1. Determine all pairs of integers $(a, b), \quad b \neq 1$ for which $a=\frac{4 b^{2}+5 b+10}{b+1}$.
Solution. With elementary transformations, we obtain

$$
a=\frac{4 b^{2}+5 b+10}{b+1}=\frac{4 b(b+1)+b+1+9}{b+1}=4 b+1+\frac{9}{b+1}
$$

which means that $a$ is an integer if and only if $\frac{9}{b+1}$ is an integer, that is, if and only if $b+1 \in\{1,3,-3,9,-9\}$. According to this:

$$
\begin{aligned}
& b+1=1 \quad \Rightarrow b=0, \quad a=10 \\
& b+1=3 \quad \Rightarrow b=2, \quad a=12 \\
& b+1=-3 \quad \Rightarrow b=-4, \quad a=-18 \\
& b+1=9 \quad \Rightarrow b=8, \quad a=74 \\
& b+1=-9 \quad \Rightarrow b=-10, \quad a=-40
\end{aligned}
$$