The geometric analogy in Ferran's solution doesn't quite do it (since it does not cover negative cases), but his computation allows another quick means. If (n+1)^3 + 8 must be a perfect cube, then this means we require for a solution two perfect cubes that differ by 8. There are only two possibilities: -8 and 0, and 0 and 8. So (n+1)^3 must equal either -8 or 0, so n must equal either -2 or 0. In the first case (-2), n^2 + 3 = 7 is not a perfect cube, in the second case (0), neither n + 3 = 3 nor n^2 + 3 = 3 is a perfect cube.

Posted by RoyCook on 2003-09-19 13:44:28