Problem 2. Find all pairs of integers $(a, b)$ for which $a=\frac{4 b-5}{b-2}$
Solution. The given expression can be transformed as follows:

$$
a=\frac{4 b-5}{b-2}=\frac{4(b-2)+8-5}{b-2}=\frac{4(b-2)}{b-2}+\frac{3}{b-2}=4+\frac{3}{b-2}
$$

The number $a$ will be an integer if and only if $\frac{3}{b-2}$ is an integer, which is true if and only if $b-2$ is a divisor of 3, i.e., if and only if $b-2 \in\{1,-1,3,-3\}$.

Accordingly, there are four possibilities, namely:

- For $b-2=1$ we get $b=3, a=4+3=7$,
- For $b-2=-1$ we get $b=1, a=4-3=1$,
- For $b-2=3$ we get $b=5, a=4+1=5$ and
- For $b-2=-3$ we get $b=-1, a=4-1=3$.

The required pairs are: $(7,3),(1,1),(5,5)$ and $(3,-1)$.