Find all integers $n \geqslant 2$ for which there exists a real number $r>0$ and $n$ real numbers $a_{1}<a_{2}<\cdots<a_{n}$ such that the $n(n-1) / 2$ differences $a_{j}-a_{i}$ obtained when $1 \leqslant i<j \leqslant n$ are equal, up to order, to the numbers $r^{1}, r^{2}, \ldots, r^{n(n-1) / 2}$.
In such a problem, it is essential to start by studying small cases.
$\triangleright$ When $n=2$, it suffices to choose $a_{1}=0$ and $a_{2}=r=1$. Therefore, $n=2$ works.
$\triangleright$ When $n=3$, it suffices to choose $a_{1}=0, a_{2}=r=\phi$ and $a_{3}=\phi^{3}$, where $\phi=(1+\sqrt{5}) / 2$ is the golden ratio and satisfies the equation $\phi^{2}=1+\phi$. Indeed, we then have $a_{2}-a_{1}=\phi$, $a_{3}-a_{2}=\phi\left(\phi^{2}-1\right)=\phi^{2}$, and $a_{3}-a_{1}=\phi^{3}$, as illustrated on the left below. Therefore, $n=3$ works.
$\triangleright$ When $n=4$, it suffices to choose $a_{1}=0, a_{2}=\lambda, a_{3}=\lambda^{4}, a_{4}=\lambda^{6}$ and $r=\lambda$, where $\lambda$ is a positive root of the polynomial $P(X)=X^{3}-X-1$, for which $P(1)0$ and $a_{1}a_{2}-a_{1}$, we even know that $r>1$. From now on, for each integer $k \leqslant t$, we associate the interval $\mathcal{I}(k)=\{i, i+1, \ldots, j-1\}$ for which $a_{j}-a_{i}=r^{k}$. When we split an interval $\mathcal{I}(k)$ into two subintervals $\mathcal{I}(\ell)$ and $\mathcal{I}\left(\ell^{\prime}\right)$, we immediately see that $r^{k}=r^{\ell}+r^{\ell^{\prime}}$.
Suppose now that there exists an integer $k$ for which the set $\mathcal{I}(k) \cup \mathcal{I}(k+1)$ is not an interval. By replacing each number $a_{i}$ with $-a_{i}$ if necessary, there exist four integers $u<v<w<x$ such that $r^{k}=a_{v}-a_{u}$ and $r^{k+1}=a_{x}-a_{w}$. Let $x, y$, and $z$ be the integers such that $r^{x}=a_{w}-a_{v}$, $r^{y}=a_{w}-a_{u}$, and $r^{z}=a_{x}-a_{v}$. Since $r^{y}=r^{k}+r^{x}<r^{k+1}+r^{x}=r^{z}$, we know that $z \geqslant y+1$, so $r^{k+1}+r^{x}=r^{z} \geqslant r^{y+1}=r^{k+1}+r^{x+1}$, which is absurd. Thus, the set $\mathcal{I}(k) \cup \mathcal{I}(k+1)$ is always an interval.
![](https://cdn.mathpix.com/cropped/2024_12_06_ea9148e0135587c25e52g-11.jpg?height=250&width=679&top_left_y=2228&top_left_x=694)

Moreover, the inverse bijection $\mathcal{I}^{-1}$ of $\mathcal{I}$ is increasing: when $\mathcal{I}(k) \subseteq \mathcal{I}(\ell)$, we know that $k \leqslant \ell$. Therefore, and by replacing each term $a_{i}$ with $-a_{i}$ if necessary, the intervals $\mathcal{I}(1)$ and $\mathcal{I}(2)$ are adjacent singletons, of the form $\{k\}$ and $\{k+1\}$.

Let $\ell$ be the largest integer for which $\mathcal{I}(\ell+1)=\{k+\ell\}$. The interval $\mathcal{I}(\ell+2)$ can contain, among its possible strict subintervals, only the singletons $\{k\},\{k+1\}, \ldots,\{k+\ell\}$. Furthermore, $\mathcal{I}(\ell+2)$ does not coincide with any of the singletons $\{k\},\{k+1\}, \ldots,\{k+\ell+1\}$. Since $\mathcal{I}(\ell+2) \cup \{k+\ell\}$ is an interval, $\mathcal{I}(\ell+2)$ contains one of the three integers $k+\ell-1, k+\ell$, and $k+\ell+1$. Therefore, there exists an integer $a$, equal to $\ell-2$ or $\ell-1$, such that $\mathcal{I}(\ell+2)=\{k+a, k+a+1\}$. We then denote $\ell^{\prime}$ the integer $\mathcal{I}^{-1}(\{k, k+1\})$. Since $r^{1}+r^{2}=r^{\ell^{\prime}} \geqslant r^{\ell}=r^{a+1}+r^{a+2}$, we know that $a=0$. Thus, $\mathcal{I}(\ell+2)=\{k, k+1\}$ and $\ell$ is equal to 1 or 2.
$\triangleright$ If $\ell=1$, then $r+r^{2}=r^{3}$. This situation is currently analogous to the case $n=2$.
Furthermore, the interval $\mathcal{I}(4)$ is a singleton $\{k-1\}$ or $\{k+2\}$, and its disjoint union with $\mathcal{I}(3)$ forms an interval $\mathcal{I}(m)$ equal to $\{k-1, k, k+1\}$ or $\{k, k+1, k+2\}$, for which $r^{m}=r^{3}+r^{4}=r^{2}\left(r+r^{2}\right)=r^{5}$. But then $\mathcal{I}(5)$ has 6 subintervals, which is impossible.
$\triangleright$ If $\ell=2$, then $r+r^{2}=r^{4}$. We immediately deduce that $r^{2}+r^{3}=r^{5}$ and $r^{3}+r^{4}=r^{6}$, so $\mathcal{I}(5)=\{k+1, k+2\}$ and $\mathcal{I}(6)=\{k, k+1, k+2\}$. This situation is currently analogous to the case $n=3$.
Thus, the interval $\mathcal{I}(7)$ is a singleton $\{k-1\}$ or $\{k+3\}$, and its disjoint union with $\mathcal{I}(6)$ forms an interval $\mathcal{I}(m)$ equal to $\{k-1, k, k+1, k+2\}$ or $\{k, k+1, k+2, k+3\}$, for which $r^{m}=r^{6}+r^{7}=r^{5}\left(r+r^{2}\right)=r^{9}$. Consequently, $\mathcal{I}(9)$ has 10 subintervals, which is impossible.
![](https://cdn.mathpix.com/cropped/2024_12_06_ea9148e0135587c25e52g-12.jpg?height=438&width=1715&top_left_y=1232&top_left_x=179)

In conclusion, no integer $n \geqslant 5$ works, and the solutions are therefore $n=2, n=3$, and $n=4$.