Let $a$ and $b$ be strictly positive integers such that $ab$ divides $a^{2}+b^{2}+1$. Calculate the quotient.
As before, we denote $k=\frac{a^{2}+b^{2}+1}{a b}$ and consider a pair $(\alpha, \beta)$ of minimal sum among the pairs $(a, b)$ satisfying this equality. If $\alpha=\beta$, then $\alpha^{2}$ divides $2 \alpha^{2}+1$ so $\alpha^{2}=1$ and $k=3$. Otherwise, we can assume without loss of generality that $\alpha>\beta \geq 0$. We solve the equation $x^{2}+\beta^{2}+1=k x \beta$. The solutions are $\alpha$ and

$$
\alpha^{\prime}=k \beta-\alpha=\frac{\beta^{2}+1}{\alpha}
$$

$\alpha^{\prime}$ is a strictly positive integer (otherwise, $\alpha^{\prime 2}-k \beta \alpha^{\prime}+\beta^{2}+1 \geq 1>0$) and $\alpha^{\prime}<\alpha$ (since $\left.\beta^{2}+1 \leq \alpha^{2}\right):$ contradiction.