Consider the infinite grid in the fourth quadrant (that is, the grid extends infinitely rightwards and downwards). For notation sake, each cell in the grid will be written as its top-left most coordinate (e.g. (0,0) represents the cell adjacent to the origin, (1,0) represents the cell to its right, etc.).
For our game, we start with one coin placed in each cell (0,0), (1,0), and (0,-1), as:
A step consists of removing one coin and replacing it with a coin in its right square and a coin in its lower square. However, we can only do this if there is no coin in either such square. For example, we can make a step with the coin in (1,0) (placing coins in (2,0) and (1,-1) as:
but now, we can NOT, for example, make a step with the coin in (0,-1), since the coin in (1,-1) now blocks it.
Define "homebase" to be the four cells {(0,0), (1,0), (0,-1),(1,-1)}. Show that after any number of steps, there will always be at least one coin in the homebase.
I suppose we are looking for an invariant (the classic one being alternative coloring) but I am at a loss for what to do.
For simplicity's sake, suppose the coordinates are nonnegative integers.
Let the value of each square $(x, y)$ be $\left(\frac12\right)^{x+y}$. Consider the combined values of each square that has a coin on it, this turns out to be the invariant.
Every time a coin on $(x, y)$ splits, it becomes two coins on $(x+1, y)$ and $(x, y+1)$. Note that the value of $(x, y)$ is equal to the sum of the values of $(x+1, y)$ and $(x, y+1)$.
Can you continue from here? In particular, by finding the combined value of the entire board?