This question was part of the problem set for PROMYS Europe 2020, a maths (or math, just in case I made anyone unhappy) camp held at Oxford every year. It was the one which I couldn't do, and I am still puzzled by it to this day: 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 (m, n) 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 ≥ 1, the kth 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? P.S. The deadline has already passed, so please do not think that I am cheating! The camp was actually cancelled due to obvious reasons, which sucks, because I really looked forward to going.
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  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 (𝑚,𝑛) 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 $$\sum_{k=5}^\infty\frac k{2^{k-1}}=\frac34&lt;1,$$ 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.
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I think it should be theoretically possible, so I'm going to leave the question open in case someone comes along in future with an answer. What is the difference between an abstract method and a virtual method? What is the difference between an abstract method and a virtual method? In which cases is it recommended to use abstract or virtual methods? Which one is the best approach? What happens to the human inside the titan when the titan dies? Just to be clear I'm NOT talking about those who where swallowed.       We know that titans used to be humans and then somehow they transformed into titans.   When they kill a titan it completely Vaporizes and nothing is left. So what happens to the human inside them?     Could it be like when Eren was testing his limits and each time he fused more with the titan body? If so, why could Ymir live in titan form for over 60 years and still transform back into human? What is the C# equivalent of NaN or IsNumeric? 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So how is gravity accomplished? "Yesterday" time specification shown when it was actually the day before yesterday I'm trying to show that for $b&gt;1$, $x&gt;0$ and $x$ irrational, that $$\sup \{b^p:p\in Q,\;0&lt;p&lt;x\} = \inf\{b^q:q\in Q,\;x&lt;q\}$$ I know this follows immediately if we define $b^x=\exp(x\log(b))$ and use the fact that the exponential function is strictly increasing and continuous over $\mathbb R$, but I'm not allowed to do that. In fact, I only have that $b^{m/n}=(b^{1/n})^m$, where $b^{1/n}$ means the positive $n^{th}$ root of $b$; $b^x$ is not yet defined for irrational $x$.  I know that the supremum and infimum both exist because if $b&gt;1$ and $p&lt;q$, then $b^p&lt;b^q$. Also, by assuming that the inf $&lt;$ sup, I got a contradiction. The part I'm stuck on is deriving a contradiction from assuming that sup $&lt;$ inf. Does anybody have any ideas?  Thank you. Steganography - How many bytes of information can you hide in a $1024 \times 768$ photo? Proof of inequality using AM-GM inequality prove that $x^2 + y^2 + z^2 \ge xy + xz + yz$  I tried proving it the way I prove $x^2 + y^2 \ge 2xy$. But it just doesn't seem to work. I tried using AM-GM. Should I be using GM-HM or AM-HM? I really don't know what the problem is and I'd appreciate any help and a full proof explanation. Thanks. For which $a \in \mathbb{C} -\ \{0,1\}$ does the equation $f(z) = f(az) $ has a non constant solution $f$ with $f$ being analytical in a neighborhood of $0$.  My attempt:  First, we can see that any such solution must satisfy:  $f(z)=f(a^kz)$ for all $k \in \mathbb{N} $.  If $|a|&lt;1$:  The series $z_{k} = a^k$ converges to 0 which is an accumulation point, and $f(z_i)=f(z_j)$ for all $i, j\in \mathbb{N} $. Thus $f$ must be constant.  If $|a|=1$:  For all $a \neq 1$ , $f$ must be constant on any circle around $0$, so again $f$ must be constant.  My quesions are:   Am I correct with my consclusions?   Also, I'm stuck in the case where $|a|&gt;1$. 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