One Dimension Up
As shown above, a square can be dissected into squares of
distinct
integer side lengths.
Is it possible to dissect a cube into cubes of distinct integer side lengths?
Clarification:
The number in each square is the side length of that square.
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6 solutions
Moderator note:
Regarding "the smallest square in a square dissection is interior":
Suppose it was not, and touching the edge. Then all three sides must be adjacent to larger squares. While it's possible to do this with two sides, the third side could only be adjacent to a square of the same size, and each square is supposed to have a unique side length.
When in the corner, a similar issue happens except only one side can be adjacent to a larger square.
Ok, I get it now. You start at a face of the big cube, so there is only one way each side adjacent cubes can 'stick out'. Then, once you determine that the free face is covered by smaller cubes, you can take this free face as your next iteration as it is a nontrivial square disection. Iterating like this yields ever smaller cubes. Nice proof. Very elegant!
I don't think it even matters if the smallest square is interior, the proof works anyways
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You need this to make sure that at each step you have the cube's face completely surrounded.
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Right. We're building a "well" of cubes, and each time we have to ensure that the height of the well is tall enough. We do so by building a well on top of the previous well.
If not, say that the smallest cube had length 100, and that the surrounding cubes (somehow) had length smaller than 110. Then, we might use a cube of side length 10 placed against the edge to help us step out of the well.
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@Calvin Lin – Well, a cube of side length 10 isn't quite possible, but regardless, it's not really realistic.
@Calvin Lin – All it needs is a square hole of size equal to the face of an existing cube. Since we can't use another cube of the same size, we have to cover that square with a collection of smaller squares. But with cubes, we already know that those cubes will create an even smaller square hole.
This is beautiful.
Doesn't the small cube have two free faces not touched by the cubes extending the adjacent squares? If you let these free faces be at the tip and bottom, you can imagine that some side adjacent cubes might be flush with the top of the little cube, and extend downward, whilst the other adjacent cubes could be flush with the bottom and extend up. Prima facie, I don't think it follows that the cubes adjacent to the top or bottom face if the little cube must be smaller. What have I missed?
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The "trick" is to consider what happens at any "face" of the cube we are trying to fill. We can arrange cubes to fill this (such as in the diagram in the question) but the faces of all these cubes are flush against the exterior of the bigger cube while the interior contains a square hole at the smallest of these cubes which has to be covered by a collection of smaller squares.
I like this. I guessed it involved looking for a "smallest" cube but it isn't until you examine how things operate on a square face that it all slots into place. Ironically the solution is staring you in the face when looking at the square disection.
The second paragraph needs a bit more rigor. "We know now that it [the smallest square] must lie at the face interior [dimension conflation], and it is surrounded by bigger cubes [at least the ones also flush with the face are bigger], corresponding to the square neighbors of the face dissection. This means that all the cubes adjacent to the remaining face [what is 'remaining face'?] must be of smaller size and they induce a square dissection of that face. "
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I revised that paragraph. Hope it is clearer now.
This hardly constitutes a proof, but rather a really good hunch why it's not likely possible. It's already a difficult problem to even find instances of a square being divided into squares of distinct side lengths, as exemplified by the example given. Let's assume there exists a cube that's divided into cubes of distinct side lengths. Then any orthogonal cross-section of this remarkable cube would yield a square divided into squares of distinct side length! An amazing super-cluster of such squares, easily dozens of distinct examples, all packed together in a single cube! My instinct said "nah", and so I picked "no".
Edit: A start on a proof would be to consider a series of orthogonal cross-sections from one side of the cube to the other, which changes at integer distances from either. Because of the distinct side lengths of the smaller cube pieces, generally speaking the difference between two successive cross-sections can only be a single, or few, squares, i.e., most of the squares stay put unchanged in size and location, while some change. Obviously it is impossible for just one square to change to just another one square "of a different distinct size", so we are left with considering how a small number of squares of different sizes can change to another small number of squares of different sizes while preserving the shape and dimensions of the perimeter of such squares. On the face of it, it's almost impossible to even do with just a few squares. The smallest known dissections of either a perfect square or a rectangle into squares of distinct sizes involves about a dozen of them, and these small examples do not permit an alternate dissection.
IMO The proof that the 3-D case is not possible is much easier than finding the solution to the 2-D case. I had to look the 2-D case up on the internet.
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Well, we can start with one of the faces of the main cube, which is already a square dissected into smaller squares of distinct sizes. There's going to be a smallest square that is necessarily surrounded by four larger ones. What happens when you examine the cross-section just past the other end of that smallest square? Obviously it'll still be surrounded by those same four cubes (as well as the rest of the cubes which sides make up the original face) that are larger. Is this the proof that you are thinking of?
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That's starting on the right path. Just have to be careful with tracking the arguments and seeing if the cube is eventually allowed to pop out of the well.
[This is not a solution. You can use this to discuss ideas.]
What techniques can one use to find such a dissection of a square?
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It's not an easy problem. It took mathematicians 20+++ years to find an example. (Martin Gardner posed it in 1958, and others possibly brought it up earlier. It was solved in 1978)
umm..can we make a rectangular prism using distinct cubes? even by using infinitely small cubes
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The usage of "infinitely small cubes" suggests that you want to allow for non-integer side lengths, in which case the argument that I was thinking of doesn't immediately apply.
I believe the answer is no, but do not have a rigorous proof as yet.
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What are "infinitely small cubes"?
My solution's line of reasoning is essentially the same as that in wikipedia . It can be used to show that different size cube dissection is impossible for any rectangular cuboid.
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@Eugene Alterman – Eugene, your "infinite descent" argument assumes that we have finitely many cubes. For example, it "might" be possible that we use cubes with side lengths 2 k 1 to construct the infinite sequence of ever smaller cubes.
I agree with Mehdi that the question as currently phrased, doesn't immediately allow for any finiteness condition as yet. Adding in the condition of "integer lengths" or "finitely many cubes" would help resolve that. As such, I've edited that into the problem.
but the question didn't state that the side lengths has to be integers and the dissects are finite
I wonder if this is related in some way to Fermat's last theorem, it can be done for 2 dimensions but no higher. Something like this: The overall square "z^2" is made from the biggest component square "y^2" and the sum of smaller squares which sum to "x^2". "x^2" then divides up yet further and so on. This can't happen for a cube (or higher dimensions). The flaw is that "x^2" does not have to be a perfect square.
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As mentioned in Ray's solution by the community, there is most likely no relation.
It is not clear how the "many squares" and the "distinct squares" come into play. Yes, it is true that both problems only have solutions when n = 2 , but relating a square to a dimension would likely require a lot more machinery.
As a suggestion for another problem, suppose we relax the constraint on distinct cubes and allow up to n copies of any of the smaller cubes. Can we determine the smallest value of n which allows us to construct a larger cube?
Trivially we know we can do this if n=8, but can we solve for n=7, 6,...,2?
someone had mentioned this question has a solution if we consider infinitely smaller cubes and this would not work because eventually we would have cubes smaller than a single integer. Although if we build the cube from the inside out starting with a cube of side length 1 and adding larger cubes to it to find the smallest cube that can be composed this way. considering the large cube can extend to an infinitely long side length then there must be a solution. Or am I missing something?
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I don't even think it is possible with an infinitely large cube being filled with an infinite number of finite and infinitely small cubes.
When considering the solution to the square problem and considering what happens when we project those squares into cubes, there is a significant difference between the size of the larger cube and the smallest exposed face. Even if this were only 10:1, that smaller cube is still only 10% of the distance towards the opposite face of the cube.
You may not get a cube. As an example, consider the Fibonacci Spiral , in which we place squares of side length f i next to the previous square. Each time, we get a rectangle, and can continue this infinitely, without ever getting a square.
Each cube has 12 pairs of 2 adjacent sides. Or many pairs if inside ones. If u take out any cube of side x1 from original cube of side p... U can not take out 11*(x1^3 ) unit cubes.. if x1 < p/2. And when u cut out cubes from any sq-side.. it will leave an uneven 3-d figure at the inside.. if all x<i>-s of the partition set of p with different numbers.. are less than p/2.
Even if they are not.. If p^2 = S<p> ... for any of the faces... with each term in the set S<p> ... being squares.. in x-y plane. And cut all cubes out of them out.. They will leave a z axis figure with different either x or different y-coordinates. A stepped 3-d figure. And those co-ordinates can not be taken again. So.. no contiguous blocks of any dimension available from p^3 - {S<p> cubes} . Without getting smaller than already taken small dimension. It can ger as small as 1 .. not any more less. There are a total of p-1 integers available for making each side.
(P-1)^3 = P^3 - (3P^2 -3P +1) =>( P-1)^3 + (3P^2 -3P +1) = P^3
Now... the second term in the ledt hand side of the above equation is not a cubic form of any number.
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Hm, I'm not quite sure what you're trying to say here.
It is possible to find distinct cubes that sum up to another cube, so a volume analysis would not work. You do seem to be going for the "it can get as small as 1, but not any more less" argument, but it's not clear to me how you justify it especially given the "random" selection that you took.
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ya.. i was not very concrete in explaining it formally and it took a little while for me to get back what i had in mind to say. i meant when one starts getting out the cubes of different numbers, from a cubic number, one may start at random by scooping out an x^3 volume from the given cubic figure. Now, what remains after the x^3 is taken out from the cube, is not a cube or can not be expressed in terms of cubes... using recurssion. x can get as small as 1 for integers. or as big as the side of the given cube itself.
Does it have something to do with Fermat's last theorem?
Maybe it does.
But certainly it could be solved without relying on it. Also, Pi Han is correct.
No.
The question asked "whether a cube can be dissected into 2 or more cubes." , not "whether a perfect cube can be expressed as the sum of precisely 2 perfect cubes."
To clarify: a "cube" represents a 3D figure, whereas a "perfect cube" represents an integer.
It's like squaring a rectangle. The first squared rectangle was discovered in 1925 by Polish mathematician Zbigniew Moroñ. In particular, Moroñ found a 33 x 32 rectangle that can ve tiled with nine different squares with lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18. He also discovered a 65 x 47 rectangle tiled with 10 square tiles with lengths 3, 5, 6, 11, 17, 19, 22, 23, 24, and 25. For years, mathematicians claimed that perfect square dissections of squares were impossible to construct.
But how does this answer the question of "Can we dissect a cube into distinct cubes of integer lengths?" ?
This is a Wikipedia's reference : https://en.wikipedia.org/wiki/Squaring the square#Cubing the cube
It doesnt say, that 2 or more squares of the same sidelengh cant be placed besides each other. If this is allowed, it is obviously possible to dissect a cube into smaller cubes. So I took yes and wondered, why this wasnt correct. There was no absolut information, it just said "dissected like this".
First of all, observe that the smallest square in a square dissection is interior, i.e. it cannot lie at an edge or in a corner of the square being dissected. Indeed, assuming one neighbor of a corner square or two neighbors of a non-corner edge square have larger sizes, there must be more neighbors of that square, and they must have smaller sizes.
Now assume that there exists such a dissection of a cube. Take any of the cube's faces and consider all the cubes that are adjacent to it. They induce a square dissection of that face. Consider the smallest square of that dissection. As we already know, it is located in the interior of the face. Consider the corresponding cube. One of its faces lies on the face of the original cube, and four other faces are adjacent to bigger cubes that surround the remaining face. The cubes adjacent to that face must be of smaller size, and they induce its square dissection.
Proceeding repeatedly in this manner we obtain an infinite sequence of ever smaller cubes, which is in contradiction with the requirement that the dissection should be finite.