $\ge 1$ and covers the big square by itself.
For $n=5$, there must be exactly one small square that does not cover a vertex of the original square. Only few configurations of neighbourship are possible, and none of the works.
When a square gets divided in to more squares there are (divisions)*(divisions) new squares. If you divide a square two times you get 4 squares.