If $\left(\alpha\,a^n + \beta\, \dfrac{(-1)^n}{a^n}\right)$ is always an integer for $a$ integer, does it force $\beta=0$? In order to find a shorter proof for this thread : 
Find all functions $f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)]$
Rem: just to be clear, $\mathbb N^*$ stands for $\mathbb N\setminus\{0\}$

I'm interested in whether it is possible to establish directly that
If $\ \forall n>0$ then $F_n=\left(\alpha\,a^n + \beta\, \dfrac{(-1)^n}{a^n}\right)\in\mathbb N^*\quad$ with $(\alpha,\beta)\in\mathbb R^2$ and $a\in\mathbb N^*$ 
$\implies (a=1) \text{ or }(\beta=0)$.

A possible additional condition could be to set $F_1=a$ also.
This seems expected because intuitively $\beta$ has to be divisible by any $a^n$, but when trying to prove it, I only manage to show that $\alpha,\beta$ are rational, but cannot find the decisive blow.
I'm sure I'm missing something obvious, but I can't see it...
 A: Note that $a \alpha = F_1 + \frac{\beta}{a},$ which implies that $$F_n = F_1 a^{n-1} + \beta \left( \frac{a^{2n-2} +(-1)^n}{a^n} \right).$$ Since both $F_n$ and $F_1 a^{n-1}$ are naturals for every $n$, we must have that for every $n$, $$\beta  \left( \frac{a^{2n-2} + (-1)^n}{a^n} \right) \in \mathbb{Z}.$$ In particular, this would hold for all odd $n$. Now note that for every $a, n >1 \in \mathbb{N},$ $a^n$ is relatively prime to $a^{2n-2} - 1.$ Indeed, if $d$ divides both $a^n$ and $a^{2n-2} - 1,$ then it must also divide $a^{n-2} a^n - (a^{2n-2} - 1) = 1,$ forcing $d = 1$. 
Thus, there are infinitely many $n$ such that $\beta /a^n$ is an integer. But, since $a>1,$ for large enough $n$ we have $|\beta/ a^{n}| <1, $ forcing $\beta = 0$.
A: No.
It is fairly common.
An example is the
Fibonacci numbers.
(see https://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio)
They satisfy
$F_n
=\dfrac1{\sqrt{5}}(\phi^n-(-\phi)^{-n})
$
where
$\phi = \dfrac{1+\sqrt{5}}{2}
$.
This is because the
$\phi$ satisfies
$x^2-x-1 = 0$.
This will happen for
recurrences of the form
$a_{n+2} = ka_{n+1}+a_n$.
