*b_n has to be infinite I think

Hamel wrote: *b_n has to be infinite I think Yes, its infinite

Set the geom progression to be $(br^n)_{n\ge 0}$ with $b = b_1 \in \mathbb Z$, then there are $a,c$ such that $\frac{br^n - c}{a} \in \mathbb Z\forall n$. Define $c_k := \frac{b}{a}(1-r^k)$, then $$ \implies c_k r^n = \frac{br^n - c}{a} - \frac{br^{n+k}-c}{a} \in \mathbb Z \forall n \in \mathbb N \cup \{0\}, k \in \mathbb N.$$Thus $r=\frac{c_kr^{n+1}}{c_kr^n} \in \mathbb Q$, so $c_k \in \mathbb Q \forall k$. Fix $k$, and suppose $\nu_q(r) < 0$ for some prime $q$. Then $\nu_q(c_kr^n) \to -\infty$ as $n \to \infty$, contradiction to $c_kr^n$ always being an integer. Thus $\nu_q(r) \ge 0 \forall q $ prime, so $r \in \mathbb Z$. YUH.

stroller wrote: Fix $k$, and suppose $\nu_q(r) < 0$ for some prime $q$. Then $\nu_q(c_kr^n) \to -\infty$ as $n \to \infty$, contradiction to $c_kr^n$ always being an integer. Thus $\nu_q(r) \ge 0 \forall q $ prime, so $r \in \mathbb Z$. Can you (or someone, who understands) explain theorem/technique/trick and the notation behind $\nu_q(r)$. Thanks in advance!

Asking for help, but i helped myself, so redacted