# Rationals polynomials with integers values

I have asked this question here (*), but there are no answer.

Let $$q \in \mathbb N \cap [2,+\infty[$$ and $$P \in \mathbb Q[x]$$ with $$\forall k \in [0,\deg(P)] \cap \mathbb N, P(q^k) \in \mathbb Z$$.

Is it true that $$\forall k \in \mathbb N, P(q^k) \in \mathbb Z$$ ?

The answer is yes, because the values $$P(q^k),k\in\mathbb N$$ satisfy a linear recurrence with integer coefficients, namely the one with characteristic polynomial $$(x-1)(x-q)\dots(x-q^{\deg P})$$ (because $$k\mapsto q^{nk}$$ satisfies it for $$n\leq\deg P$$, and $$P(q^k)$$ is a linear combination of these). This recurrence has order $$\deg P+1$$, therefore if we know this many consecutive coefficients are integers, it follows by induction all of them are.
• why the coeff of $P(q^{\deg(P)+1})$ is $1$ and the others coeff is integer ? Feb 12 at 12:43
• Expanding the polynomial $\prod_{i=0}^{\deg P}(x-q^{\deg P})=x^d-a_1x^{d-1}-\dots-a_d$, the recurrence is $x_k=a_1x_{k-1}+\dots+a_dx_{k-d}$. Feb 12 at 12:47