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 $p | ⌊k^n⋅α⌋$ ?

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)


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 $$\alpha=0.13113311133311113333\dots$$

  • $\begingroup$ 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 $n$ ? $\endgroup$ – apple Jun 2 at 7:47
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    $\begingroup$ I only considered the second orange box in your post, because this was introduced by "so my question is". Please only ask one question in each MO post. That is, open a new MO post for this other question. $\endgroup$ – GH from MO Jun 2 at 14:05

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.

  • $\begingroup$ Thank you. But are you sure that $\alpha$ is always irrational ? $\endgroup$ – apple Jun 2 at 8:32
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    $\begingroup$ @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. $\endgroup$ – Dmitry Krachun Jun 2 at 10:43
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    $\begingroup$ @DmitryKrachun Thanks. But what if $p>k$? $\endgroup$ – apple Jun 2 at 10:46

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