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Squaring the Square -> Cubing the Rectangular Prism

  1. Feb 20, 2012 #1

    It is known that simple perfect squared squares exist, with the lowest one being of order 21. It is also known that no simple perfect cubed cube exists, for the smallest cube would, again, require cubes atop it that form another squared square, which would require yet another, and so on.

    However, can it be shown that a simple perfect cubed rectangular prism exists? Or, that merely a compound perfect cubed rectangular prism exists?
  2. jcsd
  3. Feb 20, 2012 #2


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    Hey TheFerruccio

    What do you mean by the term 'rectangular'? By this do you mean that if you start with a rectangular cube with certain dimensions you generate the same kind of effect as the squared square but with the idea that you get a perfect rectangular prism with the same dimensions?
  4. Feb 20, 2012 #3
    I mean, given a rectangular prism of dimensions [tex]x_1, x_2, x_3[/tex], where [tex]x_1 \ge x_2 \ge x_3[/tex] can it be perfectly and finitely filled with unique cubes of edge length [tex]x_n[/tex], where [tex]x_{n+1} < x_{n}[/tex] and [tex]x_n < x_3 \forall n > 3[/tex]
  5. Feb 20, 2012 #4


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    Are there any other restrictions? Do the values have to be integer or rational numbers of edge size or can they be real numbers of any kind?
  6. Feb 20, 2012 #5
    They do not have to be integer values, and they do not have to be rational. I'm scratching my head over imaginary number lengths. Since edge lengths are imaginary, I can't imagine being able to span a real number (say, the length of the edge of the prism) with imaginary components, unless you simply had some positive and some negative, but that would have no effect on the real side, anyway.

    I suppose I didn't make a restriction for the realness of the larger prism, though. I suppose it doesn't have to have real number edge lengths. But, for now, let's use the restriction that all edge lengths must be real numbers.

    By the way, this is not a problem I was given. I just was asking myself this. For instance, while a VERY finite number of squared squares exist, so many more squared rectangles exist. It's a looser constraint. However, while it's shown that NO cubed cube exists, I've seen no assertion that no cubed rectangle exists. The proof of a cubed cube breaks down when you expand the restriction to all rectangular prisms, and not just cubes, since you're now able to square rectangles.
  7. Feb 20, 2012 #6


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    My intuitive guess is that 'yes' its possible but I can't really give a solid argument for the proof, so take that as you will.

    One thing that might help is to investigate the bin packing problem in three dimensions since this corresponds exactly with your query.

    This problem is used for example to find an optimal way of packing trucks with boxes for shipping so that the trailor packs in as much as possible given the different box sizes.

    Although your problem is a bit more constrained, I think you will get some good ideas from looking at these because it corresponds pretty well to your type of problem.
  8. Feb 21, 2012 #7
    I think I just answered my own question: no.

    The bottom rectangle will be cubed, which demands a case of squaring the square, so, by the same proof as above, you cannot cube the rectangular prism.

    Modified question: can you space fill a rectangular prism with smaller prisms of unique arbitrary dimensions, where no two edge lengths match between prisms? The restriction is that every edge length is unique to each prism.
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