Belarus — Final Round

It is evident that all terms of the sequence are positive. Find $a_4$. Since $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, we have
$$a_4 = \frac{a_3 a_2 + 7}{a_1} = 13. \quad (1)$$
By condition,
$$\begin{aligned} a_{n+1} &= \frac{a_n a_{n-1} + 7}{a_{n-2}} = [7 = a_n a_{n-3} - a_{n-1} a_{n-2}] = \\ &= \frac{a_n a_{n-1} + a_n a_{n-3} - a_{n-1} a_{n-2}}{a_{n-2}} = a_n \frac{a_{n-1} + a_{n-3}}{a_{n-2}} - a_{n-1}, \quad n \ge 4. \end{aligned} \quad (2)$$
If we prove that all numbers $\frac{a_{n-1} + a_{n-3}}{a_{n-2}}$, $n \ge 4$, are integers, then we obtain
that all numbers $a_n$, $n \ge 5$, are integers, since the numbers $a_1, a_2, a_3, a_4$ are integers. Let $b_n = (a_{n-1} + a_{n-3})/a_{n-2}$, $n \ge 4$. In particular
$$b_4 = \frac{a_3 + a_1}{a_2} = \frac{3+1}{2} = 2 \quad \text{and} \quad b_5 = \frac{a_4 + a_2}{a_3} = \frac{13+2}{3} = 5. \quad (3)$$
Show that the sequence $(b_n)$, $n \ge 4$, is periodic with 2 as its period, i.e. $b_n = b_{n-2}$ for all $n \ge 6$. From definition of the sequences $(b_n)$ and $(a_n)$ it follows that
$$\begin{aligned} b_n &= \frac{a_{n-1} + a_{n-3}}{a_{n-2}} = \frac{\frac{a_{n-2} a_{n-3} + 7}{a_{n-4}} + a_{n-3}}{a_{n-2}} = \\ &= \frac{a_{n-2} a_{n-3} + 7 + a_{n-3} a_{n-4}}{a_{n-4} a_{n-2}} = [7 = a_{n-2} a_{n-5} - a_{n-3} a_{n-4}] = \\ &= \frac{a_{n-2} a_{n-3} + (a_{n-2} a_{n-5} - a_{n-3} a_{n-4}) + a_{n-3} a_{n-4}}{a_{n-4} a_{n-2}} = \\ &= \frac{a_{n-2} a_{n-3} + a_{n-2} a_{n-5}}{a_{n-4} a_{n-2}} = \frac{a_{n-3} + a_{n-5}}{a_{n-4}} = b_{n-2}, \quad n \ge 6. \end{aligned}$$
Therefore, taking into account (3), we have $b_{2k}=b_4=2$ and $b_{2k+1}=b_5=5$ for all $k=2,3,\dots$. Since (see (2)) $a_{n+1} = a_n b_n - a_{n-1}$ for all $n \ge 4$, and all numbers $b_n$, $n \ge 4$, and numbers $a_1, a_2, a_3, a_4$ are integers, it follows that all terms of the sequence $(a_n)$ are integers.