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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 $\forall k \in \mathbb N, P(q^k) \in \mathbb Z$ ?

(*) : https://artofproblemsolving.com/community/c6h3250782

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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.

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  • $\begingroup$ why the coeff of $P(q^{\deg(P)+1})$ is $1$ and the others coeff is integer ? $\endgroup$
    – Dattier
    Commented Feb 12 at 12:43
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    $\begingroup$ Expanding the polynomial $\prod_{i=0}^{\deg P}(x-q^{\deg P})=x^d-a_1x^{d-1}-\dots-a_d$, the recurrence is $x_k=a_1x_{k-1}+\dots+a_dx_{k-d}$. $\endgroup$
    – Wojowu
    Commented Feb 12 at 12:47

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