This was proposed by me! In fact, the hunter wins on any (connected) graph with bounded degree. We're putting this result on arxiv after the IMO.

Remark: the 1D case is very helpful Claim: we can find the direction of the rabbit at every move. Proof: consider a first coloring where $f(x,y)\in \{0,1,2,3,4\}, f(x,y)\equiv 2x+y (\bmod\; 5)$. Then going up, right, left, down correspond to the residue increasing by $1,2,3,4$ mod 5, respectively. In the second coloring, the points where $x\ge 0, y\ge 0$ is colored with color 1, points where $x\ge 0, y<0$ has color 2, $x<0,y<0$ has color 3 and the remaining have color 4. In the third set of coloring, color with 2 colors $(x,y)$ 1 only if $|x|+|y|\in S$ where $S=\{ s_1<s_2<\cdots< \}$ where $s_i=2^{s_{i-1}}$, and color $(x,y)$ 2 otherwise. Claim: eventually the hunter can find $|x|+|y|$ Proof: let $R_i=\{(x,y) | |x|+|y| \in (c_{i-1},c_i)\}$ Observe the following: it takes at least $c_{i+1}-c_i$ turns to move from a point $(x,y)$ where $|x|+|y|=c_i$ to a point where $|x|+|y|=c_{i+1}$. On the other hand, the rabbit must move out of the set of points $(x,y)$ such that $|x|+|y|\le c_i$ in at most $4c_i^2$ moves since it cannot revisit the same square. As we also keep track of how many steps in which direction the rabbit takes (ie if the rabbit's starting position is $(x_0,y_0)$ we know it is $(x_0+\Delta x, y_0+\Delta y)$ for known values of $\Delta x, \Delta y$. Pick $N=|x_0|+|y_0|$, then if the rabbit goes from $c_N$ to $c_{N+1}$, I can tell by knowing when rabbit reports 1's and a relative change of the position from a point at $|x|+|y|=c_N$ and $|x|+|y|=c_{N+1}$, which has $0.99 c_{N+1} < |\Delta' x|+ |\Delta' y|< 1.01 c_{N+1}$ Here $\Delta'$ denotes the displacement from one point $(x,y)$ red with $|x|+|y|\le c_N$ to a point where $(x,y)$ is red and for size reasons, $c_{N}<|x|+|y|<c_{N+2}$, indicating $|x|+|y|=c_{N+1}$. Invoke a fourth coloring where $(x,y)$ has color 1 iff $|x|+\lfloor \frac{|y|}{2} \rfloor=c_j$. WLOG at a point P where $(x,y)$ has color 1 in the third coloring, the second coloring of $(x,y)$ returns 1 (i.e. $x,y\ge 0$) since the other cases for the second color are similar. This means $x+y=c_t$ for some $t$. After a while it will cross $|x+\Delta' x|+ \lfloor \frac{|y+\Delta' y|}{2} \rfloor =c_u$ where $u>t$ and $\Delta' x, \Delta' y$ are known to the hunter. We can find the signs of $|x+\Delta' x|, |y+\Delta' y|$ from the second coloring. We can now solve a system of linear algebraic equation to determine the unique $P(x,y)$ that the rabbit was on, and we use the first coloring to backtrack the original position of the rabbit. Note: CAN IMO team points out a simpler construction is 8 cases on the signs of $y,x, |y|-|x|$ and consider squares of the form $\max\{ |y|, |x|\}=c_n$ instead. Remark: How dare they make rabbit lose. The rabbit should go hunting for the hunter, because the rabbit is secretly a goddess.

Very good problem, this and the hopping frogs are my favourite

We claim that the hunter has a winning strategy. Take $N=10^{665}$. Each digit of the number of carrots on each cell will give the hunter information. The first two digits will let the hunter know how the rabbit moves every turn. Let the first digit be either $1$, $2$, or $3$, and it is equivalent to the $x$-coordinate modulo $3$. Similarly, let the second digit be $1$, $2$, or $3$, and it is equivalent to the $y$-coordinate modulo $3$. The first digit allows the hunter to know if the rabbit moved left or right, and the second digit allows the hunter to know if the rabbit moved up or down. Therefore, it suffices to find the location of the rabbit at any time. Now, let $a_i$ be the sequence $1$, $0$, $1$, $0$, $0$, $1$, $0$, $0$, $0$, $1$, $\ldots$, where the $i$th appearance of $1$ is followed by exactly $i$ $0$s. Let $b_i=1-a_i$ be the sequence when $1$ and $0$ are swapped. Let the third digit be $a_i$ if $i$ is the $x$-coordinate of the cell and $i>0$, and let the third digit be $b_{1-i}$ if $i\leq0$. Similarly, define the fourth digit to be $a_i$ if $i$ is the $y$-coordinate and $i\geq0$ and $b_{1-i}$ otherwise. Since the rabbit cannot go to a cell more than once, at least one of its $x$ or $y$ coordinates must be unbounded. The hunter can use the direction the rabbit moves to see the part of the sequence the rabbit is at by looking at the third or fourth digit. Once the hunter sees a subsequence consisting of a digit, followed by multiple appearances of another digit, followed by another appearance of the first digit, the hunter can uniquely identify what one of the rabbit's coordinates are. Then, if the cell's coordinates are $(x,y)$, then let the fifth digit be $a_{|x|+|y|+1}$ and let the sixth digit be either $0$, $1$, or $2$ and is equivalent to $|x|+|y|$ modulo $3$. Let the set of points where $|x|+|y|$ is constant be called a "ring." Then, the sixth digit tells the hunter whether the rabbit moved to a ring closer to the origin or farther away from the origin. This allows the hunter to identify a contiguous subsequence of $a$ that corresponds to the ring the rabbit is on. Since $|x|+|y|$ must be unbounded, the hunter will know a sufficiently large subsequence of $a$, such allows the hunter to identify the value of $|x|+|y|$. Now, the hunter should use the $7$th and $8$th digits such that the $7$th digit is $1$ if the $x$ coordinate is $0$, $2$ if $x>0$, and $0$ if $x<0$. Similarly, the $8$th digit is $1$ if $y=0$, $2$ if $y>0$, and $0$ if $y<0$. This allows the hunter to determine the signs of $x$ and $Y$. This gives the hunter enough information to determine $|x|+|y|$ and one of $x$ and $y$. Knowing the sign allows the hunter to get the value of either $x+y$ or $x-y$, so the hunter can solve this equation to find $x$ and $y$. Finally, since the hunter knows the rabbit's moves, the hunter can determine the rabbit's starting location. The other $657$ digits are unnecessary and the hunter can make them any value, which does not affect the hunter's strategy. A table of the digits for small $x$ and $y$ are shown below. Digit 1$\qquad\qquad\qquad\qquad\qquad$ Digit 2$\qquad\qquad\qquad\qquad\qquad$ Digit 3 12312312312$\qquad\qquad\qquad\qquad$ 22222222222$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 11111111111$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 33333333333$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 22222222222$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 11111111111$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 33333333333$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 22222222222$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 11111111111$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 33333333333$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 22222222222$\qquad\qquad\qquad\qquad$ 01101010100 12312312312$\qquad\qquad\qquad\qquad$ 11111111111$\qquad\qquad\qquad\qquad$ 01101010100 Digit 4$\qquad\qquad\qquad\qquad\qquad$ Digit 5$\qquad\qquad\qquad\qquad\qquad$ Digit 6 00000000000$\qquad\qquad\qquad\qquad$ 01000100010$\qquad\qquad\qquad\qquad$ 10210201201 00000000000$\qquad\qquad\qquad\qquad$ 10001010001$\qquad\qquad\qquad\qquad$ 02102120120 11111111111$\qquad\qquad\qquad\qquad$ 00010001000$\qquad\qquad\qquad\qquad$ 21021012012 00000000000$\qquad\qquad\qquad\qquad$ 00100100100$\qquad\qquad\qquad\qquad$ 10210201201 11111111111$\qquad\qquad\qquad\qquad$ 01001010010$\qquad\qquad\qquad\qquad$ 02102120120 00000000000$\qquad\qquad\qquad\qquad$ 10010101001$\qquad\qquad\qquad\qquad$ 21021012012 11111111111$\qquad\qquad\qquad\qquad$ 01001010010$\qquad\qquad\qquad\qquad$ 02102120120 00000000000$\qquad\qquad\qquad\qquad$ 00100100100$\qquad\qquad\qquad\qquad$ 10210201201 11111111111$\qquad\qquad\qquad\qquad$ 00010001000$\qquad\qquad\qquad\qquad$ 21021012012 11111111111$\qquad\qquad\qquad\qquad$ 10001010001$\qquad\qquad\qquad\qquad$ 02102120120 00000000000$\qquad\qquad\qquad\qquad$ 01000100010$\qquad\qquad\qquad\qquad$ 10210201201 Digit 7$\qquad\qquad\qquad\qquad\qquad$ Digit 8 00000122222$\qquad\qquad\qquad\qquad$ 22222222222 00000122222$\qquad\qquad\qquad\qquad$ 22222222222 00000122222$\qquad\qquad\qquad\qquad$ 22222222222 00000122222$\qquad\qquad\qquad\qquad$ 22222222222 00000122222$\qquad\qquad\qquad\qquad$ 22222222222 00000122222$\qquad\qquad\qquad\qquad$ 11111111111 00000122222$\qquad\qquad\qquad\qquad$ 00000000000 00000122222$\qquad\qquad\qquad\qquad$ 00000000000 00000122222$\qquad\qquad\qquad\qquad$ 00000000000 00000122222$\qquad\qquad\qquad\qquad$ 00000000000 00000122222$\qquad\qquad\qquad\qquad$ 00000000000

DottedCaculator wrote: We claim that the hunter has a winning strategy. Take $N=10^{665}$. During the American training camp one of the students found a strategy using only eight (!) colors. (I'll let the student post that construction if he wants to.) There was some talk of trying to prove that eight colors is best possible but I don't think this managed to be proved.

It is possible to use $40$ colors. It suffices to consider a finite collection of finite colorings, since then we can consider their cartesian product. First coloring. Use $5$ colors. Color $(x,y)$ by the residue of $x+2y$ modulo $5$. This coloring tells unambiguously which one of the four directions the rabbit goes at each step. Now we define a subset $$S\doteqdot\{ -n^2\mid n\in\mathbb Z_{\ge 0} \} \cup \{ n(n+1)\mid n\in\mathbb Z_{> 0} \}$$of $\mathbb Z$. Note that $S$ is unbounded in both directions and the directed differences between any two consecutive elements of $S$ are pairwise different: Each positive integer is achieved exactly once. We will define $3$ more colorings in white and black. Second coloring. Color $(x,y)$ black if and only if $x\in S$. Third coloring. Color $(x,y)$ black if and only if $y\in S$. Fourth coloring. Color $(x,y)$ black if and only if $x+y\in S$. Assume for contradiction that the rabbit moves indefinitely. Note that if $(x,y)$ is the cell's rabbit, then the values of $x,y,x+y$ change either by $0$, $1$ or $-1$. Moreover, let $A$ be the set of cells where the rabbit visits. Consider the function $f(x,y)=x$, $g(x,y)=y$ and $h(x,y)=x+y$. Hence $f(A)$, $g(A)$ and $h(A)$ are discretely contiguous and at least two of $f(A)$, $g(A)$ and $h(A)$ are unbounded. Assume for instance that the $x$-value is unbounded. Consider two consecutive moments where we visited two black cells of the second coloring, where the total walk between them is not zero in the $x$-direction. We can guarantee that since $f(A)$ is unbounded and we know the direction at each step. Then we know the directed difference of the $x$-values. Since the differences between any two consecutive elements of $S$ are pairwise different, we know the actual $x$-values and not only their directed difference. Hence we also know the initial $x$-value. Similarly for $y,x+y$. So we know the initial values of at least two of $x,y,x+y$, but then we know all of them since $x+y-(x+y)=0$.

Same as above. The answer is $\boxed{\text{yes}}.$ For convenience, we assume the hunter pre-designates a square as the origin. Consider these colorings of the grid. Color $(x,y)$ with $2x+y\pmod{5}.$ Color $(x,y)$ with $1$ if $x=\pm 2^k$ for some $k,$ and with $2$ otherwise. Color $(x,y)$ with $1$ if $y=\pm 2^k$ for some $k,$ and with $2$ otherwise. Color $(x,y)$ with $1$ if $x-y=\pm 2^k$ for some $k,$ and with $2$ otherwise. Observe the following. The first coloring can be used to determine the directions of the rabbit's moves. If the rabbit takes all sufficiently large or small $x$ values, the second coloring can be used to determine the rabbit's initial $x$-coordinate. If the rabbit takes all sufficiently large or small $y$ values, the third coloring can be used to determine the rabbit's initial $y$-cooridnate. If the rabbit takes only finitely many $x$ values or $y$ values, then it takes all sufficiently large or small $x-y$ values, so the fourth coloring can be used to determine the initial value of $x-y.$ Hence the hunter can pinpoint the rabbit's starting position given all four colorings. To finish, just note that the first coloring corresponds to an integer modulo $5$ and the other three colorings correspond to an integer modulo $8$ (using base $2$), so by CRT $40$ colors suffice.

40 color construction is incoming, I want to see if I can improve this number Define $T_n$ to be the $n$'th triangle number. We say that we $overlay$ two colorings if we assign one color to each pair of colors any square can have. Overlay the following colorings. $\mathcal{A}$. $(x,y)$ is assigned red if $x= \pm T_n$ and blue othwerwise $\mathcal{B}$. $(x,y)$ is assigned red if $y= \pm T_n$ and blue otherwise $\mathcal{C}$. $(x,y)$ is assigned red if $x+y= \pm T_n$ and blue otherwise $\mathcal{D}$. The famous $5$ coloring where each cell's orthogonal neightbors are all of different colors By coloring $\mathcal{D}$ we can know the directions of the path of the rabbit. Suppose that the $y$ coordinate of the rabbit is bounded from top and bottom. Clearly, $x$ must be unbounded in some direction. After a sufficient number of jumps, using the size of gaps between colors in $\mathcal{A}$, we can deduce the rabbit's current $x$-coordinate and reverse engineer the starting position with the paths acquired with $\mathcal{D}$ to get the $x$-coordinate of the starting position. Now, do the same thing except with the diagonals in $\mathcal{C}$. The intersection of a diagonal and a row is a single square, which must be our starting square. The same applies to a bounded $x$-coordinate. But, if $x$ and $y$ are both unbounded, we can use the same gap size trick to find the $x$-and $y$-coordiante of the starting position. Thus, the hunter can always deduce the rabbit's position in $2 \times 2 \times 2 \times 5=40$ colors.

The hunter can color the cells in a way that enables him to guess the starting point of the rabbit, providing the rabbit does not end up in a dead end. It can be done with at most 20 colors. We consider the colors as consisting of some layers, i.e. any color will be viewed as a tuple $ (c_1,c_2,c_3)$ where each variable (layer) $ c_i,i=1,2,3$ can take a finite number of colors. Each layer has a purpose. The first one is for orientation - to know the directions of each move the rabbit makes (up,down,right, left). The second layer will ensure there are no two paths with identical sequence of colors that start from different cells within a fixed frame around the origin. The third layer is to give us the information about the region (frame) within which the rabbit starts. This structure of colors is only for convenience, clearly this configuration can be encoded with one dimensional colors. The first layer $ c_1$ takes values $ 1,2,3,4,5$ and its purpose is to control the direction of the rabbit. We put these values on each row starting as $ 1,2,3,4,5,1,\dots$ and on the next row starting from $ 3$, on the next one - from $ 5$, then from $ 2,4,1$ and so on. This coloring allows us to translate the sequence of colors into directions, say, up, left, down, down, right, etc. So, if we know the starting cell, this layer helps us to recover rabbit's trajectory. It also means that two equal sequences of the first layer color correspond to trajectories such that one of them can be obtained by translating the other with the vector determined by their starting points. The second layer of colors will allow the hunter to locate the starting position of the rabbit, providing he knows the rabbit started in a fixed frame. The second layer will consist of only two colors. Suppose the hunter knows the approximate location the rabbits started from, i.e. he knows that the rabbit started in a cell inside some (big) square (frame), say $ Q$ centered at the origin. Suppose all the cells inside a bigger square $ Q'\supset Q$ (also centered at the origin) have their second layer already colored. We claim that the hunter can consecutively color the second layer of each cell in a bigger square $ Q''\supset Q'$ in a way that there are not two paths starting from different cells in $ Q$ and leaving $ Q''$ that have the same sequence of colors. Let us fix two cells $ q_1,q_2\in Q, q_1\ne q_2$ and $ v$ be the vector $ q_2-q_1$. For any cell $ q'$ outside $ Q'$, but having a common side with $ Q'$ we consider the cells $ r:=q'+ v, s:=q'-v$. They cannot be both in $ Q'$ (then q' would have been inside $ Q'$ since $ Q'$ is convex). So, in case one of them is outside $ Q'$, say $ r$, and $ s$ is inside this square, we set the second layer of $ q'$ to differ the second layer of $ s$, and the second layer of $ r$ be thae same as that of $ s,$ that is, we color them alternatively. If both $ r,s $ are outside $ Q'$ we also color the second layers of $ r,q',s$ alternatively. We do this procedure for all $ q'$ outside $ Q'$ but with a common side with $ Q'$. Suppose now, there exists two paths, say, $ P_1,P_2$ that start from $ q_1$ and $ q_2$ respectively, and with the same sequences of colors. Suppose, $ q'$ is the first cell that one of them hits outside the frame $ Q'$. At this moment the other path hits one of the cells $ q'\pm v$. But, we took care $ q'$ and $ q'\pm v$ have different second layer color. Thus, there are no paths starting from $ q_1$ and $ q_2$ respectively, that leave $ Q'$ and have the same sequence of colors. Further, we color somehow the second layer of those cells that have a common side with $ Q'$ and are still uncolored, and proceed with some other pair of cells $ q_1,q_2\in Q$ and with one unit bigger than $ Q'$ frame which we again denote as $ Q'.$ In the end, we expand the coloring to a frame $ Q''$ and are sure that there are no two paths that start inside $ Q,$ leave $ Q''$ and have the same sequence of colors. It remains to invent some sign that would show us the approximate location the rabbit started from. That's the third layer that comes to help. This layer consists of three colors, say, white, black and neutral. Place a strip, two cells wide, symmetrical to the origin, and color the third layer of its inner cells white and its outer cells black. Thus, we will know if the rabbit starts within this strip - it will cross the border as white, black pattern. Now, we set infinitely many concentric strips like that of rapidly increasing sizes. It's enough to ensure that the shortest path between two consecutive strips is longer than the longest path between the previous strips. We color the third layer of all other cells neutrally. In this way, examining the sequence ot the third layer's color, we can determine the frame from which the rabbit started. So, the number of needed colors is $ 5\cdot 2\cdot 3=30$. But looking closely to the third layer one may see that in order to determine the frame the rabbit starts in, we actually only need strips one cell wide, colored, say, white, and all the other cells colored neutrally. Hence we can decrease the number of total colors to $ 5\cdot 2\cdot 2=20$. Remark. A few more comments on my blog.

v_Enhance wrote: During the American training camp one of the students found a strategy using only eight (!) colors. (I'll let the student post that construction if he wants to.) There was some talk of trying to prove that eight colors is best possible but I don't think this managed to be proved. I think I found a construction with eight colours. Main idea is from dgrozev's construction above. Outline. We will call the eight colours $0W, 0R, 0B, 1W, 1R, 1B, 2, 3$. The construction will have three main steps. 1. Each cell will be assigned a number $0,1,2$ or $3$ which will allow us to almost reconstruct the trajectory of the rabbit. 2. Some of the $0$s and $1$s will be painted white, forming a frame which the rabbit must eventually cross, allowing us to restrict the rabbit's position to a finite set of cells. 3. The remaining $0$s and $1$s will be painted red and blue so that no pair of starting points have a trajectory with identical colours. Step 1. We first assign each cell $(x,y)$ the number $x+y^2-y$ (mod 4). Note $y^2-y\equiv\begin{cases}0 \text{ if } y\equiv 0,1 \text{ (mod 4)} \\ 2 \text{ if } y\equiv 2,3 \text{ (mod 4)} \end{cases}$ so the grid will look like $$\begin{matrix} 3&0&1&2&3&0&1\\ 3&0&1&2&3&0&1\\ 1&2&3&0&1&2&3\\ 1&2&3&0&1&2&3\\ 3&0&1&2&3&0&1\\ 3&0&1&2&3&0&1\\ 1&2&3&0&1&2&3\\ \end{matrix}$$ We can almost reconstruct the trajectory (up/down/left/right moves) of the rabbit from the sequence of numbers alone. If the number increased or decreased by $1$ (mod 4), we know the rabbit moved right or left respectively. Otherwise, the rabbit moved up or down. We will be unable to deduce the direction of the first vertical move. However, we can always deduce if each subsequent vertical move is in the same or opposite direction as the previous: if the number didn't change for both moves or changed by $2$ (mod 4) for both moves, we know that they are in opposite directions. Otherwise, they are in the same direction. Therefore, for any sequence of numbers, there are only two possible trajectories for the rabbit which differ only by a vertical reflection. Step 2. We define a block $B(x,y)$ to be the four cells $(2x,2y),(2x+1,2y),(2x,2y+1),(2x+1,2y+1)$. Also define a frame of size $k\geq 0$ to be the union of all blocks $B(x,y)$ with $|x|+|y|=2k$. It is clear that all frames only contain $0$s and $1$s. The rabbit also must step on the frame to go from the inside to the outside of a frame. For each $k\geq 2$, we paint the frames $2^k-1,2^k,2^k+1$ red, white and blue respectively. We can always tell when the rabbit crosses a white frame and in which direction. If the rabbit visits a red block immediately before (after) a consecutive sequence of white blocks, it started (ended) inside the frame. If it was a blue block, it started/ended outside the frame. When the rabbit eventually exits two more white frames than it enters, say in less than $2^k$ moves, we know that the rabbit definitely started inside the frame $2^k$, since it did not have enough moves to exit two white frames outside. This restricts the rabbit's initial position to a finite number of cells. Step 3. Suppose we know that the rabbit started from a finite set of cells $Q$. It suffices to paint the remaining cells such that for any $q_1,q_2\in Q$, a) A rabbit starting from $q_1$ and a shadow rabbit starting from $q_2$ moving in the same trajectory will produce a different sequence of colours. b) A rabbit starting from $q_1$ and a shadow rabbit starting from $q_2$ moving in vertically reflected trajectories will produce a different sequence of colours. We tackle (a) first. Fix $q_1,q_2\in Q$ and let $v$ be the vector $q_2-q_1$. We select large enough frames $\{k-a,k-a+1,\dots,k+a\}$ such that they are all not painted yet, and for every cell $q'$ in frame $k$, $q'+v$ is selected or is a $2$ or a $3$. Consider the graph on the selected cells. Two cells are connected if their difference is $\pm v$. Clearly each component is a finite chain so we paint each chain red and blue alternately. When the rabbit starting from $q_1$ eventually steps onto frame $k$, the shadow rabbit will be on a cell with a different colour or number. We can ensure (b) almost identically. Define the vector conjugate of $v=\begin{pmatrix}x\\y\end{pmatrix}$ to be $\overline{v}=\begin{pmatrix}x\\-y\end{pmatrix}$. Now if the rabbit starting from $q_1$ moved to $q'$, the shadow rabbit will be at $q_2+\overline{q'-q_1}$. We can once again select large enough frames $\{k-a,k-a+1,\dots,k+a\}$ such that they are all not painted yet, and for every cell $q'$ in frame $k$, $q_2+\overline{q'-q_1}$ is selected or is a $2$ or a $3$. A similar graph on the selected cells will produce components which are either finite chains, 2-cycles or 1-cycles. Note that the 1-cycle can only happen when $q_1$ and $q_2$ have the same $x$-coordinate and $q'$ lies on the perpendicular bisector of $q_1$ and $q_2$. However, any path from $q_1$ to $q'$ will have a different number sequence from the corresponding path from $q_2$ to $q'$. We can therefore ignore this case and paint each component red and blue alternately. When the rabbit starting from $q_1$ eventually steps onto frame $k$, the shadow rabbit will be on a cell with a different colour or number. Repeating the above process for every pair of cells $q_1,q_2\in Q$ ensures that we can catch the rabbit given it started in $Q$. We carry out this process sequentially for $Q$ being the interior of frames $4,8,16,32,\dots$. Finally, painting all remaining $0$s and $1$s red produces a construction with only eight colours. $\Box$

We claim that the answer is positive. Consider the following colorings, which can be united in one: 1. Tracking of rabbit's path form: \begin{tabular}{ccc} \text{ } \text{ } \text{ } \text{ } \text{ } 4\\ \text{ } \text{ } \text{ } \text{ } 1 2\\ \text{ } \text{ } \text{ } 4 5 6\\ \text{ } \text{ } 1 2 3 1\\ \text{ } 4 5 6 4 5\\ 1 2 3 1 2 3 \end{tabular}Paint as above (colors are continued horizontally and vertically). Then on each step we know the direction of rabbit's move. Clearly until rabbit gets stuck at least $2$ of variables $x,y,x+y$ (rabbit's coordinates and their sum) are unbounded. We present colorings which allow to find unbounded variable. 2. Finding x-coordinate: Paint all columns black and white such that all distances between adjacent black columns are distinct. If $x-$coordinate is unbounded, this coloring allows to find this coordinate after rabbit intersect black column for the second time. 3. Finding y-coordinate: Analogous to the previous one. 4. Finding x+y: Color diagonals $x+y=const$ black and white such that the gaps between adjacent black diagonals are distinct. Now knowing two of variables $x,y,x+y$ we find both $x,y.$ Then using first coloring we may found the starting cell of rabbit.

v_Enhance wrote: DottedCaculator wrote: We claim that the hunter has a winning strategy. Take $N=10^{665}$. During the American training camp one of the students found a strategy using only eight (!) colors. (I'll let the student post that construction if he wants to.) There was some talk of trying to prove that eight colors is best possible but I don't think this managed to be proved. In Argentina we found a construction with only six colours but we were unable to reduce it further. Of course four colours are impossible so the only open question is for five colours.

By CRT we are able to independently "layer" colorings as long as they have a finite number of colors. Define an arbitrary origin and impose coordinates. Claim: We are able to determine the direction of every one of the rabbit's move. Proof: Color by $2x+y \pmod 5$. Now note that moving up, left, right, and down increase the residue by distinct numbers. This means that if we can determine the x-coordinate of any point on the rabbit's path, and the y-coordinate of any point on the rabbit's path, we can determine its starting location. Note that for either the x or the y direction (or both), the rabbit must be reaching infinitely large or small coordinates; suppose it is the y-direction. Claim: We are able to determine the rabbit's y-coordinates for the entirety of their path. Proof: For each $x$, color every pair $(x,y)$ with $y = k^3$ for some integer k red. By definition the rabbit must hit infinitely many such red lines. But between any two consecutive hits, we are able to determine the y-displacement of the rabbit, so through finite differences we can determine exactly which red lines were hit. Now, if the x-coordinates of the rabbit is also unbounded, then by mirroring the previous coloring we are finished. If not, suppose that the x-coordinates of the rabbit are bounded. Now: Claim: We are able to determine the rabbit's x-coordinates for the entirety of their path. Proof: For each $x$, color every pair $(x,y)$ with $y = k^3+x$ for some integer k red. Since the x-coordinates of the rabbit are bounded, the rabbit must hit infinitely many red lines. As we already know all of the y-coordinates at each hit and the change in x-coordinate between every time the red diagonal is hit, we can again apply finite differences to calculate $k$ for some point on the rabbit's path, which in turn tells us $x$, done! Remarks: Really cool problem! The key idea is to realize that we can actually track the rabbit's position by examining finite differences; this is motivated by looking at the 1D case. After that, the only remaining trick to adjust for the 2D case is if the x-coordinates are bounded; we would then want a similar coloring that also tracks the x-coordinate as well as the y-coordinate.

BrunZo, can you please post your solution using 6 colors?

Replace colorings with number assignments. We can easily "combine" numberings by using "variable base numbers". For example if I had a three numberings using $n_1,n_2,n_3$ colors each and I wanted to combine them, a cell which was colored in $a,b,c$ respectively would become colored with $a+n_1b+n_1n_2c$; it is clear that we can extract the three original numberings this way. Pick an origin arbitrarily and number the point $(x,y)$ with $x+2y \pmod{5}$. This clearly allows us to deduce the direction of the rabbit's every move. Now introduce another numbering: number the points with $x$-value equal to $\tfrac{n(n+1)(2n+1)}{6}$ for $n \geq 0$, or $\tfrac{n(n+1)(2n+1)}{3}$ for $n<0$, with a $1$, and $0$ otherwise. Since gaps between subsequent "boundaries" (rows) of ones are either distinct squares (for $n$ nonnegative) or distinct twice-squares (for $n$ negative), all these gaps are distinct. Do the same for the $y$-values with a different numbering, the same for the $x+y$-values with another numbering, and the $x-y$-value with a final numbering. Focus on only the $x$-value and $y$-value based numberings, which divide the grid into bounded rectangles of cells labeled with two zeroes. If the rabbit does not want to lose because it can't move at some point, it will have to cross infinitely many of the boundaries in some direction (where at least one of the two numbers is a $1$), so by pigeonhole WLOG have it cross a horizontal boundary moving upwards infinitely many times. Then, the hunter will be able to tell whenever the rabbit moves upwards and steps onto a cell with $x$-based number $1$, corresponding to crossing upwards. At some point, this will happen twice on two adjacent boundaries, such that we don't cross any other horizontal boundaries in between (crossing vertical boundaries is fine), and based on the number of up/down moves between these crossings we can clearly deduce the $x$-coordinate of the rabbit when this happens, so at every point in time after this the hunter is aware of the absolute $x$-coordinate of the rabbit. If we instead focus on the $x+y$-value and $x-y$-value numberings, a similar argument works, so at some point the hunter will be able to deduce the value of either $x+y$ or $x-y$. Given this and $x$, the hunter is then able to find the exact position of the rabbit, so thy may then work backwards and determine the starting position exactly. $\blacksquare$ Remark: This coloring achieves $5\cdot 2^4=80$ colors. The $x-y$ is not actually necessary—if we use the three forms of boundaries, the rabbit must cross at least two of the types infinitely many times, and from that we can deduce its position. However, using $x+y$ and $x-y$ allows us to save a bit of time because we can just repeat the argument for $x$ and $y$. Aside from this optimization (which gets us down to $40$), I am pretty sure we can actually combine the $x$ and $y$ based colorings, and simply consider the direction the rabbit moved in before stepping onto a boundary to figure out if it is an $x$-boundary or a $y$-boundary, which gets us down to $20$.

Note that if we have two colorings, we can overlay them to make another coloring by bijecting a set of ordered pairs of colors to a set of colors. Thus, we can overlay as many colorings as we want. We'll prove that we can determine the invisible rabbit's original location in four overlaid colorings. For our first coloring, we color the squares with colors $\pmod 5$ such that each downward step decreases color by $1$, each upward steps increases color by $1$, each leftward step decreases color by $2$ and rightward step increases color by $2$. In this way, the hunter will be able to determine the rabbit's trojectory. Thus, if the hunter determine's the rabbit's current location, then the hunter can backtrack to determine the rabbit's initial location. In fact, if the hunter can determine one of these for any point in time: $x$, $y$, $x+y$, where the rabbit's position is $(x,y)$, then the hunter can backtrack to determine the rabbit's initial location. Our next three colorings will color based on $x$, $y$, and $x+y$ to help the hunter determine the location on the board. $~$ Consider the following color, only on the $x$-coordinate: color $x$ if and only if $|x|$ is a perfect square. Suppose the $x$-coordinate is unbounded. Assume that the $x$-coordinate can get infinitely big. Then, we may only consider moves in which the rabbit's $x$-coordinates reaches a new high. This makes the rabbit's moves become an infinite sequence going only rightward. This sequence hits a colored square at variable intervals. Let two consecutive interval lengths be $m$ and $n$. Because of the way we colored, after two uncolored intervals are traversed, the hunter knows the rabbit's $x$-coordinate. $~$ Similarly color the $y$-coordinate and $x+y$-coordinate. For $x+y$, consider both up and right as a move in the positive direction, and for down and left consider it to be in the negative direction. If $y$-coordinate is unbounded, the hunter can know the rabbit's $y$-coordinate. If either the $x$ or the $y$-coordinate is bounded, because of the unboundedness of the other variable, $x+y$ is unbounded, so the hunter can determine $x+y$ as well, thus determining $x$ and $y$, as desired.

Toss the configuration on the coordinate plane. We can think of the colors used by the hunter as a Cartesian product of some color sets. These color sets are defined as follows. Color $1$: $2x+y \pmod{5}$. Color $2$: if $|x|$ is a triangular number, then $1$, otherwise $0$. Color $3$: if $|y|$ is a triangular number, then $1$, otherwise $0$. Color $4$: if $|x-y|$ is a triangular number, then $1$, otherwise $0$. This will give a total of $5 \cdot 2^3 = 40 < 2021$ colors, so we just need to show that using these four color sets, the hunter can uniquely determine the rabbit's initial position at some point. First of all, we can determine the direction the rabbit has taken in a given move uniquely by Color $1$; each direction corresponds to a unique delta modulo $5$ that is one of $\{1, 2, 3, 4\}$. This finishes off the use of Color $1$. Next, we look at Colors $2$ to $4$. Note that eventually, the rabbit will hit two distinct lines in at least one of Color $2$ and Color $3$. If two distinct lines of both Colors $2$ and $3$ are hit, then we are done, since the hunter can use the delta in the coordinates to determine the current position and trace back to the initial position (via Color $1$). Otherwise, WLOG only Color $3$ has two distinct lines hit, and eventually Color $2$ is always $0$. Since no point can be revisited, the rabbit eventually hits two distinct diagonals in Color $4$. (Or else, the rabbit will be stuck inside a parallelogram-shaped region forever.) Thus, the hunter can determine $y$ using the horizontal lines and $x$ using the diagonals after that, and trace back to the initial position via Color $1$. Note: the hunter can determine if two distinct lines are hit in Colors $2$ through $4$ by checking when those colors become $1$ and if the difference between the $x$- and $y$-deltas is nonzero.

We claim there exists a winning strategy for the hunter. Let $f_k(n)$ denote the $2^k$ digit of $n$ in base $2$. Let $R_k(n)$ denote the remainder when $n$ is divided by $k$. Define the following sets: $X_i=\{(x, y): x\equiv i\pmod{3}\}$ for $i\in\{0, 1, 2\}$ $A_x=\{(0, y): y\in \mathbb{Z}\}$ $B_x=\{(2^k, y): k\in \mathbb{Z}_{\ge 0}\}$ $C_x=\{(-2^k, y): k\in \mathbb{Z}_{\ge 0}\}$ $D_x=\{(\pm 4^k, y): k\in \mathbb{Z}_{\ge 0}\}$ $E_x=\bigcup_{k=0}^\infty \{(\pm x, y): 2^{2k-1}< x<2^{2k+1}, y\equiv 0\pmod{2^{2k}}\}$ $F_x=\bigcup_{k=0}^\infty \{(\pm x, y): 2^{2k}\le x<2^{2k+2}, y\equiv 0\pmod{2^{2k+1}}\}$ $G_x=\bigcup_{k=0}^\infty \{(\pm (2^k+a), y): a<2^k, f_{R_k(y)}(a)=1\}$ Define analogous sets with $x$ and $y$ swapped. Claim: The hunter can always determine the direction the rabbit moved. Proof: The rabbit moving right is equivalent to it moving from $X_i$ to $X_{i+1}$. Similarly, the hunter can tell when it moves left, up, or down. Claim: If the rabbit goes to any cell in $A_x$, $B_x$, or $C_x$, then the hunter can determine its $x$ coordinate. Proof: The hunter always knows the rabbit's position relative to its starting position. So, it suffices to find the rabbit's $x$ coordinate at any time. If the rabbit goes through $A_0$, its $x$ coordinate is $0$ at that point. Assume it doesn't. WLOG assume it goes through $B_x$. Assume it goes through $2$ columns in $B_x$. Based on the distance between them, the hunter can determine the rabbit's $x$ coordinate. Assume it goes through only $1$ column in $B_x$. Assume that column is $x=2^k$. Based on $D_x$, we can determine whether $k$ is odd or even. Because the rabbit is trapped between two columns, it must go arbitrarily far in the $y$ direction. If $k$ is even, look at the distance between rows in $E_x$. If $k$ is odd, look at the distance between rows in $F_x$. This distance will be equal to $2^k$. From this, the hunter can determine $k$, which determines the rabbit's $x$ position. Claim: If the rabbit stays does not go to any cell in $A_x$, $B_x$, or $C_x$, then the hunter can determine its $x$ coordinate. Proof: WLOG assume the rabbit's $x$ coordinate stays strictly between $2^k$ and $2^{k+1}$. Based on gaps between rows in $E_x$ and $F_x$, the hunter can determine $k$. The rabbit moves arbitrarily far in the $y$ direction. Therefore, it must go to a cell in $A_y$, $B_y$, or $C_y$. Thus, the hunter can determine its $y$ coordinate. Let $2^k+x_0$ be the rabbit's starting $x$ coordinate. Assume the rabbit passes through cells $(2^k+x_0+a_0, y_0), (2^k+x_0+a_1, y_1),\dots, (2^k+x_0+a_{k-1}, y_{k-1})$ such that $y_i\equiv i\pmod{k}$. Then, the hunter knows the values of $f_{i}(x_0+a_i)$ for each $0\le i<k$ based on whether each cell is in $G_x$. This is enough information to determine $x_0$. Combining these claims, the hunter can always determine the rabbit's $x$ coordinate. Similarly, the hunter can always determine the rabbit's $y$ coordinate.

Gone hehe

I will use a number of colourings, first we colour each square with coordinate $x, y$ with $f(x, y)=2x-y\pmod 5$, thus going down reduces the residue by $1$, going up increases it by $1$, $1$, going left increases it by $2$ and going right decreases it by $2$, thus we know the direction the rabbit travels in each turn. For the next two colourings, they will be indetical but one will colour the coloumns identically and one will colour the rows identically. For this colourings I will explain it for the rows, choose a row colour all its square with $1$, than in the up direction colour one square with $0$ than 2 squares with 1, 1 square with 0 than 4 squares with 1 etc, in the downwards direction colour one square with 0 than 3 squares with 1 than one square with 0 than 5 squares with 1 etc, use the same colouring but different colours for the across. For the final 2 colouring we have them identical but different colours for up and across. I will show for across the colouring. Assign each row a prime, and choose a coloumn that will have value 0, now number each square in each adjacent coloumn to the left $1$, $2$, \dots, and to the right $-1$, $-2$, \dots, now for a row colour a square if its divisible by the prime assigned to the row otherwise leave it uncoloured. Thus from the first 2 colouring after the direction determining, if it moves an unbounded amount on its path in the up and accoss directions we can determine where it started. If its limited in one direction we can determine where it started as we can look at the difference in the number of squares between coloured squares and this set is finite and we know where they are so we know where it started. \end{proof}