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.

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

Clearly the factors k, k+1, and 2k+1 are mutually coprime.  In view 
of divisibility requirements modulo 9, this equation implies the existence
of integers a,b,c such that one of the following four sets of conditions 

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

The above 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
                    c^3 + (-1)^3 = 2a^3
                    b^3 + (-1)^3 = 2a^3
                    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 the 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]. 

Incidentally, 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.

Having shown there is no solution to the original problem, we might change
the problem slightly by saying that, when unpacking the box we'll stack 
the rounds in a square pyramid, but we will place one round into the 
barrel.  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.

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