IMO 2000 SL N2

Origin: FRA | Category: Number Theory

Problem

For a positive integer n, let d(n) be the number of all positive divisors of n. Find all positive integers n such that d(n)3 = 4n.

Solution

Let n = p\alpha1 1 p\alpha2 2 \cdot \cdot \cdot p\alphak k be the factorization of n onto primes (p1 < p2 < \cdot \cdot \cdot < pk). Since 4n is a perfect cube, we deduce that p1 = 2 and \alpha1 = 3\beta1 + 1, \alpha2 = 3\beta2, . . . , \alphak = 3\betak for some integers \betai \geq0. Using d(n) = (\alpha1 + 1) \cdot (\alpha2 + 1) \cdot \cdot \cdot (\alphak + 1) we can rewrite the equation d(n)3 = 4n as

(3\beta1 + 2) \cdot (3\beta2 + 1) \cdot \cdot \cdot (3\betak + 1) = 2\beta1+1p\beta2 2 \cdot \cdot \cdot p\betak k . Since d(n) is not divisible by 3, it follows that pi \geq5 for i \geq2. Thus the above equation is equivalent to 3\beta1 + 2 2\beta1+1

p\beta2 3\beta2 + 1 \cdot \cdot \cdot p\betak k 3\betak + 1. (1) For i \geq2 we have p\betai i \geq(1 + 4)\betai \geq1 + 4\betai; hence (1) implies that 3\beta1+2 2\beta1+1 \geq1, which leads to \beta1 \leq2. For \beta1 = 0 or \beta1 = 2 we have that 3\beta1+2 2\beta1+1 = 1, and therefore \beta2 = \cdot \cdot \cdot = \betak = 0. This yields the solutions n = 2 and n = 27 = 128. For \beta1 = 1 the left-hand side of (1) equals 5 4. On the other hand, if pi > 5 or \betai > 1, then p\betai i 3\betai+1 > 5 4, which is impossible. We conclude that p2 = 5 and k = 2, so n = 2000. Hence the solutions for n are 2, 128, and 2000.