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Algebra Difficulty 8.7 Shortlist Find the answer

Find all positive integers n⩾2n \geqslant 2 for which there exist nn real numbers a1<⋯<ana_1<\cdots<a_n and a real number r>0r>0 such that the 12n(n−1)\tfrac{1}{2}n(n-1) differences aj−aia_j-a_i for 1⩽i<j⩽n1 \leqslant i<j \leqslant n are equal, in some order, to the numbers r1,r2,…,r12n(n−1)r^1,r^2,\ldots,r^{\frac{1}{2}n(n-1)}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

To solve the problem, we need to find all positive integers n⩾2 n \geqslant 2 for which there exist n n real numbers a1<a2<⋯<an a_1 < a_2 < \cdots < a_n and a real number r>0 r > 0 such that the differences aj−ai a_j - a_i for 1⩽i<j⩽n 1 \leqslant i < j \leqslant n are exactly the numbers r1,r2,…,r12n(n−1) r^1, r^2, \ldots, r^{\frac{1}{2}n(n-1)} .

### Step 1: Understanding the Problem

The total number of differences aj−ai a_j - a_i with 1⩽i<j⩽n 1 \leqslant i < j \leqslant n is 12n(n−1)\frac{1}{2}n(n-1) . These differences need to correspond, in some order, to the powers of r r from r1 r^1 to r12n(n−1) r^{\frac{1}{2}n(n-1)} .

### Step 2: Analysis for Small Values of n n

Let's analyze the possibility for different values of n n starting from small integers.

#### Case n=2 n = 2 :
- We have only one difference a2−a1=r1 a_2 - a_1 = r^1 .
- This condition can be satisfied with r=a2−a1>0 r = a_2 - a_1 > 0 .

#### Case n=3 n = 3 :
- We need three differences: a2−a1 a_2 - a_1 , a3−a1 a_3 - a_1 , a3−a2 a_3 - a_2 .
- We reconcile these as r1,r2,r3 r^1, r^2, r^3 . Define the differences as:
a2−a1=r1, a3−a2=r2, a3−a1=a3−a2+a2−a1=r1+r2=r3. a_2 - a_1 = r^1,\, a_3 - a_2 = r^2,\, a_3 - a_1 = a_3 - a_2 + a_2 - a_1 = r^1 + r^2 = r^3.
- The differences can indeed be r,r2,r+r2 r, r^2, r + r^2 , satisfying the requirements.

#### Case n=4 n = 4 :
- We need six differences: a2−a1 a_2 - a_1 , a3−a1 a_3 - a_1 , a4−a1 a_4 - a_1 , a3−a2 a_3 - a_2 , a4−a2 a_4 - a_2 , a4−a3 a_4 - a_3 .
- These differences need to cover the set {r1,r2,r3,r4,r5,r6} \{ r^1, r^2, r^3, r^4, r^5, r^6 \} .
- One possible assignment can be leveraging differences as sums of sequential powers and finding construction:
a2−a1=r, a3−a2=r2, a4−a3=r3 a_2 - a_1 = r, \, a_3 - a_2 = r^2, \, a_4 - a_3 = r^3
a3−a1=r+r2, a4−a2=r2+r3, a4−a1=r+r2+r3, a_3 - a_1 = r + r^2, \, a_4 - a_2 = r^2 + r^3, \, a_4 - a_1 = r + r^2 + r^3,
which matches the necessary powers of r r .

### Step 3: Larger n n

For n≥5 n \geq 5 , consider the differences exceeding each subsequent hoop does not easily allow a matching construction due to the rapidly increasing number of differences compared to available assignment sums of powers. Thus, it becomes difficult to maintain the sequence matched exactly to required power arrangements, particularly for consecutive additions.

### Conclusion

Based on the analysis and successful assignments, the values of n n that satisfy the conditions are n∈{2,3,4} n \in \{2, 3, 4\} . Therefore, the answer is:

n∈{2,3,4} \boxed{n \in \{2, 3, 4\}}

This completes the solution process for the problem.

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