Solve the following equation in natural numbers:

$$
\frac{x-1}{1+(x-1) y}+\frac{y-1}{2 y-1}=\frac{x}{x+1} \text {. }
$$
Since $x \geq 1$ and $y \geq 1$, all three denominators are positive. Multiplying (1) by their product, we obtain an equivalent equation. By multiplying and rearranging,

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

This equation holds if and only if

$$
\begin{aligned}
& x y-x-y=0, \quad \text { or equivalently } \\
& (x-1)(y-1)=1
\end{aligned}
$$

The factors on the left side are non-negative integers, so the equation holds only if both factors are equal to 1, which means the only solution is $x=y=2$.

Sparing László (Szombathely, Tolbuhin úti Ált. Isk., 8. o. t.)

Remark. If we were to look for solutions of (1) among the integers, the factors on the left side of (2) could also be equal to $-1$, from which we would obtain the additional root $x=y=0$.