Does anyone know if the following problem has ever been studied?



> 
> Let $a$ and $b$ be two real numbers and consider the polynomial: $$p\_n(x,y)=\sum\_{i=0}^{n-1}x^{n-1-i}y^{i}$$
>  where $n$ is a positive integer. 
> 
> 
> Does there exist a value of $k$ such that if $p\_n(a,b)$ is an integer for $k$ consecutive values of $n$ then $p\_n(a,b)$ is an integer for every $n$? If so what is the minimum value of $k$?  
> 
>  What happens if we change the 'integer' condition to 'rational' values at the previous question? 
> 
> 
> 


It's not difficult to establish some recurrence relation among the values of $p\_n$ but none of them seem to be promising.  

I would like to know any reference for this problem or how it could be solved.  

Any help would be appreciated. 



$p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$ is always an integer
I will do the rational case and assume $a,b\neq 0$ otherwise the problem is trivial. You just need four consecutive values. Note that $p\_n(a,b)=\cfrac{a^n-b^n}{a-b}$.
Say you have $p\_k$, $p\_{k+1}$, $p\_{k+2}$, $p\_{k+3}$ are all rational.
Note that $p\_{k+1}^2-p\_kp\_{k+2}=(ab)^k$ and $p\_{k+2}^2-p\_{k+1}p\_{k+3}=(ab)^{k+1}$. 
Thus $ab$ is rational.Now $p\_{k+1}(a+b)=p\_{k+2}+abp\_{k}$ so it follows that $a+b$ is also rational or $p\_{k+1}=0$. But similarly $p\_{k+2}(a+b)=p\_{k+3}+abp\_{k+1}$ so if $p\_{k+2}=0$ then $a=b$ and again the problem is trivial.
Thus we have $a+b,ab \in \mathbb{Q}$ and now note that $p\_n(a,b)$ is a symmetric polynomial so it can be expressed as a polynomial with rational coefficients in $a+b,ab$ so it is always rational.


