1. Determine all pairs of natural numbers for which the difference between their LCM and GCD is 15.
Solution. Let $x$ and $y$ be the pair of natural numbers for which the difference between their LCM and GCD is 15. Let $d=\operatorname{GCD}(x, y)$ and $s=\operatorname{LCM}(x, y)$. Then, there exist natural numbers $a$ and $b$ such that $x=d a, y=d b, \operatorname{GCD}(a, b)=1$ and $s=d a b$. From the condition of the problem, we have that $s-d=15$, and if we substitute $s=d a b$ we get

$$
d a b-d=15 \Rightarrow d(a b-1)=15 \Rightarrow d \in\{1,3,5,15\}
$$

1) For $d=1$, we have

$$
a b-1=15 \Rightarrow a b=16 \Rightarrow a=1, b=16 \Rightarrow x=1, y=16
$$

2) For $d=3$, we have

$a b-1=5 \Rightarrow a b=6 \Rightarrow a=1, b=6$ or $a=2, b=3 \Rightarrow x=3, y=18$ or $x=6, y=9$,

3) For $d=5$, we have

$$
a b-1=3 \Rightarrow a b=4 \Rightarrow a=1, b=4 \Rightarrow x=5, y=20
$$

4) For $d=15$, we have

$$
a b-1=1 \Rightarrow a b=2 \Rightarrow a=1, b=2 \Rightarrow x=15, y=30
$$

Thus, the required pairs of natural numbers are: 1 and 16, 3 and 18, 6 and 9, 5 and 20, 15 and 30.