## cube + cube + cube = cube

The following identity is a bit isolated in the arithmetics of natural integers $$3^3+4^3+5^3=6^3.$$ Let $K_6$ be a cube whose side has length $6$. We view it as the union of $216$ elementary unit cubes. We wish to cut it into $N$ connected components, each one being a union of elementary unit cubes, such that these components can be assembled so as to form three cubes of sizes $3,4$ and $5$. Of course, the latter are made simultaneously: a component may not be used in two cubes. There is a solution with $9$ pieces.

What is the minimal number $N$ of pieces into which to cut $K_6$ ?

About connectedness: a piece is connected if it is a union of elementary cubes whose centers are the nodes of a connected graph with arrows of unit length parallel to the coordinate axes.

Edit. Several comments ask for a reference for the $8$-pieces puzzle, mentionned at first in the question. Actually, $8$ was a mistake, as the solution I know constists of $9$ pieces. The only one that I have is the photograph in François's answer below. Yet it is not very informative, so let me give you additional informations (I manipulated the puzzle a couple weeks ago). There is a $2$-cube (middle) and a $3$-cube (right). At left, the $4$-cube is not complete, as two elementary cubes are missing at the end of an edge. Of course, one could not have both a $3$-cube and a $4$-cube in a $6$-cube. So you can imagine how the $3$-cube and the imperfect $4$-cube match (two possibilities). Other rather symmetric pieces are a $1\times1\times2$ (it fills the imperfect $4$-cube when you build the $3$-, $4$- and $5$-cubes) and a $1\times2\times3$. Two other pieces have only a planar symmetry, whereas the last one has no symmetry at all.

Here is a photography of the cut mentionned above.

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(+1) Regarding your definition of connectedness: Do you need the "arrows" to be of unit length? Otherwise, for example, two unit cubes along the same edge of $K_6$ but on opposite faces of it would be "connected" by an arrow of length 5. – John Bentin Jan 24 2011 at 12:31
A few years ago Raphaël Cerf played around similar problems for so-called polyominoes. This was in relation with the metastability of the 3D Ising model (no less), see combinatorics.org/Volume_3/Abstracts/v3i1r27.html. So he might know the answer. – Didier Piau Jan 24 2011 at 13:01
@Gerhard The solution to a version of the dissection problem is known: one cannot turn a cube into the regular simplex of the same volume through a finite number of cuts by hyperplanes, since these have different so called Dehn invariants. This holds in every dimension at least 3. In dimension 2, the area is a complete invariant, meaning that every polygons of the same area are equivalent, and the only nontrivial step of the proof is that any rectangle $L\times\ell$ can be turned into the square with side $\sqrt{L\ell}$. – Didier Piau Jan 25 2011 at 6:50
@Denis Where can one find the solution for N=8? Is it known if there are more then one solution for N=8 ? – fastforward Jan 29 2011 at 23:32
Here is an interesting link about the analogous problem for pythagorean triples : mathafou.free.fr/pbg/sol110d.html It turns out that for small pythagorean triples, a puzzle with only 4 pieces can be constructed. The smallest triples for which the minimal number of pieces doesn't appear to be known are (20,21,29) and (28,45,53). – François Brunault Mar 13 2011 at 9:31

JHI's elegant lower bound of $8$ on $N$ is achieved by an explicit dissection. I show my construction below; you might want to try to find a solution yourself before proceeding $-$ it makes for a neat puzzle. There may well be other ways to do it.

If somebody can make a "$3$-dimensional" graphic or picture of the $8$-piece dissection, you're welcome to add it by editing my answer. My diagrams are two-dimensional, labeling each piece with its height. Fortunately the dissection is simple enough for this to be possible; in particular, the eight pieces comprise four boxes and four L-shaped prisms. This also made it possible to find the solution using just pencil and paper on an otherwise uneventful international flight.

Begin by cutting the $6 \times 6 \times 6$ cube top to bottom into three pieces, as shown in top view in the first square diagram. Then cut each piece horizontally in two, dividing AB into $3+3$,$\phantom.$ C into $4+2$, and D into $5+1$. Each AB piece is then further subdivided into a box B and an L-shaped prism A. The second diagram shows (say) the bottom layer of four pieces, and the third diagram shows the top. Note that the AB subdivisions are not quite the same.

Pieces with the same color will come together to form a smaller cube. The larger C piece is a $4$-cube, and the two A pieces form a $3$-cube as shown. It remains to construct the $5$-cube from the remaining five pieces. The last two diagrams show the bottom and top of the $5$-cube.

The two $5$'s are the larger B piece, rotated to span the entire height of the cube, and the thick D piece. The thin D piece completes the bottom, with width $1$. The top is filled by the thinner C piece and the smaller B, both rotated to height 4. QEF

I guess that a physical model won't make for a hard puzzle to reconstitute into either one or three cubes (e.g. the AB, C, and D parts of the $6$-cube are independent) but would still make a nice physical model of the identity $3^3 + 4^3 + 5^3 = 6^3$.

This dissection is specific to the solution $(a,b,c;d)=(3,4,5;6)$ of the Diophantine equation $a^3+b^3+c^3=d^3$; I don't know whether an $8$-piece dissection is possible for any other solution. JHI's analysis shows that one can never get below $8$, and in some cases even that's not possible: if $a<b<c$ and $a+c<d$ then there's at least one corner of the $d$-cube, say $(1,1,1)$, that contributes to the $a$-cube, but then any cell $(x,y,z)$ with $\max(x,y,z) = a+1$ cannot connect to any corner. This first happens for $(a,b,c;d) = (6,32,33;41)$.

What's the minimal dissection for the "taxicab" identity $1^3 + 12^3 = 9^3 + 10^3$? JHI's corner-cutting argument shows that at least nine pieces are needed.

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Beautiful! I'm especially impressed that you found a way to explain it in a convincing way by using 2-D figures. – Denis Serre May 5 at 20:06
Actually, I should like to accept your answer. Unfortunately, I accepted already JHI's. – Denis Serre May 5 at 20:07
@DenisSerre, I think you can change your accepted answer. Just click the check mark for Noam's answer. – Joel Reyes Noche May 6 at 4:21
@Joel. Yes, I can! – Denis Serre May 6 at 5:25
You should credit Jaap Scherphuis of jaapsch.net/puzzles who told me this argument. – JHI May 8 at 18:23

8 is the least.

You can't have a piece of length 6, thus no two corners of the 6x6x6 cube can be part of the same piece, the cube has 8 corners, we need then a minimum of 8 pieces.

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Bravo! Gerhard "Ask Me About System Design" Paseman, 2011.01.30 – Gerhard Paseman Jan 30 2011 at 22:38

The solution mentioned in the question consists in fact of $9$ pieces, as shown in the following picture I took yesterday. I don't know whether a solution with $8$ pieces exists.

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