Such a configuration only exists for $\fbox{n = 2,3,4}$. Construction For $n=2$, just take $r=1$ and $a_1, a_2$ such that $a_2-a_1=1$. For $n=3$, we take $r = \varphi$, so that $r^2 = 1+r$. Let $a_2 - a_1 = r$ and $a_3-a_2 = r^2$, so that $a_3 - a_1 = r+r^2 = r(1+r) = r^3$. For $n=4$, we take $r>0$ such that $r^3 = 1+r$, which exists by the intermediate value theorem. Let $a_2-a_1 = r$, $a_3-a_2 = r^2$ and $a_4-a_3 = r^3$, so that \begin{align*} a_3 - a_1 &= r+r^2 = r(1+r) = r^4,\\ a_4 - a_2 &= r^2+r^3 = r^2(1+r) = r^5,\\ a_4 - a_1 &= (a_4 - a_3) + (a_3 - a_1) = r^3 + r^4 = r^3(1+r) = r^6. \end{align*} Proof that $n>4$ do not work Now suppose a configuration as in the problem exists for a certain $n>4$. For the sake of readability, we write $S = \tfrac12 n(n-1)$. By scaling the sequence by a factor $r^{-(1+S)}$, we obtain a sequence for $r^{-1}$. Thus, we can assume w.l.o.g. that $r>1$, and hence that $r^i < r^j$ for $i<j$. Note that there are $n-1$ differences of the form $a_{i+1} - a_i$, so there must be an exponent $b\leq n$ such that $r^b = a_j - a_i$ with $j - i > 1$. By writing $a_j - a_i = (a_j-a_{i+1}) + (a_{i+1} - a_i)$, we see that $r^b = r^c + r^d$ for certain distinct $c,d$. This yields \[ r^n \geq r^b = r^c + r^d \geq r+r^2, \]hence $1+r \leq r^{n-1}$. Observe that $a_n-a_1$ is the largest difference, so $a_n-a_1$ must be $r^S$. For $1 < i < n$, we have $a_n = a_1 = (a_n - a_i) + (a_i - a_1)$, so $r^S$ can be written as $r^b + r^c$ with $b<c$ in at least $n-2$ different ways. If $c \leq S-n+1$, then $r^b + r^c \leq r^{S-n} + r^{S-n+1} = r^{S-n}(1+r) = r^{S-n}\cdot r^{n-1} = r^{S-1} < r^S$, contradiction. Thus, we have that $S-n+2 \leq c \leq S-1$. This leaves exactly $n-2$ possibilities for $c$, so all of then must occur. In particular, we have $r^b + r^{S-n+2} = r^S$ for a certain $b \leq S-n+1$. Now we see that: \[ r^S = r^b + r^{S-n+2} \leq r^{S-n+1} + r^{S-n+2} = r^{S-n+1}(1+r) \leq r^{S-n+1}\cdot r^{n-1} = r^S. \]We must have equality everywhere, so in particular, $1+r = r^{n-1}$. Now we see that, if $b\leq n-1$, then $r^b \leq r^{n-1} < r^n = r+r^2$, so $r^b$ cannot be of the form $a_j-a_i$ with $j-i > 1$. This means that the $n-1$ differences $a_{i+1}-a_i$ must be $r^1, r^2, \ldots, r^{n-1}$ in some order. In particular, we have $r^1 + r^2 + \cdots + r^{n-1} = a_n - a_1 = r^S$. Lemma. For $1\leq k \leq n-2$, we have $\sum_{j=0}^k r^j \leq r^{n-k}\cdot\sum_{j=0}^{k-1} r^j$, with equality iff $k=1$ or $k=n-2$. Proof of Lemma. We have \begin{align*} \sum_{j=0}^k r^j = r^{n-1} + \sum_{j=2}^k r^j \leq r^{n-1} + r^{n-2-k}\cdot\sum_{j=2}^k r^j = r^{n-1} + \sum_{j=n-k}^{n-2} r^j = \sum_{j=n-k}^{n-1} r^j = r^{n-k}\cdot\sum_{j=0}^{k-1}r^j. \end{align*}We have equality iff either $r^{n-2-k} = 1$, i.e., $k=n-2$, or the sum $\sum_{j=2}^k r^j$ is empty, i.e., $k=1$. This proves the Lemma. By applying the Lemma for $k=n-2, n-3, \ldots, 1$, we see that: \begin{align*} r^1 + r^2 + \cdots + r^{n-1} &= r\cdot (1+r+\cdots + r^{n-2})\\ &\leq r\cdot r^2\cdot (1+r+\cdots+r^{n-3})\\ &\leq \cdots\\ &\leq r\cdot r^2 \cdot \cdots\cdot r^{n-2}\cdot(1+r)\\ &\leq r\cdot r^2 \cdot \cdots\cdot r^{n-2}\cdot r^{n-1} \cdot 1\\ &= r^S. \end{align*}We have equality only in the first and last of these $n-2$ inequalities. Since $n-2>2$, at least one of the inequalities is strict, which is a contradiction. We conclude that the sequence cannot exist for $n>4$.

Solved with pikapika007, OronSH. The answer is $\boxed{n \in \{2,3,4\}}$. To see that $n=2$ works, take $r = 1$, $a_2 = 1, a_1 = 0$. To see that $n=3$ works, take $r = \phi = \frac{1 + \sqrt{5}}{2}$, $a_1 = 0, a_2 = \phi, a_3 = \phi^2 + \phi$. This works because $a_2 - a_1 = \phi, a_3 - a_2 = \phi^2$, and $a_3 - a_1 = \phi(\phi + 1) = \phi^3$. To see that $n=4$ works, let $r$ be a root of $x^3 - x - 1$ in the interval $(1,2)$ (which exists by IVT), $a_1 = 0, a_2 = r, a_3 = r + r^2, a_4 = r + r^2 + r^3$. This works because $a_2 - a_1 = r, a_3 - a_2 = r^2, a_4 - a_3 = r^3, a_3 - a_1 = r + r^2 = r(r+1) = r^4, a_4 - a_2 = r^2 + r^3 = r^2(r+1) = r^5$ and $a_4 - a_1 = r(r^2 + r + 1) = r(r^3 + r^2) = r^3(r + 1) = r^6$. Now it remains to show that $n > 4$ fail. Suppose some $n>4 $ worked. Let $k = \frac{n(n-1)}{2}$. If $r$ works, then we can multiply the sequence by $r^{-1-k}$ so that $\frac{1}{r}$ also works, so we can wlog assume that $r >1 $. Claim: $r^{n-1} \le r + 1$. Proof: Note that $a_n - a_1$ is the largest difference, so it must be equal to $r^k$. We can write $r^k$ as the sum of $r^i + r^j$ for $0<i < j<k$ in $n-2$ different ways because \[a_n - a_1 = (a_n - a_2) + (a_2 - a_1) = (a_n - a_3) + (a_3 - a_1) = \cdots = (a_n - a_{n-1}) + (a_{n-1} - a_1)\]This implies that there are two $i,j$ with $r^k = r^i + r^j$ such that $i,j$ are both at most $k - n + 2$ since we have $n-2$ different sums and there are only $n-3$ numbers from $k - n + 3$ to $k-1$, and each number cannot be used in a sum more than once (because $r^a \ne r^b\forall a\ne b$). Therefore using these $i,j$, we find that $r^k \le r^{k - n + 1} + r^{k - n + 2}$, so dividing both sides by $r^{k - n + 1}$, we get $r^{n - 1} \le r + 1$, as desired $\square$. Claim: $r^{n-1} \ge r + 1$. Proof: Suppose this was not the case. Notice that if $j > i + 1$, then $a_j - a_i$ can be written as $r^m + r^n$ for some distinct integers $0<m , n < k$ since $a_j - a_i = (a_j - a_{i+1}) + (a_{i+1} - a_i)$. The number of values of $a_{i+1} - a_i$ is equal to $n-1$. These are the only powers of $r$ from $r^1$ to $r^k$ that can't be written as the sum of two other distinct powers of $r$ greater than $1$. If something can be written as the sum of two other distinct powers of $r$ (not equal to $1$), then it must be at least $r^2 + r$. However, we have \[r<r^2<\cdots < r^n = r\cdot r^{n-1} < r^2 + r,\]this means that there are $n$ powers of $r$ (between $r$ and $r^k$) that cannot be written as $r^m + r^n$ for integers $m,n>0$, but this is a contradiction as there are only $n-1$ values of $a_{i+1} - a_i$ for some $i$. $\square$ Therefore by these two claims, we have $r^{n-1} = r + 1$. Now, we know that $r, r^2, \ldots, r^{n-1}$ cannot be written as the sum of two distinct powers of $r$, so they must be in some order equal to the set $\{(a_2 - a_1, a_3- a_2, \ldots, a_n - a_{n-1}\}$. This implies that their sum is equal to $a_n - a_1 = r^k$. Therefore we have the equations \[r^{n+1} = r + 1 \text{ and } r + r^2 + \cdots + r^{n-1} = r^k\] Two ways to finish here: Way 1: For $n > 4$, note that \begin{align*} r + r^2 + \cdots + r^{n-1} = r(1 + r + \cdots + r^{n-2}) \\ = r(r^2 + r^3 + \cdots + r^{n-1}) = r^3(1 + r + \cdots + r^{n-3} )\\ = r^3(r^2 + r^3 + \cdots + r^{n-3} + r^{n-1}) \le r^3( r^3 + r^4 + \cdots + r^{n-1}) \\ = r^6(1 + r + \cdots + r^{n-4}) \le r^6( r^4 + r^5 + \cdots + r^{n-1}) \\ \cdots \le r^{(n-2)(n-1)/2} (1 + r) = r^k \\ \end{align*} Now, the inequality $r^3(r^2 + r^3 + \cdots + r^{n-3} + r^{n-1}) \le r^3( r^3 + r^4 + \cdots + r^{n-1})$ is strict because $r^{n-2} - r^2 >0$, so we have \[ r + r^2 + \cdots + r^{n-1} < r^k ,\]contradiction. Way 2: The proof for $n>8$ is as follows: We have $1 = r^{n-1} - r = (r^{n-1} - r^{n-2}) + (r^{n-2} - r^{n-3}) + \cdots + (r^2 - r)$. Now, $(r^{n-1} - r^{n-2})$ is the largest term in the sum, so it must be greater than $\frac{1}{n-2}$. Multiplying both sides by $r^{\frac{n^2 - 3n}{2}}$ gives \[ r^{\frac{n^2 - n - 2}{2}} - r^{\frac{n^2 - n - 4}{2}} > \frac{r^{\frac{n^2 - 3n}{2}} } {n-2} > \frac{r^{\frac{n^2 - 4n + 3}{2}}}{n-2} = \frac{(r+1)^{\frac{n-3}{2}}}{n-2} > 1\]since $r+1>2$ and $2^{\frac{n-3}{2}} > n - 2$. Therefore, we have $r^{k - 1} - r^{k-2} >1 $, so multiplying both sides by $\frac{r}{r-1}$ gives that \[r^{k-1} > \frac{r}{r-1} = \frac{r^{n-1} - 1}{r-1} = 1 + r + \cdots + r^{n-2}\]Multiplying both sides by $r$ gives that \[ r + r^2 + \cdots + r^{n-1} < r^k ,\]contradiction. For $n = 5$, we have \[r + r^2 + r^3 + r^4 = r^5(r^2 + 1) < r^5 (r^2 + r) < r^{10} ,\]absurd. For $n = 6$, we have \[r + r^2 + r^3 + r^4 + r^5 = r^8(r^2 + 1) < r^8 \cdot r^6 < r^{15}\]absurd. For $n=7$, \begin{align*} r + r^2 + \cdots + r^6 \\ = r^7(1 + r^2 + r^4) < r^7( r^2 + r + r^4) \\ = r^7(r^7 + r^4) = r^11(r^3 + 1) \\ < r^{11}(r^3 + r^2) < r^{19} < r^{21} ,\\ \end{align*}absurd. For $n = 8$, \begin{align*} r + r^2 + \cdots + r^7 = r^{10}(1 + r^2 + r^4 ) \\ < r^{10}(r^7 + r^4) \\ = r^{14}( r^3 + 1) < r^{14} (r^3 + r^2) < r^{28}, \\ \end{align*}absurd. Therefore, all $n>4$ fail.

Let $x\geq1$ be the common ratio. Then, by scaling, we can assume without loss of generality $a_j-a_i=x^c$ for $0\leq c<\binom n2$. If $n=2$, $a_1=0$, $a_2=1$ works. If $n=3$, $a_1=0$, $a_2=x$, $a_3=x^2=x+1$ works if $x$ satisfies $x^2=x+1$. If $n=4$, let $x^3=x+1$. Then, $x^5-x^4=x^3+x^2-x^2-x=1$ and $x^5=x^3+x^2$, so $a_1=0$, $a_2=1$, $a_3=x^3=x+1$, $a_4=x^5=x^4+1=x^3+x^2$ works. Now, suppose $n\geq5$. If $a_{i+1}-a_i=1$, then we get $(a_{i+1}-a_i)+(a_i-a_j)=(a_{i+1}-a_j)$, or $x^a+1=x^b$. There are $n-2$ distinct pairs $(a,b)$, and all values of $b-a$ must be distinct since $x^{-a}=x^{b-a}-1$. We have $0<a,b<\binom n2$, so $x^{b-a}=1+\frac1{x^a}\leq\frac{x+1}x$. Since there are at least $n-2$ distinct values of $b-a$, we get $x^{n-1}\leq x+1$. We have $x_{p+q}-x_p=(x_{p+q}-x_{p+1})+(x_{p+1}-x_p)\geq1+x$ for all $q\geq2$. Since one of $1$, $x$, $\ldots$, $x^{n-1}$ corresponds to $x_{p+q}-x_p$ for some $q\geq2$, we get $x^{n-1}\geq x+1$. Therefore, we must have $x^{n-1}=x+1$, so equality must hold, implying $a_2-a_1$, $a_3-a_2$, $\ldots$, $a_n-a_{n-1}$ are $1$, $x$, $\ldots$, $x^{n-2}$ in some order. We also have that $a_{i+2}-a_{i+1}=x$ or $a_i-a_{i-1}=x$, so assume without loss of generality $a_{i+2}-a_{i+1}=x$. Therefore, $1+x+\ldots+x^{n-2}=x^{\binom n2-1}$, so $$x^{\binom n2}-x^{\binom n2-1}=x^{n-1}-1=x.$$This implies either $x_2-x_1=1$ or $x_n-x_{n-1}=1$. Assume without loss of generality $x_2-x_1=1$. Now, since $(a_{j+2}-a_{j+1})+(a_{j+1}-a_j)=a_{j+2}-a_j$, we get $x^a+x^b=x^c$ for $n-2$ pairs $a,b\leq n-2$, and all of the $c$ are distinct. Since $x^{2n-3}=(1+x)\left(1+\frac1x\right)>4$ and $x^{n-2}<2$, we get $n-1\leq c\leq2n-4$, so each value of $c$ must appear exactly once. However, $x^k+x^{k-1}=x^{k+n-2}$, so we must have $x_{k+1}-x_k=x^{k-1}$ for all $k$ by induction. Now, $x^{2n-3}=x_i-x_j$ for some $i>j$. Since $x^{2n-3}>x^{n-2}+x^{n-3}$, we must have $x^{2n-3}=1+x+x^2$, so \begin{align*} x+x^2+x^3&=x^{2n-2}\\ &=(1+x)^2\\ &=1+2x+x^2. \end{align*}Therefore, $x^3=x+1$, implying $n-1=3$, or $n=4$, contradiction. Therefore, the only integers $n$ satisfying the equation are $\boxed{n=2}$, $\boxed{n=3}$, and $\boxed{n=4}$.

Great for showersolving. Denote each difference via a closed interval on the real line. $n = 2, 3, 4$ are trivial via construction on $1, x^2 = x+1, x^3 = x+1$, so assume $n \geq 5$. Easy lemma: If two intervals are consecutive in the geometric sequence then they are not disjoint. Denote the lone differences by $a, b, c, d, \ldots$. Notice that WLOG the first two in geometric are $a, b$ then they are consecutive so $a+b$ is the first sum of $\geq 2$ elements to appear. If two elements appear before it: $a, b, a+b, c, a+b+c$ (as $a+b, c$ are tangent) is the sequence but five things must appear before $a+b+c$ so contradiction. If three elements appear before it: $a, b, c, a+b, b+c, a+b+c, d, ?, a+b+c+d$ since $a+b+c, d$ are tangent but nine things must appear before $a+b+c+d$ so contradiction. If at least four elements appear before it: $a, b, c, d, \ldots, ?, a+b$ then since $d, e, f, \ldots, ?$ are not tangent to $a, b$ it follows that $?, a+b$ are disjoint, giving contradiction. Thus by contradiction in all cases we finish.

@Inconsistent Interesting solution! I don’t quite get why the sequence in the first case is a, b, a+b, c, a+b+c and not a, b, a+b, c, b+c , a+b+c as we would then have five elements before a+b+c. I think I am misunderstanding something so could you maybe please explain that to me?

It is possible for $n\le 4$. For $n=2$ just take any two numbers. For $n=3$ take $0,\varphi,\varphi+\varphi^2$ which works because $\varphi+\varphi^2=\varphi^3$. For $n=4$ let $r^3=r+1$ then take $0,r,r+r^2,r+r^2+r^3$ which also works. Now, we show that $n\ge 5$ fails. $~$ Let $M=\frac{1}{2}n(n-1)$. If $a_1,a_2,\dots,a_n$ is such a sequence, then we can scale everything by $r^{-M-1}$ to get another sequence that works with $\tfrac1r$. Thus, WLOG assume $r>1$. $~$ Consider $a_n-a_1$. This must be the largest difference so it is $r^M$. Note that $a_n-a_1=(a_n-a_{k})+(a_{k}-a_1)$. Let $r^{x_k}=a_n-a_{k}$ and $r^{y_k}=a_k-a_1$. Note that $x_2,x_3,\dots,x_{n-1},y_2,y_3,\dots,y_{n-1}$ must all be distinct. Since there are $n-2$ pairs $(x_i,y_i)$ and $n-3$ numbers in the set $\{M-1,M-2,\dots,M-n+3\}$, there must exist a number $i$ such that $x_i,y_i\le M-n+2$. Hence, \[r^M=r^{x_i}+r^{y_i}\le r^{M-n+2}+r^{M-n+1}\]which implies $r^{n-1}\le +1$. $~$ Now, note that if $r^{n-1}>r+1$, then the minimum number that can represented as a sum of two different powers of $r$ is $r+r^2>r^n$, so the numbers $r,r^2,\dots, r^n$ all cannot be represented as a sum of two different powers of $r$. For $a_j-a_i$, note that if $j-i>1$ then it can be represented as $(a_j-a_{i+1})-(a_{i+1}-a_i)$ so all $n$ numbers $r,r^2,\dots,r^n$ must be equal to differences $a_{i+1}-a_i$. There are only $n-1$ numbers $a_{i+1}-a_i$, contradiction. Thus, $r^{n-1}=r+1$. Evidently, this also necessarily implies that the differences $a_{i+1}-a_i$ for $i=1,2,\dots,n-1$ are of the form $r,r^2,\dots,r^{n-1}$. In particular, \[r^M=r+r^2+\dots+r^{n-1}\]Now, for $n=5$ it is clear that $r+r^2+r^3+r^4=r^3+r^4+r^5=r^5+r^7$ while $r^{10}=r^6+r^7$ so contradiction. Note that \[r^M-r^{M-1}=r^{n-1}-1=r\]So $r^{M-1}=r^{M-2}+1$. We claim that for $n\ge 6$ that if $r_1,r_2>1$ are roots of $r^{M-1}-r^{M-2}-1$ and $r^{n-1}-r-1$. We will show that $r_1\le s=1+\frac{1}{2n-6}\le r_2$. $~$ Note that by calculus both polynomials are increasing for $r>1$ so it simply suffices to show that $s^{n-1}\ge s+1$ and $s^{M-1}\le s^{M-2}+1$. For the second one, it suffices to show $s^{M-1}\le 2n-6$ and for the first one, $s^{n-1}\ge 2+\tfrac{1}{2n-6}$, neither of which are too hard.

goofy goobers this solution is slightly lucky i guess (at the part where i used $r^{N-1}$ but i guess it's still decently motivated) The answer is $n=2,3,4$. For $n=2,3$ this is obvious. For $n=4$ it'd be enough to find $r$ such that \[a_4-a_3=r^3\]\[a_3-a_2=r^2\]\[a_2-a_1=r^1\]\[a_4-a_2=r^5\]\[a_3-a_1=r^4\]\[a_4-a_1=r^6\]simultaneously hold. Put another way we need $r^2+r^3=r^5$, $r^1+r^2=r^4$, and finally $r^3+r^4=r^6$. This is fine if $1+r=r^3$ which clearly has a solution in $r>0$ by IVT. WLOG $r>1$. Let's first show that $n=5$ fails. Evidently $a_5-a_1=r^{10}$. Consider $a_5-a_2$ and $a_4-a_1$. Clearly one of these is $r^9$. (At this point using a pyramid to represent differences is very useful.) The other one is then either $r^7$ or $r^8$ by Pigeonhole (there aren't enough valid spaces for $r^7$ and $r^8$ otherwise). MALD it probably fails Now we apply goofy tactics. Notice $N=\frac{n(n-1)}{2}$ then $r^{N-1}=(r^1+\dots+r^{n-1})-r^s$ for some $s$, but also $r^0+\dots+r^{n-2}=r^{N-1}$, so we can just say $r^{n-1}-r^s=1$. In particular $r^{n-1}>2$. Let $X=r^{n-1}$. We're basically almost done, realize that \[\frac{r}{r-1}\cdot (X-1)=X^{n/2}<\frac{\sqrt[n-1]{2}}{\sqrt[n-1]{2}-1}\cdot (X-1)\]so then \[\frac{X^{n/2}}{X-1}<\frac{\sqrt[n-1]{2}}{\sqrt[n-1]{2}-1}\]and the LHS has a critical value at $X=\frac{n}{n-2}<2$ so it's monotonically increasing so then \[2^{n/2}<\frac{\sqrt[n-1]{2}}{\sqrt[n-1]{2}-1}\]\[2^{n/2}-1<\frac{1}{\sqrt[n-1]{2}-1}\]\[\sqrt[n-1]{2}-1<\frac{1}{2^{n/2}-1}\]\[\sqrt[n-1]{2}<\frac{2^{n/2}}{2^{n/2}-1}\]\[2<\left(\frac{2^{n/2}}{2^{n/2}-1}\right)^{n-1}\]then $n\le 5$. Yay.

Splendid Problem! The answer is $n \in \{2,3,4\}$ Construction: In $n=2$ nothing to do, $n=3$ $r^2=r+1, a_1=1, a_2=1+r, a_3 = 1+r+r^2$ works for $n=4$, take real $r > 0$ such that $r^3=r+1$ (which exists by ,say IVT) then $a_4-a_1=r^6,a_4-a_2=r^5,a_4-a_3 = r^3, a_3-a_2 = r^2, a_3-a_1 = r^4, a_2-a_1 = r$ works (easy to show). Disproving for $n>4$ case: There are two key claims in this part of proof, The first key claim is to notice $r^{n-1}=r+1$ (which is inspired by the nature of constructions). Let $K = \frac{n(n-1)}{2}$ for the sake of convenience. Key Claim 1: $r^{n-1}=r+1$

The second key claim is also inspired by some wishful thinking(trying to somehow repeat the equality arguement used in the proof of claim 1). Key Claim 2: $\sum_{i=0}^{k} r^i \leq r^{n-k} \sum_{i=0}^{k-1} r^i$ with equality holding iff $k=0$ or $k=n-2$

Now repeatedly apply claim 2 to get , $r+r^2+ \cdots r^{n-1}= r(1+r+r^2+\cdots r^{n-2}) \leq r^S$ so equality must hold everywhere which is a contradiction for $n > 4$. $\blacksquare$ Remark:This is a excellent algebra problem which needs algebraic intuition to solve, definitely better than A6 and A7, which are fake algebra

Hopefully this is correct. Solved with an affirmation. Answer: $n=2, 3, 4$ The construction is the same as others. Suppose that there exists a configuration as in the statement for $n$. Observe that multiplying the elements of our sequence by $r^{-\frac{1}{2}n(n-1)-1}$ results in a sequence viable for $\frac{1}{r}$. Hence, we may assume that $r > 1$. Since $r$ is the smallest number among the differences, there is a positive integer $m$ such that $a_{m+1}-a_m=r$. Because there are $n-2$ differences where adding $a_{m+1}-a_m$ to it produces another difference, there must exist $n-2$ distinct pairs of integers $(p_1; q_1), \dots, (p_{n-2};q_{n-2})$ such that $r^{p_i}+r=r^{q_i}$ for all $i$. Notice that $r^{p_i}+r=r^{q_i} \implies r=r^{p_i}(r^{q_i-p_i}-1)$ implies that $q_i-p_i \neq q_j-p_j$ for all $i \neq j$. Consequently, there is a positive integer $d$ such that $q_d-p_d \geq n-2$. Therefore, we have $r^{p_d}+r=r^{q_d} \geq r^{p_d+n-2} $ which implies that $1+r \geq r + \frac{r}{r^{p_d-1}} \geq r^{n-1}$. Now, we have $r^a+r^b=r^a(r^{b-a} + 1) \geq r^n$ for all $a > b$. This forces $\{x_1, x_2, \dots, x_{n-1}\} = \{r^1, r^2, \dots r^{n-1}\}$ since the sum of any two $x_i$s is larger than all of the $r_i$s. Then there must exist a pair of positive integers $(a, b)$ such that $r^a+r^b=r^n$ which implies that $1+r \leq r^{n-1}$. Therefore, $1+r=r^{n-1}$. Since $r^\frac{1}{2}n(n-1)$ is the largest number among the differences, we must have $r^{\frac{1}{2}n(n-1)}=a_n-a_1=x_1+x_2+ \dots x_{n-1} = r + r^2 + \dots r^{n-1}$, which implies that $(1+r)^n=(r+r^2+ \dots r^{n-1})^2$. Let $y_i$ denote the coefficient behind $r_i$ in $(r+r^2+ \dots r^{n-1})^2$. Observe that $y_i=2n-1-i$ whevever $i > n-1$ and $y_i=i-1$ whenever $i \leq n-1$. Swapping the multiple $r^{n-1}$ with $1+r$ for all of the $r_i$s where $i \geq n$, we obtain $(r+r^2+ \dots r^{n-1})^2=(y_1+y_n)r^1+(y_2+y_n+y_{n+1})r^2+ \dots +(y_{n-1}+y_{2n-3}+y_{2n-2})r^{n-1}+y_{2n-2}r^n$. Notice that the coefficients are always smaller than $3n-2$ and we can manually check that $y_1+y^n = n-1 < n$ and $y_{2n-2}=1$. On the other hand, it is well known that $(1+r)^n=\sum\limits_{i=0}^n \binom{n}{i}r_i$. When $n > 4$, $\binom{n}{i} \geq 3n$ for all $2 \leq i < n-2$. Looking at the coefficients of both of these identities implies that $(1+r)^n > (r+r^2+ \dots r^{n-1})^2$, a contradiction. $\blacksquare$

Solution using minimal polynomials[Definetly isnt the expected solution]:- Answer is $n=2,3,4$ . Construction is just hit and trial pretty much same as previous solutions . So i will just focus on contradicting $n \geq 5$ . Case 1: $r<1$ Proof: Just multiply all $a_i$ by $\frac{1}{r^{\frac{n(n-1)}{2}+1}}$ to convert this to case 2. Case 2: $r>1$ Its pretty easy to see $r$ must be an algebric integer , so lets say $P(x)$ is the minimal polynomial of $r$ over $\mathbb{Z}[X]$ . let $d=deg P$. I establish a key irreducibility lemma : Lemma 1: For any positive integer $n$ the polynomial $x^n+x-1$ is reducible iff $n \equiv 5$(mod $6$) and $x^2-x+1$ is a factor then and $\frac{x^n+x-1}{x^2-x+1}$ is irreducible over $\mathbb{Q}$ . Proof: Refer to the appendixsection [example A.2] of the following paper :- https://kconrad.math.uconn.edu/blurbs/ringtheory/irredselmerpoly.pdf some gaps are however there u can easily fill them yourself . Lemma 2: There exist a positive integer $m$ such that $P(x)=x^{m+1}-x^{m}-1$ . or $P(x)=x^3-x-1$ Proof: Assume the contrary Start with the fact that $a_n-a_1= r^{\frac{n(n-1)}{2}}$ its easy to see $r^{\frac{n(n-1)}{2}-1} \in \{a_{n-1}-a_{1},a_{n}-a_{2}\}$ Assume WLOG its $a_{n-1}-a_{1}$ , so $r^{\frac{n(n-1)}{2}-c}=r^{\frac{n(n-1)}{2}-1-c}+1$ for some natural number $c$ So $P(x)|x^{\frac{n(n-1)}{2}-c}-x^{\frac{n(n-1)}{2}-1-c}-1$ now consider the reciprocal polynomial [which is redducible] and lemma 1 to conclude $P(x)=x^2-x+1$ or $P(x)=x^3-x-1$[u can verify this by observing the pattern in example A.2 among the quiotents when divided by $x^2+x+1$] obviously later isnt possible , so $P(x)=x^2-x+1$ however $x^2-x+1$ has no real roots so that also isnt an option Case 1: $P(x)=x^{m+1}-x^{m}-1$ Claim: $m \leq n-1$ . Proof: Assume the contrary observe $r,r^2,....,r^m$ are all in $a_2-a_1,a_3-a_2,....,a_n-a_{n-1}$ that obviously implies the claim. Now pick some $i$ such that $a_{i+1}-a_{i}=r$ . Now consider the paritions $a_1,a_2,...a_{i}$ and $a_{i+1},a_{i+2},...,a_n$ . Claim: There exist distinct natural numbers $i_1,i_2$ chosen from the same parition such that $a_{i_2}-a_{i_1} \geq r^{m+2}$ . Proof: Just consider the total number of ways we can choose such two pair of numbers with $i_2>i_1$ basic high school combi and observe there are atleast $n$ such pairs[the actual lower bound is a quadratic] which obviously implys the main statement as that implies one pair ammounted to a value atleast $r^{n+1}$ [$1$ is already hijacked by $a_{i+1}-a_{i}$ so the $+1$] Subcase 1: $i_1,i_2$ are chosen from the first parition. clearly $i_2>i_1$ observe $a_{i+1}-a_{i_1}=a_{i+1}-a_{i}+a_{i}-a_{i_1}=r+a_{i}-a_{i_1}$ Say $a_{i}-a_{i_1}=r^{A}$ and $a_{i+1}-a_{i_1}=r^{B}$ clearly $A \geq m+2$ so $1 \geq r^{m+2}-r^{m+1}=r$ contradiction ! . For the other parition the proof is similar i aint writing it again Case 2: $P(x)=x^3-x-1$ Just strengeniz the previous bounds a bit this case is easily done

My solution is a bit different from above solutions Assume such $r>0$ exists for $n\geq5$ WLOG, assume $r>1$ and $a_1=0$ for convenience, as we can replace $(a_i)$ with $(a_i+d)$ and $(r^{-\frac{n(n-1)}{2}-1}a_i)$ Claim 1: $a_n-a_1=a_n=r^{\frac{n(n-1)}{2}}$ (Trivial) Claim 2: $r^{n-1}>2$ Proof: Assume for contradiction $r^{n-1}\leq 2$ We see that if $r^m=a_i-a_j, i-j\geq2$ then $r^{m-1}>2$ as $r^{m-1}=r^{u_1-1}+r^{u_2-1}>2$ for some $u_1,u_2$ So for all $1\leq m\leq n$, there exists an $i$ that $a_{i+1}-a_i=r^m$ (contradiction) So $r^{n-1}>2$ Claim 3: $r^{n-2}<2$ Assume for contradiction $r^{n-2}\geq2$ $i$ is the index that $a_{i+1}-a_i=r$ We have for some $k'<k$: $r^k=a_n-a_i=a_n-a_{i+1}+a_{i+1}-a_i=r^{k'}+r\leq \frac{a_n-a_i}{r}+r$ So $r^{k-1}(r-1)\leq r$, or $r^{k-2}(r-1)\leq 1$ $\ell$ is the index that $a_{i+1}-a_1=a_{i+1}=r^{\ell}$, we also have $r^{\ell-2}(r-1)\leq 1$ Claim: The set ${a_s-a_j,i+1\leq j<s\leq n}=A_1$ and ${a_s-a_j,1\leq j<s\leq i}=A_2$ are disjoint and don't have $r$, moreover, $\max{r^k,r^{\ell}}=\max A_1,A_2$ The latter is trivial, so we will prove the former $r<b_1<...<b_u<r^k$ are all elements of $A_1$, $r<c_1<...<c_v<r^{\ell}$ are all elements of $A_2$ in which $u=\frac{(n-i)(n-i+1)}{2}-1,v=\frac{i(i+1)}{2}-1$ Hence $\max{k,l}\geq u+v+3=\frac{(n-i)(n-i+1)+i(i+1)}{2}+1\geq \frac{n(n+2)}{4}+1$ But $r^{\max{k,l}-2}(r-1)\leq 1$, so we have $r^{\frac{n(n+2)}{4}-1}(r-1)\leq1$ From assumption, we have $\displaystyle 2^{\frac{1}{n-2}}\leq 1+2^{-\frac{n^2+2n-4}{4(n-2)}}$ Or $2\leq (1+2^{-\frac{n^2+2n-4}{4(n-2)}})^{n-2}$, or $ln(1+2^{-\frac{n^2+2n-4}{4(n-2)}})\geq \frac{ln2}{n-2}$ But $ln(1+x)\leq x $ for all $x>0$ so we need $\frac{n-2}{ln2}<2^{\frac{n^2+2n-4}{4(n-2)}}$ to reach a contradiction, which is true and not hard to prove, however I'm lazy to prove it here sorry:) So the assumption is false, we have $r^{n-2}<2<r^{n-1}$ If $r^{n-1}=a_{j_2}-a_{j_1}, j_2-j_1\geq2$, then $a_{j_2}=r^{k_2},a_{j_1}=r^{k_1}$ for some $k_2>k_1\geq1$, then $r^{n-2}>2$, contradiction Therefore $\{a_{i+1}-a_i,1\leq i \leq n-1\}=\{r,r^2,..,r^{\frac{n(n-1)}{2}}\}$, so $r^{\frac{n(n-1)}{2}}=r.\frac{r^{n-1}-1}{r-1}$ We easily see that $a_n-a_2$ or $a_{n-1}$ equals $r^{\frac{n(n-1)}{2}-1}$ Case 1: $a_n-a_2=r^{\frac{n(n-1)}{2}-1}$ For $a_2=r^d$, we have $r^d=r^{\frac{n(n-1)}{2}-1}(r-1)=r^{n-1}-1$, so $r^{n-1}=r^d+1$ If $d\geq2, r^{n-2}=r^{d-1}+\frac{1}{r}\geq r+\frac{1}{r}\geq2$, contradiction So $d=1$, or $r^{n-1}=r+1,r^{\frac{n(n-1)}{2}-2}(r-1)=1$ So $r^n=r+r^2$, we must have $r^n=a_3,a_3-a_2=r^2$ $a_n-a_3$ or $a_{n-1}$ equals $r^{\frac{n(n-1)}{2}-2}$ If $a_n-a_3=r^{\frac{n(n-1)}{2}-2}$, then $r^n=a_3=r^{\frac{n(n-1)}{2}-2}(r^2-1)=r+1$, contradiction So $a_{n-1}=r^{\frac{n(n-1)}{2}-2}$ Easy to see $r^{\frac{n(n-1)}{2}-3}\in\{a_n-a_3,a_{n-1}-a_2,a_{n-2}\}$ If $a_{n-2}=r^{\frac{n(n-1)}{2}-3}$, then $a_{n-1}-a_{n-2}=r^{\frac{n(n-1)}{2}-3}(r-1)=\frac{1}{r}$, contradiction If $a_{n-1}-a_2=r^{\frac{n(n-1)}{2}-3}$, then $a_2=\frac{1}{r}$, contradiction So $a_n-a_3=r^{\frac{n(n-1)}{2}-3}$, so $a_3=r^n=r^{\frac{n(n-1)}{2}-3}(r^3-1)=\frac{r^2+r+1}{r}$, so $1+r+r^2=r^{n+1}=r^2(r+1)=r^3+r^2$, so $r^3=r+1$, or $n=4$ (contradiction) Case 2: $a_{n-1}=r^{\frac{n(n-1)}{2}-1}$: Doing the same as above, we will see that this also leads to contradiction Therefore $n\leq 4$, in this case we can show there exists $r>1$ that satisfies the condition like above posts