Problem 2.6 Determine the non-zero natural numbers $a$ and $b$ for which $\frac{a+1}{b}$ and $\frac{b+2}{a}$ are simultaneously natural numbers.
Solution $\frac{a+1}{b}$ being a non-zero natural number implies $b \mid a+1$, from which $b \leq a+1$ and hence $b+2 \leq a+3$.

Now $\frac{b+2}{a} \leq \frac{a+3}{a}=1+\frac{3}{a} \leq 4$. Since $\frac{b+2}{a}$ is a non-zero natural number, it follows that $\frac{b+2}{a} \in\{1,2,3,4\}$. $3 \text{p}$

If $\frac{b+2}{a}=1$, then from $b+2=a$ we get $b=a-2$. Then $\frac{a+1}{b}$ being a non-zero natural number implies $\frac{a+1}{a-2}=1+\frac{3}{a-2}$ is a non-zero natural number, from which $a \in\{3,5\}$. We obtain the solutions $a=3, b=1$ and $a=5, b=3$.

If $\frac{b+2}{a}=2$, then from $b+2=2a$ we get $b=2a-2$. Then $\frac{a+1}{b}$ is a non-zero natural number, or $\frac{2a+2}{2a-2}=1+\frac{4}{2a-2}$ is an even natural number, from which $a \in\{2,3\}$. We obtain the solution $a=3, b=4$; the variant $a=2, b=2$ does not satisfy the initial conditions.

If $\frac{b+2}{a}=3$, then from $b+2=3a$ we get $b=3a-2$. Then $\frac{a+1}{b}$ is a non-zero natural number, or $\frac{3a+3}{3a-2}=1+\frac{5}{3a-2}$ is a natural number, from which $a \in\{1\}$. We obtain the solution $a=1, b=1$.

If $\frac{b+2}{a}=4$, then $b+2=4a$ results in $b=4a-2$. Then $\frac{a+1}{b}$ is a non-zero natural number, or $\frac{4a+4}{3a-2}=1+\frac{6}{4a-2}$ is an even natural number, from which $a \in\{1,2\}$. We obtain the solution $a=1, b=2$; the variant $a=2, b=6 \text{ does not}$ satisfy the initial conditions.