I'm investigating when the sum of n consecutive cubes equals a cube, i.e., for what n-values does

Sum{i=0..n-1} (k+i)^3 = k^3 + (k+1)^3 + ... + (k+n-1)^3 = Y^3

have nontrivial solutions (k,Y) for k and Y > 0. I have found (using programs) that if this equation has non-trivial solutions, n is not squarefree (for n > 3). Now I'm trying to prove that n > 3 cannot be squarefree. Here is a [link](https://www.dropbox.com/s/v1snspj6ysq82g6/MathCubes_.pdf?dl=0) to my proof and what I've done so far but I've reached a wall. I have three equations that I believe contradict each other (I am almost certain they contradict each other). I just can't see how they contradict each other and I might need a new set of eyes to look at it. The three equations are given in the link but, if you like, I've put them below. I'm trying to show the following:

For natural numbers x, y, k and integer d > 1,

1. d^2*y = 2k + dx - 1
2. xy(d^4*y^2+d^2*x^2-1) = cube
3. x | k*(k-1)

cannot all be true. Please let me know if you have any questions or suggestions for me! Thanks in advance!