5. A knight is placed in each cell of a chessboard. What is the minimum number of knights that can be removed from the board so that no knight remains that attacks exactly four other knights? (A knight attacks the squares that are one square away horizontally and two squares away vertically, or vice versa.)
Answer: 8 knights.

Solution. First, we will show that no fewer than 8 knights need to be removed. On the left diagram, all knights that attack exactly 4 squares of the board are marked (for convenience, they are highlighted in different colors). Let's call such knights bad. To stop a knight from attacking four others, one must remove either this knight or one of those it attacks. We will show that even to get rid of the bad black knights, one must free no fewer than 8 squares. On the middle diagram, the circles mark the squares under attack by the bad black knights. Three bad black knights in the upper left corner attack four squares marked with the symbol $\odot$. If only one of these squares is freed, one of the black knights will remain bad. Therefore, for this trio of knights, at least two squares need to be cleared. Thus, for all four trios, at least $4 \cdot 2 = 8$ squares need to be freed.

Now, let's provide an example showing that 8 knights are sufficient. On the right diagram, the knights that need to be removed from the board are marked.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-07.jpg?height=304&width=1272&top_left_y=434&top_left_x=450)