$S_n$ is an integer for all integers $n$ 
Let $a$ be a non-zero real number. For each integer $n$, we define $S_n = a^n + a^{-n}$. Prove that if for some integer $k$, the sums $S_k$ and $S_{k+1}$ are integers, then the sums $S_n$ are integers for all integers $n$.

We have $S_{k} = a^k+\frac{1}{a^k} = m_1$ and $S_{k+1} = a^{k+1}+\frac{1}{a^{k+1}} = m_2$ where $m_1,m_2 \in \mathbb{Z}$.  Thus raising $S_k$ and $S_{k+1}$ to any positive power results in an integer. Is there a way I can prove the statement from this?

Partial Result: For any integers $p,q$, we have
$$
S_p S_q = (a^p + a^{-p})(a^{q} + a^{-q}) = 
(a^{p+q} + a^{-(p+q)}) + (a^{p-q} + a^{-(p-q)}) =
S_{p+q} + S_{p-q}
$$
and conversely, whenever $x$ and $y$ are either both odd or both even with $x > y$, we have
$$
S_{x} + S_y = S_{(x+y)/2}S_{(x-y)/2}
$$
If $q = (2k + 1)p$, we have
$$
\frac{S_q}{S_p} = \sum_{i=0}^{2k} (-1)^i a^{2(k-i)p} = \sum_{i=0}^k (-1)^kS_{ip} 
$$
