In how many ways can two different squares be chosen from an $8 \times 8$ chessboard such that the midpoint of the segment connecting their centers also coincides with the center of a square?
Place the chessboard in a coordinate system as shown in the figure. Let the origin be the center of the bottom-left square, and choose the side length of a chessboard square as the unit on the axes.

Let $P_{1}(a ; b), P_{2}(c ; d)$ be the centers of two arbitrarily chosen squares. It is known that the coordinates of the midpoint of the segment $P_{1} P_{2}$ are: $F\left(\frac{a+c}{2} ; \frac{b+d}{2}\right) \cdot F$ will be the center of a square if $\frac{a+c}{2}$ and $\frac{b+d}{2}$ are integers, i.e., if $a$ and $c$, as well as $b$ and $d$, have the same parity.

The coordinates of the centers can be divided into 4 groups based on parity:

| even; | even |
| :--- | :--- |
| even; | odd |
| odd; | even |
| odd; | odd. |

For the coordinates of the midpoint to be integers, we can only pair a point with another point from the same group.

Each group contains 16 centers (there are 64 centers in total), from which

$$
\binom{16}{2}=\frac{16 \cdot 15}{2}=120
$$

pairs can be formed. Therefore, from the 4 groups, we can select a total of $4 \cdot 120=480$ pairs of points such that the midpoint of the segment connecting them also falls on the center of a square.

![](https://cdn.mathpix.com/cropped/2024_05_02_aa4e7009215311399502g-1.jpg?height=507&width=507&top_left_y=1031&top_left_x=804)