Equal to their greatest common divisor and least common multiple profit. Find such natural numbers.



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Let the required natural numbers be $x$ and $y$, and let $m$ be their greatest common divisor. We can set $x = ma$ and $y = mb$, where $m, a, b$ are natural numbers and $a, b$ are coprime.

Then their least common multiple must be $mab$. Therefore, according to the problem statement, we have
$$
ma \times mb - (ma + mb) = m + mab.
$$

Simplifying, we get $(m-1)ab = 1 + a + b$.
(*)
$$
\therefore m-1 = \frac{1}{ab} + \frac{1}{a} + \frac{1}{b} \leq 3,
$$

which means $1 < m \leq 4$.
Thus, $m$ can take the values 2, 3, 4.
(1) When $m = 2$, (*) becomes $ab - a - b = 1$, i.e., $(a-1)(b-1) = 2$.
Solving, we get $\left\{\begin{array}{l}a=2, \\ b=3\end{array}\right.$ or $\left\{\begin{array}{l}a=3 \\ b=2 .\end{array}\right.$
Therefore, the required natural numbers in this case are 4 and 6.
(2) When $m = 3$, (*) becomes $2ab - a - b = 1$,

which simplifies to $(2a-1)(2b-1) = 3$.
Solving, we get $\left\{\begin{array}{l}a=1 \\ b=2\end{array}\right.$ or $\left\{\begin{array}{l}a=2, \\ b=1 .\end{array}\right.$
Therefore, the required natural numbers in this case are 3 and 6.
(3) When $m = 4$, (*) becomes $3ab - a - b = 1$. From this, we get $a = b = 1$.
Therefore, the required natural numbers in this case are 4 and 4.