$S_n$ 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 stuff:
Lemma: If $b+b^{-1}$ is an integer then $b^n+b^{-n}$ is an integer for all $n$.
The proof is via induction.
Notice that by Newton's theorem (and symmetry of binomial coefficients) $(b+b^{-1})^n=\sum\limits_{i=0}^{(n-1)/2}\binom{n}{i}(b^ib^{-(n-i)}+b^{n-i}{b^{-i}})+A$ (where $A=0$ if $n$ is odd and $A=\binom{n}{n/2}$ otherwise).
Notice every summand except for $b^n+b^{-n}$ is an integer by the inductive hypothesis, and $(b+b^{-1})^n$ is also an integer. We conclude $b^n+b^{-n}$ i an integer.

Setting $b=a^k$ and $a^{k+1}$ we have $a^{nk}+a^{-nk}$ and $a^{n(k+1)}+a^{-n(k+1)}$ are integers for all $n$.

Solution: Let $T_n$ be the polynomial defined by $T_0(x)=2,T_1(x)=x,T_{n+1}=xT_n(x)-T_{n-1}(x)$. We then have $T_n(a+a^{-1})=a^n+a^{-n}$, it follows from the recursion  that if two consecutive values $T_k(x),T_{k+1}(x)$ are integer then $T_n(x)$ is an integer for all $x$.
