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.

  • $\begingroup$ 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 ... $\endgroup$ – darij grinberg Nov 17 '18 at 16:19
  • $\begingroup$ ... 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. $\endgroup$ – 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.

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    $\begingroup$ Just found the reference to the problem. It is a problem from AMM, namely Problem E2998 by Clark Kimberling. The problem is stated for the integer version and again four consecutive values suffice. $\endgroup$ – Vlad Matei Nov 17 '18 at 17:44
  • $\begingroup$ Thanks! Do you know if the result is still true for $k=3$? $\endgroup$ – jack Nov 17 '18 at 19:03
  • $\begingroup$ 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 $l\geq 3$. $\endgroup$ – Vlad Matei Nov 19 '18 at 20:05
  • $\begingroup$ 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. $\endgroup$ – Vlad Matei Nov 19 '18 at 20:08

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