I have tried this recreational problem for a long time now and I haven't been able to find any approach that would take me to the solution. Any hints or tricks to use would be appreciated.
On a chessboard with infinite rows and columns 3 pieces are placed on tiles (1, 1), (1, 2) and (2, 1). A permitted move is to remove a piece on position (i, j) and place two new ones on (i+1, j) and (i, j+1) if empty. Is it possible to achieve a configuration in which tiles (1, 1), (1, 2), (2, 1) and (2, 2) are empty?
This is also known as Pebbling a Chessboard.
It can't be done. Their proof is as follows:
Give the bottom left piece a weight (value) of $1$. Every time we remove it and place two new ones, give them half the weight each. The total weight will remain invariant (will not change), and will remain $1+\frac12+\frac12=2$.
The sum of all possible weights (whole board) is $4$. This means that in order to empty out the first three pieces, we need to occupy every other square on the board.
But that would require an infinite number of moves, hence it is impossible.