4. A horse stands at the corner of a chessboard, a white square. With each jump, the horse can move either two squares horizontally and one vertically or two vertically and one horizontally (like a knight moves). The horse earns two carrots every time it lands on a black square, but it must pay a carrot in rent to rabbit who owns the chessboard for every move it makes. When the horse reaches the square on which it began, it can leave. What is the maximum number of carrots the horse can earn without touching any square more than twice?
Solution: $\square$
The horse must alternate white and black squares, and it ends on the same square where it started. Thus it lands on the same number of black squares (b) as white squares $(w)$. Thus, its net earnings will be $2 b-(b+w)=b-w=0$ carrots, regardless of its path.