Task 2. Determine all natural numbers $a$ and $b$ for which the numbers $x=\frac{2 a+b}{3 b+2}, y=\frac{3 b+2}{8}$, and $z=\frac{8}{2 a+b}$ are natural numbers.
Solution. It is not hard to verify that the equality $x y z=1$ is satisfied, and since $x, y, z \in \mathbb{N}$, we get $x=y=z=1$. Therefore, from $y=1$ we get $\frac{3 b+2}{8}=1$, i.e., $3 b+2=8$. From the last equation, we have $b=2$. On the other hand, from $z=1$ we get $\frac{8}{2 a+b}=1$, i.e., $2 a+b=8$. From the obtained equation, we have $a=3$. Now it is not hard to see that $x=1$ (for the obtained values of $a$ and $b$).

Thus, the required numbers are $a=3$ and $b=2$.