For which natural numbers does the equation

$$
x^{3}-y^{3}=x y+61
$$

hold?
The right side of the equation is positive, so $x>y>0$. On the left side,

$$
x^{3}-y^{3}=(x-y)^{3}+3xy(x-y)>(x-y)^{3}+xy
$$

thus $(x-y)^{3}<61$.

This means that $0<x-y<4$.

According to the equation and identity (2), from here,

$$
xy=\frac{61-(x-y)^{3}}{3(x-y)-1}
$$

If $x-y=2$ or $x-y=3$, then the right side of (3) is not an integer $\left(\frac{53}{5}\right.$, or $\left.\frac{34}{8}\right)$. If $x-y=1$, then from (3) we get $xy=30$.

Thus, $x$ and $-y$ are the roots of the equation

$$
t^{2}-t-30=0
$$

Factoring this:

$$
t^{2}-t-30=(t+5)(t-6)
$$

so the roots are -5 and 6. Since $x$ and $y$ are positive numbers, the only solution is $x=6$ and $y=5$.

Marton Tamás (Paks, Paks Nuclear Technical Secondary School, 1st year)