$\begin{array}{l} \dfrac{{{a^2} + 1}}{{ab - 1}} \in \mathbb{Z}\\ \Rightarrow {a^2} + 1 \vdots ab - 1\\ \Rightarrow \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right) \vdots ab - 1\\ \Rightarrow {a^2}{b^2} + {a^2} + {b^2} + 1 \vdots ab - 1\\ \Rightarrow {a^2}{b^2} - 2ab + 1 + {a^2} + {b^2} + 2ab \vdots ab - 1\\ \Rightarrow {\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2} \vdots ab - 1\\ \Rightarrow {\left( {a + b} \right)^2} \vdots ab - 1 \Rightarrow a + b \vdots ab - 1 \end{array}$
Lại có:
$\begin{array}{l} {a^2} + 1 - \left( {{b^2} + 1} \right) = {a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) \vdots ab - 1\\ \Rightarrow {b^2} + 1 \vdots ab - 1\\ \Rightarrow \dfrac{{{b^2} + 1}}{{ab - 1}} \in \mathbb{Z} \end{array}$
