Czech and Slovak Mathematical Olympiad

Since all terms $a_i$ are positive integers, we have
$$a_5 = a_1a_2 + a_2a_3 + a_3a_4 - 1 \ge 1 + 1 + 1 - 1 = 2, \text{ and therefore } a_5 \ne 1.$$
The number $a_5$ is thus divisible by at least one prime number.
For every $n \ge 4$, the following holds
$$\begin{align*} a_n &= (a_1a_2 + a_2a_3 + \dots + a_{n-3}a_{n-2}) + a_{n-2}a_{n-1} - 1 = \\ &= (a_{n-1} + 1) + a_{n-2}a_{n-1} - 1 = \\ &= a_{n-1}(a_{n-2} + 1), \end{align*}$$
and therefore $a_{n-1} \mid a_n$. Let $p$ be any prime divisor of $a_5$, then the relation $a_{n-1} \mid a_n$ implies $p \mid a_6$, from where $p \mid a_7$, and so on. By mathematical induction we get $p \mid a_n$ for every $n \ge 5$. Thus, the prime $p$ divides infinitely many terms of the given sequence. This completes the proof of part a).

Let $\mathcal{P}$ denote the set of all primes that divide infinitely many terms of the sequence. Suppose that the set $\mathcal{P}$ is finite, i.e. $\mathcal{P} = \{p_1, \dots, p_k\}$ for a suitable $k$. Obviously, for every $i \in \{1, 2, \dots, k\}$ we find such a term $a_{n_i}$ that is divisible by $p_i$ and $n_i \ge 5$. Due to the relation $a_{n-1} \mid a_n$ (proved earlier for each $n \ge 4$) we have $p_i \mid a_n$ for all $n \ge n_i$. If we now denote $N = \max(n_1, \dots, n_k)$, then $a_N$ is divisible by all primes $p_1, \dots, p_k$. Therefore, $a_N+1 > 1$ is not divisible by any primes from $\mathcal{P}$, so there must be a prime $q \notin \mathcal{P}$ satisfying $q \mid a_N + 1$. This prime $q$ is then also a divisor of the number $a_{N+2} = a_{N+1}(a_N + 1)$, so $q \mid a_n$ holds for each $n \ge N+2$, and therefore $q \in \mathcal{P}$. Thus, we get a contradiction, that proves the statement in part b).