Folklore

Find all pairs of natural numbers $(a, b)$ for which the equality $\operatorname{LCM}(a, b)-\operatorname{GCD}(a, b)={ }^{a b} / 5$ holds.

#
Let's use the fact that the LCM $(a, b)$ is divisible by the GCD $(a, b)$, and the identity LCM $(a, b) \cdot \operatorname{GCD}(a, b)=a b$. Let GCD $(a, b)=n$, then LCM $(a, b)=k n$ (where $k$ and $n$ are natural numbers). Then $5(k n-n)=k n^{2}$, or $k(5-n)=5$. This equation has a unique solution: $k=5, n=4$. Therefore, GCD $(a, b)=4, \operatorname{LCM}(a, b)=20, a b=80$. Both numbers are not less than 4, and one is a multiple of 5, i.e., not less than 20. Hence, the answer.

## Answer

$\{4,20\}$.