## Solution

Two of the seven fields within the figure on the left in Figure 1 can be covered simultaneously with an L-tetromino. These fields therefore have different colors, in particular $n \geq 7$. Assume that $n=7$ is possible and color the fields in the area as in Figure 1. The two fields to the right and left below the area must now have the colors 1 and 3, because otherwise two fields of the same color could be covered with an L-tetromino. The color of the two other fields on the right in Figure 1 is determined in the same way. Now there is no color left for the field with the marking, contradiction. An example for $n=8$ can be found in Figure 2 .
![](https://cdn.mathpix.com/cropped/2024_04_28_5496cd1b274f6d47390cg-03.jpg?height=564&width=282&top_left_y=237&top_left_x=570)

Figure 1: $n \geq 8$ applies.

![](https://cdn.mathpix.com/cropped/2024_04_28_5496cd1b274f6d47390cg-03.jpg?height=494&width=505&top_left_y=273&top_left_x=1095)

Figure 2: $n=8$ is sufficient.
3. the plane is divided into unit squares. Each square is to be colored with one of $n$ colors, so that the following applies: If four squares can be covered with an L-tetromino, then these squares have four different colors (the L-tetromino may be rotated and mirrored). Determine the smallest value of $n$ for which this is possible.