Solve the following equation in the set of natural numbers:

$$
[a, b]+(a, b)+a+b=a \cdot b
$$

(Here $[a, b]$ denotes the least common multiple of $a$ and $b$, while $(a, b)$ denotes their greatest common divisor.)
Let the greatest common divisor of $a$ and $b$ be denoted by $A$, and let $a = a_1 \cdot A$, $b = b_1 \cdot A$. Then the least common multiple is $A \cdot a_1 \cdot b_1$, which can be verified by examining the prime factorization of the numbers: if $a = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots \cdots p_r^{\alpha_r}$, $b = p_1^{\beta_1} \cdot p_2^{\beta_2} \cdots \cdots p_r^{\beta_r} (\alpha_i \geq 0, \beta_i \geq 0)$, then

$$
\begin{array}{cc}
A = (a, b) = p_1^{\min(\alpha_1, \beta_1)} \cdots \cdots p_r^{\min(\alpha_r, \beta_r)}, & [a, b] = p_1^{\max(\alpha_1, \beta_1)} \cdots \cdots p_r^{\max(\alpha_r, \beta_r)} \\
a_1 = p_1^{\alpha_1 - \min(\alpha_1, \beta_1)} \cdots \cdots p_r^{\alpha_r - \min(\alpha_r, \beta_r)}, & b_1 = p_1^{\beta_1 - \min(\alpha_1, \beta_1)} \cdots \cdots p_r^{\beta_r - \min(\alpha_r, \beta_r)}
\end{array}
$$

that is,

$$
A \cdot a_1 \cdot b_1 = p_1^{\alpha_1 + \beta_1 - \min(\alpha_1, \beta_1)} \cdots p_r^{\alpha_r + \beta_r - \min(\alpha_r, \beta_r)} = p_1^{\max(\alpha_1, \beta_1)} \cdots p_r^{\max(\alpha_r, \beta_r)}
$$

since for any $\alpha, \beta$, $\alpha + \beta - \min(\alpha, \beta) = \max(\alpha, \beta)$.

Substituting the above into the equation and simplifying:

$$
A \cdot a_1 \cdot b_1 + A + A \cdot a_1 + A \cdot b_1 = A \cdot a_1 \cdot A \cdot b_1, \quad A \cdot (a_1 b_1 + a_1 + b_1 + 1) = A \cdot (A \cdot a_1 \cdot b_1), \quad A \cdot (a_1 + 1) \cdot (b_1 + 1) = A \cdot (A \cdot a_1 \cdot b_1)
$$

First, let's consider the case $A = 0$. If $a \neq 0$ and $b \neq 0$, then $A \neq 0$, so at least one of $a$ and $b$ must be zero. By definition, $(a, 0) = a$ if $a \neq 0$, so the only remaining case is $a = b = 0$. The number $(0, 0)$ is not usually defined; if we do define it, then $(0, 0) = 0$ in which case $a = b = 0$ is a solution, while $(0, 0) \neq 0$ in which case it is not. Whether or not $0$ is considered a natural number does not affect the essential part (and the grading) of the solution, regardless of whether the solution $a = b = 0$ is missing or present.

Thus, we can assume $A \neq 0$, and we can simplify by dividing by $A$:

$$
(a_1 + 1) \cdot (b_1 + 1) = A \cdot a_1 \cdot b_1
$$

It is clear that the cases $a_1 = 0$ and $b_1 = 0$ do not provide solutions, so the following rearrangement is permissible:

$$
A = \frac{a_1 + 1}{a_1} \cdot \frac{b_1 + 1}{b_1}
$$

Since $1 \leq a_1$ and $1 \leq b_1$, we have

$$
1 < \frac{a_1 + 1}{a_1} \leq 2, \quad 1 < \frac{b_1 + 1}{b_1} \leq 2
$$

Thus, only $A = 2, 3, 4$ are possible.

1. Case: $A = 2$. From (1), we get

$$
a_1 b_1 + a_1 + b_1 + 1 = 2 a_1 b_1
$$

Adding 2 to both sides,

$$
2 = a_1 b_1 - a_1 - b_1 + 1 = (a_1 - 1)(b_1 - 1)
$$

This is only possible if $a_1 - 1 = 1$ and $b_1 - 1 = 2$, or vice versa. In the former case, $a_1 = 2$, $b_1 = 3$, so $a = 4$, $b = 6$; in the latter case, $a = 6$, $b = 4$.

2. Case: $A = 3$. From (1), we get

$$
\begin{gathered}
3 a_1 b_1 = a_1 b_1 + a_1 + b_1 + 1 \\
2 a_1 b_1 - a_1 - b_1 - 1 = 0 \\
4 a_1 b_1 - 2 a_1 - 2 b_1 - 2 = 0 \\
(2 a_1 - 1)(2 b_1 - 1) = 3
\end{gathered}
$$

Then necessarily $2 a_1 - 1 = 3$ and $2 b_1 - 1 = 1$, or vice versa, from which $a = 6$, $b = 3$, or $a = 3$, $b = 6$.

3. Case: $A = 4$. Then

$$
\begin{gathered}
4 a_1 b_1 = a_1 b_1 + a_1 + b_1 + 1 \\
3 a_1 b_1 - a_1 - b_1 - 1 = 0 \\
9 a_1 b_1 - 3 a_1 - 3 b_1 - 3 = 0 \\
(3 a_1 - 1)(3 b_1 - 1) = 4
\end{gathered}
$$

Since $3 a_1 - 1 = 4$ and $3 b_1 - 1 = 1$ are not possible, we must have $3 a_1 - 1 = 3 b_1 - 1 = 2$, from which $a = b = 4$.

We have found five pairs of numbers, and it is easy to verify that each of them is indeed a solution; the solutions are:

$$
(3, 6), \quad (4, 6), \quad (4, 4), \quad (6, 4), \quad (6, 3)
$$

Reviczky Ágnes (München, Gymnasium Der Armen Schulschwestern, II. o.t.)