In other words, if you have a pile of cannonballs in a square pyramid, can you rearrange them as a cube instead, or do you have to shell someone first? Or, instead, does $$n(n+1)(2n+1)=6x^3$$ have any nontrivial integer solutions?

The analogous problem where you want to put the cannonballs in a square instead of a cube is know to have only one nontrivial solution: $$\sum_{n=1}^{24}n^2=70^2$$

It can be seen that $n,n+1,2n+1$ are pairwise coprime by using Euclid's algorithm. Consider the six cases of the rest $n \pmod 6$:



Since the product of 3 coprime integers is a cube each is a cube too. Let $y=a^3$, $6y+1=b^3$, $12y+1=c^3$. A solution is the equivalent to a solution to the system $$b^3-6a^3 = c^3 - 12 a^3 = 1$$


Applying the same argument and substitution: $$2b^3-a^3=3c^3-2a^3=1$$









So if it is proven that none of these systems of "cubic Pell equations" has a nontrivial solution, the theorem is proven. I have a suspicion a quick proof would require an algebraic number theory atom bomb.


1 Answer 1


All six cases boils down to whether one can find integer solutions for Skolem's equation $$x^3 + dy^3 = 1$$

for $d = 2$ and $4$. According to my reference, there is a theorem:

(Skolem) - If $d \in \mathbb{Z}$ is given with $d \ne 0$, there exists at most one pair $(x,y) \in \mathbb{Z} \times \mathbb{Z}$ with $y \ne 0$ such that $x^3 + dy^3 = 1$

For $d = 2$, the only $v \ne 0$ solution is $(u,v) = (-1,1)$ and there is no $v \ne 0$ solution for $d = 4$.
As a corollary of this, there is no $n > 1$ solution for the equation:

$$m^3 = \frac{n(n+1)(2n+1)}{6}$$

Please consult reference below for more details.


  • Henri Cohen, Number Theory Volume I, Tools and Diophantine Equations,
    $\S 6.4.7$ Skolem's Equations $x^3 + dy^3 = 1$.

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