Consider the 81 points where the 9x9 lines dividing the chessboard intersect, and let's call them grid points. How many squares are there whose all four vertices are grid points? How many remain if the center of the board cannot be a vertex of a square?
I. The sides of a square that meets the condition can be parallel to the sides of the chessboard; if they are not, we can draw lines parallel to the sides of the board through its vertices, enclosing it in a square whose sides are parallel to the board's sides—let's call these base squares—such that the original vertices lie on the sides of the base square. We will count, on the one hand, the squares that can be inscribed in a single base square in the manner described, and on the other hand, the base squares that can be placed on the board.

![](https://cdn.mathpix.com/cropped/2024_05_02_61b775bd26cd05fb2474g-1.jpg?height=472&width=455&top_left_y=390&top_left_x=824)

In a base square, we can inscribe as many squares as there are internal grid points on one side of the square. To this, we also count the base square itself, so we get as many squares as the length of the side of the base square, with the unit being the side of one square on the chessboard (since a square can be drawn from each unit segment of the side, starting from its initial point, and the first one can be paired with the base square itself).

Now let's see where the lower left vertex of a base square with side length $c$ can be placed. This can clearly be on the left side of the board, or on any of the grid lines up to the $(8-c)$-th one from there. (If, for example, $c=6$, then the lower left vertex can be 0, 1, or 2 units away from the left edge of the board.) Similarly, the vertex in question can be up to $(8-c)$ units away from the bottom edge of the board. Thus, this vertex can be any of the grid points of an $(8-c)$-sided base square placed in the lower left corner of the board. This square has $(8-c+1)=9-c$ grid points on each side, so it contains $(9-c)^2$ grid points. A base square with side length $c$ can thus be placed on the chessboard in $(9-c)^2$ different ways such that its vertices are grid points. (For example, for $c=6$, $(9-6)^2=3^2=9$ ways.)

In each of these, $c$ squares can be inscribed (including the base square itself), so we get a total of $(9-c)^2$ squares that are inscribed in a square with side length $c$. Summing this for $c=1,2,\ldots,8$ gives us the number of squares we are looking for. This is

$$
8^2 + 7^2 \cdot 2 + 6^2 \cdot 3 + 5^2 \cdot 4 + 4^2 \cdot 5 + 3^2 \cdot 6 + 2^2 \cdot 7 + 8 = 540
$$

II. Now we will count the squares to be excluded, which have one vertex at the center $K$ of the board. Such a square is inscribed in a base square whose one side passes through $K-n$, and thus the side of the base square is at most 4 units, because the distance from $K$ to the edges of the board is 4 units. If we place one of the grid points on the boundary of a base square with side length $c$ at $K$ and inscribe the square whose one vertex is this point, we get an excluded square, and by placing different grid points of the base square at $K$, we get different squares. Thus, there are $4c$ excluded squares inscribed in a base square with side length $c$, because there are $4c$ grid points on the boundary of the base square. Summing this for $c=1,2,3,4$ gives the number of excluded squares

$$
4 \cdot (1+2+3+4) = 40
$$

and thus there are 500 squares whose vertices are grid points of the chessboard different from the center $K$.

Aranka Siket (Makó, József A. g. II. o. t.)