2002 IMO Problems/Problem 4

Problem: Let be an integer and let be all of its positive divisors in increasing order. Show that

Solution

We proceed with two parts:

Part 1: Proof of Inequality

We have: \[ \sum_{i=1}^{r-1} d_i d_{i+1} = d_1d_2 + d_2d_3 + \cdots + d_{r-1}d_r \] For each term \( d_i d_{i+1} \), note that \( d_{i+1} \leq \frac{n}{d_i} \) (since divisors come in pairs). Thus: \[ d_i d_{i+1} \leq d_i \cdot \frac{n}{d_i} = n \] Summing over all \( r-1 \) terms: \[ \sum_{i=1}^{r-1} d_i d_{i+1} \leq (r-1) \cdot n < n^2 \] The last inequality holds because \( r \) (the number of divisors) is at most \( 2\sqrt{n} \), making \( (r-1)n \leq 2n^{3/2} < n^2 \) for \( n > 4 \). Smaller cases can be verified directly.

Part 2: Equality Condition

Equality occurs **only** when \( n \) is prime: - For prime \( n \), the divisors are \( 1 \) and \( n \), so:

- For composite \( n \), there exists at least one additional divisor \( d_2 \) (where \( 1 < d_2 < n \)), making the sum strictly larger than \( n \) but still less than \( n^2 \).

See Also

2002 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
All IMO Problems and Solutions