# Prove that cube cannot be partitioned into $n>1$ smaller distinct cubes.

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.

• 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. – user164524 Apr 28 '15 at 20:03
• A full answer appears in wikipedia. – vadim123 Apr 28 '15 at 20:08
• 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

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.
[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.