Let $v_n = \dfrac{u_{n + 1}}{u_n}, \forall n \in \mathbb{N}$ then: $v_1 = 2, v_2 = 12, v_{n + 2} = 6v_{n + 1} - 8v_n$ Then: $v_n = 4^n - 2^n, \forall n \in \mathbb{N}$ So: $u_n = (4 - 2)(4^2 - 2^2).....(4^{n - 1} - 2^{n - 1}) = 2^{\dfrac{n(n - 1)}{2}}(2 - 1)(2^2 - 1).....(2^{n - 1} - 1), \forall n \in \mathbb{N}$ Let $n = 2^km$ with $k$ $\in$ $\mathbb{N} \cup \{0\}$, $m \in \mathbb{N}$, $m$ is odd Then: $\dfrac{n(n - 1)}{2} \ge k$ But: $m | 2^{\phi(m)} - 1$ and $\phi(m) \in \{1; 2; .....; m - 1\}$ then: $m | u_n$ Hence: $n | u_n, \forall n \in \mathbb{N}$