Vasilev N.B.

On a plane, there is an $8 \times 8$ square divided into $1 \times 1$ cells. It is covered with right-angled isosceles triangles (two triangles cover one cell). There are 64 black and 64 white triangles. We consider "proper" coverings - such that any two triangles sharing a side are of different colors. How many proper coverings exist?
The first method. In one cell, a black triangle can be placed in four ways (two ways to choose the diagonal and two ways to choose which of the triangles will be black). It is easy to verify that if the arrangement of triangles in the cells on the main diagonal is known, the position of the triangles in the remaining cells is uniquely determined. Therefore, the number of correct coverings is $4^{8}=2^{16}$.

The second method. The top-left cell can be covered in four ways. The adjacent cell to the right can be covered in two ways, the next adjacent cell in two ways, and so on. We get that the top row can be covered in $2^{2} \cdot 2^{7}=2^{9}$ ways. The remaining seven cells in the left column can similarly be covered in $2^{7}$ ways. In total, we get $2^{16}$ ways. If three cells in a $2 \times 2$ square are already covered, the fourth cell is uniquely determined, as is easy to verify. Therefore, the entire remaining part of the board is uniquely covered.

## Answer

$2^{16}$ coverings.