Flip the Board

Let's say we wanted to make all the black squares white. All $8$ of the black squares marked with a red $X$ in the picture above need to be flipped at least once, but none of them share the same row or column. Since a double-flip will flip over an entire row AND an entire column, it is possible for one double-flip to flip $2$ of the red $X$ squares at most. Since there are $8$ red $X$ squares, at least $\boxed{4}$ double-flips are needed.

One example of a solution using $4$ double-flips is if all of the blue $O$ squares are double-flipped. This works because all the white squares have an even number of $O$ s in their row and column, so they are flipped an even number of times and stay on white. All the black squares have an odd number in their row and column, so they are flipped an odd number of times and go to white.

Let's number the rows and columns from $1$ to $8$ , starting in the top left. Then the checkerboard can be thought of having been created from a blank white board by flipping all the even-numbered rows and columns (which are black except where they intersect). Undoing these flips, e. g. by applying flips at $(2,2)$ , $(4,4)$ , $(6,6)$ and $(8,8)$ , recovers the all-white board.

Consider instead an n by n checkerboard covered with tiles which are black on one side and white on the other. It is easy to see that for n = 1, we need 0 double-flips; for n = 2, we need 1 double-flip; for n = 3 or n = 4, we need 2 double-flips....

More generally, starting with the tile in the upper left corner, a single double-flip produces a 2 by 2 square of black tiles. Furthermore, the top 2 rows have identical patterns, with a 2 by 1 white column adjacent to the black square; similarly for the two columns on the far left. And the remaining (n-2) by (n-2) board has the original pattern; call it the "new board". We perform a double-flip to the tile in the upper left corner of the new board to continue.

Using mathematical induction, we can see that if n = 2k, then we shall need k double-flips to get an entirely black board. When n = 2k - 1, we still need k double-flips. In particular, for an 8 by 8 board, we need 4 flips.

Bcuz 8/2=4

The interesting point is that there are more than one solution... 96, to be precise.

We have the following 4Χ4 grid:

B2 B4 B6 B8

D2 D4 D6 D8

F2 F4 F6 F8

H2 H4 H6 H8

We choose 4 squares,in a way that they are all in different rows and columns of the grid.

For example:

B4, D8, F2, H6

Every different combination is a solution to our problem. There are 24 of them.

Same thing for this grid:

A1 A3 A5 A7

C1 C3 C5 C7

E1 E3 E5 E7

G1 G3 G5 G7

So, we have a total of 48 solutions for the one color. Same way for the other color, so the number of solutions becomes 96.