Is it true that there always exists a positive integer $n$ such that $p | ⌊k^n⋅α⌋$

Let $$k,M$$ be positive integers such that $$k−1$$ is not squarefree. Prove that there exist a positive real number $$α$$, such that $$⌊α⋅k^n⌋$$ and M are coprime for any positive integer $$n$$.

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

However, if $$α$$ must be irrational, then I have a feeling that there are no such $$α$$ 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 $$p | ⌊k^n⋅α⌋$$ ?

(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)

No, it is not true. For a counterexample, take $$p=2$$, $$k=10$$, and $$\alpha=0.13113311133311113333\dots$$
• Thank you for your answer. However,if M is not a prime number, will there exist a positive irrational number $\alpha$ such that $gcd(\lfloor \alpha \cdot k^n \rfloor , M)=1$ for every positive integer $n$ ? Jun 2, 2019 at 7:47
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.
• Thank you. But are you sure that $\alpha$ is always irrational ? Jun 2, 2019 at 8:32
• @apple If $k\geq 2$ you have at least two choices at each step so there are uncountably many numbers that can be constructed this way. Hence, a lot of them will be irrational. Jun 2, 2019 at 10:43
• @DmitryKrachun Thanks. But what if $p>k$? Jun 2, 2019 at 10:46