Both 31 and 32 cannot be shown to be a sum of three cubes.
Prove that there is an infinite number of such "twins" (successive integers).

A cube may be {0,1,8}mod3^2.

Hence the sum of 1, 2, or 3 cubes is {0,1,2,3,6,7,8}mod3^2.

In any 9 numbers there will be a successive pair worth {4,5}mod3^2 which cannot be the sum of 3 or less cubes.

 

Posted by broll on 2012-11-27 00:34:28