Prove that every positive rational number can be written as (a3 + b3)/(c3 + d3) for some positive integers a, b, c, d.
Solution
Note that (m + n)3 + (2m - n)3 = 9m(m2 - mn + n2) and (m + n)3 + (2n - m)3 = 9n(m2 - mn + n2). So a = m + n, b = 2m - n, c = m + n, d = 2n - m gives m/n. We require a, b, c, d positive. Obviously a and c are positive, but b and d are only positive if 2m > n and 2n > m, in other words if m/n belongs to the open interval (1/2, 2).
For k > 4, we have (k+1)3/k3 < 2. So consider the sequence 53/43, 63/43, 73/43, ... it obviously tends to infinity and the ratio between successive terms is less than 2. So at least one member belongs to each of the intervals (2, 4), (4, 8), (8, 16), ... . Now any positive rational h < 1/2 must belong to some interval [1/2n+1, 1/2n], but we can find an integer k such that k3/43 belongs to (2n, 2n+1) so h k3/43 belongs to (1/2, 2) and hence can be written as (a3 + b3)/(c3 + d3). So h = ( (4a)3 + (4b)3)/( (kc)3 + (kd)3). Similarly for any rational h > 2.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2002