China Mathematical Olympiad

Call a set $S$ "good" if it satisfies the property as stated in the problem.

(1) If $S$ has only one element, $S$ is "good".

(2) Now we can assume that $S$ contains at least two elements. Let $$d=\min \{|m-n|:m,n\in S,m\ne n\}.$$ Then there is an integer $a$, such that $a+d,a+2d\in S$. Note that $$a+4d=3(a+2d)-2(a+d)\in S,$$ $$a-d=3(a+d)-2(a+2d)\in S,$$ $$a+5d=3(a+d)-2(a-d)\in S,$$ $$a-2d=3(a+2d)-2(a+4d)\in S.$$ So we have proved that if $$a+d,a+2d\in S,$$ then $$a-2d,a-d,a+4d,a+5d\in S.$$ Continuing this procedure, we can deduce that $$\{a+kd\mid k\in \mathbb{Z},3\nmid k\}\subseteq S.$$ Let $S_0=\{a+kd\mid k\in \mathbb{Z},3\nmid k\}$, it is easy to verify that $S_0$ is "good".

(3) Now we have proved that $S_0\subseteq S$. If $S\ne S_0$, pick a number $b\in S\setminus S_0$, then there exists an integer $l$, such that $a+ld\le b<a+(l+1)d$. Since at least one of $l$ and $l+1$ is indivisible by 3, we know that at least one of $a+ld,a+(l+1)d$ is contained in $S_0$. If $a+ld\in S_0$, note that $0\le b-(a+ld)<d$, by the definition of $d$, we must have $b=a+ld$. If $a+(l+1)d\in S_0$, note that $$0<|a+(l+1)d-b|\le d,$$ by the definition of $d$, we also have $b=a+ld$. In both cases, we have proved that there is a number $b\in S\setminus S_0$ of the form $b=a+ld$. This $l$ must be divisible by 3, hence $$a+ld,a+(l+1)d,a+(l+2)d\in S,$$ $$a+(l-2)d,a+(l-1)d\in S.$$ So $$a+(l+3)d=3(a+(l+1)d)-2(a+ld)\in S,$$ $$a+(l-3)d=3(a+(l-1)d)-2(a+ld)\in S.$$ Continuing this procedure, we have $a+(l+3j)d\in S$, $j\in \mathbb{Z}$, which implies that $\{a+kd\mid k\in \mathbb{Z}\}\subseteq S$. We claim that $S=\{a+kd\mid k\in \mathbb{Z}\}$, since for any number $x\notin \{a+kd\mid k\in \mathbb{Z}\}$, there is an element in $y\in \{a+kd\mid k\in \mathbb{Z}\}$ such that $0<|x-y|<d$. By the definition of $d$, we must have $x\notin S$.

So $S=\{a+kd\mid k\in \mathbb{Z}\}$, and it is easy to verify that such $S$ is "good".

We conclude that there are three classes of "good" sets: (1) $S=\{a\}$; (2) $S=\{a+kd\mid k\in \mathbb{Z},3\nmid k\}$; (3) $S=\{a+kd\mid k\in \mathbb{Z}\}$, here $a,d\in \mathbb{Z},d>0$.