TITLE: A special chess board
QUESTION [11 upvotes]: Consider a special $n\times n$ chess board such that each $1\times 1$ square is ether black or white.
We  know that every $1\times 1$ square has exactly one white adjacent square (two squares are adjacent, if they share exactly one side). For example the figure shows a possible coloring for $n=4$.
Example:

a) Prove that if $n$ is an odd number, then the coloring is not possible!
b) Prove that if $n$ is an even number, then the coloring is possible!
c) Find the number of white cells if $n$ is an even number!
These problems are really beautiful, try them!

REPLY [3 votes]: Here is the answer for a):
Consider the $2\times 2$ squares from the bottom left corner to the top right corner. Since $n$ is odd, the top right $1\times 1$ square is not in either of the $2\times 2$ squares. Write an A into each square of the main diagonal (from the bottom left to the top right corner), and a B into each square of the adjacent diagonals. 
The bottom left $1\times 1$ square’s adjacent white square is in the bottom left $2\times 2$ square in a B square. Hence the other A now has an adjacent white square. Therefore in the next $2\times 2$ square the bottom left A’s adjacent white square must be one of the Bs in that $2\times 2$ square. The othet A now has an adjacent white square. And so on... Untill we arrive to the top right $1\times 1$ square. We get that it can’t have an adjacent white square, because in that case the A under it would have two white adjacent squares.