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Let $\alpha>0$ and define $S(\alpha)=\{\lfloor n \alpha \rfloor: n\in\Bbb Z^+ \}$. (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 $\frac{1}{\alpha}+\frac{1}{\beta}=1$.

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.

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This was on the 1995 Putnam exam, according to a comment on an answer to OP's question at mathoverflow.net/questions/151533/… –  Gerry Myerson Dec 12 '13 at 5:24
    
1995 Putnam Problem B6 asks to prove the three sets cannot partition $\mathbb{Z}^{+}$, and this has easier solution. –  i707107 Dec 13 '13 at 10:00

1 Answer 1

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.

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