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:

n=6y

y(6y+1)(12y+1)=x^3

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 = 1n=6y+1

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

n=6y+2

c^3-4a^3=6b^3-c^3=1

n=6y+3

c^3-6a^3=4b^3-c^3=1

n=6y+4

b^3-2a^3=2b^3-3c^3=1

n=6y+5

c^3-2a^3=12b^3-c^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.

**Answer**

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.

**References**

- Henri Cohen,
*Number Theory Volume I, Tools and Diophantine Equations*,

\S 6.4.7*Skolem’s Equations x^3 + dy^3 = 1*.

**Attribution***Source : Link , Question Author : Sophie , Answer Author : achille hui*