On an infinite chessboard, how many squares are there from which a knight can reach a designated square in exactly $n$ moves - but not in fewer?
In his article "A special geometry" (KML IX. vol. 3-4, p. 71), Kárteszi Ferenc proves that on an infinite chessboard, starting from a square, the squares reachable with $n (n \geq 5)$ knight moves are regularly arranged, specifically, they are located precisely in the $(n-1)$-th and $n$-th rings. Using this theorem, we solve the problem by counting the squares in the combined $(n-1)$-th and $n$-th rings that are labeled $n$, i.e., the squares reachable with $n$ knight moves.

Since the configuration has two perpendicular axes of symmetry, it is sufficient to examine one quarter of it. We can choose the thick, arrow-marked step line as one boundary of the quarter to be examined; the other boundary of the quarter is obtained by rotating the adopted boundary line $90^{\circ}$ around the center of the square labeled 0. This results in the doubly arrowed step line.

![](https://cdn.mathpix.com/cropped/2024_05_02_642c09ac902ed42005eeg-1.jpg?height=753&width=921&top_left_y=480&top_left_x=602)

Divide the quarter of the configuration to be examined into the two parts shown in the figure. First, examine the part labeled $A$. The outer row of the $n$-th ring consists of $2n+1$ squares, and the two outermost squares of this row are always labeled $n$. This implies that in the outer and inner rows of the $n$-th ring, and in the outer and inner rows of the $(n-1)$-th ring, there are $(n+1), n, (n-1), (n-2)$ squares labeled $n$, respectively. Therefore, the part labeled $A$ of the combined $(n-1)$-th and $n$-th rings of the quarter configuration contains

$$
(n+1) + n + (n-1) + (n-2) = 4n - 2
$$

squares labeled $n$.

In the part labeled $B$, the squares labeled $n$ are arranged in 3 equal-length, parallel diagonals in the $n$-th and $(n-1)$-th rings, specifically, in the $n$-th ring, the two outermost and in the $(n-1)$-th ring, the middle diagonal. Since the middle diagonal in the $n$-th ring has $n$ squares, and the two outer diagonals have $(n-1)$ squares, the number of squares labeled $n$ in the part labeled $B$ is $3(n-1)$.

The total number of squares labeled $n$ in parts $A$ and $B$ is therefore

$$
4n - 2 + 3(n-1) = 7n - 5
$$

The total number of squares labeled $n$, i.e., exactly $n$ knight moves away, in the entire configuration (for $n \geq 5$) is

$$
S_{n} = 4(7n - 5) = 28n - 20
$$

For $n < 5$, direct counting yields

$$
S_{1} = 8, S_{2} = 32, S_{3} = 68, S_{4} = 96
$$

Imre Csiszár (Bp. I., Petőfi g. III. o. t.)