This question has been asked before, at least twice by people who were trying to cheat on the PROMYS Europe 2020 Application, and once in a question closed because the OP showed no effort. I haven't been able to solve it, so I'm going to to ask it again. PROMYS applications have closed, and I hope the effort shown below is enough to keep this question from being closed.
The Mathematical Forest is grown in a two-dimensional plane, where trees can only grow on points with integer coordinates. To start with, there are no trees at all. The foresters plant the first tree at $(0,0)$. Each year, they carry out tree planting according to the following rule. If there is a tree on the point (,) but there are no trees on the points $(m+1,n)$ and $(m,n+1)$, then they can choose to remove the tree on $(m,n)$ and plant new trees on the points $(m,n+1)$ and $(m+1,n)$. For an integer $k\geq1$, the $k$th diagonal consists of all points $(m,n)$ with $m+n=k−1$. Is it possible for the foresters to arrange their planting so that eventually there are no trees on the first $2$ diagonals? What about the first $3$ diagonals? $4$ diagonals? Can you generalize?
It's easy to get to a position with no trees on the first two diagonals. (It only takes $4$ plantings.) I can prove it's impossible to reach a position with no trees on the first $4$ diagonals, and I believe it's impossible to reach a position with no trees on the first $3$ diagonals, but I can't prove it.
For the $k=4$ case, I used a pagoda function, as in a peg solitaire problem. For $k=1,2,3,\dots$ define the potential of a tree on the $k$th diagonal as $\frac1{2^{k-1}}$, and the potential of the forest as the sum of the potentials of all the trees in it. When a tree of potential $\frac1{2^{k-1}}$ is removed, it is replaced by $2$ trees of potential $\frac1{2^k}$ so the potential of the forest never changes. Initially, the potential of the forest is $1$.
Since there are $k$ trees on diagonal $k$, the potential of all points beyond the fourth diagonal is $$\sum_{k=5}^\infty\frac k{2^{k-1}}=\frac34<1,$$ so it is impossible that there are no trees on the first $4$ diagonals.
The potential of the fourth diagonal is $\frac12$, so this argument doesn't show that it's impossible to have no trees on the first $3$ diagonals. I've done a lot of experimenting, though, and I believe that the statement is true. (I even wrote a little computer game so I could experiment quickly.) The best I've been able to do is to get down to one man on the third diagonal.
The black circles are trees eligible for removal, and the gray circles are other trees. You can see that there is a "traffic jam" in front of the lone tree on diagonal $3$, and there seems to be no way to clear it.
I've been trying to come up with an argument that reflects this traffic jam, but I haven't been at all successful. We know that after $n$ removals there are $n+1$ trees, and I've been trying to prove somehow that they can't all be sufficiently far from the origin to allow all trees to get past diagonal $3$, but I haven't come close.
In the diagram above, if the tree at $(1,1)$ were moved to $(0,2)$, and the tree at $(1,2)$ were moved to $(2,0)$, then it would be possible to remove the tree at $(0,2)$, emptying the second diagonal. In the fictitious position, we still have one tree on diagonal $2$ and two on diagonal $3$.
So, I think an argument along the lines I was trying cannot just be concerned with the distance from the origin; it must somehow take into account how the trees interfere with one another. I haven't been able to do this.
I'd be grateful for solutions, hints, or counterexamples.
I wasn't really sure what tags to attach. Please add any that seem appropriate.
You're just missing 1 more observation.
Reveal as much as you need.
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But this requires infinitely many years, so it cannot be done.