It is enough to solve the problem when all the numbers in $S$ are powers of 2; taking logs; we get the following problem. Quote: Let $S$ be a set of $n$ nonnegative integers. Prove that they may be ordered $a_1, \dots, a_n$ such that $a_i - a_j \neq a_j - a_k$ for all $i < j < k$. It is enough to solve the case where $S = \{0, \dots, 2^m - 1\}$ (pick $m$ large and remove the unwanted numbers). This is easy by induction on $m$: base cases are vacuous, and if $a_0, \dots, a_{2^m - 1}$ is a valid arrangement for $m$ it can be easily seen that \[2a_0, \dots, 2a_{2^m - 1}, 2a_0 + 1, \dots, 2a_{2^m - 1} + 1\]is a valid arrangement for $m+1$. Done.

CantonMathGuy wrote: It is enough to solve the problem when all the numbers in $S$ are powers of 2; This is true, but not that trivial IMHO I guess what you wanted to do was the following. (1) Look at the prime $2$. "Pack up" the integers with the same $v_2$ and order the packages as done above. (2) Now the "bad cases" where $f(j)^2 = f(i)f(k)$ must happen when $f(i), f(j), f(k)$ belongs to the same package. (3) Now, look at the next prime. Pack them up again, and sort them as done above. Repeat. This terminates finitely, and it removes all "bad cases" as there are no duplicate elements in $S$.

A solution at the contest We use induction on $n$. Assume that there does not exist some bijection $f$ for some $n$ and $S$. Let $S$ be the one with minimum $\sum_{a \in S} a$ among these $S$. Construct a graph $G=(V,E)$ with $V=S$, $E=\{(x,y) | xy=z^2 \text{ for some } z \in S \}$ If $G$ is not connected, then use the induction hypothesis on the connected components and combine them. Or, if $G$ is connected, then take any $x,y \in S$ and consider a path(with non-repeated vertices) $x e_1 z_1 e_2 z_2 \cdots z_{t-1} e_t y$, where $e_i$ satisfies $e_i^2=z_iz_{i+1}$. Then, $xz_1^2z_2^2 \cdots z_{t-1}^2y=(e_1 \cdots e_t)^2$, so $xy=\frac{(e_1 \cdots e_t)^2}{(z_1z_2 \cdots z_{t-1})^2}$. Note that $xy$ is a positive integer, so it is a perfect square. So for every $x,y \in S$, $xy$ is a perfect square. Hence there exists some $a$ such that $\sqrt{\frac{x}{a}}$ is a integer for all $x \in S$, and replacing $S$ by $\{\sqrt{\frac{x}{a}} | x \in S \}$ contracts the minimality of $S$.

Does this hold for all distinct real numbers?

Sketch : We use this lemma: Let $n$ is positive integer and $X=[0,1,2,...n]$. And element of $X$ may be ordered $a_1, \dots, a_n$ such that $a_i - a_j \neq a_j - a_k$ for all $i < j < k$. We take prime numbers $p_1, p_2, .... $ which divides any number in $S$. At first, we select $p_1=p$. We will divide $S$ to $X_0, X_1 ,... $ $X_k=[p^k*m | gcd(m,p)=1] $ And we can rearrange $X_0, X_1 ,... $ by (lemma). In $X_k$ , select $p_2$. And for $X_k$, divide and rearrange. ....... Sorry for my poor Eng

The quotation in #2 also holds for any distinct complex numbers.

liekkas wrote: The quotation in #2 also holds for any distinct complex numbers. Could you share your proof?

Let $n$ complex numbers be $z_1,\cdots,z_n$. Consider the maximal linearly independent group, WLOG $z_1,\cdots,z_k$, by the definition, for any $1 \le i \le k$, there exist rational numbers $q_{ij}$ such that $z_i=\sum_{j=1}^{k} q_{ij}z_j$. From $z_1,\cdots,z_k$ are linear independent, we conclude that $z_t+z_s=2z_r$, $1 \le t <r<s \le k$ is equivalent to $q_{tj}+q_{sj}=2q_{rj}$, $\forall 1 \le j \le k$. So it suffices to have at least one $j$ that fails the equation. Now we use induction to show this. $k=1$ is just the case of integers (because we can multiply an integer to all rationals to make them integers). Suppose it is true for $k-1$, consider $k$. Let $z_i$ corresponding to k-tuple $(q_{i1},\cdots,q_{ik})$. Let $c_1,\cdots,c_m$ be all the possible values in $q_{11},\cdots,q_{n1}$.And $D_j=\{(q_{i2},\cdots,q_{ik} \mid q_{i1}=c_j \}$. First we arrange $c_1,\cdots,c_m$ as $k=1$ case, and WLOG this arrangement is still $c_1,\cdots,c_m$. By the induction hypothesis, we can arrange $D_1,\cdots,D_m$ as wanted, then we extend it to $k-th$ division. Now if for some $1 \le t <r<s \le k$ one have $q_{tj}+q_{sj}=2q_{rj}$. Note that if $t,s$ are in same $D_j$, by the induction hypothesis, it fails for some $2 \le j \le k$. If not, it fails for $j=1$. The induction is complete.

Let $S$ be a set of $n$ nonnegative integers. Prove that they may be ordered $a_1, \dots, a_n$ such that $a_i - a_j \neq a_j - a_k$ for all $i < j < k$. After this step, you can induct! Let the first half of your sequence be odd and the second half even. Then, all arithmetic sequences have to be all odd or all even. In either case, if you position the numbers, (2x, 2y, 2z) or (2x+1, 2y+1, 2z+1) such that x, y, z do not form an arithmetic sequence (which we know can do by induction), we are done yay!

it has something similar with Iranian TST 2015 second exam p5......something really silimar.