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Prove that cube cannot be partitioned into $n>1$ cubes, such that each of them has different side length.

I believe tallhis is not hard problem, but I just do not have an idea how to start. I tried to prove it by contradiction. I assumed that large cube is tiled with a $n>1$ cubes such that at all of small cubes have different side length. Then I tried to use induction.

For $n=2$ it is trivial that it is impossible because in that case one side of a small cube must be twice of other side. For $n=3$ it is impossible because wherever we place first small cube, the remaining area cannot be a cuboid so we cannot partition the remaining area into $2$ cubes. For $n=4$ I do not see an easy way to prove it.

I do not believe that induction is the best method, but any solution will be appreciated.

Mike Pierce
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    I don't understand? For instance, you can partition a cube into 8 equal sub-cubes. – Mankind Apr 28 '15 at 20:02
  • What does tile a cube with cubes mean? – Henry Apr 28 '15 at 20:02
  • @HowDoIMath. This is a typo. I edited question. –  Apr 28 '15 at 20:03
  • @Henry. That means partition of cube into small cubes such that there is no cube ovarlap and remaining space. –  Apr 28 '15 at 20:08
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    A full answer appears in wikipedia. – vadim123 Apr 28 '15 at 20:08
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    I recall seeing a proof. First you can use the fact that any given face must be a square that is decomposed into smaller distinct squares. Then you consider the smallest square on the face. It must be a hole because the surrounding squares are larger, therefore part of taller cubes. The lower face must then be a squared square as well and you get into infinite descent. – Ross Millikan Apr 28 '15 at 20:09

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As user vadim123 said in the comments, a full answer to this question appears on this page on Wikipedia.

[Given a cube $C$] suppose that there is such a dissection. Make a face of $C$ its horizontal base. The base is divided into a perfect squared rectangle $R$ by the cubes which rest on it. Each corner square of $R$ has a smaller adjacent edge square, and $R$'s smallest edge square is adjacent to smaller squares not on the edge. Therefore, the smallest square $s_1$ in $R$ is surrounded by larger, and therefore higher, cubes on all four sides. Hence the upper face of the cube on $s_1$ is divided into a perfect squared square by the cubes which rest on it. Let $s_2$ be the smallest square in this dissection. The sequence of squares $s_1, s_2, \dotsc\;$ is infinite and the corresponding cubes are infinite in number. This contradicts our original supposition.

If a $4$-dimensional hypercube could be perfectly hypercubed then its 'faces' would be perfect cubed cubes; this is impossible. Similarly, there is no solution for all cubes of higher dimensions.

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Mike Pierce
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Suppose it is possible to represent a cube of length k(say),using smaller cubes. Assume the maximum number of smaller cubes needed to represent the larger cube to be n. Then the total volume , $ k^3 = x_1^3 + x_2^3 ... +x_n^3$ call this (1). From (1) we can find $x_1^3$ by multiplying the whole expression(1) by $x_1^3/k^3$ if we substitute $x_1^3$ back in (1) we get $k^3 = (x_1^2/k)^3 + (x_1.x_2/k)^3 ...+ x_2^3 + ...+x_n^3 $ a representation of more than n elements contradicting maximal n. If there is a mistake please tell.

nrynn
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  • One important thing is that smaller cubes must be distinct. Maybe some elements in your last expression are the same. –  Jun 19 '16 at 11:58
  • you are correct.. Can we instead take the representation among the remaining representation which has a minimal element x1(say) then contradicting there is another minimal which is x1 ^2/k ? – nrynn Jun 19 '16 at 12:16
  • Maybe it is possible to prove it in some way using your approach, but very simple and short proof can be found on this page. –  Jun 19 '16 at 16:45