One In The Chamber

A well-known puzzle, attributed to Lucas, involves stacking cannon 
balls in a square-based pyramid.   The apex has 1^2 ball(s), the 
second layer down has 2^2 balls, the third layer 3^2 balls, and so
on.  The classic question asks "How many layers are necessary so that 
the total number of balls is a perfect square?"   What if we alter
this question to ask for the number of layers necessary to give a
perfect cube?  This might seem more natural, assuming the balls come
packed in cubical crates.

Be careful - it's a trick question.  When unpacking the box we'll 
stack the rounds in a square pyramid, but one round goes in the 
barrel.  (This is commonly overlooked.)  Thus we need a cube that 
equals a pyramid PLUS 1.  If we ship the rounds in a cubical box of 
26^3 = 17576, we can put one round in the chamber and stack the other 
17575 in a square-based pyramid of 37 layers.

The sum of the first k squares is (k)(k+1)(2k+1)/6, so if this is
to equal the cube of some integer N we have

                 (k)(k+1)(2k+1) = 6N^3

and clearly the factors k, k+1, and 2k+1 are mutually coprime.  In
view of mod 9 requirements, this implies the existence of integers
a,b,c such that one of the following four sets of conditions apply:

              k = 6a^3      k+1 = b^3      2k+1 = c^3
          or
              k = a^3       k+1 = 6b^3     2k+1 = c^3
          or
              k = 2a^3      k+1 = b^3      2k+1 = 3c^3
          or
              k = a^3       k+1 = 2b^3     2k+1 = 3c^3

Therefore, since (k) + (k+1) + (-(2k+1)) = 0, we're led to an 
equation that is either of the form

               x^3 + y^3 + 6z^3 = 0
or else
               x^3 + 2y^3 + 3z^3 = 0

These equations are closely related to each other and, in general, 
solutions to them are quite rare, even without the additional 
requirements that the components be of the form k, k+1, and 2k+1.  
In fact, it seems that Legendre once (mistakenly) stated there 
were no integer solutions of x^3 + y^3 = 6z^3.  For more on this
problem, and what Legendre might have been thinking, see the
article The 450 Pound Problem.

I should add that these conditions preclude the existence of any 
solution other than the trivial ones.  This follows because the 
four possibilities listed above lead, respectively, to

                    c^3 + (+1)^3 = 2b^3
               or
                    c^3 + (-1)^3 = 2a^3
               or
                    b^3 + (-1)^3 = 2a^3
               or
                    a^3 + (+1)^3 = 2b^3
              
and, as Euler proved, the only integer solutions of x^3 + y^3 = 2z^3
are with x=y.  Thus only integers solutions of k(k+1)(2k+1)=6N^3 are
[k=1,N=1], [k=0,N=0], [k=-1,N=0], and [k=-2,N=-1].

By the way, the classical problem of stacking the balls in a square-
based pyramid has essentially the same answer as stacking the balls 
in a triangular-based pyramid, except multiplied by a factor.