IMO 2002/B1 solution

IMO 2002

Problem B1

The positive divisors of the integer n > 1 are d₁ < d₂ < ... < d_k, so that d₁ = 1, d_k = n. Let d = d₁d₂ + d₂d₃ + ... + d_k-1d_k. Show that d < n² and find all n for which d divides n².

Solution

d_k+1-m <= n/m. So d < n²(1/(1.2) + 1/(2.3) + 1/(3.4) + ... ). The inequality is certainly strict because d has only finitely many terms. But 1/(1.2) + 1/(2.3) + 1/(3.4) + ... = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... = 1. So d < n².

Obviously d divides n² for n prime. Suppose n is composite. Let p be the smallest prime dividing n. Then d > n²/p. But the smallest divisor of n² apart from 1 is p, so if d divides n², then d ≤ n²/p. So d cannot divide n² for n composite.

43rd IMO 2002