67th Romanian Mathematical Olympiad

Since the fractions $f_1=\frac{a+1}{b}$ and $f_2=\frac{b+2}{a}$ are positive integers, $a+1\ge b$ and $b+2\ge a$, hence $a+1\ge b\ge a-2$. This leaves the cases:

1. $b=a+1$: then $f_2=1+\frac{3}{a}\in \mathbb{N}$, whence $a=1,b=2$ or $a=3,b=4$, which are convenient values.

2. $b=a$: then $f_1=1+\frac{1}{a}\in \mathbb{N}$, whence $a=1,b=1$, which are convenient values.

3. $b=a-1$: then $f_2=1+\frac{1}{a}\in \mathbb{N}$, whence $a=1,b=1$.

4. $b=a-2$: then $f_1=1+\frac{3}{b}\in \mathbb{N}$, whence $a=3,b=1$ or $a=5,b=3$, which are convenient values.