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Answered 2007-11-23 19:51:42

This is a cracking question. [I'm assuming that 'integral side lengths' is intended also to mean integral positions. That is to say, the position vectors of the cubes' vertices should consist only of integer components...such that we are effectively drawing our cube in a 3D grid. If this was not a condition, the answer would be infinity.] So then: first, there are 53 = 125 1x1x1 cubes. Then 43 = 64 2x2x2cubes. 33 = 27 3x3x3 cubes. 23 = 8 4x4x4 cubes. And, stunningly, 13 = 1 5x5x5 cube. So far we have 225 cubes. That would not be interesting. But we have only counted the cubes with edges parallel to the edges of the main cube. Suppose there are some more cubes formed by diagonal lines? Since edge lengths and vertex positions both have to be integral, and we are working in 3 dimensions, we are actually looking for Pythagorean quadruples - integer solutions to a2 + b2 + c2 = d2 Up to now we have only used trivial solutions like this: 52 + 02 + 02 = 52 Our edges have only moved in one dimension. We might consider edges that move in two dimensions, using the smallest Pythagorean triple: 32 + 42 + 02 = 52 But diagonal edges of length five are clearly not going to fit. So introduce the third component, and we find this: 12 + 22 + 22 = 32 By using diagonal edges it is possible to constuct a few more valid 3x3x3 cubes. It turns out, I think, that four of these can be formed. So we have 229 cubes in total.

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