Solve the equation $[x] \cdot [y] = x + y$ on the set of positive real number pairs. (Here $[x]$ denotes the integer part of the number $x$, that is, the largest integer not greater than $x$.)
We denote the fractional part of $x$ and $y$ by $\{x\}$ and $\{y\}$, respectively. Then our equation is:

$$
[x] \cdot [y] = [x] + [y] + \{x\} + \{y\}
$$

The left side is an integer, so the right side must also be an integer. By definition, $0 \leq \{x\}, \{y\} \leq 1$, so $\{x\} + \{y\} = 0$ or $\{x\} + \{y\} = 1$. In the case $\{x\} + \{y\} = 0$, we have $[x] \cdot [y] = [x] + [y]$, from which

$$
([x] - 1)([y] - 1) = 1
$$

Thus, $[x] - 1 = 1$ and $[y] - 1 = 1$, or $[x] - 1 = -1$ and $[y] - 1 = -1$. According to our condition $\{x\} + \{y\} = 0$, $x$ and $y$ are integers, so $x = 2$ and $y = 2$ or $x = 0$ and $y = 0$. According to the problem's conditions, $x$ and $y$ are positive, so only the solution $x = y = 2$ is valid.

In the case $\{x\} + \{y\} = 1$, we have $[x] \cdot [y] = [x] + [y] + 1$, from which

$$
([x] - 1)([y] - 1) = 2
$$

Thus, $[x] - 1 = 1$ and $[y] - 1 = 2$, or $[x] - 1 = 2$ and $[y] - 1 = 1$, or $[x] - 1 = -1$ and $[y] - 1 = -2$, or $[x] - 1 = -2$ and $[y] - 1 = -1$. From the third and fourth cases, $[y] = -1$ or $[x] = -1$ would follow, so $x$ or $y$ would not be positive.

In the first two cases, $[x] = 2$ and $[y] = 3$, or $[x] = 3$ and $[y] = 2$, so in both cases $[x] + [y] = 5$. We also know that $\{x\} + \{y\} = 1$, so $x + y = 6$.

It is easy to see that if $2 \leq x < 4$, $x \neq 3$ and $y = 6 - x$, then $(x, y)$ is a solution, and there are no other solutions.

Thus, $x = y = 2$ and $2 \leq x < 4$, $x \neq 3$, $y = 6 - x$ give all positive solutions to the equation.

Based on the work of Péter Stoll (Miskolc, Földes F. Gymnasium, 2nd grade)

Note. The integer solutions of equations of the form $x y = a x + b y + c$ (where $a, b, c$ are integers) can similarly be obtained from the equation $(x - b)(y - a) = a b + c$ by factorizing $a b + c$.