Do the squares have to have integer sides? Because nine squares, each of side 5/3, is a perfectly fine solution.
This isn't doable with integer square sides, as follows:
You cannot have a square of size 4x4, because that leaves 9, which only divides 4/4/1, 4/1/1/1/1/1 or 1/1/1/1/1/1/1/1/1 for solutions with 4, 7 and 10 smaller squares respectively. Each time you replace a 2x 2 square with 4 1x1 squares, you increase the number of squares by 3.
You can only have 1 square of size 3x3 for fit reasons. That leaves 16, which can be split into 4, 7, 10, 13 or 16 squares, for solutions with 5,8,11,14 or 17 squares in total.
So the largest square is 2x2, and by the same approach as before that will generate solutions with 25, 22, 19, 16, 13, 10 or 7 squares. None of those have 9 squares.
no subject
This isn't doable with integer square sides, as follows:
You cannot have a square of size 4x4, because that leaves 9, which only divides 4/4/1, 4/1/1/1/1/1 or 1/1/1/1/1/1/1/1/1 for solutions with 4, 7 and 10 smaller squares respectively. Each time you replace a 2x 2 square with 4 1x1 squares, you increase the number of squares by 3.
You can only have 1 square of size 3x3 for fit reasons. That leaves 16, which can be split into 4, 7, 10, 13 or 16 squares, for solutions with 5,8,11,14 or 17 squares in total.
So the largest square is 2x2, and by the same approach as before that will generate solutions with 25, 22, 19, 16, 13, 10 or 7 squares. None of those have 9 squares.
So there is no 9-square split with integer sides.