sqing wrote: For which real numbers $k > 1$ does there exist a bounded set of positive real numbers $S$ with at least $3$ elements such that $$k(a - b)\in S$$for all $a,b\in S $ with $a > b?$ Remark: A set of positive real numbers $S$ is bounded if there exists a positive real number $M$ such that $x < M$ for all $x \in S.$ 1) The set has no accumulation point If the set has such an accumulation point, we can always find $a-b$ positive and as small as we want and since $k(a-b)\in S$, we get $\inf(S)=0$ Since bounded, $\exists M=\sup(S)$ Then choosing $a$ as close as we want of $M$ and $b$ as close as we want of $0$, we get a member of S as close as we want of $kM$, which is impossible since $k>1$ and $M=\sup(S)$ Q.E.D 2) $|S|$ is finite This is an immediate consequence of "no accumulation point" and "bounded" Q.E.D 3) Only solution with $|S|=3$ and $k=2$.* or $k=\frac{1+\sqrt 5}2$ Let $0<a_1<a_2<a_3<... <a_n$ the elements of $S$ where $|S|=n\ge 3$ finite. So the $n-1$ numbers $k(a_2-a_1),k(a_3-a_1),k(a_4-a_1),...,k(a_n-a_1)$ are distinct, increasing and members of $S$. And so they are $a_1,a_2,..,a_{i-1},a_{i+1},a_{i+2},...,a_n$ for some $i\in\{1,2,....,n\}$ If $i\le n-2$, then the two last numbers are : $k(a_{n-1}-a_1)=a_{n-1}$ and $k(a_n-a_1)=a_n$ Which is impossible since this would mean $a_{n-1}=a_n=\frac{ka_1}{a-1}$ So $i=n-1$ or $i=n$ As a consequence, we have : $k(a_2-a_1)=a_1$ $k(a_3-a_1)=a_2$ ... $k(a_{n-1}-a_1)=a_{n-2}$ And either $k(a_{n}-a_1)=a_{n-1}$, either $k(a_{n}-a_1)=a_{n}$ And so $(a_1,a_2,....,a_{n-1})=\left(a_1,a_1(1+\frac 1k),a_1(1+\frac 1k+\frac 1{k^2}),...,a_1(1+\frac 1k+...+\frac 1{k^{n-2}})\right)$ And either $a_n=a_1(1+\frac 1k+...+\frac 1{k^{n-1}})$, either $a_n=\frac{ka_1}{k-1}$ If $n>3$, taking in above $k(a_3-a_2)$, we get $\frac{a_1}k\in(a_1,a_2)$ and so not in $S$ If $n=3$ and $(a_1,a_2,a_3)=(a_1,a_1(1+\frac 1k),a_1(1+\frac 1k+\frac 1{k^2})$, then again $k(a_3-a_2)\notin S$ If $n=3$ and $(a_1,a_2,a_3)=(a_1,a_1(1+\frac 1k),\frac{ka_1}{k-1})$, then $k(a_3-a_2)=\frac{a_1}{k-1}$ And so : either $\frac{a_1}{k-1}=a_1$ which means $k=2$ either $\frac{a_1}{k-1}=a_1(1+\frac 1k)$ which means $k=\frac{1+\sqrt 5}2$ And so two solutions : $\boxed{k=2}$ and for example $S=\{2,3,4\}$ $\boxed{k=\frac{1+\sqrt 5}2}$ and for example $S=\{2,1+\sqrt 5,3+\sqrt 5\}$

Thank you very much.

At first we can prove that S is finite set. To make contradiction, we will verify that the infimum of s is obtainable. Donate $inf:S = m$ , which mean that for any $\epsilon$ , there exist $x \in S$ , such that $x-m < \epsilon$. Suppose m is not obtainable (may say $m \not \in S$) , this is equivalent that there is an infinite strictly declining sequence $\{a_n\}_{n\geq 0}$ , such that $a_i \in S$ together with $a_i-m<\frac{1}{2^i}$. By the condition $k(a_i-a_{i+1})\in S$, together with the apparent fact that the value of progression $\{a_i-a_{i+1}\}$ can be arbitrarily close to $0^+$ thus we have $m = 0$ , which is contradiction because if we set the superb of S as $c$ and we have an element $u$ in S such that $0\leq c-u < \epsilon$ and $0<v<\epsilon $ , and as the condition goes , $k(u-v)\in S$ , yet $k(u-v) >k (c-2\epsilon)$. If we set $\epsilon < \frac{k-1}{2k}c$ , we have $k(u-v) > c$ , contradicting! So S has a precise bottom. On top of this, since regarding any two consecutive element(in amount) , their difference must be greater than $\frac{m}{k}$ and $S$ is a bounded set , we have $S$ is a finite set. Assume $S = \{a_1<a_2<a_3...<a_n\} , n\geq 3$ Pay attention that the difference set of $S$ contain at least $\{a_2-a_1,a_3-a_1,a_n-a_1\}$ , which means that merely one element of $S$ cannot be presented in the way as $k(a_i-a_1)(2\leq i\leq n)$ It is that there is an integer $r$(we call $r=n+1$ the same meaning as r is not exist) , such that $k(a_2-a_1)=a_1$ $k(a_3-a_1)=a_2$ $\cdot \cdot \cdot$ $k(a_{r-1}-a_1)=a_{r-2}$ $k(a_r-a_1)=a_r$ $\cdot \cdot\cdot$ $k(a_n-a_1)=a_n$ Observe that for any $k\geq r,$ $a_k =\frac{k}{k-1}a_1$ , a constant. So $r=n \ or=n+1$ $1) r \geq 3$ It is easily got that $a_i=\frac{k^{i-1}+k^{i-2}+...+1}{k^{i-1}}a_1$ but $a_3-a_2=\frac{k+1}{k^2}a_1<a_2$ , thus $k+1=k^2$, implying $k=\frac{1+\sqrt{5}}{2}$. It is feasible, by setting $S=\{ 2,1+\sqrt5 , 3+\sqrt5\}$ $2) r = 2$ we have $a_2=\frac{k+1}{k}a_1 , a_3 = \frac{k}{k-1}a_1$ , easy to deduce $k=2 \ or\ k =\frac{1+\sqrt{5}}{2}$ For $k=2$ , there is an example $\{2,4,6\}$