From:  (David Moews)
Date:           11 Mar 1998 19:29:42 -0800
Newsgroups:     rec.puzzles,sci.math
Subject:        Re: No CUBE of cubes.

In article <6e7eb1$2am$>,
Bill Taylor <> wrote:
|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?

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