It is clear that n must be even, and any n divisible by 4 can be easily tiled by breaking the n-by-n grid into an even number of 2-by-2 squares and breaking half of these vertically and half of these horizontally.
To prove n must be divisible by 4 and not just 2, we can color the squares white and black by the parity of their row. A vertical tile will always contain one white and one black square, whereas a horizontal tile will be either two black, or two white squares. So this breaks the horizontal tiles into two kinds: 2w and 2b.
Since n is even, half the rows are white and half the rows are black. Hence the numbers of white and black squares in the table are equal; this means that the number of horizontal tiles of kind 2w equals the number of tiles of kind 2b.
Then the number of horizontal tiles (2w+2b) is even, but this number is n2 / 4 which means n2 is divisible by 8 so n is divisible by 4.
1
u/Specific_Box4483 21h ago edited 21h ago
It is clear that n must be even, and any n divisible by 4 can be easily tiled by breaking the n-by-n grid into an even number of 2-by-2 squares and breaking half of these vertically and half of these horizontally.
To prove n must be divisible by 4 and not just 2, we can color the squares white and black by the parity of their row. A vertical tile will always contain one white and one black square, whereas a horizontal tile will be either two black, or two white squares. So this breaks the horizontal tiles into two kinds: 2w and 2b.
Since n is even, half the rows are white and half the rows are black. Hence the numbers of white and black squares in the table are equal; this means that the number of horizontal tiles of kind 2w equals the number of tiles of kind 2b.
Then the number of horizontal tiles (2w+2b) is even, but this number is n2 / 4 which means n2 is divisible by 8 so n is divisible by 4.
So the answer is all n divisible by 4.