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

@minhkhanhvu5180

2 жыл бұрын

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

@and_the_first_last

2 жыл бұрын

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

@zmaj12321

2 жыл бұрын

@@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

2 жыл бұрын

The classic proof by “just trust me bro”

@TheRenegade...

2 жыл бұрын

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

"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

2 жыл бұрын

Fermat moment lol

@Rudxain

2 жыл бұрын

@@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

Жыл бұрын

Rubik's cubes

@YezzirGamer

23 күн бұрын

depends on the size of the cubes

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

2 жыл бұрын

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

@gamekiller0123

2 жыл бұрын

@@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

2 жыл бұрын

@@hecko-yes wtf that's brilliant

@sullivanbell2397

2 жыл бұрын

That'd be awesome

@rustyruinz

2 жыл бұрын

I would square at it at all times

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.

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

@somedudewatchingyoutube9163

2 жыл бұрын

It’s asmr but better

@RTOF

2 жыл бұрын

@@somedudewatchingyoutube9163 yes

@BamsyTheSergal

2 жыл бұрын

picking items up after breaking a chest

@RTOF

2 жыл бұрын

@@BamsyTheSergal really sounds like it

@sofia.eris.bauhaus

2 жыл бұрын

kzread.info/dash/bejne/anh3x7ich7TWiLw.html

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

@carykh

2 жыл бұрын

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

2 жыл бұрын

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

@leanna5733

2 жыл бұрын

e

@sage5296

2 жыл бұрын

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

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

@somedudewatchingyoutube9163

2 жыл бұрын

It’s sounds like guns reloading

@TotallyNotSushii

2 жыл бұрын

@@somedudewatchingyoutube9163 kinda does ngl

@Okaz_

2 жыл бұрын

It’s so satisfying

@iateyourbread

2 жыл бұрын

asmr squaring squares

"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

2 жыл бұрын

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

2 жыл бұрын

Proof by intimidation.

@LewisBavin

2 жыл бұрын

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

@lyricalcarpenter

2 жыл бұрын

@@nooneinparticular3370 in conclusion, AAAAAAAAAAAAAAAAAAAAA

@Rudxain

2 жыл бұрын

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

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.

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

@-FutureTaken-

2 жыл бұрын

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

@SlyHikari03

2 жыл бұрын

Minecraft too

@thepizzaguy8477

2 жыл бұрын

pov you bombed the quarry

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

2 жыл бұрын

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

2 жыл бұрын

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

@DreadKyller

2 жыл бұрын

@@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_

2 жыл бұрын

@@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

2 жыл бұрын

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?

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

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

7 ай бұрын

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).

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.

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

@somedudewatchingyoutube9163

2 жыл бұрын

That’s like half a 10 months

@iabgunner8701

2 жыл бұрын

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

@normalman9123

2 жыл бұрын

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

@cheeseburgermonkey7104

Жыл бұрын

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

why is this so calming??

@Rolagle

2 жыл бұрын

Dunno

This video gives Fermat's Last Theorem vibes.

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

2 жыл бұрын

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

2 жыл бұрын

@@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.

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.

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

I really like these mathematical videos. Great video!

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

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

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.

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

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

i want a cubed cube puzzle

@magica3526

2 жыл бұрын

did you watch even 2 minutes in?

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

Ngl a jigsaw puzzle of this would be kinda fun

I need more of this

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

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

@brandonsaffell4100

2 жыл бұрын

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

@KatzRool

2 жыл бұрын

what the heck is a 1.5-cube lmao

@brandonsaffell4100

2 жыл бұрын

@@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

2 жыл бұрын

@@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

Жыл бұрын

@@drdca8263 is this English

Bros uploading now. Thank you!

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

2 жыл бұрын

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

2 жыл бұрын

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

@drdca8263

2 жыл бұрын

@@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

2 жыл бұрын

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

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

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.

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

IDK why but reading the title gave me an aneurysm

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.

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

@shayne3005

2 жыл бұрын

So I'm not the only one lol

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

this is relaxing to watch

Alt title: Cary talking while weird bubble sounds play

I love when the 50 square appears

carys voice at night is so calming

I love ur vids!!

Love you Cary 😭 you and Michael-my childhood kings

epic sound design

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

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

finally a new video

what about hyper cubed hyper cubes

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?

This would work as a corridor crew satisfying render.

nice new upload!

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

LOOKS VERY NICE AND NICE EXPLANATION

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

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

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

Hearing your old self feels super nostalgic.

Now we need to try the tessaracted tessaract

Getting a lot of cube content recently

This looks cool

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).

Coincedental: numberphile video for this appears!

funtastic animation much good

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

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.

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

And I was just watching other carykh videos, what timing

"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!"

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

Neat animation

1+ number to each square = 1 bfdi bubble dying

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

This has so much Fermat energy to it

It would be cool someday to find a real solution to cube dissection!

@drdca8263

2 жыл бұрын

I think he has shown that it is impossible? (not publicly though)

Noice he's back

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

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

carykh do you remembered ALGICOSATHLON

Good video

2011 cary sounds like he is pitched up a few semitones

I could watch hours of this.

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.

Top Tier ASMR

wow so satisfying

cary it's been six months

Would a cubed cube be possible if the center was not required to be filled? A cubed box?

I like the sounds the squares make lol

When you make the full video about this, you could submit it to the SoME2 (Summer of Math Exposition 2) by 3Blue1Brown

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

Cary's Last Theorem

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

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.

Happy birthday gelatin

is it just me, or does this channel sneak into my recommended and I don’t realize this is the same channel as the time travel video?

I really like that bubble pop sound fx used for the cubes. Where can I find it at?

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