A5. Find all positive integers $n \geqslant 2$ for which there exist $n$ real numbers $a_{1}0$ 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 $r^{1}, r^{2}, \ldots, r^{\frac{1}{2} n(n-1)}$.
Answer: $n \in\{2,3,4\}$.
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 $1^{3}-1-10\right)$ 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 $a_{1}1$ 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 $r^{b-(n-1)}-r^{3} / r^{n-1}1$ and $f(r)-r^{n}$ is convex, we have
$$
r^{6-1}-r^{b-2}>r^{b-2}-r^{b-3}>\ldots>r^{6-(n-3)}-r^{b-(n-2)} \text {, }
$$
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
$$
r^{a_{n-3}} \cdot(r-1)=r^{a_{n}-2}-r^{\alpha_{n-3}}=r^{b-(n-3)}-r^{b-(n-2)}-r^{b-(n-2)} \cdot(r-1),
$$
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:
$$
r^{b}=r^{b-(n-2)}+r^{b-(n-2)-1} \text { and } r^{b}-r^{b-(n-3)}+r^{b-(n-2)-3} \text {, }
$$
which can be rewritten as
$$
r^{n-1}-r+1 \text { and } r^{n+1}-r^{4}+1 .
$$

Simple algebra now gives
$$
r^{4}+1=r^{n+1}-r^{n-1} \cdot r^{2}-r^{3}+r^{2} \rightarrow(r-1)\left(r^{3}-r-1\right)=0 .
$$

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