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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 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!

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  • $\begingroup$ Maybe it would be easier to consider $\left(\frac{b(b+1)}{2}\right)^{2} - \left(\frac{a(a+1)}{2}\right)^{2}.$ $\endgroup$
    – Geoff Robinson
    Commented Mar 7, 2015 at 8:50
  • $\begingroup$ I've tried this. Unfortunately, you get back to the same equation I have... $\endgroup$
    – Derek Orr
    Commented Mar 7, 2015 at 16:03
  • $\begingroup$ In case anyone was wondering, Derek has already recorded this sequence in the OEIS. oeis.org/A240970 $\endgroup$
    – Timothy Chow
    Commented Mar 7, 2015 at 22:37
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    $\begingroup$ Derek, does your system of three equations have any known solutions if some of the variables are allowed to be rational numbers rather than integers? $\endgroup$
    – Timothy Chow
    Commented Mar 7, 2015 at 22:44
  • $\begingroup$ There is some discussion at mathpages.com/home/kmath147.htm $\endgroup$
    – Gerry Myerson
    Commented Sep 14, 2022 at 1:16

2 Answers 2

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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

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    $\begingroup$ Forgot to mention, but it seems that 25201 is squarefree. pipo $\endgroup$
    – pipo
    Commented Mar 8, 2015 at 18:34
  • $\begingroup$ $25201=11\times29\times79$. $\endgroup$
    – Gerry Myerson
    Commented Mar 8, 2015 at 22:22
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    $\begingroup$ I've confirmed pipo's example (Y = 1764070), so this answers Derek's question. Perhaps Derek (or someone else) can say what solution to Derek's system of three equations pipo's example corresponds to. $\endgroup$
    – Timothy Chow
    Commented Mar 8, 2015 at 23:03
  • $\begingroup$ @TimothyChow : I believe this corresponds to a solution $x=2291$, $y=980$, $d=11$, and $k=46690$. $\endgroup$
    – Vladimir Dotsenko
    Commented Mar 9, 2015 at 0:07
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    $\begingroup$ @pipo how did you find these? It seems {48841, 967190} and {277729, 711785} are incorrect. $\endgroup$
    – Derek Orr
    Commented Mar 10, 2015 at 1:10
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The question was explored on sci.net 20 years ago ; see here. To summarize what was said then (the answer by Dave Rusin) : We have identified several types of solutions:

  1. The trivial solutions -- $m=(1-k)/2$ (so $N=0$) for odd $k$, and $m=-k/2$ or $-k/2+1$ for even $k$
  2. The torsion solutions -- $m=(u^3-2u^2-4u-4)(u-1)/6$ when $k=u^3$
  3. Other elements in the subgroup generated by 1. and 2. (Conjecturally only $k=4$, $m=11$)
  4. Other elements of rank-1 curves (None such? see Sect. VI)
  5. Other generators of rank >1 curves (and linear combinations thereof) as for $k=3, 20, 25, 49, 99, 153, 288$
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    $\begingroup$ Thanks but I've looked at this link many times before. Rusin doesn't prove that n cannot be squarefree. $\endgroup$
    – Derek Orr
    Commented Mar 7, 2015 at 16:02

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