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 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. 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 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 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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