Problem 3. Let the numbers $x=\frac{a}{b-2}$ and $y=\frac{3b-6}{a-3}$. Determine the pairs $(a, b)$ of natural numbers such that $x$ and $y$ are simultaneously natural numbers.
## Solution. 1p by default

| If $x$ and $y$ are natural numbers, then $x \cdot y=\frac{a}{b-2} \cdot \frac{3 b-6}{a-3}$ is also a natural number. | $\mathbf{1 p}$ |
| :--- | :--- |
| Therefore, $x \cdot y=\frac{3 a}{a-3}=\frac{3(a-3)+9}{a-3}=3+\frac{9}{a-3} \in \mathbb{N} \Leftrightarrow a-3 \in\{1,3,9\} \Leftrightarrow$ | $\mathbf{2 p}$ |
| $a \in\{4,6,12\}$. | $1 \mathbf{p}$ |
| If $a=4 \Rightarrow \frac{4}{b-2} \in \mathbb{N}, \frac{3 b-6}{1} \in \mathbb{N} \Leftrightarrow b \in\{3,4,6\}$  $\Rightarrow(a, b) \in\{(4,3),(4,4),(4,6)\}$ | 1p |
| If $a=6 \Rightarrow \frac{6}{b-2} \in \mathbb{N}, \frac{3 b-6}{3} \in \mathbb{N} \Leftrightarrow b \in\{3,4,5,8\}$  $\Rightarrow(a, b) \in\{(6,3),(6,4),(6,5),(6,8)\}$ | 1p |
| If $a=12 \Rightarrow \frac{12}{b-2} \in \mathbb{N}, \frac{3 b-6}{9} \in \mathbb{N} \Leftrightarrow b \in\{5,8,14\}$  $\Rightarrow(a, b) \in\{(12,5),(12,8),(12,14)\}$ |  |