# $p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$ is always an integer

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.

• Interesting question. All I can say for now is that if $k \geq 2$, then the "integer" version will follow from the "rational" version, because for each $g \geq 1$, each of the $p_n\left(a,b\right)$ is integral over the ring $\mathbb{Z}\left[p_g\left(a,b\right), p_{g+1}\left(a,b\right)\right]$. Better yet, the ring of all symmetric polynomials in $a$ and $b$ is a finitely generated free $\mathbb{Z}\left[p_g\left(a,b\right), p_{g+1}\left(a,b\right)\right]$-module when $a$ and $b$ are indeterminates. This is a particular case of ... – darij grinberg Nov 17 '18 at 16:19
• ... Theorem 1 in my A quotient of the ring of symmetric functions generalizing quantum cohomology, since the $p_n$ are just the complete homogeneous symmetric polynomials $h_n$ in the two variables $a$ and $b$. But the "rational" version doesn't follow from this argument. – darij grinberg Nov 17 '18 at 16:21

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.
• Thanks! Do you know if the result is still true for $k=3$? – jack Nov 17 '18 at 19:03
• I would bet that yes. Here is a way to go about it, though its more heuristic than a proof. From above you get that $ab=\sqrt[l]{m}$, $m\in\mathbb{Q}$ is not an $l$ power where $l|k$ and $a+b=p+q\sqrt[l]{m}$. Suppose $l\geq2$. Then expressing $\cfrac{a^k-b^k}{a-b}$ and $\cfrac{a^{k+1}-b^{k+1}}{a-b}$ in terms of $a+b$ and $ab$ and using the fact that $\sqrt[l]{m},\ldots, \sqrt[l]{m^{l-1}}$ are algebraically independent over $\mathbb{Q}$ you get two curves on which $(p,q)$ is a rational point if $l=2$, or at least $4$ if $l\geq 3$. – Vlad Matei Nov 19 '18 at 20:05
• If you vary $m$ also you end up with an intersection of two surfaces if $l=2$, and at least four if $l\geq 3$. Bombieri Lang conjecture gives finitness of rational points so this why I believe it should be rare, or even impossible if you have more than four surfaces. – Vlad Matei Nov 19 '18 at 20:08