Existence of $m \in M \subsetneq \mathbb{N}$ such that $p \nmid a+bm$, $p$ prime

Question

Let $f(x)=a+bx$, with $a,b \in \mathbb{N}-\{0,1\}$ and $\gcd(a,b)=1$.

Let $p$ be a prime number.

$\{f(n)\}_{n \in \mathbb{N}} \subset \mathbb{N}$, so we can ask if $p$ divides $f(m)$ for some fixed $m \in \mathbb{N}$.

Let $M$ be an infinite proper subset of $\mathbb{N}$.

Is it possible to determine if there exists $m \in M$ such that $p \nmid f(m)$?

Notice that if $M=\mathbb{N}$, then the answer is positive; just take large enough $m$ such that $f(m)$ is a prime number larger than $p$ (this can be done by Dirichlet's theorem on arithmetic progressions), and then clearly $p \nmid f(m)$.

Any hints and comments are welcome!

Answer 1

Rewriting your statement, you want so see whether there is $m \in M$ so that

$$ m \equiv -a \cdot b^{-1} \pmod{p} $$

where $b^{-1}$ is the inverse of $b$ in $\mathbb{Z}_p$. Let $k = a\cdot b^{-1} \in \{0, \dots, p-1\}$. If there does not exist $m \in M$ so that $p \not| f(m)$, then $m \equiv k \pmod{p}$ for all $m\in M$, or equivalently, $$ M \subseteq k + p\mathbb{N}_0. $$ Moreover, this characterizes the desired property.