TITLE: $p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$ is always an integer
QUESTION [6 upvotes]: 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.

REPLY [11 votes]: 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.