Why was this visual proof missed for 400 years? (Fermat's two square theorem)

[@caspermadlener4191]

This proof is so beautiful that I wrote an entire essay about numbers as the sum of two squares. When the essay was "finished" (I admit that it wasn't), I sent it to the main competition for this type of math essays in the Netherlands, and it got third place. Also, because I heavily studied the subject in my spare time and Olympiad training, I got really good at this type of number theory. When I participated at the IMO in Oslo this year (second time), I solved question 3 with full points, which was about this type of number theory. I got a perfect score on the first day, and scored 7+5+4=16 points on the second day, for a total of 37 points! GOLD! 19th place worldwide! Relative best for my country ever! I really don't know if I would have gotten this score without this proof, so thank you so much for making this video. I hope that you are going to inspire lots of other people as well!

[@PC_Simo]

*_WOW!_* That’s really impressive 😮👏🏻! *_CONGRATULATIONS!_* 🥳😃👍🏻

[@Avighna]

You are a legend

[@jannegrey]

This is great! Congratulations. And I hope that maybe you will be one inspiring people as well!

[@gilberttheisen9270]

25/3/2024. La preuve se démontre en 4 lignes. Niveau: classe de 4e en France ! Plus, revoir sa copie; Bon courage.

[@mikemthify]

This proof was discovered by Roger Heath-Brown in 1971, and was later condensed into the one sentence version by Don Zagier. It's one of two proofs of this theorem found in the wonderful book "Proofs from THE BOOK" 6th ed by Martin Aigner and Günter M. Ziegler in chapter 4.

[@Mathologer]

Thanks for that. I bought the book when it came out (1ed.). Loved it then. Looks like I should have a look at the most recent edition. Who knows what other gems have found their way in there :)

[@seanziewonzie]

I love Zagier's sentence, even without the windmills. It serves as a great exercise in reading proofs. If I ever teach one of those "intro to proofs" class, I would assign the task of deciphering it as some sort of class discussion for the day.

[@Macieks300]

@mikemthify Roger Heath-Brown was 19 in 1971. Could you post some sources?

[@mikemthify]

@@Macieks300 page 21 of the book I mentioned. As a source it cites: D. R. Heath-Brown: Fermat's two squares theorem, Invariant (1984), 2-5. latex version, with appendix on history, January 2008, at eprints.maths.ox.ac.uk/677/1/invariant.pdf The URL is archived at: web.archive.org/web/20110606154228/eprints.maths.ox.ac.uk/677/1/invariant.pdf

[@Macieks300]

@@mikemthify He said "My original notes date from 1971." I don't know if that means he came up with the proof then but if he did he really would've been 19 and that just blows my mind.

[@ChrisSeltzer]

Videos like this make me marvel at the internet. Growing up I could never have access to content like this but now I can watch a brilliant mathematical mind explain fascinating concepts to me. this channel is an example that should give everyone faith in the future of humanity.

[@jakegerke7188]

I never made it past geometry in public school, and yet I was able to follow most of this well, and appreciate how beautiful this proof really is. I chalk that up not only to your ability to explain things in various ways, but also to just how clean and professionally edited this video was. Well done. You have yourself a new fan. (Or... a new windmill.)

[@Mathologer]

That's great :)

[@OKEKOBEB]

I don't know what I am doing on this video but that last bit of your comment is better than the proof

[@FourthDerivative]

"The proof is left as an exercise for the reader" -Fermat

[@pianoclassico718]

Fermat on every ''theorem'' and conjecture

[@emojidinosaur7300]

lol

[@maxwellsequation4887]

Yeye

[@melancholiaenshrinesalltriumph]

I have a proof but they go to another school and you wouldn't know them

[@troyterry5759]

To be fair, he had us pretty exercised over his theorem for many, many years...

[@vsevolodvoronov7526]

Thanks for the video! When I first heard about this proof, I asked Alexander Spivak who invented the visual version. And he said that there was no other source, it was his own idea. Because we don't know anybody who came up with this before 2007, it's almost certainly that he was the first. Unbelievable, but the Zagier's proof (and the previous proof by Heath-Brown) had appeared without any connection to geometry.

[@Mathologer]

I actually had a link to a writeup by Spivak and I dug up an e-mail address. Sadly he never replied to my e-mail asking him whether he discovered the windmills (neither did Don Zagier) :(

[@vsevolodvoronov7526]

@@Mathologer I have a few friends in common with him, and it was easier for me.

[@rainjar]

@@vsevolodvoronov7526 No harm in him replying now?

[@Vanechki]

@@Mathologer I'm sorry! I haven't seen your letter. (Most likely it was sent to spam automatically or you did not write to the email address I use.) I generally answer everyone, and I love and respect your channel, so of course I would answer you immediately. The proof was invented by me no later than 2005. After some time, the opportunity arose to insert it into my book "Arithmetic-2". Thanks for making the video. They did it very well. Almost a million views. Thank you!

[@Mathologer]

@@Vanechki Glad you got to see it after all and that you like it :)

[@raynmanshorts9275]

Fermat: "Hey, here's this cool thing about numbers." Mathematicians: "Amazing! Can you prove it?" Fermat: "I already did." Mathematicians: "Wow! Can we see it?" Fermat: "Hmmm... nah."

[@Fingerblasterstudios]

Fermat: *dies*

[@archiebellega956]

Fermat : I'm sorry but I run out of space to write the stuff anyway bye Everyone : ... you can just get another paper

[@justpaulo]

Fermat's👻: Aaaahh, now let's just sit and enjoy their struggle !

[@gfhrtshergheghegewgewgew1730]

perhaps fermat chose to let other people work on the problem than to just spoonfeed the proofs for them, so as to not spoil the pursuit of mathematics for people. since he knew he was able to prove it he can reasonably assume that anyone else could be able to as well

[@chickenduckhappy]

When it came to granting access to his proofs, he seems to have been slightly on the egg plant side of behaviors 🤔 On the other hand, he also was an extremely strict judge, so maybe he wanted people to demonstrate their ability to grok things on their own while watching with a frown 🙂

[@scooldrood]

"4k+1, now can you see the patter on the left?" "Yeah 😄, 4k-1!" "4k+3! "😑"

[@McDaldo]

Is there a reason that it is notated as 4k+3 in stead of 4k-1?

[@MuffinsAPlenty]

@@McDaldo There is nothing wrong with using 4k-1 instead of 4k+3. An integer is 1 less than a multiple of 4 if and only if it is 3 more than a multiple of 4. So 4k-1 and 4k+3 describe the same sets of integers. The arguments/proofs in this video would work exactly as well using 4k-1 as it does using 4k+3. So why does Mathologer use 4k+3? Because of modular arithmetic! In modular arithmetic, we work with the _remainders._ So if you were asked, "what is 7 modulo 4; in other words, what is the remainder when you divide 7 by 4?" you would probably answer with "3", not with "-1". And Mathologer's next video (after this one), uses modular arithmetic, so feel free to check it out: watch?v=X63MWZIN3gM

[@Alexgaby15Channel]

@@McDaldo it's because when you do modulo the remainder of 7 / 4 is 3 not -1. Because of this is more standardized to use 4k 4k+1 4k+2 4k+3 and not things like 4k-1 or 4k+4

[@anniecenter]

MuffinsAPlenty Thank you so much for answering this. This makes so much sense

[@redpanda2961]

@@MuffinsAPlenty Isn't it a trade-off of domain consistency for the consistency of modular arithmetic? For 4k+1 --> k>=1 but for 4k+3 --> k>=0.

[@chirayu_jain]

So elegant. At 19:17, I understood where this proof is going, that is the happiest moment of your video when I understand where the proof is going 😃

[@captainpints]

Chirayu Jain Nice!

[@blackpenredpen]

I agree! That kind of feeling is just amazing!

[@nisargbhavsar25]

Hi Chairayu

[@MrVerece]

Jo didn't realize it was 19 minutes of Math already at that moment

[@chirayu_jain]

@@blackpenredpen you are too here!!!! 😮

[@admiralbananas]

20:55 I had to immediately upvote here. I love when a proof concludes and it all comes together and makes sense. I wish that visuals were more commonplace in math papers (and in maths in general), because I feel like less people would feel like math is something they'll never be able to understand. Great video, very easy to follow, very enlightening!

[@johnnysparkleface3096]

So far I'm utterly hopeless. Your eureka moment went right by me, I don't see how anything fits together. I was completely lost every inch of the way. I believe there are people who just CAN NOT understand math no matter how gifted the teacher. And I HATE that I am one of those people, because I think I'd really like math if I could just catch on.

[@admiralbananas]

@@johnnysparkleface3096 That's ok! Even though this video is aimed at being a simple proof, it is still somewhat advanced to be able to grasp. Don't beat yourself up, there is always plenty of math for you to enjoy that you'd be able to digest, not matter your skill level.

[@PC_Simo]

I love that eureka-moment, as well; and this proof and video certainly delivers. 👍🏻

[@benjaminmiddaugh2729]

I love the structure of this video. The moment when I understood how the visual proof would go (just before we moved to visual representations of it) is why I watch videos like this.

[@bjdiament]

Thank you, Mathologer for your wonderful videos! David Wells's survey sadly omits Cantor's diagonalization, which, in my opinion, belongs no lower than position 2 on his list of most beautiful proofs. Cantor's proof is also the granddaddy (through Goedel) of Turing's proof of the undecidability of the halting problem (which also sends chills down my spine whenever I read it), and which ushered in the field of computer science.

[@Mathologer]

Speaking of omissions. What about Pythagoras's theorem ? :)

[@muskyoxes]

I kept hearing "a 4k+1 prime" and wondered how or if the primality mattered. It's amazing how late, and how crucially, it finally comes into play.

[@programmer4047]

Where in the proof it mattered? Can you give me timestamp? I still don't understand why it has to be a 4k+1 prime.

[@muskyoxes]

@@programmer4047 20:07 primality comes in

[@serkanmuhcu1270]

I like, that 3blue1brown is also a patron

[@dikephobia]

Yes. I love that "3lue1brown" is a "patreon."

[@MrYAY100]

Shirt = To infinity and beyond?

[@dimitrispapadakis2122]

why is > beyond?

[@MrYAY100]

@@dimitrispapadakis2122 Im thinking it refers to a number greater than infinity (>inf). In other words beyond infinity

[@adama7752]

@@dimitrispapadakis2122 because it's not >=

[@livedandletdie]

2 Infinity or greater than. And is after all the multiplicative function.

[@linyenchin6773]

There is no "beyond" the boundless aka infinite...Buzz Lightyear was stoned on "star command," a powerful strain of marijuana...

[@ghostrng]

It is good to see that mathloger is back online...

[@JasonEwton]

wooo!!! Yes! Re-subbed!

[@tomkopolt1619]

And to see all the comments restored aswell!

[@heydudeyahbro5492]

Yes! In the era of 2+2=4 is racist!

[@KnakuanaRka]

Why was he offline?

[@jonsey3645]

I am numerically challenged. I have a bachelor's degree in nursing and have never passed algebra...(please don't ask). I am addicted to your channel and genuinely understand the pleasure that you exhibit from elegant solutions. Thank you for this long undiscovered pleasure that you have introduced me to.

[@АлександрЯкунин-т5й]

I am very very fascinated by 1) How hardworking you are with all these presentations 2) How kind, positive and interested in math you are. It's perfect that you make these videos, it literally makes me much happier because i fall in love with math more and more. P. S. Sorry for my english, it's not my language.

[@Mathologer]

Glad you like the videos. It's a lot of work but it's also very rewarding to then get comments like this that show people really appreciate what I am doing :)

[@TommasoGianiorio]

Euler's formula for polyhedra can easily reach #1 if you realise it's actually d0-d1+d2-d3+d4...dn=1 where di is the number of i-dimensional objects that form an n-dimensional polyhedron

[@csDiablo1]

Out of curiosity: are you sure about the right side? I am certainly no expert in this particular subject, but having an odd number there seems.... Well... Odd 😁😁 Jokes aside though, this is kinda new form of knowledge for me and I want to see where you got this from :)

[@rmsgrey]

@@csDiablo1 It checks out for the familiar 3D case - V-E+F-1=1 (the last 1 on the left is the body itself). In 2D, it can be rewritten as V=E (the shape and the constant 1 on the right cancel).

[@zemoxian]

I noticed that pattern in high school when playing with polytopes. Never tried to prove it though. I think I also noticed that the n-1 dimensional surface of an n-dimensional sphere is the derivative of its hyper-volume. I think that might have been an assumption on my part given that it’s true for the first couple of examples. I did integrate hyperspheres and derive a formula for n-dimensional spheres. It’s interesting that you get an extra factor of pi at every even dimension. I’ve wondered if that has anything to do with the number of independent axes of rotation you can have. I feel like I should study math again. Don’t think I could derive that formula now.

[@TommasoGianiorio]

@@zemoxian I think there is a recent video of 3B1B exactly on that extra Pi

[@TommasoGianiorio]

@@csDiablo1 yeah, absolutely sure! It's easy to see that that sum equals 1 in the case of a n-dimensional tetrahedron for example. If you didn't know, the n-th row of Pascal's triangle describes the number of i-dimensional objects that form an n-dimensional tetrahedron ( for example, a 3-dimensional pyramid has 4 V 6E 4 F and 1 Pyramid, 4-6-4-1) and the 1 left over in the equation is the first 1 in Pascal's rows (it is another well-known result that the alternating sums of the numbers in the rows equals zero)

[@luisbenites4825]

You guys really outdid yourselves with the presentation of this visual proof. Nice addition of the uniqueness proof. Spectacular job!

[@15silverblade]

Okay, this is actually one of the most beautiful things I've seen in math.

[@AntonBourbon]

I've seen many beautiful 4K videos on KZbin, but out of *4k+1* videos, this is definitely the best :)

[@tejing2001]

I really love the graphical intuition added onto that one sentence proof. It makes it a lot clearer WHY that function is an involution and has exactly 1 fixed point. Also, you misspoke. At 28:54 you said "b squared" instead of "c squared." >.< Gotta be tough to get through that stuff without any mistakes. At least it's clear what you meant cause of the written equations.

[@ThePharphis]

A great reason for redundancy in information given!

[@Saki630]

It was me, I discovered this proof back in grade school when making arts & crafts. I wrote a note in my journal of discovering the proof, but I had to also go back and watch Power Rangers.

[@michaelwoodhams7866]

When you do Euler's polyhedron formula, here is an interesting bit you could include. For any polyhedron*, the angular deficits at the vertices sum to 720 degrees (4 pi steradians.) This can be very quickly proved via Euler's polyhedron formula, using for a polygon sum-of-angles = 180 x number-of-vertices - 360. The appeal is that this is about a 30 second proof. For example, consider a square pyramid with regular triangles. The 'top' vertex has 4 triangles, so the deficit is (360 - 4x60)=120 degrees. The other four vertices have a square and two triangles so the deficit is (360-90-2x60)=150. The sum of the deficits is 4x150+120=720. I expect (I haven't looked into it) that this is a special case of a theorem which says integrate-curvature-over-a-topologically-spherical-surface = 4 pi, and in turn gives surface area of a unit sphere = 4 pi. And probably integrate-curvature-over-any-surface = 4 pi (1 - number of holes in surface) * Not self-intersecting, topologically equivalent to a sphere.

[@TheOneThreeSeven]

A year ago I left a comment on one of these video's saying I was so inspired I was going to make my own math education you tube video's. I have something very special for everyone coming very soon, it's a free software project that I created while working on a tool to make animations for my video's and is almost ready to be released. I just published the first video on my channel, check it out!

[@MrAlRats]

You've got to learn to use apostrophes correctly!

[@yt-sh]

The numbers what do they mean?

[@Machu_channel]

I wanna learn python. Make the UI of the software user friendly. I wanna try the software. I saw your video and that was great.

[@elonmusk501]

I had subscribe your channel

[@TheOneThreeSeven]

Hi everyone! I just published a 30 minute video on my channel where I show a proof of the fundamental theorem of algebra at the end, but it's about a lot more than just that. Enjoy!

[@koenth2359]

The simple part: any odd number n that can be written as the sum of two squares must be the sum of an even square a^2 and an odd square b^2. Now a^2=0 (mod 4) and b^2=1 (mod 4), so that n must be 1 (mod 4).

[@Shadow81989]

For an easier understanding I'd like to add that every odd b^2 can be expressed as (x+1)^2, with x being an even number. Now obviously that makes b^2 equal to x^2 + 2x + 1. As x is even, both x^2 and 2x are always divisible by 4, so any b^2 must be of the form 4k+1. (therefore obviously any a^2 + b^2 with a being even and b being odd has to be of the form 4k+1 as well...)

[@Shadow81989]

@Šimon Rada good point! I changed to the good old "x" to avoid confusion with the original "a".

[@Shadow81989]

@Šimon Rada yes, that was part of the first statement (not mine): "any odd number n *that can be written as the sum of two squares* [...must be of the form 4k+1]" :-)

[@dhritajitkalia2653]

Beautiful beautiful explanations. Every student deserves a professor like you

[@peterjamesfoote3964]

Thank you for presenting this. Haven’t had a math class in more than 40 years but I did have formal logic which helped a bit when following this video. If I had seen this in high school I might have had a whole different career path.

[@nilshoppenstedt6073]

WOW! Definitiv eines der besten Mathe-Videos auf KZbin! Und auch sehr schön aufbereitet und präsentiert!

This is really beautiful. It's even more beautiful than the theorem itself, which was hard to beat.

[@pierineri]

Thank you for this fantastic video! Note that the footprint-preserving involution defined in 18:01 does not need the special form of the prime p, and in fact the conclusion in 20:30 is: the footprint-preserving involution has exactly one fixed point if p=4k+1, and none if p=4k+3. Thus the number of windmills is odd if p=4k+1 and even if p=4k+3. The argument in Chapter 6 also still works if we do not assume the form of the prime p, but the conclusion reads: "there is at most one way of writing p as a sum of two squares". So if we like this video actually also includes the trivial case 4k+3: p=4k+1: odd number of windmills, exactly one fixed point of yz, p writes uniquely as a sum of two squares. p=4k+3: even number of windmills, no fixed points of yz, p is not a sum of two squares.

[@BoringExtrovert]

7:02 yes it can. It's sufficient to look at the last two digits of a number to check if it's divisible by 4 since 4 divides 100. The last two digits were 81 which is one above a multiple of four.

[@maulaucraw1209]

Thank you kind sir

[@Gulyus]

Yes, but can you prove it is prime : P That would be the issue in this case.

[@BoringExtrovert]

@@maulaucraw1209 😆😆

[@johnny_eth]

7:00 yes it can. The number ends in 81. That's a multiple of 4 + 1.

[@keyboard_toucher]

To elaborate a bit, 6513...46381 = 6513...46300 + 81. The number on the left obviously has no remainder when divided by 4 (being a multiple of 100), leaving only 81 to be considered.

[@incoralium9211]

@@keyboard_toucher Thx captain abvious, but "multiple of 4 depends of last 2 digit " is a tool given at school before the age of ten, just like " sum up digits of a number to know if you can divide it by 3 "

[@davidmeijer1645]

"Step back and squint your eyes." Brilliant guide to this insight!

[@nanamacapagal8342]

32:20 Any odd number can be written as x² - y². We first factor x² + y² as usual, leaving us with: k = x² - y² k = (x + y)(x - y) We want to get rid of the y term and cancel it into 1 so that k can simply be represented as 2x + 1 (or in this case, 2x - 1). To do this we set y = x - 1. The rest of the computation is as follows: k = (x + (x - 1))(x - (x - 1)) k = (2x - 1)(1) k = 2x - 1 Therefore every odd number can be written as the difference of two squares by using consecutive x and y. 32:30 All odd primes have a unique way of being represented as a difference of two squares. We have already proved above that all odd numbers can be represented as the difference of two squares regardless of whether or not the numbers themselves are prime. To prove that there are no other possible choices for prime numbers we may look at the difference of squares a bit closer. The expression x² - y² can be factored into (x + y)(x - y). In this case any composite number ab (in this case, 15) can be expressed multiple ways because we can write it as 1*ab (1*15) or a*b (3*5), both of which can be converted into difference of squares, one for each pair of factors. 1*15 = (8-7)(8+7) = 8² - 7² 3*5 = (4-1)(4+1) = 4² - 1² 1*ab = (((ab+1)/2) - ((ab-1)/2))(((ab+1)/2) + ((ab-1)/2) = ((ab+1)/2)² - ((ab-1)/2)² a*b = ((a+b)/2 - (b-a)/2)((a+b)/2 + (b-a)/2) = ((a+b)/2)² - ((b-a)/2)² However, there is only one factorization for any prime p, namely: 1*p Therefore, since we can only factor primes in one way, there must also be exactly one way to represent p as a difference of two squares.

[@divyadulmini374]

Thank you very much..I wrote this to my assignment in university.Thank you.Thank you.❤️❤️❤️❤️

[@Wyverald]

You can write these proofs much more succintly. 1) Any odd number can be written as 2k+1. Obviously 2k+1 = (k+1)^2 - k^2, so 2k+1 can be written as the difference of two squares. 2) Given the above, we know that for any integer k there always exist integers p, q such that 2k+1 = p^2 - q^2 = (p+q)(p-q). Both (p+q) and (p-q) must be odd, since 2k+1 is odd. So if 2k+1 is also prime, one of (p+q) and (p-q) must be 1 -- it's obvious that it's the latter.

[@swingardium706]

The movie "Fermat's Room" is indeed excellent, I'm glad it's getting a bit of publicity!

[@morphx666]

Just finished watching it... thanks for the recommendation!

[@guillermogil3391]

I see eye to eye with you! Totally!

[@SimonBuchanNz]

Huh, I vaguely remember watching it a while ago and sort of liking it, but not thinking it especially awesome? I should rewatch it I guess?

[@mitjamastnak9206]

Awsome video! In the x^2-y^2 problem at the end, all solutions divisible by 4 are also possible (if you assume that x and y are coprime then you can get all odd numbers as well as numbers divisible by 8).

[@jezzag9739]

This is great. You're a good teacher and I appreciate the time you spent making it

[@Luxaray2000]

Great video. I actually had a project in my number theory class to verify the one sentence proof. Very fun, but this is way more enlightening.

[@seiggrainhart4719]

I'm surprised there aren't more comments about how your shirt literally says "To infinity and beyond" in math geek. At least, I think it does?

[@Pointlesschan]

Yeah I noticed his shirt too

[@georgm3257]

Thank you for this great video. A long time ago I heard that Zagier did a one-sentence-proof without knowing what it was until two weeks ago. I did a bit of thinking on my own and want to share what I found (probably not as the first one) because it might be interesting. In his original paper Zagier states that his proof is not constructive. In itself both involutions (the trivial t:(x,y,z) --> (x,z,y) and the zagier-involution z as discribed in the video) don't give many new solutions starting from a given one. But combined they lead from the trivial solution to the critical, from the fixpoint of the zagier-involution F := (1,1,k) to the fixpoint of the trivial involution t. Proof (sry no latex here): Let n be the smallest integer with (z*t)^n(F) = F. So t*(z*t)^(n-1)(F) = F (multiply by z on both sides). And therefore (t*z)^m * t * (z*t)^m (F) = F with m = (n-1)/2. Bringing (t*z)^m to the other side proofs that (z*t)^m (F) is a (the) fixpoint of the trivial involution, ie a critical solution. Note that n is always odd, assuming n is even results in a contradiction: If n is even we have t*(z*t)^k * z * (t*z)^k * t(F) = F with k=(n-2)/2. So again we see that (t*z)^k*t(F) is a fixpoint, this time of z, and therefore equals F. Multiplying by z gives us (z*t)^(k+1)(F) = F contradicting the choice of n.

[@leoneschle3112]

Minecraft villager be like: 5:30

[@ploopybear]

when the paper is worth 2 emeralds

[@draketungsten74]

Speaking of Minecraft... 33:13 first PayPal supporter 🤔

[@SathvickSatish]

Drake Tungsten notch agrees 😂

[@Narinjas]

I wonder if there is a villager sound expansion mod that includes this take of the sound in the variety or if it will have it included now.

[@squibble311]

mathologer is a gamer confirmed

[@iridium9512]

Wow. That's a lot to take in. I get the idea, but I feel like to truly get an intuitive grasp, I would need to take some time to think it all over. Amazingly well explained. Well done!

[@Mathologer]

In his 1940 book “A Mathematician’s apology” the mathematical superstar G.H. Hardy writes: “Another famous and beautiful theorem is Fermat’s ‘two square’ theorem... All the primes of the first class” [i.e. 1 mod 4] ... “can be expressed as the sum of two integral squares... This is Fermat’s theorem, which is ranked, very justly, as one of the finest of arithmetic. Unfortunately, there is no proof within the comprehension of anybody but a fairly expert mathematician.” My mission in today’s video is to present to you a beautiful visual proof of Fermat’s theorem that hardly anybody seems to know about, a proof that I think just about anybody should be able to appreciate. Fingers crossed :) Please let me know how well this proof worked for you. And here is a very nice song that goes well with today’s video: kzbin.info/www/bejne/p3y5apWBYqh1jtE Added a couple of hours after the video went live: One of the things that I find really rewarding about making these videos is all the great feedback here in the comments. Here are a few of the most noteworthy observations so far: -Based on feedback by one of you it looks like it was the Russian math teacher and math olympiad coach Alexander Spivak discovered the windmill interpretation of Zagier's proof; see also the link in the description of this video. -Challenge 1 at the very end should be (of course :) be: an integer can be written as a difference of two squares if and only if it is odd or a multiple of 4. -one of you actually some primality testing to make sure that that 100 digit number is really a prime. Based on those tests it's looking good that this is indeed the case :) -one of you actually found this !!! 6513516734600035718300327211250928237178281758494417357560086828416863929270451437126021949850746381 = 16120430216983125661219096041413890639183535175875^2 + 79080013051462081144097259373611263341866969255266^2 - a nice insight about the windmill proof for Pythagoras's theorem is that you can shift the two tilings with respect to each other and you get different dissection proofs this way. Particularly nice ones result when you place the vertices of the large square at the centres of the smaller squares :) -proving that there is only one straight square cross: observe that the five pieces of the cross can be lined up into a long rectangles one of whose short side is x. Since the area of the rectangle is the prime p, x has to be 1. Very pretty :) -Mathologer videos covering the various ticked beautiful theorems: e^i pi=-1 : kzbin.info/www/bejne/Y5XLeaWdYrCVgJI (there are actually a couple of videos in which I talk about this but this is the main one) infinitely many primes was mentioned a couple of times already. This video has a really fun proof off the beaten track:kzbin.info/www/bejne/gnfahHyagbiiqas pi^2/6: again mentioned a couple of times but this one here is the main video: kzbin.info/www/bejne/r4HPZ2eunsSNkKM root 2 is irrational: one of the videos in which I present a proof: kzbin.info/www/bejne/nGLcdXiug6Z4g8k pi is transcendental: kzbin.info/www/bejne/b5jOkGujhtqYqMk And actually there is one more on the list, Brower's fixed-point theorem that is a corollary of of what I do in this video: kzbin.info/www/bejne/baSQioBjoMh-g6c -When you start with the 11k windmill and then alternate swapping yz and the footprint construction, you'll start cycling through different windmill solutions and will eventually reach one of the solutions we are really interested in. Zagier et al talk about this in an article "New Looks at Old Number Theory" www.jstor.org/stable/10.4169/amer.math.monthly.120.03.243?seq=1

[@EagerLearner23]

4k+1, 4k-1

[@andlabs]

"Very nice song" is a link back to this video Prof. Hardy's life appears to be increasingly anticlimactic. Always overshadowed or outdone, it seems.

[@madhuragrawal5685]

Windmill summary is 404ing

[@kenhaley4]

The link to the "very nice song" is incorrect. It simply links right back to this video.

[@Mathologer]

@@kenhaley4 Fixed the link :)

[@evanparsons123]

I watched this video on Christmas morning 2020. At the risk of goading, this is a stunning video and I'm tremendously grateful for it.

[@ciscoortega9789]

I gasped out loud when he pointed out that the windmills pair up with each other. That was amazing

[@shatter6012]

Cisco Ortega what does gasped out mean

[@thomassabino5440]

@@shatter6012 audibly

[@shatter6012]

@@thomassabino5440 oh thanks now it makes sense

[@kenhaley4]

Brilliant! I admit I didn't follow every step of all this on first viewing, but I know there's nothing there beyond my ability to understand. I will watch it again (maybe several times), because it's easy to see that it's truly beautiful!

[@johnchessant3012]

This is the proof found in "Proofs from the Book"! Don Zagier condensed this into one (not easily understood) sentence.

[@Mathologer]

Must be a more recent edition than the one on my bookshelf :) Maybe also have a look at the links in the description of this video :)

[@peterdriscoll4070]

I gotta admit, that was an awsome proof. Not long-winded, just windmilled.

[@tamirerez2547]

2^2+ i^2=3

[@Mathologer]

Veeery funny :)

[@JMairboeck]

Wait a minute, does that mean that if we extend the domain of x and y into the complex numbers, it works for any (real) prime? 4^2+(3i)^2=7, for example

[@willnewman9783]

@@JMairboeck Yes. As he mentions at the end of the video, any odd number can be written as x^2-y^2. So any odd prime p has p=x^2-y^2=x^2+(iy)^2

[@tamirerez2547]

Yes Joachim. looks like. And so 6^2 + 5i^2 = 11 Or we can simply say that ANY PRIME NUMBER CAN BE WRITEN AS a^2 + b^2 or a^2 - b^2 (and we dont need imaginary numbers) 10^2 - 9^2 = 19 12^2 - 11^2=23. 16^2 - 15^2=31 Only now I notice: 10+9=19 12+11=23 16+15=31

[@jerberus5563]

He says 4k+3, and that's equivalent to 4k-1.

[@SoleaGalilei]

I remember the Numberphile video and I'm amazed that such a simpler proof is available now! Thanks for sharing it.

[@CC-hx8gj]

>One person assigned each theorem a score of 0, with the comment, “Maths is a tool. Art has beauty”; that response was excluded from the averages listed below, as was another that awarded very many zeros, four who left many blanks, and two who awarded numerous 10s. lol

[@mathsfermattest6494]

Challenge 2: p =x^2 - y^2=a^2 - b^2 p = (x+y)(x-y) = (a+b)(a-b) That implies x-y = a-b = 1 So x+y = a+b = p So we get x=a by those equation s that implies y=b □

[@chicohigs]

p=x^2-y^2=(x+y)(x-y) => if p-prime, then x=y-1 => p=2x+1 (proof of the unique)

[@Dusk-MTG]

I'm studying mathematics right now nad I really love integer numbers, they have so many interesting properties and you really need to stretch your mind to find them. I find calculus, topology, geometry and all that stuff seemingly complicated, but actually easy (the proofs are very often similar), but number theory is always fascinating. At first glance it may seem the easiest part of mathematics, but it's probably the hardest one to understand deeply.

[@GreenMeansGOF]

Mathologer’s Theorem: π is the sum of two squares. 21:19

[@heliy_25]

Impossible. For a degree greater than 2 .

[@hugo3222]

It's actually a simple corollary of the theorem that a circle cannot be transformed into *one* square.

[@federico6416]

guys relax, he was referring to the fact that he pronounced "P" as π (pie)

[@heliy_25]

@@federico6416 😜

[@amirilan4435]

Wow, the windmill proof is such a beautiful proof, amazing.

[@eliyasne9695]

20:16 This is brilliant! That's the very reason this theorem is about primes.

[@ts4gv]

eliya sne It's crazy to think about it that way, but you're totally right. The proof wasn't very "primey" until that key moment.

[@sighthoundman]

But, because of the famous identity, known to the ancient Greeks, any number that is a product only of primes of the form 4k + 1 (and possibly including 2) will also be a sum of two squares. Things get more complicated if you allow primes of the form 4k + 3. The simplest way to describe it (YMMV) is that in the Gaussian integers (that's numbers that can be written in the form a + bi, where a and b are integers), primes are exactly the numbers that are either of the form a + bi where (a + bi)(a - bi) = a^2 + b^2 = p (prime in the [regular] integers) or p prime in the integers, with p = 4k + 3. That's one reason this theorem is important. It tells us how to factor complex integers.

[@Neptunade]

Handy little elegant trick. Quite intuitive, i was a step or two ahead as the explanation was on-going.

[@Quwertyn007]

5:38 All primes that can be written as a sum of two squares are primes

[@Mathologer]

:)

[@doctorwhouse3881]

I got that... and then remembered what I was watching and felt silly.

[@FrankHarwald]

@@Mathologer I mean, it's not wrong, is it? ;)

[@mcris7727]

This reminds us of the old saying that mathematics is a giant truism (or tautology) that reduces to something like 1+1=2. In Physics, Dirac said: The world of elementary particles would be much more scarce if not for so many imaginative physicists.

[@MrEvilNES]

(p^q)->p , yes

[@DarrelFrancis]

Beautiful proof, beautifully explained!

[@myrthryn]

I have the most excellent documentation of who came up with the windmill interpretation of this proof, but there isn't enough space to place it into this youtube comment.

[@alexandersanchez9138]

Man, this channel is awesome. Keep up the great work!

[@MK-13337]

21:15 "and therefore pi is a sum of two squares" 🤔 now that is some mathologer magic I missed in between the lines

[@Mathologer]

Time to watch it one more time. Double the fun :)

[@phiefer3]

The crux of it is that he had x^2 + 4y^2, and 4y^2 is the same as (2y)^2, so that's a square, and x^2 is obviously a square number, so that's the sum of two squares. In fact, this is how he started out this section of the video, go back to 10:38 and watch that bit. He starts out by defining p this way (since this is what he was trying to show), then he split the y^2 into y(y) and replaced one of the y with z to make a more general formula, and then from there he proved that there is always a case where y and z are equal.

[@MK-13337]

I'm a mathematician myself so I know how the proof works. "pi" in my comment is not a typo since it *sounds* like he says that pi (3.1415....) is the sum of two squares 🤔 Technically true if we don't consider integer squares

[@davidr2421]

Yeah it does sound like he's saying "pi is the sum of two squares", but I assume he just mispronounced "p"?

[@loganstrong5426]

I'm wondering if the original proof uses the function pi(n), referring to the nth prime number? He swapped out for p, but misspoke once after all his research.

[@tbabubba32682]

I love the friendly rivalry between you and numberphile. I also love your visualizations.

[@benjaminbrady2385]

6:59 the primes of the form 4k + 1 can be written as the sum of two integer squares. We only need to check the last two digits to determine a numbers modulo 4. This yields 81 which is 20*4 + 1 ⚀

[@Sir_Isaac_Newton_]

This is wrong. Here's why: Although what you claim might be correct in most scenarios, it isn't in this one specifically; the fact that 4k is divisible by 2 and the 1 is prime* means that the aforementioned theorem cannot be extrapolated unto said value. In other words, the theorem doesn't "fit" for the equivallence we are trying to prove.

[@pengin6035]

You are a godsent angel, I've had my mouth open the whole video, I wish I could subscribe twice

[@shoam2103]

Typical Fermat. Claiming he has proofs but not delivering. *Unlike* Mathologer of course 😜

[@raghunanangecuni7034]

Mathologer is the most useful math channel . I like your explanation sir What a way that you explain. Thank You sir

[@_abdul]

21:20 And therefore pi is a Sum of Two Square. That Excitement Nearly Killed me.

[@Mathologer]

:)

[@mjcard]

Me2

[@ldd6098]

I'm 40 now and just recently got recently interested in (some) mathematics. Thanks for these videos.

[@vj_henke]

I have a question regarding 32:19, the challenge at the end. You claim that the existence of integers x,y with x^2 - y^2 = n (> 0, for simplicity) leads to n being odd. As i found the counter example x = 4, y=2 and therefore n=16 - 4 = 12 being not odd , I probably misunderstood you. Any help is kindly taken. Greetings from Germany.

[@Mathologer]

Yes, well spotted, of course that statement is wrong. The correct statement is: an integer can be written as a difference of two squares if and only if it is odd or a multiple of 4 :)

[@davidvose2475]

I'm not a proper mathematician, but this proof is intriguing and satisfying - because it is elegant and I can follow. Thank you

[@alexanderboscan2087]

Videos are back :D

[@Vanechki]

I am happy to see my proof in English. Thank you!

[@randompuppy789]

This man is straight up a beast.

[@tonyschofield4489]

Thankyou for reigniting my fascination with Maths.

[@clermeil]

I wish everyone talking about the harder sciences (physics, chemistry, etc.) and math spoke English in a German accent. It seems appropriate. Additionally, biologists should speak English with a British or American accent and Philosophers should speak English with a French accent. Am I crazy?

[@SkyOverEvrythng]

With all due respect to Descartes, Pascal, & Voltaire, and Levi-Strauss, Barthes, Baudrillard, & Derrida… I think Hegel, Kant, Leibniz, Nietzsche, Schopenhauer, Heidegger, & Wittgenstein might take issue with you about philosophy with a French accent.

[@ScorieDivine]

I beg to differ. French accent should go to maths, no small part of classic physics and chemistry (until 1800-1850), and a good chunk of medicine. Germans should inherit philosophy and chemistry (especially from 1850 to 1950). Americans get either everything or nothing, from 1950 to our day, depending on how you look at it. As for Brits, fuck em.

[@portolan4454]

The Spanish have a saying Speak to your loved one in French; speak to your customer in English; speak to your horse in German; but- speak to your God in Castellano.

[@pixequil]

I used to do my math homework in Myriad Pro so I'm happy to see you using that font for math

[@spacemanspiff2137]

TheOneThreeSeven. I love the fine structure of his name

[@LukeSumIpsePatremTe]

I think that he's one 37 year old man who likes math. I know it goes deeper, but that's my impression.

[@richardfarrer5616]

That number is a constant surprise to me.

[@TheOneThreeSeven]

HA!! You nailed it =) If I get enough subscribers on my new channel I just launched yesterday I will do a username backstory reveal

[@dlevi67]

And he drives an Alfa?

[@TheOneThreeSeven]

@@LukeSumIpsePatremTe lmao this is actually how old I am =)

[@fuuryuuSKK]

Also, the fact that the wind/mills pair up in two different ways, and the y=z and x=y=1 solutions are the odd ones out respectively, means that you can start at the trivial x=y=1 solution, and recursively calculate the other half of each pair, alternating pairings. We will arrive at y=z eventually, because since y=z and x=y are different for all p>5 (in which case we already have the solution), and all that are not the odd one out are paired up, we will always get a windmill that we have not yet encountered when taking the paired windmill of the previous step. Since p is finite, this means that the number of windmills is also finite, and we will eventually either exhaust the set (getting the y=z case) or get the y=z case early. While thinking it through more, it occured to me that what the previous paragraph does is just taking strips of length k and winding four of them around a 1². This then led me to find a path for skipping straight to the end, by considering that this means that the wings of the paired windmill of the will be shorter than the center, and thus the center will be o, the largest odd square smaller than p. Then it is simply a matter of calculating (p-o)/4 for y, and determine the paired windmill of this to get the square decomposition of p. Simplifying more, the even term of the decomposition is thus 2y=(p-o)/2.

[@kallewirsch2263]

"If there are any parts of this video that you struggled with, just ask" Yes. Where do you get your T-shirts from? It took me a few seconds to figure this one out, but hey - you've got a friend :-) Edit: Greetings from AUstria to Australia

[@shoam2103]

Haven't figured out. Can you tell? 🤔 Asking for a friend..

[@Mathologer]

I get my t-shirts from all over the world. The one in this video I actually made myself :)

[@52flyingbicycles]

Mathologer does it mean “to infinity and beyond”?

[@shoam2103]

We are struggling with figuring out what the symbols mean 😅

[@KCML82]

First thing I noticed in the video hehe. Nice Toy Story reference :p

[@yahccs1]

Wow this makes sense and is a great visual way to see when sums of squares work. Similar to the patterns I've found (with primes) looking into integer triangles using Pythagoras, just out of curiosity. When I saw the thumbnail image for the video I thought it looked like using LEGO to solve maths puzzles! A difference between squares is just turning the equation around from a^2+b^2=c^2 to b^2=c^2-a^2 which can be factorised to (c-a)(c+a) so if b is prime b^2 only divides by b and 1 so since a and c are different, c-a and c+a are 2 different factors of b^2, so c-a=1, c=1+a and c+a = b^2 so a triangle with a prime on one side needs the other 2 sides to be separated by 1. Since c is the longest (hypoteneuse), a must be the second largest and b the smallest. If the equation a^2+b^2=c^2 must have the smallest number first (written as an "a, b, c" triangle) then a would be the prime and b cannot be prime. So it's better to say a^2 is the prime squared and (c-b)=1 and (c+b) = a^2 so c = average of (c-b) and (c+b) = (a^2+1)/2 and b = half the difference between them = (a^2-1)/2 I thought this was a neat way of getting Pythagoras integer solutions where the hypoteneuse is one more than the larger of the 2 squares. It doesn't work for a=2, but for odd primes a^2+1 and a^2-1 are always even so are still integers when dividing by 2. They can be integers for any odd number 'a', but if a is an odd number that is not prime, this would not be the only solution but (c-b) and (c+b) could have other values -other ways of combining factors to make a^2. If 'a' is a prime (p) squared then a^2 has factors 1, p, p^2 and p^3 so (c-b) might be p and (c+b) would be p^3 - another interesting result. Anyway I think I got as far as 526 integer solutions with 100 unique side ratios, unless I missed some! It was good to plot 'a' against 'b' and see how they spread out. Also interesting to have a bar chart of how many times each side ratio appeared in the whole set, the 3,4,5 triangle appeared 139 times and the 5, 12, 13 triangle 46 times, and I found 49 with side ratios that only occurred once up to the 100th which was (400, 561, 689). I did them in order of 'b' because each value of 'b' sets a minimum and maximum value of 'a' from the square root of ((b+1)^2-b^2) to b-1, which means a limited number of 'a's to investigate. If we only want unique side ratios and not a multiple of them a, b and c must not have any common factors so they can't all be even, nor have 2 even and 1 odd because the two even squares cannot add or subtract to make an odd square. They can't all be odd either because the difference between or sum of two odd squares is even so one of them is even. 'c' cannot be an even number because if c^2 divides by 4 and a and b must both be odd, so their squares have to add up to an even number that doesn't divide by 4. OK most of this is probably really obvious to maths experts out there but I have forgotten so much and wanted to get back into it so I could understand enough to finally get round to reading that book I inherited with my Dad's stuff - on Fermat's last theorem. Maths videos online have been so helpful, that I was able to understand quite a lot of it - although I had no idea what 'modular forms' meant.

[@hakeemnaa]

7:05 yes if a number has last two digits which can be divided by 4, the whole number can be because 100 can be divided by 4 so any multiple of 100 can be, like 83500 and you can check by delete all the other digits like 83516 it will be 83500+16 83500 can be divided and you have to check 16 now with the prime number it end with 81 which is 80+1 4(20)+1 :) the rest don't not matter because they can be divided by 4 any way

[@Mathologer]

That's it and that's the answer I was expecting :) I was actually quite surprised by this answer by ben1996123: 6513516734600035718300327211250928237178281758494417357560086828416863929270451437126021949850746381 = 16120430216983125661219096041413890639183535175875^2 + 79080013051462081144097259373611263341866969255266^2

[@chayansarma4443]

Lovely explanation and illustrations.Really a nice proof.

[@ImranMoezKhan]

Fermat was where "The proof is left as an exercise" started.

[@Mathologer]

:)

[@nboisen]

Brilliant. And explained with amazing clarity!

[@Jack-vm1fg]

Makes me wonder just how much of mathematics can be reduced to stuff that's easier to understand.

[@ts4gv]

THIS VIDEO IS FANTASTIC!!! THANK YOU

[@denny141196]

I’m pretty sure I got the 4k+3 proof. Might need corrections: 1st claim: to get an odd number as the sum of two numbers, they must have opposite parity (one even one odd). Proof: by exhaustion. Even+even=even, odd+odd=even. 2nd claim: The square of a number has the same parity as the number itself. Proof: (2k)^2 = 4k^2 = 2(2k^2). (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1. Therefore, for an odd number to be the sum of two squares, it must be an even number squared plus an odd number squared. Consider (2k)^2 + (2m+1)^2. Using previous working, this is equal to 4k^2 + 4m^2 + 4m + 1 = 4(k^2 + m^2 + m) + 1. Therefore, the sum of the squares of two numbers with opposite parity is always one more than a multiple of 4. There is no other way to get an odd number as the sum of two squares, so getting any number of the form 4k+3 is impossible.

[@ViceroyoftheDiptera]

Yes, but you have not proved the hard part: that a number of the form 4K+1 can indeed be written as the of two squares.

[@martinepstein9826]

Amazing video as always! I see some commenters sharing their favorite theorems. In the theme of counting how many objects can be created in a certain way I recently learned about Kurotowski's closure-complement problem. It asks: given any subset of any topological space, by taking successive closures and complements how many different sets can be created? The answer turns out to be 14 ! What a strange number. It seems too high, but if you smush together enough weird subsets of R you can achieve it.

[@hugo3222]

At 24:37, instead of cutting the tiles, why not consider the whole plane, which is covered by "equally many" blue+green and red squares. Of course, one has to consider a proper limit, but it's still easier to see what's going on than with the cut-and-rearrange procedure.

[@subhabratabasak5681]

hey!! your videos are really helpful ..please keep uploading such stuff. please do not stop.

[@JERMAG07]

I see Mathologer's new upload. I just literally drop anything else I do, and watch. Cat video after this, maybe? :)

[@123mailashish]

Marvellous!!! U r an excellent teacher. U know the nuances of voice modulation while teaching. Excellent write up.

[@allmycircuits8850]

behold: new Amazon Prime service translating in 4k+1 resolution :)

[@ts4gv]

AllMyCircuits Nice one dude

[@danielwimmer4698]

But you don't need prime to get 4k.

[@danielwimmer4698]

I guess, you can't get 4k with prime would have been better at least considering that it is more accurate and you don't need prime to get to 1 (mod 4) either. Oh, well I didn't think of it at the time.

[@linuxgaminginfullhd60fps10]

I really appreciate the work you are doing. I wouldn't find(look for) this nice proof on my own and if you didn't post the video I would spent this limited time I had today on something useless... Your videos boost my inspiration and thus make me feel better. Keep going!

[@Mathologer]

That's great :)

[@Censeo]

Mathologer making complicated math available for amateurs since 2016

[@goebelmasse]

Every odd number can be written as a difference of two squares in at least one way. Trivial proof: n² equals the sum of the first n odd numbers 1, 3, 5, 7, …, (2n-1), which is easily shown by induction and very easy to visualize too. Just try squared paper and see. 😎 So it is easy to see 2n-1 equals n² - (n-1)². For any odd prime number p = 2n-1 this is the only way to equal a difference of two squares. Proof by contradiction: Let's assume p is a difference of two squares n² - (n-m)² with m > 1 and m < n Then p = n² - (n-m)² p = (n + (n - m)) * (n - (n - m)) p = (2n - m) * m p is prime and p | m means m = p or m = 1, and the last one contradicting m > 1. But if m = p, then 2n - m = 1 and therefore m = 2n - 1, contradicting m < n. So there is exactly one way to express any odd prime number p as a difference of two squares. This was so much easier than Fermat's theorem. Even possible to me. 😉 (Excuse my English, but it is better than most people's German.)

[@ankeunruh7364]

Math teaching in the GDR included proofs. What you wrote here, is what my math teacher showed as an example of "proof by contradiction". Afterwards we had to prove the Fundamental Theorem of Arithmetic.