3. Determine all pairs of natural numbers \(a\) and \(b\) for which

\[
2[a, b] + 3(a, b) = a b
\]

where \([a, b]\) denotes the least common multiple and \((a, b)\) denotes the greatest common divisor of the natural numbers \(a\) and \(b\).

(Jaroslav Švrček)
SOLUTION. Using the known equality $m \cdot n=[m, n] \cdot(m, n)$, which holds for any natural numbers $m$ and $n$, we can transform the given equation into the form

$$
2[a, b]+3(a, b)=[a, b] \cdot(a, b)
$$

For simplicity, let $u=[a, b]$ and $v=(a, b)$, where $u$, $v$ are natural numbers, for which it is clear that $u \geq v$, since the least common multiple of numbers $a$ and $b$ is certainly not less than any of their common divisors. We thus need to solve the equation $2 u+3 v=u v$, which we rearrange to the form

$$
u v-2 u-3 v+6=6 \text { or }(u-3)(v-2)=6 \text {, }
$$

where $u, v$ are natural numbers satisfying the condition $u \geq v$ and also the condition $v \mid u$ (a common divisor divides the common multiple).

For $v=1$, $u$ does not yield a natural number, so it must be that $v-2 \geq 0$, and thus also $u-3 \geq 0$.

The number 6 can be written as the product of two non-negative numbers in four ways:

$\triangleright 6=6 \cdot 1$, so $u=9, v=3$. Since $(a, b)=3$, we can write $a=3 \alpha, b=3 \beta$, where $(\alpha, \beta)=1$, thus $9=[a, b]=3[\alpha, \beta]$, which gives $\{\alpha, \beta\}=\{1,3\}$ or $\{a, b\}=\{3,9\}$.

$\triangleright 6=3 \cdot 2$, so $u=6, v=4$, but then the condition $v \mid u$ is not satisfied.

$\triangleright 6=2 \cdot 3$, so $u=5, v=5$ and $a b=[a, b](a, b)=u v=5 \cdot 5$. If the greatest common divisor of numbers $a$ and $b$ equals their least common multiple, then $a=b=u=v$, i.e., $a=b=5$.

$\triangleright 6=1 \cdot 6$, so $u=4, v=8$, but then the condition $u \geq v$ is not satisfied.

The solutions are the ordered pairs $(a, b) \in\{(5,5),(3,9),(9,3)\}$.