In how many ways can 32 knights be placed on an $8 \times 8$ chessboard so that no two attack each other?
Solution. Let's examine the order of the fields such that by moving through them, we step on each field exactly once, in knight's moves, and return to the starting field.

| 15 | 34 | 39 | 4 | 17 | 32 | 49 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 40 | 3 | 16 | 33 | 38 | 5 | 18 | 31 |
| 35 | 14 | 37 | 56 | 53 | 50 | 7 | 48 |
| 2 | 41 | 54 | 51 | 60 | 57 | 30 | 19 |
| 13 | 36 | 59 | 62 | 55 | 52 | 47 | 8 |
| 42 | 1 | 24 | 11 | 58 | 61 | 20 | 29 |
| 25 | 12 | 63 | 44 | 27 | 22 | 9 | 46 |
| 0 | 43 | 26 | 23 | 10 | 45 | 28 | 21 |

If we place knights on the chessboard such that they do not attack each other, then no two consecutive numbered fields can have a piece, so the difference between the numbers of any two knights must be at least 2. However, we must place a knight on every other field, otherwise, we cannot place 32 pieces. Therefore, we either place a knight on every even-numbered field or on every odd-numbered field: either on the light or the dark squares, ensuring that they do not attack each other.

Thus, there are two ways to place 32 knights on the chessboard such that no two attack each other.