For convenience, let $\alpha=\frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}$. Then \begin{align*} S_n &= \frac{1}{\sqrt{5}}\Big[\Big(1+C_n^1\alpha + C_n^2\alpha^2+\cdots+C_n^n\alpha^n\Big)\\ &\quad -\Big(1+C_n^1\beta+C_n^2\beta^2+\cdots+C_n^n\beta^n\Big)\Big]\\ &=\frac{1}{\sqrt{5}}\Big[(\alpha+1)^n-(\beta+1)^n\Big]\\ &=\frac{1}{\sqrt{5}}\Big[\Big(\frac{3+\sqrt{5}}{2}\Big)^n-\Big(\frac{3-\sqrt{5}}{2}\Big)^n\Big] \end{align*} Therefore the two roots to sequence $\{S_n\}$'s characteristic equation are $\frac{3\pm\sqrt{5}}{2}$. This means the equation is $$t^2 -3t+1=0$$ Accordingly, the recursion is \begin{equation} S_{n+2}=3S_{n+1}-S_n\qquad(n=1, 2, 3, \cdots) \end{equation} The two initial values are: \begin{align*} S_1 &= \frac{1}{\sqrt{5}}\Big[\Big(\frac{3+\sqrt{5}}{3}\Big)-\Big(\frac{3-\sqrt{5}}{3}\Big)\Big]=1\\ S_2 &=\frac{1}{\sqrt{5}}\Big[\Big(\frac{3+\sqrt{5}}{3}\Big)^2-\Big(\frac{3-\sqrt{5}}{3}\Big)^2\Big]=3 \end{align*} Because both $S_1$ and $S_2$ are integers, by recursion \myJustRef{eq_s8}, all terms in $\{S_n\}$ are integers. Meanwhile it is easy to see that \begin{equation} 5S_{n+1}=15S_n - 5S_{n-1}\qquad(n=2, 3, 4, \cdots) \end{equation} Subtracting \myJustRef{eq_s82} from \myJustRef{eq_s8} gives $$S_{n+2}=8S_{n+1}-16S_n+5S_{n-1}\qquad (n=2, 3, 4, \cdots)$$ or $$S_{n+3}=8S_{n+2}-16S_{n+1}+5S_{n}\qquad (n=1, 2, 3, \cdots)$$ Now it is clear from this recursion that $S_{n+3}$ is divisible by 8 if and only if $S_n$ is divisible by 8. Because $S_1 = 1$, $S_2=3$, $S_3=3\times 3-1=8$, we find $S_n$ is divisible by 8 if and only if $n=\boxed{3k}$ where $k$ is a positive integer.