At most how many rooks can be placed on an $8 \times 8$ chessboard so that each one is attacked by at most one other?
The rows and columns of a chessboard, referred to collectively as lines, allow us to establish the following:

1. At most two rooks can be in one line;
2. two mutually attacking rooks occupy 3 lines of the board in the sense that no more rooks can be placed on these three lines according to the condition;
3. if a rook is not attacked by any other, it occupies two lines.

Thus, if we denote the number of rook pairs by \( P \) and the number of "lonely" rooks by \( M \), then \( 3P + 2M \leq 16 \), since the chessboard has a total of 16 lines. The number of rooks that can be placed on the board is therefore

\[
2P + M \leq 2P + \frac{4}{3}M = \frac{2}{3}(3P + 2M) \leq \frac{2}{3} \cdot 16 < 11
\]

that is, at most ten, and ten can indeed be placed. One possible arrangement is shown in the diagram.

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Gábor Megyesi (Szeged, Juhász Gy. Teacher Training College 1st Year Practice Elementary School, 8th grade)

Remarks. 1. Our line of thought can be generalized to an \( n \times n \) chessboard. On its \( 2n \) lines, we can place at most \(\left\lfloor \frac{4n}{3} \right\rfloor\) rooks in the desired manner. This many can indeed be placed, for example, by the following method:

For \( n = 1 \), we can place a single rook; for \( n = 2 \), we can place two rooks in any two squares.

For \( n \geq 3 \), place one rook pair in the first column and the third row. The 6 lines occupied by these reduce the board to an \((n-3) \times (n-3)\) board. Continue the process on this reduced board in the same way.

2. Several people referred to the 1973-74 Mathematical Competition Problems, where a solution to the problem can be found on page 57.