, 2022

First, note that if there are two coloring methods, where the first method colors point $(x,y)$ with color $c_1(x,y)$ and the second method colors point $(x,y)$ with color $c_2(x,y)$, then we can let $c(x,y) = 2^{c_1(x,y)}3^{c_2(x,y)}$, and color $(x,y)$ with color $c(x,y)$, thereby obtaining a coloring method that simultaneously achieves the functions of both coloring methods. Therefore, we will give five coloring methods below, which together allow us to determine the rabbit's position, thus achieving what the problem requires.

1. First let $c_1: \mathbb{Z}^2 \rightarrow \{1,2,3\}$ be $c_1(x,y) \equiv x(\text{mod } 3)$. This coloring tells us whether the rabbit moved left, right, or vertically.

2. Next let $c_2: \mathbb{Z}^2 \rightarrow \{1,2,3\}$ be $c_1(x,y) \equiv y(\text{mod } 3)$. This coloring tells us whether the rabbit moved up, down, or horizontally.

3. Define
$$K := \{0,2,2+2^2,2+2^2+2^3,\dots\} \cup \{0,-3,-3-3^2,-3-3^2-3^3,\dots\},$$
and let $c_3: \mathbb{Z}^2 \rightarrow \{1,2\}$ satisfy
$$c_3(x,y) = \begin{cases} 1 & x \in K, \\ 2 & x \notin K. \end{cases}$$
Note that the gaps between adjacent numbers in $K$ are all different, so combined with $c_1$, when $c_3$ equals 1 for the second time, we can determine with certainty the $x$-coordinate of the rabbit's location.

4. Similarly, let $c_4: \mathbb{Z}^2 \rightarrow \{1,2\}$ satisfy
$$c_4(x,y) = \begin{cases} 1 & y \in K, \\ 2 & y \notin K. \end{cases}$$
Then when $c_4$ equals 1 for the second time, we can determine with certainty the $y$-coordinate of the rabbit's location.

$$c_5(x, y) = \begin{cases} 1 & x + y \in K, \\ 2 & x + y \notin K. \end{cases}$$
Then when $c_5$ equals 1 for the second time, we can determine with certainty $x+y$ of the rabbit's location.

Now, suppose the rabbit is able to move every round; then among the three quantities $x,y,x+y$, at least two must be unbounded, which means that among $c_3, c_4, c_5$, at least two of them will equal 1 for the second time. In other words, we must be able to determine at least two of $x,y,x+y$, and thereby determine $(x,y)$. Combining all of the above, the hunter necessarily wins.