For n a positive integer find f(n), the number of pairs of positive integers (a, b) such that ab/(a + b) = n.
ab = n(a + b) implies a(b - n) = nb > 0, so a and b must both exceed n. Let b = n + d, then a = n(n+d)/d = n + n2/d. This is a solution iff d divides n2. So f(n) = the number of divisors of n2.
21st Putnam 1960
© John Scholes
15 Feb 2002