100 is a square which is also a sum of 4 consecutive cubes.
Are there any others?
(In reply to re: Solution - flaw?
You might not want to spend too much further time investigating this.
Let the square sought be a^2. The sum of the first n cubes is a square, ((n(n+1))/2)^2. So is the sum of the first (n-4) cubes, (((n-4)(n-3))/2)^2, say b^2. The intriguingly Pythagorean look of this is misleading, as a^2+b^2 = c^2 is just as true if a=c and b=0. Here, it's just a matter of zeroing the factors of b to allow the two remaining squares, a and c, to equal each other. Naturally the same is also true if one asks for the sum of 5 cubes, or 6 cubes, or even a trillion cubes.
Edited on June 19, 2013, 7:01 am
Posted by broll
on 2013-06-19 06:06:31