Problem 2. Determine the pairs $(a, b)$ of integers for which

$$
\frac{a+2}{b+1}+\frac{a+1}{b+2}=1+\frac{6}{a+b+1}
$$

Lucian Dragomir
Solution. Note that $b \neq-2$ and $b \neq-1$. Adding 2 to both sides of the equality, we obtain

$$
\left(\frac{a+2}{b+1}+1\right)+\left(\frac{a+1}{b+2}+1\right)=3+\frac{6}{a+b+1} \Leftrightarrow(a+b+3)\left(\frac{1}{b+1}+\frac{1}{b+2}\right)=\frac{3(a+b+3)}{a+b+1}
$$

Case 1. If $a+b+3=0$, then any pair of the form $(-3-u, u)$, with $u \in \mathbb{Z} \backslash\{-2,-1\}$ is a solution to the equation.

Case 2. If $a+b+3 \neq 0$, then $\frac{1}{b+1}+\frac{1}{b+2}=\frac{3}{a+b+1}$, from which $a=\frac{b^{2}+4 b+3}{2 b+3}$.

Since $a \in \mathbb{Z}$, it follows that $2 b+3 \mid b^{2}+4 b+3$, from which we obtain that $2 b+3 \mid 3$, i.e., $b \in\{-3,-2,-1,0\}$. The only solution in this case is $b=0, a=1$.

In conclusion, the solutions to the equation are $(1,0)$ and pairs $(-3-u, u)$, with $u \in \mathbb{Z} \backslash\{-2,-1\}$.