Say I have a polynomial $x^2 + a x + b$, where $a, b \in \mathbb{Z}$, which has two real roots. Denote these roots $\alpha, \beta$. Then Newton's identities tell us that $\alpha^k + \beta^k \in \mathbb{Z}$ for every integer $k \geq 1$.
My question is, what can we say about the value of $\alpha^k + \beta^k$ as $k$ varies? In particular, could we relate (some) of its prime factors to (the prime factors of) $a, b$ say?
Let $S_k = \alpha^k + \beta^k$ with $S_0 = 2$ and $S_1 = -a$, then:
$$ -a S_k = S_kS_1 = (\alpha^k + \beta^k)(\alpha + \beta) = \alpha^{k+1} + \beta^{k+1} + \alpha\beta(\alpha^{k-1} + \beta^{k-1})=S_{k+1} + b S_{k-1} $$ $$ \iff \;\;\;\; S_{k+1} = -a S_k - b S_{k-1} $$
Below are a few divisibility properties that follow directly from the recurrence relation above.
If $a \mid S_{k-1}$ then $a \mid S_{k+1}$. Since $a \mid S_1 = -a$, it follows by induction that $a \mid S_k$ for all odd $k$.
Let $d = \gcd(a,b)$, then $d^c \mid S_{k}, S_{k-1} \implies d^{c+1} \mid S_{k+1}$. Since $d \mid S_1 = -a$ and $d \mid S_2 = a^2 - 2b$ it follows by induction that $d^{\lceil k/2 \rceil} \mid S_k$ for all $k$.
Let $p \mid b$ be an odd prime that divides $a$ but not $b$, then $p \mid S_{k-1} \iff p \mid S_{k+1}$. Since $p \mid S_1 = -a$ and $p \not\mid S_2 = a^2 - 2b$ it follows by induction that $p \mid S_k$ iff $k$ is odd.
Let $q \mid b$ be a prime that divides $b$ but not $a$, then $q \not\mid S_k \implies q \not\mid S_{k+1}$. Since $q \not\mid S_1 = -a$ it follows by induction that $q \not\mid S_k$ for all $k$.