The 9 squares of a $3 \times 3$ board, as shown in the figure below, must be painted in such a way that in each row, each column, and each of its two diagonals, there are no squares of the same color. What is the minimum number of colors needed for this painting?

![](https://cdn.mathpix.com/cropped/2024_05_01_ffb48c2069f209c1311cg-10.jpg?height=343&width=347&top_left_y=611&top_left_x=736)

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Solution

Let's denote the colors of the main diagonal by the numbers 1, 2, and 3, as shown in the first figure.

| 1 |  | $X$ |
| :--- | :--- | :--- |
|  | 2 |  |
| $Y$ |  | 3 |

Now consider the colors of the two corner squares of the secondary diagonal, denoted by $X$ and $Y$ in the first figure. Since $X$ is in a row that contains the color 1, a diagonal with the color 2, and a column with the color 3, the color $X$ is not in the set $\{1,2,3\}$. By the same argument, the color $Y$ is also not in this set. Moreover, since $X$ and $Y$ are on the same diagonal, they must represent different colors. Therefore, we need at least $3+2=5$ colors. To ensure that 5 is the minimum number, it suffices to exhibit a configuration with this number of colors. This is done in the next figure, where each number represents a color.

| 1 | 3 | 5 |
| :--- | :--- | :--- |
| 3 | 2 | 1 |
| 4 | 1 | 3 |