31st CanMO 1999 Problem 3 Find all positive integers which equal the square of their number of positive divisors.
Let N = p a q b ... . Then the square of the number of positive divisors of N is (a + 1) 2 (b + 1) 2 ... . Any square has all its prime factors to an even power, so a, b, ... must all be even. So writing a = 2m, b = 2n etc, we have p m q n ... = (2m+1)(2n+1) ... . Since the rhs is odd, all the primes p, q, ... must be odd. A trivial induction shows that 2n + 1  1. Hence 2n + 1  1 for any odd prime p. For n = 1, 2n + 1 = 3 < p unless p = 3. So 2n+1 < p n , where p is an odd prime and n is a positive integer, except in the case n = 1, p = 3. Thus we can have at most one prime p in the factorisation of N and N = 1 or 3 2 . 31st CanMO 1999 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002