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?
It's not true if $a$ is allowed to be complex. For example, if $a$ is a root of $z^4 - z^2 + 1$ we have $a^2 + a^{-2} = 1$ and $a^3 + a^{-3} = 0$ but $a + a^{-1} = \pm \sqrt{3}$.
EDIT: Using the symmetries $a \to 1/a$ (leaving $S_k$ invariant) and $a \to -a$ (with $S_k \to (-1)^k S_k$), we may assume $a \ge 1$. Of course if $a = 1$, all $S_k = 2$ are integers, so we're reduced to the case $a > 1$. Moreover we may assume $n > 1$, since if $a + 1/a$ is an integer so are all $a^k + 1/a^k$. Of course all $S_k > 0$.
Consider the polynomials $P(z) = z^{2n} - S_n z^n + 1$ and $Q(z) = z^{2n+2} - S_{n+1} z^{n+1} + 1$. $a$ is a root of both of these. In particular it is an algebraic integer. But if $a$ is rational, it must be an integer, and then (since $a > 1$) it would be impossible to have $S_n$ an integer.
Now both $P(z)$ and $Q(z)$ are multiples of the minimal polynomial $M(z)$ of $a$, which is at least quadratic. If $z$ is a root of $P(z)$, then so is $\omega z$ where $\omega$ is any $n$'th root of $1$. Thus the roots of $P(z)$ are $a \omega$ and $a^{-1} \omega$ for the $n$'th roots $\omega$ of $1$. Similarly, the roots of $Q(z)$ are $a \zeta$ and $a^{-1} \zeta$ for the $n+1$'th roots $\zeta$ of $1$. But except for $1$, an $n$'th root of $1$ is not an $n+1$'th root of $1$, so the common roots of $P(z)$ and $Q(z)$ are $a$ and $1/a$. Thus these are the only possible roots of $M(z)$, and $M(z)$ is quadratic: in fact $M(z) = (z-a)(z-1/a)$, and thus $S_1 = a + 1/a$ must be an integer.
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 $x$.
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 $$ \frac{S_q}{S_p} = \sum_{i=0}^{2k} (-1)^i a^{2(k-i)p} = \sum_{i=0}^k (-1)^kS_{ip} $$
