Saudi Arabia booklet, 2024

Since $a_n$ is odd and is congruent to $2$ modulo $3$, it always has a prime divisor of the form $3h + 2$ for $h \in \mathbb{Z}^+$. Suppose by contradiction that the sequence $(a_n)$ has finitely many prime divisors, then the number of prime divisors of the form $3h + 2$ of the sequence is clearly also finite, let them be $p_1 < p_2 < \dots < p_k$.

Setting $m = p_1p_2\cdots p_k$ then clearly $\gcd(m, 3) = 1$, by using Euler's theorem, one can get $m \mid 3^{\varphi(m)} - 1$. We also have
$$\varphi(m) = (p_1 - 1)(p_2 - 1)\cdots(p_k - 1)$$
is a positive integer not divisible by $3$, so setting $n = \varphi(\varphi(m))$ then using Euler's theorem again, one could get
$$\varphi(m) \mid 3^{\varphi(\varphi(m))} - 1 = 3^n - 1.$$
Note that $3^x - 1 \mid 3^y - 1$ where $x, y$ are positive integers such that $x \mid y$. Hence,
$$3^{\varphi(m)} - 1 \mid 3^{3^{n-1}} - 1.$$
Therefore $m \mid 3^{3^{n-1}} - 1$. From here, it follows that
$$a_n = (3^{3^{n-1}} - 1) + 3 \equiv 3 \pmod{m}.$$
On the other hand, clearly $a_n$ will have a prime divisor of the form $3h + 2$ so there exists $i$ with $1 \le i \le k$ such that $p_i \mid a_n$. But $p_i \mid m$ which implies that $p_i \mid 3$, which is clearly a contradiction. Therefore, $(a_n)$ has infinitely many prime divisors.