23rd VMO 1985 Problem A1 Find all integer solutions to m 3 - n 3 = 2mn + 8.
(m,n) = (2,0), (0,-2) Solution Put m = n+k. Then 3n 2 k+3nk 2 +k 3 = 2n 2 +2nk+8, so (3k-2)n 2 +(3k 2 -2k)n+k 3 -8 = 0. For real solutions we require (3k 2 -2k) 2 ≥ 4(3k-2)(k 3 -8) or (3k-2)(32-2k 2 -k 3 ) ≥ 0. The first bracket is +ve for k ≥ 1, -ve for k ≤ 0, the second is +ve for k ≤ 2, -ve for k ≥ 3. Hence k = 1 or 2. If k = 1, then n 2 +n-7 = 0, which has no integer solutions. If k = 2, then 4n 2 +8n = 0, so n = 0 or -2. Thanks to Suat Namli 23rd VMO 1985 © John Scholes jscholes@kalva.demon.co.uk 12 February 2004 Last corrected/updated 12 Feb 04