Solution. When counting the different cases, we do not take the coloring of the chessboard into account. The corner of a $5 \times 5$ square should fall into the center of one of the corner squares of the chessboard, and the sides of the square should be parallel to the sides of the chessboard. This gives one possibility for placing the square, because no matter which corner square we start from, the placements can be transformed into each other by reflection, for example $((a)$ figure $)$.

Since the side of a 5-unit square covers the centers of 6 small squares, the next placement is also possible. The corner of the square placed as described above can be moved to the center of the small square adjacent to the corner square, either to the right or diagonally downward. This gives us 2 more possibilities ((b) and (c) figures).

Finally, let's examine the possibility where the sides of the $5 \times 5$ square are not parallel to the sides of the chessboard. In this case, each corner of the $5 \times 5$ square will fall into the center of the fourth square in one of the outermost columns or rows, following the same direction. To prove this, draw lines through the corners of the shape parallel to the sides of the chessboard. The resulting four right-angled triangles are clearly congruent, with hypotenuses of 5 units. The lengths of their legs are integers, so by the Pythagorean theorem, they can only be 3 and 4 units long. This gives one more solution ((d) figure). Therefore, the total number of solutions is 4.

![](https://cdn.mathpix.com/cropped/2024_05_02_23d737db1c5f89179061g-1.jpg?height=282&width=287&top_left_y=688&top_left_x=434)

$(a)$

![](https://cdn.mathpix.com/cropped/2024_05_02_23d737db1c5f89179061g-1.jpg?height=262&width=282&top_left_y=693&top_left_x=756)

$(b)$

![](https://cdn.mathpix.com/cropped/2024_05_02_23d737db1c5f89179061g-1.jpg?height=277&width=268&top_left_y=691&top_left_x=1080)

$(c)$

![](https://cdn.mathpix.com/cropped/2024_05_02_23d737db1c5f89179061g-1.jpg?height=260&width=282&top_left_y=694&top_left_x=1385)

$(d)$
In how many ways can a square with a side length of 5 units be placed on an $8 \times 8$ chessboard so that each of its vertices falls on the center of a square? (Solutions that can be transformed into each other by reflection or rotation are not considered different.)





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