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 $d_{i+1}\le \frac{n}{d_i}$ (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 $(r-1)n\le 2n^{3/2}<n^2$ 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