A "simple" divisibility problem

The given recurrence has closed form $a_n = \alpha^{2n+1} - \beta^{2n+1}$ , where $\alpha = \sqrt{2} + 1$ and $\beta = \sqrt{2} - 1$ . Assume that $2m+1$ divides $2n+1$ and let $2n+1 = k(2m+1)$ . Then, we have:

$a_n = \alpha^{k(2m+1)} - \beta^{k(2m+1)} = (\alpha^{2m+1} - \beta^{2m+1}) S = a_m S$

where $S$ is a sum of combinations of powers of $\alpha$ and $\beta$ (why?), therefore it is an algebraic integer. On the other hand, it is clearly rational, so it must be a rational integer, and therefore $a_m$ divides $a_n$ . Since $2786 = a_4$ , we have that $a_4, a_{13}, a_{22}, a_{31}...$ are all divisible by $a_4 = 2786$ .