How many knights can we place on an $8 \times 8$ chessboard so that no two attack each other? What about on a $5 \times 5$ chessboard?
On a chessboard, a knight moves from a white square to a black one, and from a black square to a white one. This means that no two knights placed on squares of the same color can attack each other. On an $8 \times 8$ chessboard, there are 32 black and 32 white squares, so 32 knights can be placed in the desired manner, for example, by placing them on the white squares. On a $5 \times 5$ chessboard, there are 12 squares of one color and 13 of the other, so at least 13 knights can be placed on this board.

![](https://cdn.mathpix.com/cropped/2024_05_02_95e87ca9b55358201fe5g-1.jpg?height=390&width=223&top_left_y=331&top_left_x=886)

| 1 | 20 | 9 | 14 | 3 |
| ---: | ---: | ---: | ---: | ---: |
| 10 | 15 | 2 | 19 | 24 |
| 21 | 8 | 25 | 4 | 13 |
| 16 | 11 | 6 | 23 | 18 |
| 7 | 22 | 17 | 12 | 5 |

Can more knights be placed on the boards under the given conditions than in the examples above?

Consider a $2 \times 4$ section of the board. The 8 squares in this section can be numbered with 4 numbers such that knights on squares with the same number attack each other. One such numbering is shown in Figure 1. Thus, more than 4 knights cannot be placed on such a section, as otherwise, there would be knights on squares with the same number, and thus attacking each other. Since an $8 \times 8$ chessboard can be constructed from 8 such $2 \times 4$ sections, more than $8 \times 4 = 32$ knights cannot be on the board. Therefore, our result is the best possible.

| 1 | 54 | 19 | 38 | 3 | 56 | 21 | 40 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 18 | 37 | 2 | 55 | 20 | 39 | 4 | 57 |
| 53 | 12 | 63 | 30 | 47 | 14 | 41 | 22 |
| 36 | 17 | 46 | 13 | 64 | 31 | 58 | 5 |
| 11 | 52 | 29 | 62 | 15 | 48 | 23 | 42 |
| 28 | 35 | 16 | 45 | 32 | 61 | 6 | 59 |
| 51 | 10 | 33 | 26 | 49 | 8 | 43 | 24 |
| 34 | 27 | 50 | 9 | 44 | 25 | 60 | 7 |

Figure 3

A $5 \times 5$ board cannot be assembled from $2 \times 4$ rectangles, so a different method is needed here. In Figure 2, we numbered the squares of the $5 \times 5$ board from 1 to 25. It is easy to verify that if a knight starts from the 1st square and always moves to the next numbered square, it will visit all squares in 25 moves. Thus, if more than 13 knights are placed on the $5 \times 5$ board, there will be at least two knights on squares with consecutive numbers, and thus, according to the property of the numbering, they will attack each other. Therefore, 13 knights can be placed on the $5 \times 5$ board in the desired manner, but more than that is not possible.

Note. The latter method also provides the correct upper bound for the $8 \times 8$ board if we use the known fact that an $8 \times 8$ chessboard can also be traversed by knight moves. We provide such a traversal in Figure 3.