Can the position of a pawn on a chessboard be determined by the following equation, where $x, y$ represent the row and column numbers of the square:

$$
x^{2}+x y-2 y^{2}=13
$$
I. solution. Solve the equation for $x$:

$$
x=\frac{-y \pm \sqrt{9 y^{2}+52}}{2}
$$

$x, y$ are positive integers, so $9 y^{2}+52$ must also be an integer, and more specifically, it must be a perfect square:

$$
9 y^{2}+52=k^{2}, \quad k^{2}-9 y^{2}=(k+3 y)(k-3 y)=52
$$

It is sufficient to take $k$ as positive, so $k+3 y$ and consequently $k-3 y$ are positive integers. Therefore, 52 must be factored into two positive integers. Factoring 52 into prime factors gives $52=2^{2} \cdot 13$. One factor will be even, so the other must also be even because the sum and difference of two integers (i.e., $k$ and $3 y$) have the same parity. Thus, only the $26 \cdot 2$ factorization is relevant. From the equations

$$
k+3 y=26, \quad k-3 y=2
$$

we get $k=14, y=4$, and the positive value of $x$ is $x=5$. Therefore, the doll can only be at the intersection of the 5th row and the 4th column, and its position is uniquely determined.

Staub Klára (Budapest, Apáczai Csere J. Gyak. G. I. o. t.)

II. solution. The left side of the given equation can be factored into two terms:

$$
x^{2}-y^{2}+x y-y^{2}=(x+y)(x-y)+y(x-y)=(x+2 y)(x-y)
$$

The factors are integers, and the first one is positive, so the 13 on the right side must also be factored into positive integers. Since 13 is a prime number, $13=13 \cdot 1$. Now, from the equations

$$
x+2 y=13, \quad x-y=1
$$

we get $x=5, y=4$. There is no other solution because $x+2 y>x-y$.

Simon György (Budapest, József A. G. I. o. t.)

Remark. In neither solution did we use the size of the chessboard. Therefore, the equation determines the position of the doll on any $n \times n$ board, provided $n \geq 5$.

Székely Gábor (Budapest, Madách I. G. II. o. t.)