Is it true that there always exists a positive integer $n$ such that $p \mid \lfloor k^n\cdot\alpha\rfloor$?

Question

Let $k,M$ be positive integers such that $k−1$ is not squarefree. Prove that there exist a positive real number $\alpha$, such that $\lfloor\alpha\cdot k^n\rfloor$ and M are coprime for any positive integer $n$.

Since $k−1$ is not squarefree, then there exists a prime number $p$ such that $p^2\mid k−1$. Choose $\alpha=N+\frac{1}{p}$, with $N$ a positive integer such that $p\cdot N+1$ is divisible by all prime factors of $M$ (except $p$ if $p\mid M$), and $N$ is not divisible by $p$. (We can choose $N$ by using Chinese Remainder Theorem.) Then for every positive integer $n$, $$\lfloor\alpha\cdot k^n\rfloor=N\cdot k^n+\lfloor\frac{k^n}{p}\rfloor=N\cdot k^n+\frac{k^n−1}{p}=\frac{k^n⋅(p⋅N+1)−1}{p}. $$ Since $p\cdot N+1$ is divisible by all prime divisors of $M$, and $\lfloor\alpha\cdot k^n\rfloor$ is not divisible by $p$, because $N\cdot k^n$ is not divisible by $p$ (we consider this in case $p\mid M$), therefore, $\lfloor\alpha\cdot k^n\rfloor$ and $M$ are coprime.

However, if $\alpha$ must be irrational, then I have a feeling that there are no such $\alpha$ that suit the problem's condition.

So my question is:

Let $p$ be a prime integer, $k$ be a positive integer and $α$ be a positive irrational number. Is it true that there always exists a positive integer $n$ such that $p \mid \lfloor k^n\cdot\alpha\rfloor$ ?

Any answers or comments will be appreciated.

(Please let me know if this question should be closed, off-topic or unclear. I may not visit this page frequently, so I may not be able to know what is going on. Sorry for this inconvenience.)

Answer 1

I will answer your question (the one in the second orange box).

No, it is not true. For a counterexample, take $p=2$, $k=10$, and $$\alpha=0.13113311133311113333\dots$$

Answer 2

In the case that $p=k=2$ there is no real, neither rational nor irrational, such that $\lfloor 2^n \alpha \rfloor$ is always odd. But for $\min (p,k) \gt 2$ where $k$ is a real number, integer or not, there are an uncountable number of integer sequences, missing all multiples of $p$ and obtainable as $\lfloor k^n \alpha \rfloor$ for some $\alpha.$ Each such sequence completely determines the corresponding $\alpha.$ Since there are uncountably many, some (really, most) are irrational.