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

$\sum\limits_{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, Y \in \mathbb{N} $. 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, d \in \mathbb{N} $ and $ d > 1,$

1. $ d^2y = 2k + dx - 1 $
2. $ xy(d^4y^2+d^2x^2-1) = cube $
3. $ x {\space} | {\space} k(k-1) $

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