XLVI OM - II - Problem 6

A square with side length $ n $ is divided into $ n^2 $ unit squares. Determine all natural numbers $ n $ for which such a square can be cut along the lines of this division into squares, each of which has a side length of 2 or 3.
If $ n $ is an even number, then a square of dimensions $ n \times n $ can be cut into $ 2 \times 2 $ squares. Let's assume, then, that $ n $ is an odd number. The square $ n \times n $ consists of $ n $ rows, with $ n $ cells (unit squares) in each row. We paint these rows alternately black and white (in a "zebra" pattern). Due to the oddness of $ n $, both the top and bottom rows will be the same color—let's assume black. The difference between the number of black cells and the number of white cells in the entire checkerboard is $ n $.
Suppose that a division into $ 2 \times 2 $ and $ 3 \times 3 $ squares is possible. Each $ 2 \times 2 $ square has two white and two black cells. Each $ 3 \times 3 $ square has three white and six black cells, or vice versa. The difference between the number of black cells and the number of white cells in a $ 2 \times 2 $ square is $ 0 $, and in a $ 3 \times 3 $ square, it is $ 3 $ or $ -3 $. By adding all these differences (corresponding to the individual squares in the division), we obtain a number divisible by $ 3 $. However, this sum is also equal to the difference between the number of black cells and the number of white cells in the entire checkerboard, which is $ n $.
Conclusion: If the considered division is possible, then the number $ n $ is divisible by $ 2 $ or by $ 3 $. Conversely, if $ n $ is divisible by $ 2 $ or by $ 3 $, then the $ n \times n $ square can be cut in the desired way (into only $ 2 \times 2 $ squares or $ 3 \times 3 $ squares).
Note 1. When $ n $ is divisible by $ 6 $, the checkerboard can, of course, be cut into both $ 2 \times 2 $ and $ 3 \times 3 $ squares. Moreover, mixed divisions are possible; the reader can verify (this is an easy exercise) that if $ n $ is a number greater than $ 6 $ and divisible by $ 2 $ or by $ 3 $, then the $ n \times n $ square can be cut into $ 2 \times 2 $ and $ 3 \times 3 $ squares, using both types.