Find all integer solutions to m3 - n3 = 2mn + 8.
Answer
(m,n) = (2,0), (0,-2)
Solution
Put m = n+k. Then 3n2k+3nk2+k3 = 2n2+2nk+8, so (3k-2)n2+(3k2-2k)n+k3-8 = 0. For real solutions we require (3k2-2k)2 ≥ 4(3k-2)(k3-8) or (3k-2)(32-2k2-k3) ≥ 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 n2+n-7 = 0, which has no integer solutions. If k = 2, then 4n2+8n = 0, so n = 0 or -2.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
12 February 2004
Last corrected/updated 12 Feb 04