Saudi Arabian Mathematical Competitions, 2012

Since $a$ divides $n$ divides $a^2 + b^2 + 1$, it follows that $a \mid b^2 + 1$. Similarly $b \mid a^2 + 1$. Thus we have $\gcd(a, b) = 1$.
We have $ab \mid a^2 + b^2 + 1$. By symmetry, we can assume $a \le b$. It follows that
$$a^2 + b^2 + 1 = kab \quad (1)$$
for some positive integer $k$. In the case $a = b$ we have $2a^2 + 1$ is a multiple of $a$, meaning $a = 1$. Consequently, $(a, b) = (1, 1)$ and $k = 3$.

Assume now $a > b$. The equation (1) considered as a quadratic equation in $a$ has a positive integer solution. Its second solution is
$$a' = kb - a = \frac{b^2 + 1}{a},$$
and it is also a positive integer. Moreover,
$$a' = \frac{b^2 + 1}{a} \le \frac{b^2 + 1}{b+1} \le b < a.$$
Thus, if equation (1) has a solution $(a, b)$ with $a > b$, then it also has another solution $(b, a')$ with a strictly smaller sum of numbers. Applying the same argument to this new solution, then to the next solution, etc., we eventually arrive at a pair $(a_0, b_0)$ that cannot be further reduced. Hence,
$$a_0 = b_0 = 1,$$
and thus $k = 3$.
Thus, starting with an arbitrary solution one can descend to the pair $(1, 1)$. Therefore, all possible solution $(a, b)$ of (1) are constructed in the infinite chain
$$(1, 1) \rightarrow (2, 1) \rightarrow (5, 2) \rightarrow (13, 5) \rightarrow (34, 13) \rightarrow \dots$$
The transfer to the next pair is by the rule $(x, y) \rightarrow (3x - y, x)$.
It can easily be observed and then proved by induction that the $k^{\text{th}}$ pair, if the pair $(1, 1)$ has number 0, consists of numbers $(F_{2k+1}, F_{2k-1})$, where $(F_i)_{i \ge 0}$ is the Fibonacci sequence, $F_0 = 0$, $F_1 = 1$, and $F_{i+2} = F_{i+1} + F_i$, $i = 0, 1, \dots$.
Finally, since $n$ is a multiple of $ab$ and is a divisor of $3ab$, it follows that either
$$n = ab \text{ or } n = 3ab.$$
Thus, the solutions are
$$n = 1, 3, \quad n = F_{2k-1}F_{2k+1}, \quad n = 3F_{2k-1}F_{2k+1}, \quad k = 1, 2, \dots$$