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 $p_1<p_2<\cdots <p_k$.

Setting $m=p_1{p}_2\cdots p_k$ then clearly $\gcd (m,3)=1$, by using Euler's theorem, one can get $m\mid 3^{\phi (m)}-1$. We also have $$\phi (m)=(p_1-1)(p_2-1)\cdots (p_k-1)$$ is a positive integer not divisible by $3$, so setting $n=\phi (\phi (m))$ then using Euler's theorem again, one could get $$\phi (m)\mid 3^{\phi (\phi (m))}-1=3^n-1.$$ Note that $3^x-1\mid 3^y-1$ where $x,y$ are positive integers such that $x\mid y$. Hence, $$3^{\phi (m)}-1\mid 3^{3^{n-1}}-1.$$ Therefore $m\mid 3^{3^{n-1}}-1$. From here, it follows that $$a_n=(3^{3^{n-1}}-1)+3\equiv 3(\mathrm{mod}m).$$ 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.