For a given $a\in \mathbb{Z}$, define $P(a)$ to be the set of all prime numbers dividing $a$. Also define $\mathcal{P}$ to be the set of all prime numbers. Let $a,b,c\in \mathbb{Z}\setminus \{0\}$ be such that neither $\frac{b}{c}$ nor $\frac{c}{b}$ is a power of $a$. Then is it true that $$\mathcal{P}\setminus\cup\_{n\in \mathbb{N}}P(ca^n-b)$$ is an infinite set?



Question about iterations not divisible by infinitely many prime numbers
Yes. This follows from a result of Corrales-Rodrigáñez and Schoof (see the paper [here](https://www.sciencedirect.com/science/article/pii/S0022314X97921144?via%3Dihub)) solving the support problem of Erdős.


In particular, suppose that there are only finitely many primes $p$ that do not divide $ca^{n} - b$ for any $n$. Let $x = a$ and $y = b/c$. If $q$ is a prime that divides $ca^{n} - b$ for some $n$ and does not divide $c$ then we have $a^{n} \equiv \frac{b}{c} \pmod{q}$. Then if $k$ is an integer, $x^{k} = a^{k} \equiv 1 \pmod{q}$ implies $y^{k} \equiv (b/c)^{k} \equiv (a^{n})^{k} \equiv 1 \pmod{q}$. In particular, $x^{k} \equiv 1 \pmod{q}$ implies that $y^{k} \equiv 1 \pmod{q}$.


Theorem 1 of the paper cited above shows that if $F$ is a number field, $x, y \in F^{\times}$ have the property that for all integers $n$ and almost all prime ideals $p$ of the ring of integers of $F$, one has $y^{n} \equiv 1 \pmod{p}$ whenever $x^{n} \equiv 1 \pmod{p}$, then $y$ is a power of $x$.


This implies that $b/c$ is a (positive or negative) power of $a$.


