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 \(\tfrac{1}{2}n(n-1)\) differences \(a_j-a_i\) for \(1 \leqslant i<j \leqslant n\) are equal, in some order, to the numbers \(r^1,r^2,\ldots,r^{\frac{1}{2}n(n-1)}\).
Answer: \(n \in\{2,3,4\}\).
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\[r^{n} \geqslant r^{e}=a_{j}-a_{i}=\left(a_{j}-a_{j-1}\right)+\left(a_{j-1}-a_{i}\right)>r+r=2r\]
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\[S=\left\{\left.r^{b-(n-2)-\frac{1}{2} i(i+1)} \right\rvert\, 1 \leqslant i \leqslant n-2\right\}\]
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\[\begin{equation*} r^{n-1}=r+1 \quad \text { and } \quad r^{n+1}=r^{4}+1 \tag{1} \end{equation*}\]
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.