2. Find all pairs of positive integers $a, b$, for which the number $b$ is divisible by the number $a$ and at the same time the number $3a+4$ is divisible by the number $b+1$.


2. Since the number $a$ divides the number $b$, we can write $b=k a$, where $k$ is a positive integer. It is therefore sufficient to find positive integers $a$ for which there exists a positive integer $k$ such that the number $3 a+4$ is a (positive) multiple of the number $k a+1(=b+1)$. From this condition, we get the inequality $k a+1 \leqq 3 a+4$, from which it follows that $k-3 \leqq(k-3) a \leqq 3$, and thus $k \leqq 6$. Moreover, for $k \geqq 3$, we already have $2(k a+1)>3 a+4$ for any $a \geqq 1$, so it must be that $k a+1=3 a+4$. Let's consider all six possibilities for the number $k$:

$k=1: a+1 \mid 3 a+4$, and since $a+1 \mid 3 a+3$, it would have to be $a+1 \mid 1$, which is not possible, as $a+1>1$.

$k=2: 2 a+1 \mid 3 a+4=(2 a+1)+(a+3)$, so $2 a+1 \mid a+3$. However, for any natural $a$, $2 \cdot(2 a+1)>a+3$, so it must be $2 a+1=a+3$ or $a=2$ and thus $b=k a=4$.

$k=3: 3 a+1=3 a+4$, which is not possible.

$k=4: 4 a+1=3 a+4$, so $a=3, b=12$.

$k=5: 5 a+1=3 a+4$, which does not satisfy any integer $a$.

$k=6: 6 a+1=3 a+4$, so $a=1, b=6$.

The solutions are the pairs $(1,6),(2,4)$, and $(3,12)$.

For a complete solution, award 6 points. For merely listing all pairs that satisfy the problem, award $1 \mathrm{~b}$.