Do cubed cubes exist?

I've been wanting to make more math videos recently, so here's one! Not counting the animations, I probably put this together in 6 hours.
Squaring squares website: www.squaring.ne...
The 2017 NumberPhile video about squared squares: • Squared Squares - Numb...
My tweet from a month earlier showing this exact same animation: / 1529068510579523584
Source that says that the 21-square squared square was discovered in 1978: mathworld.wolf...
Music:
"Wings - Touching Piano Music - Royalty Free" by Jonny Easton
• Wings - Touching Piano...

[@thegiftedspriter7427]

Ah, the classic “I have a proof for this, but not the time/space to show it here.” Fermat would be proud.

[@minhkhanhvu5180]

the proof is trivial and is left as an exercise for the viewer

[@and_the_first_last]

@@minhkhanhvu5180 Wait really? I must be bad at trivia then cause I don't have a clue what it is

[@zmaj12321]

@@and_the_first_last "Left as an exercise for the reader" is a common joke people make about mathematical works, which often leave "obvious" things for the reader to figure out themselves, even if they're not always that obvious.

[@yackamajez]

The classic proof by “just trust me bro”

[@TheRenegade...]

@@and_the_first_last trivial means "simple", not "related to trivia"

[@jackeea_]

"It is impossible to seperate a cube into many cubes. I have discovered a truly marvelous proof of this, which my time before VidCon is too narrow to contain" - carykh, 2022

[@Rudxain]

Fermat moment lol

[@Rudxain]

@@indetermite I read a comment that talked about how a variant of FLT is key to the solution, so yes, you're kinda right

[@DreamyyArt]

Rubik's cubes

[@YezzirGamer]

depends on the size of the cubes

[@ENDI8089]

​@@DreamyyArtthe cubes are all the same tho

[@smilesnack9482]

This makes me wonder if a screen saver that generated squared squares would be possible? It would pick a random squared square with a selection of 4 different colors, and tile it until the computer wakes up, or the square is filled (which would just make it start over with new colors and a new squares square.)

[@hecko-yes]

or you could make the squared square be a component square of another squared square and just keep zooming out

[@gamekiller0123]

@@hecko-yes This is one of the easiest ways to show that there are infinitely many squared squares (once you know that at least one exists).

[@Mkrabs]

@@hecko-yes wtf that's brilliant

[@soapycanthandle]

That'd be awesome

[@SNESpaghetti]

I would square at it at all times

[@creativebeetle]

I know it's a minor detail, but I always love the touch of making repetitive sound effects change pitch randomly. It goes a long way in terms of appeal, and it's often unnoticed.

[@RTOF]

It was actually satisfying watching that animation come together, the popping sounds are also kinda a bonus

[@somedudewatchingyoutube9163]

It’s asmr but better

[@RTOF]

@@somedudewatchingyoutube9163 yes

[@BamsyTheSergal]

picking items up after breaking a chest

[@RTOF]

@@BamsyTheSergal really sounds like it

[@sofia.eris.bauhaus]

kzbin.info/www/bejne/bnWmlomZjLajjrs

[@sushiirolled]

The animation was so satisfying to watch! The sounds and the squares sliding into place were really calming.

[@somedudewatchingyoutube9163]

It’s sounds like guns reloading

[@sushiirolled]

@@somedudewatchingyoutube9163 kinda does ngl

[@Okaz_]

It’s so satisfying

[@iateyourbread]

asmr squaring squares

[@YoshiFawful64]

Actually figured out a pretty simple proof Let's assume a cubed cube exists. Each of its sides must be a squared square, so let's look at one of those sides - specifically, the smallest cube on it (let's say it has side length _n_.). How do we cover its opposite face? Since it's surrounded by larger cubes, their sides create a "tunnel", thus forming a flat _n•n_ area. Since size _n_ is already taken, the only way to cover that surface would be with another, smaller squared square of cubes; however, then the same issue arises with the smallest cube in *that* array, and the problem repeats ad infinitum. Thus, a cubed cube is impossible.

[@RubyPiec]

"This can't possibly exist, you might be wondering why can't there be a perfect cube dissection if there can be perfect square dissections because a square and a cube are kind of the same thing. Well they're not" 10/10 proof

[@HanzCastroyearsago]

1^3+2^3=9 (square) 9+3^3=36 (square) 36+4^3=100 (square), So for a number to be a square and a cube, it has to be a power of 6 (as 2 and 3 are prime numbers)

[@nooneinparticular3370]

Proof by intimidation.

[@LewisBavin]

Proofs by Contradiction, Induction & Brute Force found dead in a fucking ditch

[@lyricalcarpenter]

@@nooneinparticular3370 in conclusion, AAAAAAAAAAAAAAAAAAAAA

[@Rudxain]

That reminds me of turning a sphere inside-out topologically. You can't turn a circle inside out without creating sharp bends, but a sphere can allow you to do so

[@matthew.y]

Littlewood mentions a proof for this in a book. It involves a neat infinite descent argument - a trick I rarely come across nowadays.

[@carykh]

Interesting, that sounds like the same direction I was planning on going with my video! Maybe I somehow read his book a decade ago and retained the proof, but forgot I read the book? haha

[@matthew.y]

Either way, i'm sure your visuals will add a lot of depth and intuition to it ^^

[@leanna5733]

e

[@sage5296]

Something like "So you have a lower area you need to fill in, and you can't do it with one cube since you already used that size of cube, now take the smallest cube on that layer and you're back where you started", and then therefor there would have to be a smaller positive integer for every integer, which there isn't so it's no t possible? I remember grappling with this problem some time ago, but I don't remember why

[@NYKevin100]

​@@sage5296 Apparently someone has written the full argument on Wikipedia. The short explanation is that you look for the smallest cube on one of the big cube's faces, then ask what's on the other side of that cube (another squared square), find the smallest cube in that, and it turns out that you can keep playing this game indefinitely (you'll never get to the far face of the big cube, because at each step, you're always selecting a cube which is less deep than any of its neighbors). You don't even strictly need infinite descent, just observing that the overall solution must be finite is enough.

[@KatzRool]

I love how the rate of growth was proportional to the side length. If you paid enough attention, you could probably approximate the final size of the squares before they finished popping.

[@NOL_bm]

Proof: By using any of the perfect squared squares, you could easily turn them all into cubes for a 3d view, however with a 3d view, you will notice that with the smaller cubes such as 2, there will be a hole that can only be filled with the number 2, breaking the rules, and therefore, a perfect cubed cube is impossible.

[@wightboy12345]

Exactly what I was thinking. You could claim that the surface of the smaller cube could be a square tiling problem itself, but that becomes recursive with the cubed height problem so you’re back to square one (pin not intended until I was typing it, then it became completely intended).

[@isavenewspapers8890]

I have discovered a truly remarkable proof of this theorem, but this video is too small to contain it.

[@brunoldo]

I'm a stardew valley fan and the sound of the squares growing drove me insane

[@-FutureTaken-]

That feeling when you need to withdraw 300 stone from a chest

[@SlyHikari03]

Minecraft too

[@thepizzaguy8477]

pov you bombed the quarry

[@johnbrown-jd7od]

When I was watching the animation, I assumed the condition was that the areas of the squares need to be integers (as opposed to the side lengths). However, this brings up an interesting question. If the areas had to be integers, then this would give you a bit more flexibility (as square root values for the side lengths are now possible). Therefore, I wonder if it's possible to make a cubed cube in this case? If the volumes of the cubes had to be integers, then the side lengths can have any cube root value, giving you more flexibility.

[@DreadKyller]

Unfortunately not really. You can think of any decimal representation of this problem as integers by multiplying all lengths by 10^n where n is the number of decimal places of the number with the most decimal places. So a total size of 100 with a cube taking up 31.425 could be equated to the same problem with a total size of 100,000 and a cube taking up 31,425. You may argue irrationals, but then you just have to consider the theoretical case of the question as n approaches infinity.

[@AssemblyWizard]

@@DreadKyller And how would you consider the infinite case?

[@DreadKyller]

@@AssemblyWizard Again, you can't really consider the infinite case, but you can consider the limit as it approaches infinity. Essentially applying a bit of calculus. The point of the matter is that at any given integer decimal length you can make an equivalent problem using strictly integers, this fact remains true no matter how close n gets to infinity. At infinity we'd be dealing with infinitely large integers instead of numbers with infinite decimal places. The point of the argument is that decimal places make no difference in this problem as decimal places become arbitrary when we're discussing ratios between numbers. the ration of 4 and 3.333333333... is the same as 40,000 an 33,333.33333... when dealing with ratios/percentages between numbers decimal places become more or less arbitrary as long as both sides are scaled the same. Regardless, if the proof he's going to show is related to what I expect his proof explanation in the next video will be (as Matthew in the comments stated, using Fermat's method) this isn't even important to the problem.

[@_Heb_]

@@DreadKyller I'm compelled to say that we aren't allowed to estimate like this, because this puzzle still plays in the realm of finite space. Sure, (10^n)/3 and (10^n)/3 - 1/3 are almost negligibly similar for arbitrarily large n, but in a game like this changing a side length by the tiniest amount can change the validity of a solution. However, what we can do is multiply each side length by the LCM of every denominator of every non-integer side length. That would put us in the realm of integers fairly easily, but it unfortunately only works with real side length. As far as I can tell, there's no similar method to convert square root side lengths into integers.

[@dinamosflams]

you gave me another idea. if its possible to use squares and cubes with irrational lengths (like root 2) can we also use negative and complex numbers for lengths to subtract the areas and volumes?

[@epicoanimator4302]

After such a long time (5 months), Cary has finally uploaded 2 more videos.

[@somedudewatchingyoutube9163]

That’s like half a 10 months

[@iabgunner8701]

@@somedudewatchingyoutube9163 dude, like thats a quarter of 20 months.

[@normalman9123]

@@iabgunner8701 dude, thats like 1/20th of 100 months. (8y 4m)

[@cheeseburgermonkey7104]

@@normalman9123 dude that's like a hundredth of 41 years and 8 months

[@OmegaDoesThings]

Yeah, this is definitely one of those "intuitively I know why cubed cubes are impossible, but I don't know how/recall enough maths to formalize it". It's basically something like, when you place a cube, you then have to fill all of that remaining space with other cubes. But you need unique side lengths, and the *volumes* have to sum up to the remaining space you need while satisfying both "the sum of the side lengths is equivalent to the supercube side length" *and* "the sum of the areas have to equal a given side", and as other comments have mentioned, that just boils down to some kinda Fermat problem. So then you either have to violate the uniqueness constraint, or the "it has to be a cube and not a rectangular prism" constraint. Taniyama-Shimura eat your heart out.

[@CountingCents]

This video gives Fermat's Last Theorem vibes.

[@Riiyye]

why is this so calming??

[@Rolagle]

Dunno

[@SlappleBattles]

0:31 Um, technically, you only need one square 🤓👆

[@gamingwithpapaandfriends]

and it would be side length 112

[@ScutoidStudios]

2011 CARY GOT ME SO NOSTALGIC WHAT-- Regardless, I really like this video, it's always so nice to see a new carykh video.

[@T3sl4]

Oh boy, are we soon going to add CaryKH Cubes alongside the likes of Parker Squares?

[@PoisonIvory088]

Alt title: Cary talking while weird bubble sounds play

[@Le_Codex]

I hope you'll be doing a follow-up video where you detail your proof in full details, this sounds like a really cool math problem

[@jonathanlevy9635]

I think I got a proof which is quite simple: There must be the smallest cube in this partition and it must touch at least three other cubes, but this is impossible since they will always overlap with each other. It is a trivial statement but it requires a bit more explanation which I will leave as a task for the reader.

[@Neyobe]

Love you Cary 😭 you and Michael-my childhood kings

[@masela01]

cary it's been six months

[@OjingeoChaemuchim]

A face of cubed cubes must be a squared square. Consider the smallest square of the squared square. Its face must also be adjacent to squared square. The process must go indefinitely and cannot end. QED.

[@treestomped6954]

That is an interesting idea. I feel it won't work though as cubes have different levels to how deep they are. For example, if you have a 2x2x2 cube and have the rest of the area around it fit perfectly somehow, you're still going to be left with a gap that can only be fulfilled by an additional 2x2x2 cube, which is an issue as you've already used one, thus making it impossible to solve unfortunately.

[@toatrika2443]

well thats not entirely true. you assume that one would have to start by aligning cubes on one plane which is not necessarily the case. therefore you havent proven that its impossible generally.

[@NOL_bm]

@@toatrika2443 the cubes are required to be aligned on each of the 6 faces, meaning what he is saying is exactly right I thought this through myself when trying to figure out what proof there could be, came to the same conclusion and did not see any loopholes.

[@bugorgans]

carys voice at night is so calming

[@karlkewbz]

Honestly even if this is unfinished! this still looks *very* nice :o

[@Paxep_]

IDK why but reading the title gave me an aneurysm

[@hisopio231]

Ngl a jigsaw puzzle of this would be kinda fun

[@tokiponafan4126]

I need more of this

[@PoisonIvory088]

I wish i could come to vid con 2022 and the annoying part is that i just LEFT California

[@quasistarsupernova]

I was so excited about a new video, and all I got was popping sounds and a proof that wasn't even shown.

[@ben9583_]

This is a very interesting topic! I suspect this will remain in the back of my head until the proof video comes out...

[@KatzRool]

Unfortunately hypercubing the hypercube is just as impossible as cubing the cube, and all higher dimensions fail for the same reasons.

[@brandonsaffell4100]

What about fractal dimensions? Is there any where it works, or is it strictly a 1d and 2d thing.

[@KatzRool]

what the heck is a 1.5-cube lmao

[@brandonsaffell4100]

@@KatzRool some math wizard shit for sure. Unfortunately I am not a math wizard, but a mere math scribe. Was hoping to find a proper math wizard that might be able to help me answer your question better.

[@drdca8263]

@@KatzRool this probably wont work well, but here's an idea: If we think of a n-cube with opposite sides identified, this is like the group S^n , and if you like, take the Pontryagin dual of that, you get like, Z^n , and if you think of the elements of this Z^n as being like, different momentum eigenstates, and consider what the energies of those momenta world be. The kinetic energy is ((1/(2m)) times) the square of the momentum, and so the number of energy eigenvalues (counting multiplicity) less than a given cutoff, should be approximately proportional to the volume of an n-ball with radius the square root of that cutoff. The formula for the n-volume of an n-ball with radius R is V_n(R) = pi^(n/2) / Gamma((n/2) + 1) R^n , which we can evaluate at non-integer values like n=1.5 . So, we can estimate how many energy eigenvalues (counting multiplicity) with energy less than a given cutoff, a free system where the position space is S^n would have, and like, also this gives the number of momentum eigenvalues with energy less than whatever cutoff. Then, maybe we could like, look for a group where the number of elements "near" the identity (in, some sense?) grows at that rate, this would presumably be a non-abelian group? (either that or not finitely generated?) so, trying to take the Pontryagin dual of this, wouldn't like, work quite normally, but there are generalizations to non-abelian groups, and so could maybe get some sort of like, quantum "group" out of it? idk. Some sort of like, non-commutative geometry kind of thing?

[@tinydong4586]

@@drdca8263 is this English

[@williamrutherford553]

This seems like Fermats Last Theorem would be important, especially because it's the difference between squares and cubes. But in this case, you're summing up arbitrary many cubes. Also the edges are interesting, because if you have a^3 + b^3 + ... + z^3 = n^3, the edges mean we have 6 equations like a + b + c = n, d + e = n, etc.

[@Lalovoe]

I really like these mathematical videos. Great video!

[@moth.monster]

Cary's Last Theorem

[@mar8925]

epic sound design

[@wiseSYW]

allowing duplicates, it is easy to create a parker cube of cubes

[@larry_berry]

P = NP and I have a proof for this, but I don't have time before vidcon to release the proof, so just trust me.

[@yaboinooby]

I didn’t realize a video would be sent now but I’m glad it did!

[@KylerLiam]

I legit thought this was posted 9 months ago, but it was actually posted today??? Wow, never been so early

[@zestyorangez]

Something that stands out as a problem would be the situations where you have a small square surrounded by large squares would make long tube cavities on the inside which would have to all be filled in with a limited number of smaller cubes.

[@gallium-gonzollium]

This might be the most excellent video I have gotten recommended in a while.

[@godzilla928]

omg i gonna play this animation everytime before sleep! its so relaxing and satisfying!!

[@bitzibaerlie]

The popping sound of the animation reminds me of Stardew Valley speedruns.

[@shayne3005]

So I'm not the only one lol

[@Blueskies2513]

solution: make a single 50 x 50 x 50 cube

[@arivanhouten6343]

It's been ages but I'm glad you uploaded

[@HanzCastroyearsago]

i want a cubed cube puzzle

[@magica3526]

did you watch even 2 minutes in?

[@trollster7133]

This video is presented like a 3 blue 1 brown video and I completely didn’t realize this was Cary until the audio clip

[@4zaUniv]

Bros uploading now. Thank you!

[@TAKTlmao]

This would work as a corridor crew satisfying render.

[@NotScotishYellow]

The only way for cubed cubes, is imperfection

[@Akshatdoeswhatever]

This video is very satisfying to watch the counting sound and the insert sound is so satisfying and also I’m making a animation that is inspired by tpot

[@Log-d9f]

When i saw "old cary" i looked at the channel name, and i was impressioned that cary started making math things :o

[@mrquethisway]

I love when the 50 square appears

[@sdfmnb2314]

any line is a lined line

[@SlyHikari03]

This guy straight up just said “trust me bro” when sighting his sources.

[@FrostMonolith]

Hearing your old self feels super nostalgic.

[@siyustuff213]

damn i didn’t even realise it was cary speaking until he played the clip of 2011 him speaking 💀

[@DumbMuscle]

Any orthogonal cross section of the cubed cube has to be a squared square (otherwise the cube would not be full). Consider the smallest cube on the surface, with side length s (not necessarily the smallest in the cube, but the smallest one whose surface is part of the surface of the whole cube). In the squared square that is the surface of the cube, it must be surrounded by squares of side length greater than s. Consider a cross section parallel to that surface at depth s+0.5. In that cross section, the cube s is replaced by a square gap of side s between the cubes that border the cube s at the surface (as those cubes must extend to at least depth s+1, since they are larger than s and also have a face at depth 0). That gap can only be filled by cubes smaller than s, and therefore s must be a number which allows for a squared square of side s to be formed. Now consider the smallest square of that tiling of side length s, and let that smallest square have side length s2. It cannot be at the edge of that tiling (otherwise no squared square would be possible, since it would not be possible to arrange larger squares around it to fill a square), so if we move down again by the size of that square (to depth s+s2+0.5), then we again reach a cross section with a square gap of side s2, which could only be filled by a squared square of side s2. This logic continues forever, requiring smaller squared squares each time to fill the gap left by the smallest square in the cross section one step above. However, the smallest available integer is 1, so we cannot continue making smaller squared squares forever to fill the gaps (let alone to smooth off the opposite surface of the cube eventually), and so a cubed cube is impossible.

[@XxAxoshardxX]

Coincedental: numberphile video for this appears!

[@CubicCreeper7914]

And I was just watching other carykh videos, what timing

[@dissmo706]

If you make a video about these cubed cubes, will you include other info about the squared square? Like why it’s limited to 21 squares n not under? Also why the side length 112?

[@jimgu2578]

Because that’s the smallest possible solution (I believe). Of course all squares must be in different sizes or the solution would be trivial.

[@DanatronOne]

@@jimgu2578 Alternatively, you need more than 1 square, else the solution is even more trivial :)

[@drdca8263]

@@DanatronOne more than one size of square, I assume you mean? Because 4 squares of the same size, can make a square, but this is trivial.

[@DanatronOne]

@@drdca8263 Technically speaking, a single square fulfills both conditions; Making a square, and each component square having a unique length.

[@plopsmcgee9672]

"That's all past me had to say about this problem." "And that's also all your gonna get out of current me too! Get pranked!"

[@fluxalchemist]

You may not be able to make a cubed cubed, but I'm sure I'd be able to make a lined line!

[@DuochromaticBichromes]

Do tessaracted tessaracts exist?

[@funnyhamsteronyt-pc9d]

this is relaxing to watch

[@YamamotoTV2021]

I didn’t catch the proof as to why cubed cubes don’t exist. But what about tesseracted tesseracts? Do they exist? (And of course, line-segmented line-segments exist. The simplest example is 1+2=3.)

[@zemyla]

A "side" of a tesseracted tesseract would be a cubed cube, just like the side of a cubed cube is a squared square. So cubed cubes not existing mean all higher dimensions are off limits too.

[@YamamotoTV2021]

@@zemyla That sucks. But what if we bent the rules to allow up to 7 of the same size cube? Thank you.

[@zemyla]

@@YamamotoTV2021 I'm pretty sure it's still forbidden by Fermat's Last Theorem through an infinite descent argument. Even if you had 7 cubes of size N, you'd still have to fill the last cube with a cubed cube of size N.

[@jeffreyflores3094]

Cary, stop teaching me this stuff! I’m on summer break!

[@ratweiler]

"well they're not " i died . i love you

[@Veptis]

It's like a 3Blue1Brown video where the problem is described and then a few stories get told before we get to a beautiful solution. just in multiple parts (I hope).

[@johnchessant3012]

is the proof reasonably easy or am I getting nerd sniped here?

[@nw2kr8bc3t]

Getting a lot of cube content recently

[@TheE_2K_24]

1+ number to each square = 1 bfdi bubble dying

[@PoisonIvory088]

Can we make a Circled Circle or a Triangled Triangle ?

[@yellowmarkers]

Circles do not tile the plane and definitely cannot be formed into a larger circle at all. Triangles do tile the plane and can be formed to make triangles so the question can actually be asked but I don't think it's possible for this reason: The only tessellation to make an equilateral triangle out of equilateral triangles is the one with schlafli symbol {3,6}, but to create a formation that contains no identical triangles, it is possible to merge some of the triangles together (for any arbitrary size of the large triangle). I'm pretty sure this isn't possible to do without leaving at least one rhombus (60deg polyiamond, made of triangles, which are identical and cannot be merged) behind

[@donovanmahan2901]

You can argue that it's inevitable that the smallest square to square a square would need at least one identical cube stacked on top of it to fill the hole once the squares are made into cubes. If you try to square the square hole that is left, you are still left with this issue with the smallest square in THAT square being the problem square instead. This problem persists no matter how many recursions of this process you undergo. If you were to attempt to cube a cube, squaring a square is necessary to cube a cube, therefore cubing a cube is impossible.

[@trwn87]

Please, make a video with the proof! 🥺

[@amesstarline5482]

As a fan of both Numberphile and Cary, this is fun.

[@primrose9693]

Okay, but this video is so funny, because he plays us an audio clip from younger him where he's saying it then, but going to prove it later, only for it to be Now, much later, where he's telling us, again, that he's saying it now, but going to prove it later.

[@RatKillCat]

My ”intuitive proof” (which is probably not exactly complete) is that cubes of varying side lengths create differences in height - so whenever you have a cube surrounded by larger cubes in the xy plane, it creates a hole in the z plane which can only be filled by a cube of the same height as you’ve already used, unless you square it with smaller cubes, but then the same problem will present itself again on a smaller scale, and intuitively, it will converge at a scale that’s too small. Or something like that.

[@theotherdashmelted]

Before I discovered BFDI, carykh was still a very interesting channel to me.

[@lorenzospedups]

carykh do you remembered ALGICOSATHLON

[@Person-ef4xj]

What if we remove the requirement that the side lengths must be integers and only require them to be different? Would it be possible to have a cube made of smaller cubes of different real number side lengths?

[@Релёкс84]

This has so much Fermat energy to it

[@newy_]

Your the type of mathematician i like.

[@EntergeticalakaBot]

Top Tier ASMR

[@creproductions102]

No carykh joke? I guess i might do one. Cary knows heaven

[@WDNH24]

Do hypercubed cubes exist supercubed cubes exist do elevated elevators exist

[@oliverfeuer9323]

Now we just need to know if a tesseracted tesseract exists.

[@EHMM]

finally a new video

[@nanamacapagal8342]

Few questions for the upcoming proof. 1. If three dimensions is impossible, what about higher ones? Are there solutions (if any) to the squaring the square problem? 2. What makes 0, 1, and 2 dimensions special? Obviously the whole squaring a square in 0 and 1 dimensions is trivial but it's not so obvious why a 2d square should be able to be tiled this way. Is there some required property that still holds in 2 dimensions but breaks in the third? 3. Will the proof involve the volume of the tiling cubes, or will it involve their shape and arrangement? Is it something that boils down to equations, or graphs?

[@cmentarnikPL]

I assume it has something to do with Fermatsy last theorem, as others have mentioned it can be used here. And it won't work in higher dimensions. For 3 dimensions I came with a simple proof. Faces of the cubed cube will have to be squared squares because the face has to be flat and square. Now consider the smallest cube on that face. Since all other cubes are bigger than it, on the other side of it it's a square hole at least 1 deep. Then to tile it you will again need a squared square. Then we find the smallest cube and repeat the process. This means that we can find an arbitrarily small positive integer which obviously doesn't exist. Thus by proof by contradictions cubed cube doesn't exist. And by similar logic a "3D face" of 4D cube has to be a cubed cube, so it won't work for higher dimensions

[@kraccurmomsbacc69420]

Carykh, there is a swastika in bfb 14: dont dig straight down in golf balls place.

[@CFAW2024]

LOOKS VERY NICE AND NICE EXPLANATION

[@lily_littleangel]

I can make a cubed cube. All I need is a single size 1 cube and we're done.

[@christinahamilton7676]

11 years pass, and your voice barely changes. 2011 Cary kinda sounds like 2022 Cary on helium.

[@anaunaga5471]

I feel like part of the proof is because you are basically making 6 sides that are squared squares. Its likely that its impossible to have two bounding squares (The square that contains all the internal squares) that don't have the same internal square somewhere in the construction; so in that case the opposite sides of a cube need to have the same cube somewhere, but they don't share any common edges to make that possible. To put simply. The same internal cube would need to exist on opposite sides of the big cube, which would contradict cubing the cube being possible.