Painting chess board

The task is to paint each of the $64$ squares on a chess board either blue or red. I need to find the number of distinct ways this can be done given that any $2\times 2$ square on the board has two red and two blue squares.

I've tried solving it for a $4\times 4$ board, but I am getting no where. Would appreciate any help

I'll just give some hints that will allow you to easily deduce the formula that user14111 gave in a comment. Call any pair of adjacent squares a domino, which can be horizontal or vertical. Call two dominos neighbours if their union is a $2\times2$ square. Call a domino monochromatic for a colouring if its squares have the same colour.

• If a colouring has some monochromatic domino, then so is any neighbour of it (and it has the opposite colour).
• If there is any monochromatic hoizontal domino, then there is one in each row (in the same pair of columns)
• If there is any monochromatic vertical domino, then there is one in each column (in the same pair of rows).
• Both these conditions cannot be met simultaneously.
• Therefore it suffices to count the following cases, and add up the results:
  1. There is at least one monochromatic horizontal domino
  2. There is at least one monochromatic vertical domino
  3. There are no monochromatic dominoes.
• A solution for case 1. is completely determined by the colouring of its first row, a solution for case 2. is completely determined by the colouring of its first column, and a solution for case 3. is completely determined by the colouring of its top-left corner square.

For an $m\times n$ chessboard there are $2^m+2^n-2$ ways.

Case I. There are two horizontally adjacent squares of the same color: $2^m-2$ ways.

Case II. There are two vertically adjacent squares of the same color: $2^n-2$ ways.

Case III. None of the above: $2$ ways.

Hint for Case I: There are $2^m-2$ ways to color one row so that two adjacent squares have the same color. The rest of the coloring is determined from that; colors must alternate in each column. (Note, therefore, that Cases I and II do not overlap.)