SAUDI ARABIAN IMO Booklet 2023

The answer are $n\in \{2,3,4\}$. We first show a solution for each $n\in \{2,3,4\}$. We will later show the impossibility of finding such a solution for $n\ge 5$.

For $n=2$, take for example $(a_1,a_2)=(1,3)$ and $r=2$.

For $n=3$, take the root $r>1$ of $x^3-x-1=0$ (the golden ratio) and set $(a_1,a_2,a_3)=(0,r,r+r^2)$, then $$(a_2-a_1,a_3-a_2,a_3-a_1)=(r,r^2,r+r^2=r^3).$$

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 $2^3-2-1>0$) and set $(a_1,a_2,a_3,a_4)=(0,r,r+r^2,r+r^2+r^3)$, then $$(a_2-a_1,a_3-a_2,a_4-a_3,a_3-a_1,a_4-a_2,a_4-a_1)=(r,r^2,r^3,r^4,r^5,r^6).$$

For $n\ge 5$, we will proceed by contradiction. Suppose that there exist numbers $a_1<a_2<\cdots <a_n$ 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 $e\le n$ and a difference $a_j-a_i$ with $j\ge i+2$ such that $a_j-a_i=r^e$. This implies that $$r^n\ge r^e=a_j-a_i=(a_j-a_{j-1})+(a_{j-1}-a_i)>r+r=2r,$$ 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 $$(a_5-a_4)+(a_4-a_1),(a_5-a_3)+(a_3-a_1),(a_5-a_2)+(a_2-a_1).$$ 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 $n-2,n-3,\dots ,2$. Then comparing any two such equations to get a contradiction unless $n\le 4$.

Now we go back to the full proof for any $n\ge 5$. 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: $$a_n-a_1=(a_n-a_i)+(a_i-a_1)\text{ for }i\in \{2,\dots ,n-1\}.$$ In each equation, one of the two terms on the right-hand side must be at least $\frac{1}{2}(a_n-a_1)$. But from the lemma we have $$r^{b-(n-1)}=r^b/r^{n-1}<\frac{1}{2}(a_n-a_1),$$ so there are at most $n-2$ sufficiently large elements in $\{r^k|1\le k<b\}$, namely $r^{b-1},\dots ,r^{b-(n-2)}$ (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 $$L=r^{b-1},\dots ,r^{b-(n-2)}.$$ Next we claim that the “small” terms in the $n-2$ equations must be equal to the elements in $$S=\{r^{b-(n-2)-\frac{1}{2}i(i+1)}|1\le i\le n-2\},$$ in the corresponding order (the largest “large” term with the smallest “small” term, etc.). Indeed, suppose that $$r^b=a_n-a_1=r^{b-i}+r^{{\alpha }_i}\text{ for }i\in \{1,\dots ,n-2\},$$ where $1\le {\alpha }_1<\cdots <{\alpha }_{n-2}\le b-(n-1)$. Since $r>1$ and $f(r)=r^n$ is convex, we have $$r^{b-1}-r^{b-2}>r^{b-2}-r^{b-3}>\cdots >r^{b-(n-3)}-r^{b-(n-2)},$$ implying $r^{{\alpha }_2}-r^{{\alpha }_1}>r^{{\alpha }_3}-r^{{\alpha }_2}>\cdots >r^{{\alpha }_{n-2}}-r^{{\alpha }_{n-3}}$. Convexity of the function $f(r)=r^n$ further implies $${\alpha }_2-{\alpha }_1>{\alpha }_3-{\alpha }_2>\cdots >{\alpha }_{n-2}-{\alpha }_{n-3}.$$ Note that ${\alpha }_{n-2}-{\alpha }_{n-3}\ge 2$: 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 $$\begin{array}{rl}{\alpha }_{n-2}-{\alpha }_1& =({\alpha }_{n-2}-{\alpha }_{n-3})+\cdots +({\alpha }_2-{\alpha }_1)\\ & \ge 2+3+\cdots +(n-2)=\frac{1}{2}(n-2)(n-1)-1=\frac{1}{2}n(n-3).\end{array}$$ On the other hand, from ${\alpha }_{n-2}\le b-(n-1)$ and ${\alpha }_1\ge 1$ we get $${\alpha }_{n-2}-{\alpha }_1\le b-n=\frac{1}{2}n(n-1)-n=\frac{1}{2}n(n-3),$$ implying that equalities must occur everywhere and the claim about the small terms follows. Now assuming $n-2\ge 2$, 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 $$r^{n-1}=r+1\text{ and }{r}^{n+1}=r^4+1.$$ Simple algebra now gives $$r^4+1=r^{n+1}=r^{n-1}\cdot r^2=r^3+r^2⟶(r-1)(r^3-r-1)=0.$$ Since $r\ne 1$, using (1) we conclude $r^3=r+1=r^{n-1}$, thus $n=4$, which gives a contradiction. □