100 is a square which is also a sum of 4 consecutive cubes.
Are there any others?

Let's call the cubes (x)³,(x+1)³,(x+2)³,(x+3)³
The simplified sum is 4x³+18x²+42x+36
this factors to 2(2x+3)(x²+3x+6)
this can only be a square if
2(2x+3)=(x²+3x+6) or (2x+3)=2(x²+3x+6)
The first of these is the quadratic 0=x²-x which has solutions x=0 and x=1
The second is the quadratic 2x²+4x+9=0 which has complex solutions.

So there are two squares that are the sum of 4 consecutive cubes:

0²+1³+2³+3³=36=6²
1³+2³+3³+4³=100=10²

Posted by Jer on 2013-06-18 12:11:52