Impossible Squares - Numberphile

[@numberphile]

Catch a more in-depth interview with Ben on our Numberphile Podcast: kzbin.info/www/bejne/Y6Wqn5xvhMd9jc0

[@bradensorensen966]

What if you make a slanty square... INSIDE a slanty square?

[@krishdevi6433]

Is there an equation (or equations) for the sequence/s of odd numbers that are the result of the sum of two squares? (Not including 0^2 + n^2) For example: 5 13 17 25 29 37 41 45 53...? Where you put in 'n' and it gives you the number in the sequence?

[@tarynleffler2606]

@@krishdevi6433 I would also like to know this.

[@dont5014]

WOWW YOU DON'T READ MY PROFILE PICTURE 😶😶❌❌

[@numberphile]

This was filmed BEFORE the lockdown but edited during it! :) - Brady

[@stevemattero1471]

What a great topic! Why 4k+3 and not 4k+1?

[@MrPictor]

@@stevemattero1471 Watch mathologers's videos.

[@anantkerur557]

4k+3 is equivalent to 4k+4-1 which is the same as one less than a multiple of 4 [ 4k-1]. Note that these 'k's are different

[@jojojorisjhjosef]

Illegal maths meeting.

[@yaminireddy5157]

Ahh ,i was just about to ask. :)

[@howsjimmysocool]

I was yelling at my computer asking why the area wasnt being solved using pythagoras - and then he surprised me with it being a proof for pythagoras...

[@oldcowbb]

exactly, i was about to comment "this is a wasted chance to use pythagoras "

[@eve8372]

Haha same here!

[@AngryDuck79]

Me too. Pythagoras jumped out at me about a minute in and he took eight minutes to get around to it lol

[@kezzyhko]

Yep, started writing a comment already, then decided to check if there is already such a comment

[@uncle_esau]

Was looking for this comment...

[@itwasinthispositionerinoag7414]

Rubik's cube in the background feeling all superior with its extra dimension

[@darksnowman7192]

Never thought to meet a kripperino on numberphile

[@claymournesden8705]

Continuum transfunctioner

[@praveenanookala4457]

Another agadmator fan!

[@ffynloparnell1888]

Klein bottle next to it feeling superior with its fourth dimension

[@chronyx685]

dark snowman kripp was not carried by annihilan battlemaster this game

[@chrismcdonald6195]

"We're getting square numbers because we're drawing squares." FINALLY - something on Numberphile I kinda know already!

[@caleblewis8169]

What's a square

[@rexlapis3126]

Quadrilateral

[@rameshshinde4488]

Hi

[@dont5014]

WOWW YOU DON'T READ MY PROFILE PICTURE 😶😶❌❌

[@svz5990]

​@@dont5014because your name says Don't!

[@craigmcqueen7992]

7:03 That was a real pleasure to see how such an elegant proof of Pythagoras' theorem just popped out like that.

[@EnteiFire4]

I really like the one without algebra, where you rearrange the four triangles to make two rectangles, where one on the top left corner horizontally, and the other is on the bottom right vertically. The rest of the square is made of a square of side "a" and a square of side "b".

[@svz5990]

​@@dont5014why the chicken kfc borgor are you everywhere?

[@MozartJunior22]

8:52 It's amazing how Brady has developed a mathematician's mind after all these years of doing these vidoes. This is exactly the question a mathematician would ask

[@inigo8740]

When I first found the channel, I had no idea he wasn't a maths guy, he really seemed to know. Of course after having watched many videos and having learned about the channel, I can now tell a bit he isn't originally a math guy. But you can also see he's getting a bit of a hang on it.

[@Liggliluff]

I wished we were given an answer to that question.

[@alessandrofelisi6037]

​@@Liggliluff They become "less sparse" as you go up! In fact, they tend to "fill" all the natural numbers, in a certain sense.

[@jamieg2427]

He's been filming Numberphile long enough that some of his first viewers could have gotten a masters in math twice over.

[@RipleySawzen]

@@alessandrofelisi6037 Do you have a list of these somewhere or a proof of this?

[@edwardstennett4794]

Ben Sparks is by far my favourite KZbin mathematician. His knack for explaining things in a way that's easy to understand for pretty much anyone makes maths so much more accessible. I regularly rewatch his videos - would love to see him do even more videos on Chaos.

[@fredresz7773]

Edward Stennett Man I love all of the people on this channel! ALL of them! They’re all so fun to watch and enjoy math with.

[@apothecurio]

I feel sometimes the guests can speak pretty dryly. Ben Sparks is NOT one of these guests. Not by any means whatsoever.

[@dont5014]

WOWW YOU DON'T READ MY PROFILE PICTURE

[@adamqazsedc]

He is, a math teacher

[@CursedJoker]

I'm sure that if you go and film Matt Parker long enough, he'll come up with some "kinda possible" Squares.

[@nanamacapagal8342]

3.

[@quinn7894]

*cough* *cough* classic Parker square

[@jw41538]

well done

[@caleblewis8169]

Parker squares

[@dont5014]

WOWW YOU DON'T READ MY PROFILE PICTURE 😶😶❌❌

[@midwinter78]

As soon as I saw the square-in-a-square diagram, I started yelling "that's the square on the hypotenuse!"

[@GuyNamedSean]

I realized that as well! I also roundabout found my way toward the theorem behind what numbers are and are not candidates.

[@loganstrong5426]

Same! Which made it pretty obvious to me that any square you can make has an area that is the sum of two squares.

[@criskity]

Me too, and I was wondering why he decided to go the messier route by subtracting the areas of 4 triangles. Pythagoras is right there to begin with!

[@fredresz7773]

Logan Strong Stumbling across little gems like this and the comment from GuyNamedSean above is what really deepened my love for math!

[@lamusicadepedrovicente]

yes! it got me a bit nervous they not using that to find the area

[@maxharrison9020]

PLEASE DO A PODCAST WITH THIS MAN

[@TimothyReeves]

Done recently

[@robertveith6383]

Stop yelling your post in all caps.

[@HonkeyKongLive]

Ben has this knack for taking something we all know about and hitting from a different direction and I love it.

[@jeremyrixon150]

This!

[@dont5014]

WOWW YOU DON'T READ MY PROFILE PICTURE

[@mariosonicfan2010]

"suddenly there's this deep glimpse of maths that goes way beyond what they're ready for" you make it sound like math is some ancient forbidden arcane knowledge or something

[@sergey1519]

ancient, not forbidden, maybe arcane

[@ruben307]

It is.

[@leonthethird7494]

So pretentious

[@atimholt]

Well, he’s talking about 9 year olds. I remember having those kinds of epiphanies, if only because the curriculum was geared specifically to lead down a logical path.

[@Codricmon]

The dark side of mathematics is a pathway to many abilities some consider to be... unnatural.

[@Peetzaahhh]

5:41 in and I'm bewildered how Pythagoras didn't come up EDIT: 7:24 oh it’s because he’s proving it

[@matthewlyons6544]

Peetzaahhh this was exactly what I was thinking. In my head I was shouting Pythagoras, but then realised the reason it wasn’t referenced was because it was being proved!

[@pickles974]

@Peter Attia when did you stop learning maths and how old are you? I'm asking a bunch of people in the comments because I'm assuming people who are amazed by this video must be about 11 years old or younger.

[@mhr6780]

@@pickles974 🙄

[@OwlyFisher]

@@pickles974 rude. not all of us intuit maths

[@adamqazsedc]

@@pickles974 yknow how this video isn't _just_ about The Pythagorean Theorem.

[@chrisbersabal102]

3:10 i thought i am the only who does that pen cover thing

[@hamiltonianpathondodecahed5236]

xD I didn't notice

[@sparkytheteacher]

@@hamiltonianpathondodecahed5236 neither did I...

[@fasjdays]

Thats exactly what I said too! I'm not the only one after all

[@Dude-sr4ji]

7:20 why do I feel like I just got rickrolled by Pythagoras?

[@dovahseod]

You got Pythagorasrolled

[@TangoWolf09]

Pythagorolled

[@Wigglemice]

6:42 as soon as I saw this, I was like, "Ohhh of course! That's how you visualize the Pythagorean theorem! I should have seen that sooner!" Man, I love those ah ha moments.

[@waltercisneros9535]

7:10 "We prove Pythagoras" *drops mic*

[@DouglasZwick]

At 11:20, I was really hoping he was going to circle the number on his screen with that marker.

[@stevejobs5488]

That moment made me go 😬

[@danielstephenson7558]

Ben Sparks' videos are some of the most watchable videos. The Mandelbrot videos. the Golden Ratio and the Chaos Game are among my favourites.

[@AdamBomb5794]

That one 3Blue1Brown video of which coordinates are on a circle just popped into my mind.

[@zackszekely6618]

Because you can't have an odd number of threes 😃

[@programmingpi314]

also that video hints at another way of finding if a square is possible or not.

[@RajSingh-qp9st]

8837666846

[@cfgauss71]

The real question we want answered: where does that ladder lead to?

[@Bibibosh]

Arthur Clay hahaha !!!! Nice observation

[@electronics-by-practice]

It leads to z axis

[@ChocoHearts]

It leads to Dennis, of course.

[@trueriver1950]

Clue: This was filmed just before lockdown, when Covid-awareness was rising. It's the emergency escape in case the other person coughed unexpectedly. Hold breath while outclimbing the viral aerosols and on exit breathe out before inhaling. Luckily we subsequently thought of using masks.

[@sumdumbmick]

@9:00 I love that you describe this as a problem that you personally have no intuition about. So often I see mathematicians and scientists talk about intuition as something that is universal, and so if they don't have a good intuition about something they're highly likely to write the entire human race off as having no intuition about it, which is astoundingly solipsistic really. So it deserves mention and respect that you did not fall into that pattern at all but demonstrated recognition that you are but one of many minds, and just because you lack knowledge or intuition about something does not imply that others necessarily would as well. To be perfectly clear, I also have absolutely no intuition about this particular thing, but I quite expect that some people do.

[@danjtitchener]

Occurs to me that this is similar to the infinite forest problem, when it was asked which trees could you see!

[@clockworkkirlia7475]

Oooh, I don't know that one! It sounds interesting.

[@timh.6872]

If each grid point has a line orthagonal to the plane (representing a tree trunk), and you stood near the origin, can you see the horizon? If so, how much?

[@anandsuralkar8376]

Right

[@Bibibosh]

We need part two. This guy is awesome!!!!! The idea is awesooooooome!! This channel is _________!!!!!!!!

[@dascha78]

...awes(49*o)me

[@Filipnalepa]

... Numberphile?

[@ffggddss]

"Which numbers are possible?" Ans: All the 2-square numbers; i.e., any number that's a sum of two squares. 0, 1, 2, 4, 5, 8, 9, 10, 13, 16, etc. Not 3, 6, 7, 11, 12, 14, 15, etc. Reason: once you draw one side of the square, the rest is determined (but allowing reflection across the initial side). That side must connect a pair of grid dots, the square of whose separation, s, is always a sum of two squares : s² = ∆x² + ∆y². But the square's area *is* just A = s² = ∆x² + ∆y² And of course, ∆x & ∆y are always integers. PS: The method he uses to prove Pythagoras is, I believe, due to James A. Garfield, when he was schoolteacher, before becoming 20th president of the US. PPS: The characterization of the 2-square numbers is based on characterizing primes in the ring of complex integers. [If you don't know what a mathematical ring is, don't pay it any mind - it isn't necessary; it just might help a little if you do know.] Warning: This gets a bit heavy, which is probably why it isn't in the video, so proceed at your own risk! Real primes can be sorted into 3 classes, modulo 4 [when dividing any integer by 4, the remainder is one of: 0, 1, 2, or 3; equivalently, 0, ±1, or 2]: There are no primes that are 0 mod 4. (i.e., no multiples of 4 are prime!) There's only 1 prime that's 2 mod 4; 2 itself. All others are ±1 mod 4 [I.e., 1 or 3 mod 4]. 2 can trivially be written as a sum of 2 squares: 2 = 1 + 1. Any number that is -1 mod 4, cannot, because all squares are 0 or 1 mod 4, so any sum of two of them can only be 0, 1, or 2 mod 4; never 3 == -1 mod 4. So among real primes, only 2, and the +1 mod 4 primes, can be written as a sum of 2 squares. It so happens that all +1 primes can be written as a sum of 2 squares - I'm not recalling the proof of that at this time. [I invite anyone who knows how to do that, to show it here!] So among the complex integers, the +1 primes are composite, being factorable into a product of 2 complex integers: p = a² + b² = (a + bi)(a - bi) while the -1 primes remain prime, because any product of 2 complex integers must be a conjugate pair in order for the product to be real; and such a product is necessarily a sum of 2 squares, which in turn, cannot be -1 mod 4. Now, the _coup de grace._ For complex numbers, the squared modulus [modulus = its "length"] of a product of them is the product of their squared moduli: w = u + vi; z = x + yi; wz = (ux-vy) + (uy+vx)i |w|² = u² + v² ; |z|² = x² + y² |w|²|z|² = |wz|² ; that is, (x² + y²)(u² + v²) = (ux-vy)² + (uy+vx)² . . . [This can be verified by simply expanding both sides.] Thus showing that a product of a pair of 2-square numbers is again a 2-square number. Now consider the prime factorization of any positive integer, N. Factor out any squares; that is, any prime, p, raised to a power ≥ 2, can be factored into p times an even power of p, which is thus p times a square. You now have N = one big product of squares, which itself is a square, times a product of single, distinct primes. If any of those distinct primes is -1 mod 4, N cannot be written as a sum of 2 squares; if none of them are -1 mod 4, N can be written as a sum of 2 squares. Thus, 3, 7, 11, 19, 23 cannot, being -1 primes; but neither can 6, 12, 14, 15, 21, 22, 24, 27, or 28, because of their prime factorizations. 0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, and 29 can each be written as a sum of 2 squares. Fred

[@alexcerullo3143]

ffggddss Fred

[@josevillegas5243]

Thanks for this! I think I mostly get it, except for the second to last (penultimate) paragraph: "If any of those distinct primes is -1 mod 4..." Cam you explain? So far, I get N has been factored into N = p*q*...*r*(a*b*...*c)^2 = pq...rA^2 where p,q,...r are primes and a,b,...c can be any integers, and A = a*b*...*c. Let pq...r be shorthand for p*q*...*r A^2 is a 1-square and is it trivially a 2-square since A^2 = 0^2 + A^2? Then using the fact that the product of a pair of 2-square numbers is itself 2-square, pq...rA^2 is a 2-square iff pq ..r is a 2-square? I think that's where my confusion arises because I don't know how modular classes behave under multiplication. As you mentioned the primes p,q, ..r have to be +/-1 mod 4. Does your conclusion (the penultimate paragraph) hinge on which mod 4 class the pq...r product is in? If you multiply two +1 mod 4 numbers, you get another +1 mod 4 number: (4k+1)(4j+1) = 4m+1 for some m But also if you multiply two -1 mod 4 numbers, you still get a +1 mod 4 number: (4k-1)(4j-1) = 4n+1 for some n To be exhaustive, if you multiply a -1 mod 4 number by a +1 mod 4 number, you get a -1 mod 4 number: (4k-1)(4j+1) = 4o-1 for some o So what would happen if p,q,..r had an even number of -1 mod 4 primes? E.g. if p,q,...r was just 3 and 7. Their product is 21 which is a +1 mod 4. Thanks for reading this far! Looking forward to your response and hopefully understanding what's going on. I'm really curious.

[@mufasao6776]

The Fred at the end is so funny to me. No QED, no square or symbol, just "Trust me, I'm Fred" lol

[@MrCreeper20k]

I’d be interested to see this extended into 3D. Might be a little more tedious than insightful though

[@HeavyMetalMouse]

In 3D... Assume (0,0,0) is a vertex, and lattice point (a,b,c) is a vertex (with integers a,b,c >=0). The other two vertices on the cube 'adjacent' to the origin in the other two directions would need to be of the form (x,y,z) and satisfy ax + by + cz = 0 (perpendicular to (a,b,c) and (x^2) + (y^2) + (z^2) = (a^2) + (b^2) + (c^2) and x,y,z in the Integers At this point, I'm not entirely sure what method to use to show when you can find two suitable lattice points satisfying those conditions. But if you do, then you get the other four for free, as they're just adding together the vectors, and adding integers always yields integers. If you want to know what *integer* volumes of the cubes are possible, then you also are restricting your search to cases where sqrt(a^2 + b^2 + c^2)^3 is an integer, which only happens when sqrt(a^2 + b^2 + c^2) is an integer. In which case, your solution set is some subset of the cube numbers. However, all cubed integers are, by definition, formable on lattice points (just take the orthogonal points), therefore any solution that could theoretically be formed by a 'tilted' cube must also be formed by a non-tilted cube. Therefore, if you only want integer cube volumes, the solution is a trivial "All integers of the form s^3, where s is a positive integer.", as as any tilted cube on the lattice points must have a either a volume in that set, or a non-integer volume.

[@anantkerur557]

At 4:03, You can get numbers that are of the form a²+b², so 5 = 2²+1², 9 = 3²+0², and so on, but you can't write three as such. Edit: Yes!! I never knew such a simple problem could be so intricate and advanced!

[@judychurley6623]

but you cant have a side of 0 units...

[@tomwakefield1726]

@@judychurley6623 the two numbers are the sides of the triangles which creates the slant. If you have a 0 it just means that the triangle is just a straight line, so there's no slant

[@praveenanookala4457]

Nice

[@judychurley6623]

@@tomwakefield1726 it's the length of the sides of the triangles that are squared.

[@renpnal229]

@@Seven-ez5ux The actual proof of the fact that a number can be expressed as a sum of two square if and only if its prime factorization contains no primes of the form 4k + 3 raised to an odd power.

[@stefanjoeres7149]

Is area 51 possible?

[@stydras3380]

51 is 3*17 and the power of 3 is odd, so no :0

[@MattiaConti]

President send me a message to eliminate this comment as soon as possible

[@brokenwave6125]

Area 51 was an inside job

[@Jivvi]

@@brokenwave6125 if it was an outside job, it would be area ∞-51.

[@robo3007]

@@stydras3380 Proof that mathematics was invented by the government to cover up their secrets!

[@MarcDittner]

I agree with Ben, this is also my favourite proof of Pythagoras.

[@mr.researcher2736]

Numberphile is just incredible, I love this, the best thing is that the best people explain everything

[@Cardgames4children]

It's also my favorite proof of the Pythagorean theorem. It's so simple and intuitive.

[@KayvanAbbasi]

I know James Grimes is people's (probably myself included!) most favorite on this channel, but I also love videos from Ben Sparks. Specifically, I loved his video about the bifurcation. Thank you!

[@wj11jam78]

This video was excellently done, because in the first few minutes I had essentially watched the whole thing. The information was presented in a way which meant that I could easily jump ahead, and figure out the formulas and proofs on my own, without the explanation. It made all the math behind the problem jump out at me. As soon as I saw the triangles, I knew Pythagorean theorem was coming, so I tried it out, and the whole thing solved itself. I'm not the best at math, especially algebra (though I do love geometry), so props to this guy. Really intuitive way of teaching this.

[@nymalous3428]

This was unreasonably interesting for me. I find myself compelled to make a spreadsheet with [a,b] possibilities...

[@raginghobbit3018]

That pythagoras' proof is so smooth and satisfying, I absolutely love it

[@michaels4340]

4:33 Since any square you make (after 2) would have a smallest square inside it, with four triangles around it, and triangles have an area of 1/2ab, and the areas of all four triangles would be 2ab, I'd say to make a square you'd have to be able to write the number as n^2+2m for some integers n,m

[@michaels4340]

now farther along in the video, and of course, I should have known from the triangles :P ...and I've taken number theory, lol, I've probably even done this problem

[@matron9936]

If you can’t express a number as a^2 +b^2 you can’t get a square of this area. It‘s because of the Pythagorean theorem where you get one size of the square is sqrt(a^2 +b^2 ) squaring which you‘d get the area. So the area is always a^2 +b^2 where a and b are natural numbers. Edit: Oh, I didn’t watched the video to the end. You mentioned it. Cool video :D

[@Danilego]

7:08 Oh man, I wasn't expecting that! It must be the simplest proof of pythagoras

[@mikedrewson5545]

Wow, I did not expect this to be a proof of Pythagoras. This is why math is amazing.

[@CaptainSpock1701]

I'm looking at this and the whole time I'm thinking hang on guys, why not just use Pythagoras? It's so obvious. Then "... do you realise we just prove Pythagoras?" - *Mind = Blown* Wow! Simple proof. Going around the complete oposite way as what I was expecting. Great work guys. Always love your videos!

[@whatno5090]

Another fun way to figure this out, is that you know that for any such "slanty square" lying anywhere in the real plane, you can fix one of the vertices on a lattice point and rotate the square about that point; if (and only if) somewhere along the way, one of the nearest vertices hits another lattice point, then you can do a slanty square of that area. This means that we can reduce the problem to finding lattice points on the circle of radius s, where s is the side length of the square; and s = sqrt(A). But the equation for the circle of radius r is x^2 + y^2 = r^2, so of course, this means we need to find integer solutions x^2 + y^2 = A!

[@gabrielmello3293]

As a current engineering student, the moment he tried to calculate the area of the square, I was yelling in my head "just use the damm pythagorean theorem". A few minutes later I remembered how I used to watch numberphile way back in middle school when I still didn't know the pythagorean theorem and all the heavy math I know now, and only then I could appreciate the beauty and the art in the video. This video is intended for people like 14 year old me who didn't know that much math, but absolutely loved these kinds of problems. Thank you for keeping up the making of videos that motivate and introduce math outsiders into such a beautiful discipline.

[@numberphile]

He also deliberately held back pythagoras for the reveal later on!

[@gabrielmello3293]

@@numberphile Yes, that's what I was trying to say.

[@ricardovalentin5056]

Thanks Brady, for that great time of pure mathematics.

[@JM-us3fr]

This is why understanding which primes are sums of two squares is important. 3Blue 1Brown does an excellent video on this, showing why these are the only numbers that can't be expressed in this way.

[@rosiefay7283]

This is beautiful. As you show, it has elements that can appeal to many ages. Once you know how to calculate the area of a right-angled triangle, you can calculate the area of a slanty square, and can at least collect possible and impossible areas. But there's non-trivial number theory there as well. Suggestions for further exercises: 1. Prove that if x and y are possible, so is xy. 2. Repeat the same exercise with equilateral triangles on a triangle grid. (The triangle whose sides are 1 counts as area 1.)

[@pietrocelano23]

the animation at 2:50 made me think immediately of the 3b1b video about primes and circles, so i know where this is going!

[@agr.9410]

Pietro Celano I was reminded of the exact same thing!

[@douglasjackson295]

Trigonometry... You can draw any square in which the size is the sum of two squared integers. In a Square grid if you can draw a long then you can draw that line rotated 90°, that if you can draw a line of a given line you can draw a square of size of the square of the length of the line (line length = c , Square size =c^2). Given the constraints outlined in the video (Lines must be between two points) we can make a right triangle using this line or rather we can create every line using a right triangle and this right triangle for the line to be valid must have legs of integer lengths. Thus All valid lines must be the hypotenuse of a right triangle with integer legs. Thus the length of valid lines (c) must be the square root of The quantity of The sum of the squares of two integers. Thus all squares will have the size of the sum Of the squares of two integers

[@Benlucky13]

I got really excited with the first few numbers in the string because they're adding the digits of pi after the initial 3. so 3 to 6 is '3', 6 to 7 is '1', 7 to 11 is '4', and 11 to 12 is '1'. unfortunately the pattern breaks after that, was hoping this would be another one of those odd ball "why the heck does pi show up here" strings. 3141 is still a fun coincidence, though.

[@timh.6872]

Actually, you can get to Pi from this fact! That divisibility rule he shows can be used to count how many grid points are at a distance sqrt(n) from the origin. If you add up all the counts for n from 1 to some large integer R, you approximate the area of a circle of radius R. Using that 4k+1, 4k+3 only if odd rule, you can rearrange the count into the sum 1 - 1/2 + 1/3 - 1/4 + ... times 4 R², which means the alternating sum is equal to π/4. 3Blue1Brown has a more in depth walkthrough, I think it's called "Approximating Pi with Prime Numbers", but I might be wrong there.

[@5eurosenelsuelo]

This guy is simply on another level

[@konstantinkh]

If you want to draw more square sizes on a dotted grid, all you have to do is place your grid in more dimensions. In 3d, significantly more areas are possible, such as square of area 3. And in 4 dimensions, all integer sizes are possible! (Legendre's Three and Four Square Theorems respectively.)

[@tamirerez2547]

3= 2^2 + i^2

[@anantkerur557]

That leads to a interesting question - what if we allowed Complex numbers?

[@sergey1519]

@@anantkerur557 it is easy to see that it is equivalent to just searching for a²±b²=c solutions

[@diogor379]

@@anantkerur557 My intuition would be that instead of a grid of dots we would have a space of dots to work with. The area 3 square would have to be "lifted" in space by one of its corners into the complex axis I think.

[@alansmithee419]

You can't travel a distance i on the square grid, making this not applicable.

[@alfeberlin]

@@diogor379 Intuition then fails us because even if we drew the imaginary part in a third dimension, the squares stretching into this would appear to grow with a growing imaginary part, but mathematically they should shrink.

[@notoriouswhitemoth]

@7:30 not only does it prove the Pythagorean theorem, it proves it in the same way the Pythagoreans discovered it in the first place.

[@grandexandi]

I asked this question as a comment on a Numberphile video years ago. I'm going to go ahead and presume this video was made in response to that one comment of mine, of course. In which case, thank you! I love it!

[@snowingbook]

Klein Bottle in the background : **exists and Ben doesn't mention it** Clifford Stoll : YOU HAVE SINNED, MORTAL

[@justinjustin7224]

How to draw a square of area 3 using a grid of equally spaced dots: 1. Draw the smallest square you can on the grid. 2. Define the shortest distance between dots to be sqrt(3). 3. Laugh at the problem giver for not clearly specifying units.

[@konstanty8094]

5:48 that's the long way side is sqrt(a^2+b^2) (pytagorean theorm) so the square is a^2 + b^2 7:20 oh so that's why

[@ThemJazzyBeats]

This is a lot like the "Pi hiding in prime regularities" video of 3b1b, where one of the things he does in that video is check if a number can be expressed by the sum of 2 squares

[@amoradioheadd]

me: *watches these math/geometry videos* my math homework sitting on my desk: [sadness noises]

[@xCorvus7x]

7:49 I propose that every square whose side length has a prime factor of the form 4k+3 , k being 0 or a natural number (such as 3, 7, 11, etc.), cannot be drawn as a slanty square in this grid, because those primes cannot be written as the sum of two integer squares. Since the Pythagorean Theorem scales up (3, 4, 5 are a pythagorean triple, and so are 3n, 4n, 5n where n is zero or a natural number), what holds for any such prime must also hold for integer multiples of such a prime. Edit around 11:00 : it works if that prime factor has an even power because (4k+3)^(2n) equals 4m+1 ~~and all numbers of the form 4n+1 _can_ be written as the sum of two squares~~. That means, the whole number will have factors of the form 4a times 4b+1 times 4c+2 which will not result in a number of the form 4d+3.

[@willnewman9783]

This is not quite right. 21 is 4(5)+1, and it cannot be written as the sum of two squares

[@xCorvus7x]

@@willnewman9783 Yeah, that comment is too rudimentary, an amateur's rambling. I had thought I had recognised the problem but I have missed several things. For one, as you say, that a number is of the form 4n+1 is not a sufficient criterion for determining whether it can be written as the sum of two squares. Since there are numbers of the form 4n and 4n+2 who can, it isn't even a necessary condition (unless we only look at odd numbers). Furthermore, my argumentation rests on the assumption that the pair of integer squares for any non-prime number with an odd-powered prime factor of the form 4k+3, if it exists, could be deduced simply by scaling up from the integer squares that sum to that prime of the form 4k+3. I. e. iff a number with an odd-powered prime factor of the form 4k+3 can be written as a sum of two integer squares, then that prime factor can be, too. And since the latter can't, I have supposed that neither can the former. But the 'odd-powered' part should have made me reconsider. What I have written, is only what I have regurgitated from an inspection of which primes can be written as the sum of two integer squares. To generalise that to all natural numbers, I have to do more work.

[@EebstertheGreat]

This is a direct corollary of Fermat's theorem on the sum of squares, which states that a prime number p is the sum of two integers squared x² + y² if and only if p ≡ 1 (mod 4).

[@russellthorburn9297]

So many questions popped into my brain while watching this(some of which were answered): 1. What method can I use to check to see if any number works? 2. Is there a visual pattern going on here? 3. Do the numbers that don't work get farther apart? 4. Are there any examples of where 3 in a row don't work or is 2 in a row the maximum? 5. Are there any other examples of where two integers in a row don't work? 6. Can this problem be extended into 3d space as well? This, by the way, is how math should be taught. Rather than simply dumping the pythagorean theorem on children (Here kid. Use this.) it would be far better to give them this type of problem to investigate which in turn leads them to the pythagorean theorem.

[@heynyquildriver]

i am unreasonably proud of myself for finding out that it’s a proof of pythagorean’s theorem before they talked about it.

[@pickles974]

Why? Its nothing to be proud of... people figure this out at 11 years old in school.

[@benzeh4769]

MORE BEN!!! I LOVE THIS MAN

[@robertveith6383]

Stop yelling in all caps.

[@markkaidy8741]

what would happen if you assigned the spacing between the "dots" to root 2 or root 3 what "squares" could one draw.. 3 area would work etc...what spacing would only yield primes? or what happens with triangles in real number then irrational spacing...then other shapes...maybe I will investigate myself!

[@theblackeagle48]

Square at 1:41 remembered me the last puzzle of Profesor Layton and the mysterious village lol

[@macronencer]

I was asked to prove Pythagoras' Theorem during a university entrance interview for Cambridge in 1982, and this is the way I did it! I think the interviewer liked my approach, because he said he enjoyed geometric proofs. :) I didn't pass the entrance exam so I ended up at Southampton... but I often remember this. Another interview I sat was for Nottingham, oddly enough, where many of Brady's videos are made. In THAT interview, I was asked to integrate e^x.sin(x).cos(x) while they watched, which was WAY harder and made me sweat a bit!

[@kemikao]

Nice! Now, what volume cubes can you make in a "dotted" lattice?

[@trogdorstrngbd]

My off-the-cuff conclusion is that only the cube numbers are possible. Proof (?): The volume of a cube with all integer-component edge vector is V = (a^2 + b^2 + c^2)^(3/2). Since a^2 + b^2 + c^2 and V are both integers, a^2 + b^2 + c^2 must be a square number and V a cube number (no other kind of integer can be raised to the 1.5th power and get another integer). EDIT: I'm assuming you want V to be an integer. Interestingly, this ambiguity wasn't present in the 2D case since restricting the edge vector components to be integers there automatically forced the area to be an integer as well.

[@timh.6872]

I think it's possible, and in fact, a volume of 3 is peanuts (a = b = c = 1). The trick is figuring out whether there's any discernable pattern. I suspect 4 dimensions and quaternions would be easier to work with.

[@trogdorstrngbd]

​@@timh.6872 Can you elaborate? If all sides have a length of 1, the volume is 1. If one corner is at (0,0,0) and an adjacent one at (1,1,1), the volume is 3^1.5, which isn't an integer.

[@cerwe8861]

My favorite proof for the Pythagorean Theorem ist one with a Torus. I saw it on the Dong Video "squaring a Doughnut" from Vsauce Michael. My 2. Favorite proof is the one from Garfield (the President) 'cause it's so clever.

[@jpe1]

Dbzfan _21 isn’t Garfield’s proof a generalized version of what is shown in this video? He used a trapezoid, more general case than a square. Although he broke the trapezoid down into two isosceles triangles and a scalene, not 4 right triangles and smaller square as done here, so I guess it is different... never mind.

[@cee_jay_0]

but lets ask the opposite question: for which integers can you find a pair of multiple solutions, like 0²+5² = 25 and 3²+4²=25. up to 1000 there are 6 integers, that can be written as the sum of two squares in 3 different ways, and i haven't found any number above that qith more pairs yet. and i haven't found any pattern in them either. here's the list: 325 425 650 725 845 850

[@willnewman9783]

There is a known formula given an "n" that gives how many pairs of a,b have a^2+b^2=n. 3blue1brown derives this formula in his video about pi/4=1-1/3+1/5-...

[@jhnoor9705]

For the past two years, I have been studying an area of math known as googology. Googology is basically the study of very large numbers and the notations that is used to express them. When you study googology in depth, you can see that the so-called scientific notation which we usually use to express large numbers is actually incredibly weak in comparison to many of the commonly used notations in googology such as Knuth's up-arrow notation, fast-growing hierarchy, Bird's Array Notation and Bashicu Matrix System, although it may not seems weak at all to an average person. This is mainly because we don't need numbers much larger than those that can be made using exponents in real life. For example, the mass of Sun is approximately 2*10^30 kg and the number of subatomic particles in the universe is approimately 10^80. Below I will show you the formal definition and some examples of expression in Knuth's up-arrow notation: a^^^^...^^^b with n uparrows = a^^^...^a^^^...^^a^^^...^^a... ...a^^^..^^a with (n-1) uparrows between each successive a's Where a^b is the same as a raised to the power tower of b First, we will take a review of addition, multiplication and exponentiation. We had all studied in school that addition is repeated counting, multiplication is repeated addition and exponentiation is repeated multiplication, mathematically: a+b = a+1+1+1...+1 (repeated b times) a*b = a+a+a...+a (repeated b times), and a^b = a*a*a*a...*a (repeated b times) Now, we will start with the double up-arrow operator, which is better known as tetration. a^^b (read this as "a tetrated to b") = a^a^a^a^...^a (b times) (a power tower of a's b terms high) Keep in mind that exponents are always right-associative, so a^b^c^d is the same as a^(b^(c^d)) For example, 2^^3 = 2^2^2 = 2^4 = 16 2^^4 = 2^2^2^2 = 2^2^4 = 2^16 = 65536 2^^5 = 2^2^2^2^2 = 2^2^2^4 = 2^2^16 = 2^65536 = 10^(log10(2)*65536) using rules of logarithm = approx. 2*10^19728 (a number WITH 19729 DIGITS!!!!!!) 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987 (approximately 7.6 trillion for short) 10^^^3 = 10^10^10 = 10^(10 billion) = 1 followed by ten billion zeroes Now do you see how powerful tetration is in comparison to scientific notation? But that's not the end of the story. Just like how tetration is repeated exponentiation, pentation is repeated tetration, which is normally denoted as triple up arrows (^^^) In summary, the n arrow operator is repeated (n-1) arrow operator After that, I suggest you to learn the definition of fast-growing hierarchy, which is basically like this: f_n(a) = (f_(n-1))^a(a), when n is a successor ordinal, or in other words, f_(n-1)(f_(n-1)(...(f_(n-1)(a))...)) (nested n times) When n is a limit ordinal, f_n(a) is defined as f_(n[a])(a), where n[a] is the n-th element of the fundamental sequence of the ordinal n For the definition of successor and limit ordinals, you can search it yourself in Googology Wiki Now, here's an equation for you. Given that 1/f_x(100) is the amount of DNA I inherit from my mom, try to find the ordinal x. The ordinal x here is known as my DNA ordinal Here's the approximate value of x in Bashicu Matrix System: (0,0,0)(1,1,1)(2,2,2)(3,3,3)(3,3,0)(4,4,1)(5,5,2)(6,6,2)(7,7,0)(8,8,1)(9,9,2)(10,9,2)(11,9,0)(12,10,1)(13,11,2)(13,11,2)(13,11,1)(14,12,2)(14,11,1)(15,12,2)(15,11,1)(16,12,0)(17,13,1)(18,14,2)(18,14,2)(18,14,1)(19,15,2)(19,14,1)(20,15,2)(20,14,1)(21,15,0)(22,16,1)(23,17,2)(23,17,2)(23,17,1)(24,18,2)(24,17,1)(25,18,2)(25,17,0)(26,18,1)(27,19,2)(27,19,2)(27,19,1)(28,20,2)(28,19,1)(29,20,2)(29,19,0)(30,20,1)(31,21,2)(31,21,2)(31,21,1)(32,22,2)(32,21,1)(33,22,2)(33,21,0)(34,22,1)(35,23,2)(35,23,2)(35,23,1)(36,24,2)(36,23,1)(37,24,2)(37,23,0)(38,24,1)(39,25,2)(40,25,2)(40,25,1)(41,26,2)(41,22,1)(42,23,2)(42,23,2)(42,23,1)(43,24,2)(43,23,1)(44,24,2)(44,23,0)(45,24,1)(46,25,2)(47,25,2)(47,25,1)(48,26,1)(49,27,0)(50,28,1)(51,29,2)(52,29,2)(52,29,1)(53,30,0)(54,31,1)(55,32,2)(56,32,2)(56,32,0)(57,33,1)(58,34,2)(59,34,2)(59,34,0)(60,35,1)(61,36,2)(62,36,2)(62,36,0)(63,37,1)(64,38,2)(65,38,2)(65,38,0)(66,39,1)(67,40,2)(68,40,2)(68,40,0)(69,41,1)(70,42,2)(71,42,2)(71,42,0)(72,43,1)(73,44,0)(74,45,1)(75,44,0)... ... For the definition of Bashicu Matrix System, you can search it up yourself in google So can you please help me to analyze my DNA ordinal?

[@brian23fink]

What about the opposite: slanty squares that bound, instead of being bounded by, a non-slanty square?

[@Satokaさとか]

Looking at the animation at 11:54 I think the answer to Brady's question would be more sparse, because as you go on the angles available becomes more and therefore the numbers.

[@theoriginalstoney]

Question: I noticed there's a secondary pattern happening which is not quite so obvious. If you take the "most diagonal" offset - which gives the smallest area of the inside square (so for 2x2 square, this is [1,1], for 3x3 square, this is [1,2], for 4x4 = [2,2], 5x5 = [2,3], etc. and then repeat that, you find the area of the square goes up in a very regular pattern: 1, 2 (+1), 5 (+3), 8 (+3), 13 (+5), 18 (+5), 25 (+7), 32 (+7), ... Where does that pattern arise from? I've been sitting here staring at diagonal squares and wondering...

[@patrickryckman3867]

Those squares you drew at the end that created a spiral, I bet the rate that they grow at is some metallic ratio.

[@sevenhundred77_]

Impossible challenge: solve the Riemann hypothesis

[@ayush.kumar.13907]

why not try something simpler first like the 3x+1 conjecture.

[@Unidentifying]

I proved it using the abc conjecture and mock modularity with compact non-hausdorff manifolds on gauge symmetric Fermi propagators tensored with 10 dimensional vertical tangent space in U18, but the comment space is too small to contain it.

[@themeeman]

not with that attitude

[@trueriver1950]

If you can prove it's impossible you are a better number theorist than me. Or anyone else (yet?)

[@RhejMacTavish]

Really well put together video; nice one :)

[@numberphile]

Thanks

[@helloofthebeach]

I stumbled into that Pythagorean proof on my own back when I was just starting calculus and, for all the math I did after, nothing will ever top that moment for me. I peaked early.

[@electraelpindrai1964]

yay, classic numberphile

[@laurak1545]

These mathematicians are all so nice and funny and entertaining... and most of us would never realise if it wasn't for numberphile :)

[@tonypiff]

and now i have to watch every ben sparks video.

[@ManojKumar-lhb6788]

The general solution to this is :- 1) k(m^2-n^2) 2.)2mn 3.)k(m^2+n^2). K belong to positive integer and m and n also.

[@willmunoz1638]

13:02 We spend so much time getting our bodies into shape. Me: *laughs while eating second breakfast.*

[@m.rohwer6989]

This channel is for people that think amazon prime is about numbers. I love it

[@dewaard3301]

I hope Brilliant will still be around when my boys grow up.

[@jamespalmer2620]

3 videos in a row with Bath professors! Exciting times.

[@tzisorey]

"And how big is that square?" _16 uni...._ "Don't worry, it's not a trick question" ..... _Well now I'm not sure..._

[@Joe-un1tl]

What an incredible way to show the proof of the Pythagorus theorem. I found a very non rigorous proof of it using similar triangles and an inscribed rectangle.

[@Chalseu___]

I wish I learnt Pythagoras that way, a lot more natural and intuitive imo

[@Fritzafella]

I remember watching a video by 3blue1brown giving a proof for something where youd draw a circle with its center on the origin, and only with an integer radius and refering to where these circles intercect at an (integer,integer) pair. This is essentially the same question just with a slightly differnet spin but it was different in the way that the proof did not use pythagoran or 4k-1)^(2n-1) where k and n are integers, but used the fact that certain prime numbers can be writen in a+bi form that makes them not prime , for example, 13. 13x1 is the only real number pair but in imaginary there are four related ones. And they are (3+2i)x(3-2i) you get 9+6i-6i-4i^2 or 13 (also cuz 3^2 + 2^2) but 3,7, 11 etc cannot be writen as regular primes or as complex primes either. Ill see if i can find the video cuz if you like numberphile ull like 3blue1brown.

[@electronics-by-practice]

Can you draw an equilateral triangle on that gird and what is the diameter of the circle which intersect the maximum of dots

[@thehiddenninja3428]

Before I watched it all, I figured it out. The area of a square is the square of its side-length, obviously. But If the corners are all on points, then the side length, by pythagoras's theorem, is sqrt ( a^2 + b^2), where a and b are the two whole numbers being the vertical and horizontal distance between the two points. So if we're sticking to the grid, a and b can only be whole numbers, so a^2 and b^2 can only be square numbers so the area of the square is sqrt (a^2 + b^2)^2 = a^2 + b^2 is the sum of any 2 square numbers.

[@robyngwendolynshiloh5277]

Whenever I'm stressed, I count letters on things, and I determine if the number of letters is 4K, 4K+1, 4K+2, or 4K+3 and it keeps my mind occupied. i rarely count the exact number, but only determine if its one of the 4 options

[@MrNacknime]

It's literally only about which distances of points exist, having squares around just complicates the issue. And because of pythagoras, you can have all distances of the form sqrt(a^2+b^2), a,b>=0. And then your square sizes are just squares of these numbers, so a^2+b^2.

[@chinareds54]

One small quibble... I would say a>0, b>=0, because a 0x0 is not a square; it's a point.

[@rossholst5315]

You can draw a square of 3 if you add in another axis for a z dimension. As the length of the diagonal of the cube is root 3.

[@stealthis]

11:54 spirograph. You never know where math will end up.

[@yinjohn23]

As for sparsity of numbers that is a sum of two squares, we can just look at the proportion of numbers which satisfy this criterion. Let p be any prime, and let n be a positive integer. Then, as a first step, we want to see what proportion of numbers from 1 to n satisfy this criterion, at least according to p. That is, how many numbers between 1 and n have an odd power of p in its prime factorization. To calculate this, you can ask how many numbers are divisible by p^0 but not p^1, then p^2 but not p^3 and so on. You can show that it is precisely n-floor(n/p)+floor(n/p^2)-floor(n/p^3)-... using inclusion exclusion. This is just n(1-1/(p(1+p)))+O(log_p(n)). Now you want to know the number of numbers satisfying the criterion for all p. This is just 1 - the proportion of numbers not saying the criterion for some p. Here you have to use inclusion exclusion again, except on infinitely many prime numbers.

[@NocturnalJin]

Wow. This one probably has the highest coolness-to-simplicity ratio ever.

[@nice3294]

Does this have something to do with 3blue1brown's video on pi hiding it prime regularities? His video talks about how many grid points show up on circles of square roots of numbers and the same pattern showed up and he goes into some detail about why certain numbers hit a certain amount of grid points.

[@szilardecsenyi516]

This is very exciting, which also means that there is no hypothetical booklet (at finite many rational numbers) that can produce squares corresponding to all integers.