Integer parts of multiples of irrationals Let $\alpha>0$ and define 
$$S(\alpha)=\big\{\lfloor n \alpha \rfloor: n\in\Bbb Z^+ \big\}.$$ Here $\lfloor x\rfloor$ is the integer part of $x$ and $\mathbb Z^+$ the set of positive integers.
Question. Is it possible to find $\alpha,\beta,\gamma>0$, such that
$$
S(\alpha)\cap S(\beta)= S(\beta) \cap S(\gamma) =S(\alpha)\cap S(\gamma) = \varnothing\text{?}
$$
It is well-known that the following holds
$$
S(\alpha)\cap S(\beta)=\varnothing\quad\,\text{and}\,\quad S(\alpha)\cup S(\beta)=\mathbb Z^+
$$
if and only if $\alpha$ and $\beta$ are irrational (and positive) and $\,\,\dfrac{1}{\alpha}+\dfrac{1}{\beta}=1$.
Update 1. According to a 1995 Putnam examination question, there are no $\alpha$, $\beta$ and $\gamma$ for which $S(\alpha)$, $S(\beta)$ and $S(\gamma)$ form a partition of $\mathbb Z^+$. 
Update 2. The following result holds:
$S(\alpha)\cap S(\beta)=\varnothing$ if and only $\alpha,\beta$ are irrational and if there exist positive integers $m,n$, such that
$$
\frac{m}{\alpha}+\frac{n}{\beta}=1.
$$
This is Theorem 8 in:
Th. Skolem, On certain distributions of integers in pairs with given differences, Math. Scand., 5 (1957) 57-68.
 A: Using the known fact, we see that 
$S(\alpha)\cap S(\beta) = \varnothing $
If we have 
$$\frac{k}{\alpha}+\frac{l}{\beta}  =1 $$
for some positive integers $k, l$
Conversely, we use Kronecker's theorem ( If $1$, $\alpha$ and $\beta$ are linearly independent over $\mathbb{Q}$ then the set $\{((n\alpha), (n\beta)) : n \in \mathbb{Z}^{+}\}$ is dense in $[0,1]^2$) , and detailed analysis in case when they are linearly dependent over rationals, we obtain that the condition above for $\alpha$ and $\beta$ is indeed holds when $S(\alpha)\cap S(\beta)=\varnothing$. 
On the other hand, from the link above we can also deduce that if $\alpha$ rational, then $S(\alpha)\cap S(\beta)\neq \varnothing$ regardless $\beta$. 
Hence, we have 
(Theorem) $S(\alpha)\cap S(\beta) = \varnothing $ if and only if $\alpha$ and $\beta$ are irrational numbers satisfying $$\frac{k}{\alpha}+\frac{l}{\beta}  =1 $$
for some positive integers $k, l$
If $S(\alpha)\cap S( \beta)=\varnothing$, and $S(\beta)\cap S(\gamma)=\varnothing$, then 
there are positive integers $k_1, k_2, k_3, k_4$ such that 
$$\frac{k_1}{\alpha}+\frac{k_2}{\beta}=\frac{k_3}{\beta}+\frac{k_4}{\gamma}=1$$
If $k_2\neq k_3$, then we have 
$$\frac{k_1k_3}{\alpha}-\frac{k_2k_4}{\gamma}=k_3-k_2$$
In this case, $S(\alpha)\cap S(\gamma)\neq \varnothing$. 
If $k_2=k_3$, then we have 
$$\frac{k_1}{\alpha}=\frac{k_4}{\gamma}$$
This case also, $S(\alpha)\cap S(\gamma)\neq \varnothing$. 
Hence, we have proved that for real numbers $\alpha, \beta$ and $\gamma$, 
$$
S(\alpha)\cap S(\beta)= S(\beta) \cap S(\gamma) =S(\alpha)\cap S(\gamma) = \varnothing 
$$
is impossible. 
