Problem (Algebra):
Find all positive integers $n \geqslant 2$ for which there exist $n$ real numbers $a_{1}<\cdots<a_{n}$ and a real number $r>0$ such that the $\frac{1}{2} n(n-1)$ differences $a_{j}-a_{i}$ for $1 \leqslant i<j \leqslant n$ are equal, in some order, to the numbers $r^{1}, r^{2}, \ldots, r^{\frac{1}{2} n(n-1)}$.

Solution:
We first show a solution for each $n \in\{2,3,4\}$. We will later show the impossibility of finding such a solution for $n \geqslant 5$.

For $n=2$, take for example $\left(a_{1}, a_{2}\right)=(1,3)$ and $r=2$.

For $n=3$, take the root $r>1$ of $x^{2}-x-1=0$ (the golden ratio) and set $\left(a_{1}, a_{2}, a_{3}\right)=$ $\left(0, r, r+r^{2}\right)$. Then

$$
\left(a_{2}-a_{1}, a_{3}-a_{2}, a_{3}-a_{1}\right)=\left(r, r^{2}, r+r^{2}=r^{3}\right)
$$

For $n=4$, take the root $r \in(1,2)$ of $x^{3}-x-1=0$ (such a root exists because $1^{3}-1-1<0$ and $\left.2^{3}-2-1>0\right)$ and set $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=\left(0, r, r+r^{2}, r+r^{2}+r^{3}\right)$. Then

$$
\left(a_{2}-a_{1}, a_{3}-a_{2}, a_{4}-a_{3}, a_{3}-a_{1}, a_{4}-a_{2}, a_{4}-a_{1}\right)=\left(r, r^{2}, r^{3}, r^{4}, r^{5}, r^{6}\right)
$$

For $n \geqslant 5$, we will proceed by contradiction. Suppose there exist numbers $a_{1}<\cdots<a_{n}$ and $r>1$ satisfying the conditions of the problem. We start with a lemma:

Lemma. We have $r^{n-1}>2$.

Proof. There are only $n-1$ differences $a_{j}-a_{i}$ with $j=i+1$, so there exists an exponent $e \leqslant n$ and a difference $a_{j}-a_{i}$ with $j \geqslant i+2$ such that $a_{j}-a_{i}=r^{e}$. This implies

$$
r^{n} \geqslant r^{e}=a_{j}-a_{i}=\left(a_{j}-a_{j-1}\right)+\left(a_{j-1}-a_{i}\right)>r+r=2 r
$$

thus $r^{n-1}>2$ as desired.

To illustrate the general approach, we first briefly sketch the idea behind the argument in the special case $n=5$. In this case, we clearly have $a_{5}-a_{1}=r^{10}$. Note that there are 3 ways to rewrite $a_{5}-a_{1}$ as a sum of two differences, namely

$$
\left(a_{5}-a_{4}\right)+\left(a_{4}-a_{1}\right),\left(a_{5}-a_{3}\right)+\left(a_{3}-a_{1}\right),\left(a_{5}-a_{2}\right)+\left(a_{2}-a_{1}\right) .
$$

Using the lemma above and convexity of the function $f(n)=r^{n}$, we argue that those three ways must be $r^{10}=r^{9}+r^{1}=r^{8}+r^{4}=r^{7}+r^{6}$. That is, the "large" exponents keep dropping by 1 , while the "small" exponents keep increasing by $n-2, n-3, \ldots, 2$. Comparing any two such equations, we then get a contradiction unless $n \leqslant 4$.

Now we go back to the full proof for any $n \geqslant 5$. Denote $b=\frac{1}{2} n(n-1)$. Clearly, we have $a_{n}-a_{1}=r^{b}$. Consider the $n-2$ equations of the form:

$$
a_{n}-a_{1}=\left(a_{n}-a_{i}\right)+\left(a_{i}-a_{1}\right) \text { for } i \in\{2, \ldots, n-1\}
$$

In each equation, one of the two terms on the right-hand side must be at least $\frac{1}{2}\left(a_{n}-a_{1}\right)$. But from the lemma we have $r^{b-(n-1)}=r^{b} / r^{n-1}<\frac{1}{2}\left(a_{n}-a_{1}\right)$, so there are at most $n-2$ sufficiently large elements in $\left\{r^{k} \mid 1 \leqslant k<b\right\}$, namely $r^{b-1}, \ldots, r^{b-(n-2)}$ (note that $r^{b}$ is already used for $a_{n}-a_{1}$ ). Thus, the "large" terms must be, in some order, precisely equal to elements in

$$
L=\left\{r^{b-1}, \ldots, r^{b-(n-2)}\right\}
$$

Next we claim that the "small" terms in the $n-2$ equations must be equal to the elements in

$$
S=\left\{r^{b-(n-2)-\frac{1}{2} i(i+1)} \mid 1 \leqslant i \leqslant n-2\right\}
$$



in the corresponding order (the largest "large" term with the smallest "small" term, etc.). Indeed, suppose that

$$
r^{b}=a_{n}-a_{1}=r^{b-i}+r^{\alpha_{i}} \text { for } i \in\{1, \ldots, n-2\},
$$

where $1 \leqslant \alpha_{1}<\cdots<\alpha_{n-2} \leqslant b-(n-1)$. Since $r>1$ and $f(r)=r^{n}$ is convex, we have

$$
r^{b-1}-r^{b-2}>r^{b-2}-r^{b-3}>\ldots>r^{b-(n-3)}-r^{b-(n-2)},
$$

implying

$$
r^{\alpha_{2}}-r^{\alpha_{1}}>r^{\alpha_{3}}-r^{\alpha_{2}}>\ldots>r^{\alpha_{n-2}}-r^{\alpha_{n-3}} .
$$

Convexity of $f(r)=r^{n}$ further implies

$$
\alpha_{2}-\alpha_{1}>\alpha_{3}-\alpha_{2}>\ldots>\alpha_{n-2}-\alpha_{n-3}
$$

Note that $\alpha_{n-2}-\alpha_{n-3} \geqslant 2$ : Otherwise we would have $\alpha_{n-2}-\alpha_{n-3}=1$ and thus

$$
r^{\alpha_{n-3}} \cdot(r-1)=r^{\alpha_{n-2}}-r^{\alpha_{n-3}}=r^{b-(n-3)}-r^{b-(n-2)}=r^{b-(n-2)} \cdot(r-1)
$$

implying that $\alpha_{n-3}=b-(n-2)$, a contradiction. Therefore, we have

$$
\begin{aligned}
\alpha_{n-2}-\alpha_{1} & =\left(\alpha_{n-2}-\alpha_{n-3}\right)+\cdots+\left(\alpha_{2}-\alpha_{1}\right) \\
& \geqslant 2+3+\cdots+(n-2) \\
& =\frac{1}{2}(n-2)(n-1)-1=\frac{1}{2} n(n-3) .
\end{aligned}
$$

On the other hand, from $\alpha_{n-2} \leqslant b-(n-1)$ and $\alpha_{1} \geqslant 1$ we get

$$
\alpha_{n-2}-\alpha_{1} \leqslant b-n=\frac{1}{2} n(n-1)-n=\frac{1}{2} n(n-3),
$$

implying that equalities must occur everywhere and the claim about the small terms follows.

Now, assuming $n-2 \geqslant 2$, we have the two different equations:

$$
r^{b}=r^{b-(n-2)}+r^{b-(n-2)-1} \text { and } r^{b}=r^{b-(n-3)}+r^{b-(n-2)-3}
$$

which can be rewritten as

$$
r^{n-1}=r+1 \quad \text { and } \quad r^{n+1}=r^{4}+1
\tag{1}
$$

Simple algebra now gives

$$
r^{4}+1=r^{n+1}=r^{n-1} \cdot r^{2}=r^{3}+r^{2} \Longrightarrow(r-1)\left(r^{3}-r-1\right)=0 .
$$

Since $r \neq 1$, using Equation (1) we conclude $r^{3}=r+1=r^{n-1}$, thus $n=4$, which gives a contradiction.

Final answer: 2,3,4