The puzzle was to show that a cube cannot be dissected into a finite number of smaller cubes, all of different sizes. Here’s a slightly expanded version of the proof given in *Littlewood’s Miscellany*. The approach, as with many impossibility proofs, is to assume that a solution exists and to show that this leads to a contradiction.

We start by considering the bottom face of the big cube; let’s call this face B for simplicity. The cubes standing on B induce a perfect square dissection of B, i.e. they divide it into a finite number of squares of different sizes. [Check: a finite number, because the number of cubes must be finite; unequal sizes, because the cubes are of unequal size.]

The smallest cube standing on B must stand on an internal square, i.e. not along the edge of the big cube. [Check: why? Think about the square dissection and consider whether, if the smallest square is in the corner, you can fit larger squares around it.]

The smallest cube standing on B must be surrounded by larger cubes. Therefore, the top face of this smallest cube is surrounded by “walls” formed by the faces of the surrounding cubes. Therefore, the cubes standing on this top face — let’s call it T — must induce a square dissection of T. [Check: why can’t T just be covered by a single cube?]

But now we’re back to where we started, with a square dissection problem. Take the smallest cube involved in the dissection of T, and the process repeats. At every stage we end up with a square dissection which requires smaller and smaller cubes to be added; this process never terminates and so we can never dissect the original cube into a finite number of unequal smaller cubes.

Neat, huh?

(DP)

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