12.44 Find the natural number solutions \(x, y\) for the equation \(x^{3}-y^{3}=x y+61\).
(15th All-Soviet Union Mathematical Olympiad, 1981)
【Solution】From the given equation, if $x, y$ are natural number solutions of the equation, then it must be that $x > y$.
Let $x = y + d, \quad d \geq 1, d \in \mathbb{N}$.
Then the given equation becomes
$$
\begin{array}{c}
3 d y^{2} + 3 d^{2} y + d^{3} = y^{2} + d y + 61, \\
(3 d - 1) y^{2} + d(3 d - 1) y + d^{3} = 61 .
\end{array}
$$

Thus, $d^{3} < 61$, i.e., $d = 1, 2, 3$.
(1) If $d = 1$, then the equation becomes
$$
\begin{array}{c}
2 y^{2} + 2 y - 60 = 0 . \\
y_{1} = 5, \quad y_{2} = -6 \text{ (discard). }
\end{array}
$$

Therefore, the equation has a natural number solution $x = 6, y = 5$.
(2) If $d = 2$, then the equation becomes
$$
5 y^{2} + 10 y - 53 = 0 .
$$

This equation has no integer solution $y$, so the original equation has no natural number solution.
(3) If $d = 3$, then the equation becomes
$$
8 y^{2} + 24 y - 34 = 0 .
$$

This equation also has no integer solution $y$, so the original equation has no natural number solution.
Therefore, the given equation has only one unique set of natural number solutions:
$$
\left\{\begin{array}{l}
x = 6, \\
y = 5 .
\end{array}\right.
$$