Show that a cube of integer side cannot be partitioned into a finite number of smaller cubes of integer sides, all of different sizes.
geometry/dissections/cubed.cube New
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Problem
Show that a cube of integer side cannot be partitioned into a finite number of smaller cubes of integer sides, all of different sizes. Why does the proof not also rule out a square of unequal smaller squares, which is possible? There is a standard puzzle - show it is impossible to have a cube of integer side partitioned into a finite number of other cubes, also of integer sides, and all of DIFFERENT sizes. ... Why doesn't the same proof extend to "no square of unequal smaller squares", which we know IS possible. Why can't we just look at the top edge and the smallest square there, go across, then carry on as before? ... What's specially different between 2D and 3D here? -- Bill Taylor <mathwft@math.canterbury.ac.nz>, March 1998 [ Editor: the first paragraph is the editor's summary. See also geometry/dissections/squared.square. ]
Show solution
In this posting, by an Xd X I will mean an X partitioned into a finite number of Xs of unequal sizes. The proof is as follows: Look at the bottom of the cubed cube; you will see a squared square. *The smallest square, S, in this squared square cannot be on the boundary.* Therefore, the cube with S for a face is surrounded by four larger cubes, so its opposite face abuts another squared square. We can now look at the smallest square in this squared square. It is then the face of some cube surrounded by four larger cubes, which we can look at the opposite face of, and so on: since this process can be continued indefinitely, there can be no cubed cube. The difference between 2D and 3D is in the starred statement. The smallest line in a segmented segment can be on the boundary, but the smallest square in a squared square cannot be. -- David Moews <dmoews@xraysgi.ims.uconn.edu>