Problem 1675

1. Initial Assumptions and Setup:
- Let $n$ be an odd positive integer.
- Consider an infinite square grid where each cell is filled with one of the numbers $1, 2,$ or $3$.
- The grid must satisfy the following conditions:
1. Any two cells sharing a common side do not have the same number.
2. For any $1 \times 3$ or $3 \times 1$ subgrid, the numbers filled do not contain $1, 2, 3$ in that order, regardless of the reading direction.
3. The sum of numbers in any $n \times n$ subgrid is the same.

2. Analyzing the Grid:
- We start by analyzing the placement of the number $2$ since it is the most restricted by condition 2.
- If there are no $2$'s, the grid must be a checkerboard pattern of $1$ and $3$. This can be easily eliminated by condition 3.
- If the $2$'s form a checkerboard pattern, this can also be eliminated using condition 3.

3. Lemma:
- Lemma: The 4 adjacent cells (buddies) of any cell labeled $2$ have equal labels.
- Proof:
- Label the cell with $2$ as $(0,0)$ in the coordinate plane.
- By condition 2, the two numbers to the left and right of a $2$ must be equal.
- Assume the 4 buddies are not all equal. We must have a configuration like:
$$\begin{array}{ccc} & 3 & \\ 1 & 2 & 1 \\ & 3 & \end{array}$$
- This configuration leads to a periodic pattern with period 2 along the rows and columns, implying $2$'s appear in a checkerboard fashion, which contradicts our earlier elimination. Hence, the 4 buddies of any $2$ must be equal. $\blacksquare$

4. Claim:
- Claim: Let $X = (0,0)$ be a $2$. WLOG, the 4 buddies around it are $3$. Then for any point $(a,b)$:
$$(a,b) = \begin{cases} 2 \text{ or } 3 & \text{if } a+b \text{ is odd} \\ 2 \text{ or } 1 & \text{if } a+b \text{ is even}. \end{cases}$$
- Proof:
- Consider the quadrant $a, b > 0$. We use induction on $a+b$.
- Base case $a+b = 1$ is clear.
- Suppose $a+b$ is odd. We know $(a-1,b)$ is $2$ or $1$ by induction.
- If $(a-1,b) = 2$, then $(a-2,b) = 3$, so $(a,b) = 3$.
- If $(a-1,b) = 1$, then $(a,b)$ is $2$ or $3$.
- This proves the claim for $a+b$ odd. The case for $a+b$ even is similar. $\blacksquare$

5. Coloring the Board:
- Color the board black and white like a checkerboard, with $X$ at a white square.
- The claim tells us that all points an odd taxicab distance from $X$ are $2$ or $3$, and all points an even taxicab distance from $X$ are $2$ or $1$.
- Hence, all black squares are $2$ or $3$, and all white squares are $2$ or $1$.

6. Using Condition 3:
- Define the diversity of a box $\text{div}[(a,b),(c,d)]$ as the number of $2$'s on black squares minus the number of $2$'s on white squares.
- Define $\sum[(a,b),(c,d)]$ as the sum of the numbers in the box.
- Since $[(1,0),(n+1,n)]$ has one more black square than $[(0,0),(n,n)]$, we have:
$$0 = \sum [(1,0),(n+1,n)] - \sum [(0,0),(n,n)] = 2 + \text{div}[(1,0),(n+1,n)] - \text{div}[(0,0),(n,n)].$$
- Removing the center $[(1,0),(n,n)]$ block, this is equivalent to:
$$\text{div}[(n,0),(n+1,n)] = \text{div}[(0,0),(1,n)] + 2.$$
- Let $(a_1, \ldots, a_n, a_{n+1}) = (x_{0,0}, \ldots, x_{0,n}, x_{0,n+1})$ and $(b_1, \ldots, b_n, b_{n+1}) = (x_{n,0}, \ldots, x_{n,n})$, where:
$$x_{ij} = \begin{cases} 1 & \text{if } (i,j) \text{ is 2 and black} \\ -1 & \text{if } (i,j) \text{ is 2 and white} \\ 0 & \text{if } (i,j) \text{ is not 2}. \end{cases}$$
- By the definition of diversity, we have:
$$\text{div}[(n,0),(n+1,n)] = b_1 + \cdots + b_n,$$
$$\text{div}[(0,0),(1,n)] = a_1 + \cdots + a_n.$$
- Therefore:
$$b_1 + \cdots + b_n = 2 + a_1 + \cdots + a_n.$$
- Shifting the strips up by 1, we get:
$$b_2 + \cdots + b_{n+1} = -2 + a_2 + \cdots + a_{n+1}.$$
- Subtracting these equations, we get:
$$b_{n+1} - b_1 = a_{n+1} - a_1 - 4.$$
- Since $a_i, b_i \in \{-1, 0, 1\}$ for all $i$, this implies:
$$b_{n+1} = -1, b_1 = 1, a_{n+1} = 1, a_1 = -1.$$
- Therefore, all of $(0,0), (0,n+1), (n+1,0), (n+1,n+1)$ are $2$. By similar logic, $(1,0), (1,n+1), (n+2,0), (n+2,n+1)$ are all $2$, leading to a contradiction.

$\blacksquare$