Saudi Arabian Mathematical Competitions

The recursive relation is equivalent to
$$3 a_{n+1} - 1 = \frac{1}{2} \left(9 a_{n}^{2} - 6 a_{n} + 1\right), \quad n = 1, 2, \ldots$$
Let $x_{n} = 3 a_{n} - 1$, $n = 1, 2, \ldots$, and get $x_{1} = 8 = 2^{3}$,
$$x_{n+1} = \frac{1}{2} x_{n}^{2}, \quad n = 1, 2, \ldots$$
Let $x_{n} = 2^{y_{n}}$, $y_{1} = 3$, $n = 1, 2, \ldots$, and obtain
$$y_{n+1} = 2 y_{n} - 1, \quad n = 1, 2, \ldots$$
It follows $y_{n} = 2^{n} + 1$, $n = 1, 2, \ldots$, hence $x_{n} = 2^{2^{n} + 1}$, and we get
$$a_{n} = \frac{1}{3} \left(2^{2^{n}} + 1\right), \quad n = 1, 2, \ldots$$
If $n = 3^{k}$, then
$$2^{2 \cdot 3^{k}} = 4^{3^{k}} = (3 + 1)^{3^{k}} \equiv 1 \quad (\bmod\ 3^{k+1})$$
hence
$$\left(2^{3^{k}} - 1\right)\left(2^{3^{k}} + 1\right) \equiv 0 \quad (\bmod\ 3^{k+1})$$
But, we have
$$2^{3^{k}} - 1 = (3 - 1)^{3^{k}} - 1 \equiv -2 \quad (\bmod\ 3^{k+1})$$
and we obtain
$$2^{3^{k}} + 1 \equiv 0 \quad (\bmod\ 3^{k+1})$$
hence
$$2^{3^{k}} \equiv -1 \quad (\bmod\ 3^{k+1}) .$$
It follows $2^{3^{k}} = 3^{k+1} m - 1$, for some odd integer $m$, hence
$$\begin{gathered} 2^{2^{3^{k}+1} + 1} = 2^{3^{k+1} m} + 1 = (3 - 1)^{3^{k+1} m} + 1 \\ \equiv (-1)^{3^{k+1} m} + 1 \quad (\bmod\ 3^{k+1}) \equiv 0 \quad (\bmod\ 3^{k+1}) \end{gathered}$$
We obtain $3^{k+1} \mid 2^{2^{n}} + 1$, hence $3^{k} \mid a_{n}$, that is $n \mid a_{n}$.