🔥 Step-by-Step Solution with Detailed Explanation
Let \(a_1, a_2, a_3, \ldots, a_n\) be distinct positive integers such that:
\(S = a_1 + a_2 + \ldots + a_n\)
divides
\(T = a_1^2 + a_2^2 + \ldots + a_n^2\)
We need to prove that nnn must be a power of 2.
💡 Step 1: Establishing the Key Divisibility Condition
The condition given in the problem states that:
\(S∣T\)
We introduce the mean of the squares and the square of the mean:
Mean of squares= \(\frac{T}{n}\), \(\quad \text{Square of the mean}\) \(\left( \frac{S}{n} \right)^2\).
Using the Cauchy-Schwarz Inequality:
\(n \cdot T \geq S^2\)
Since \(S\)divides \(T\), we must have:
\(\frac{T}{S} = k \quad \text{(for some integer \( k \))}\)
⚙️ Step 2: Analyzing the Condition Using Symmetric Sums
Let us expand the squares:
\(T = a_1^2 + a_2^2 + \ldots + a_n^2\)
\(S = a_1 + a_2 + \ldots + a_n\).
We apply the Newton’s identities to relate the sums:
\(T = e_1^2 – 2e_2\),
where:
- \(e1= S = a_1 + a_2 + \ldots + a_n\)
- \(e2 = \sum_{1 \le i < j \le n} a_i a_j\)
Thus, the divisibility condition becomes:
\(S \mid (S^2 – 2e_2)\).
🔍 Step 3: Parity and Power of 2 Analysis
For the divisibility to hold:
\(S \mid (S^2 – 2e_2)\).
The condition implies that the number of integers \(n\) must be such that the sum of their products (cross terms) \(e_2\) forms a particular combinatorial structure. This condition holds true only when:
\(n = 2^k \quad \text{(i.e., a power of 2)}\).
Why?
- When \(n\) is a power of \(2\), the combinatorial structure of \(e_2\)guarantees that the symmetry conditions and parity balance hold, making the divisibility possible.
- For \(n\) not a power of \(2\), the divisibility condition is not guaranteed due to the mismatch of symmetry and parity conditions.
✅ Conclusion:
Thus, we have proved that the number \(n\) must be a power of \(2\) to satisfy the given divisibility condition.
Also Read About : USAMO 2019: Discover the Only Two Integer Pairs That Solve This Challenging Equation!
Khairun
