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


$$\sum\_{i=0}^{n-1} (k+i)^3 = k^3 + (k+1)^3 + \cdots + (k+n-1)^3 = Y^3 $$


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!



Sum of consecutive cubes
Found some bigger numbers:
{n,k} as you call them:
{4913 , 11368}
{6591 , 305}
{6859 , 18171}
{8000 , 22534}
{10648 , 33558}
{12167 , 40381}
{13923 , 3010}
{14161 , 1624}
{25201 , 46690}
{33124 , 18551}
{63001 , 11170}
{48841 , 967190}
{277729 , 711785}
Most 'n' are squares, but a strange ones are 6591, 13923 and 25201


pipo


