19 Feb
2012
19 Feb
'12
9:19 a.m.
If I'm understanding: What ensures that n^3 - (n-1)^3 won't require another (n-1)^3 in its sum of cubes? --Dan << Sure -- just let n^3 - (n-1)^3 be sufficiently large as well, no? --Michael On Feb 19, 2012 10:48 AM, "Dan Asimov" <dasimov@earthlink.net> wrote: In what follows, "cube" means the cube of a positive integer. It is known that every sufficiently large integer is the sum of distinct cubes. See < http://oeis.org/A001476 >. QUESTION: Is every sufficiently large integer the sum of *more than one* distinct cubes? --Dan
________________________________________________________________________________________ It goes without saying that .