Fermat's Christmas theorem: Visualising the hidden circle in pi/4 = 1-1/3+1/5-1/7+...

NEW (Christmas 2019). Two ways to support Mathologer
Mathologer Patreon: / mathologer
Mathologer PayPal: paypal.me/mathologer
(see the Patreon page for details)
Leibniz's formula pi/4 = 1-1/3+1/5-1/7+... is one of the most iconic pi formulas. It is also one of the most surprising when you first encounter it. Why? Well, usually when we see pi we expect a circle close-by. And there is definitely no circle in sight anywhere here, just the odd numbers combining in a magical way into pi. However, if you look hard enough you can discover a huge circle at the core of this formula.
Here is a link to the relevant chapter in Hilbert and Cohn-Vossen's book Geometry and the Imagination (Google books). I am pretty sure that the idea and proof for the circle proof of the Leibniz formula that I mathologerise in this video first appeared in this book and is due to the authors: books.google.com.au/books?id=...
Here is a link to a video in which 3blue1brown about the same hidden circle in Leibniz formula:
• Pi hiding in prime reg...
And another video by him about a hidden circle in the solution to the Basel problem:
• Why is pi here? And w...
There is also a neat generalisation to what we talked about in this video to the solution of the Basel problem - in terms of the lattice points in a 4-dimensional sphere and the 4-square counterpart of the 4(good-bad) theorem. If you are interested in some details have a look at the last proof in this write-up by Robin Chapman: empslocal.ex.ac.uk/people/sta...
Links to two Numberphile videos about the one-sentence proof by Don Zagier featuring Matthias Kreck: • The Prime Problem with... (intro), • The One Sentence Proof... (the math)
Link to the original Jodocus Hondius engraving of Jodocus Hondius that Google tries to pass of as a portrait of the mathematician Albert Girard
www.swaen.com/zoomV5e.php?id=...
Thank you very much to Marty for all his help with polishing the script of the video and Karl for his idea for the 2019 Easter egg.
Today's t-shirt: google "spreadshirt pi+tree+christmas+math"
Enjoy :)

[@Mathologer]

Mathologer recently hit 500K subscribers and I would like to thank you all for your interest and your support over the years. Since I started the channel four years ago, it has pretty much been a one-man labour of love. However, maybe now is a good time to take Mathologer to the next level and hire someone to assist with editing the videos, preparing subtitles, etc. In preparation for this, I recently monetised the videos by switching on the least annoying ads on KZbin. I also just launched a Patreon page a couple of minutes ago: www.patreon.com/mathologer If you enjoy these videos and you can afford it, please consider taking out one of the Patreon memberships or making a one-time donation via PayPal: paypal.me/mathologer :) My plan is to also use this Patreon page as a platform to share more Mathologer materials with you, get into live chats, etc. Let’s see what’s possible and what makes sense here. Merry Christmas :)

[@eduardosilveira8685]

Knowing that Mathloger might expand was my best recieved Christmas gift so far

[@Sz3ntAr]

I'd love to see this channel growing big. All the immesureasurable effort put into your videos is just awesome. Merry Christmas :)

[@cmuller1441]

This channel reminds me why I love maths. The most beautiful and perfect stuff in the universe and beyond!

[@abhisheksamal1970]

It will become...1M.....very very.....soon

[@hamiltonianpathondodecahed5236]

Hey I found 2019 in your π-shirt after the 244th digit after the decimal #yay

[@MrSigmaSharp]

We dont celebrate Christmas in my country but I celebrate any Mathologer video anytime

[@mandolinic]

Hey, Mathologer is back after a short interruption of service. Nice to see you again!

[@koenth2359]

The good, the bad and the even

[@gyrgrls]

What an odd way to put it.

[@patrickchristensen96]

The disaster has been averted. Glad you’re back!

[@YuzuruA]

hope you sue the fuck out of sob... he commit several offenses and probably even criminal

[@andreguimaraes9347]

Yupp

[@cohendane4875]

Instablaster...

[@Qwerty13146]

Merry Christmas! I think the easiest way to show pi = lim N(r) / r^2 is to say that the squares centered in grid points = N(r) / r^2 >= pi * (r - 1)^2 / r^2 and since both leftmost and rightmost side of this inequality approach pi as r->inf, we get what we wanted

[@reetasingh1679]

Squeeze theorem... Gotta love the name

[@gaurangagarwal3243]

Well I call it sandwich theorem :D

[@peterw9006]

Policemen Theorem!

[@pyglik2296]

Three sequences theorem...? I was thaught boring names.

[@inigo8740]

@@peterw9006 "Théorème des Gendarmes"

[@AmitKureel]

Thanks for recognizing Indian Mathematician Madhava for his work on Calculus and infinite series relating to Pi. More than the credit, I am interested in bringing in awareness among talented Indian youth towards Maths which is currently more influenced by cash-generating Wall-street jobs and related courses.

[@JAlexCarney]

Hooray I was sure that the videos would need to be reuploaded from scratch. Very glad to see the old comments were not lost.

[@edwardc5700]

I think they were just private.

[@MisterMajister]

Let's all just forget about yesterday shall we? Great to see the videos are back and I really hope they work their situation out (IN PRIVATE)!

[@Mathologer]

At dawn somewhere deep in the woods. My personal choice of weapons is swords :)

[@lucaambrogioni]

@@Mathologer I would go for a shotgun... better safe than sorry

[@theseeker7194]

@@Mathologer just enable 2 step verification in your google account, no 'swords' needed ;)

[@inyobill]

@@Mathologer That was ugly.

[@TonyLambregts]

Way too much drama as far as I am concerned. I llike my math to avoid it.

[@AttilaAsztalos]

Mathematician looks at 3:10 - "hey, neat, the area proof of pi!" Engineer looks at 3:10 - "hey, look, nuclear reactor tiles!" ;)

[@harikrishnank.j.4954]

Madhava is an Indian (Keralite) mathematician. Actually I am watching this video miles away from his native place (Irinjalakkoda, sangamagrama as you said). Nice work sir👏

[@harikrishnank.j.4954]

@@sachinnandakumar1008 ya me too

[@Utesfan100]

Odd factors of 2020: 1,5,101,505, all good. This tells us there are 16 pairs. 2020 = 16^2 + 42^2 = 38^2 + 24^2. The two permutations and four sign choices yield all cases.

[@cheshire1]

The problem is that you can only choose two signs at a time. There are four cases involving 16, 42 and their negatives and also four cases with 38 and 24. There have to be two more pairs of positive integers whose squares add up to 2020.

[@Utesfan100]

@@cheshire1 : There are eight for each. We can also swap the order, which is what I intended by permutations. Thus the other two are 42, 16 and 24, 38. The smallest number with this pattern of dots is 5*13=65, which one can quickly verify has precisely two pairs in the first half-quadrant: (8,1) and (7,4).

[@Abhishekkumar-gu5gi]

@@Utesfan100 no there is only 4 ways for each and eight ways combined both because commutative property says that a+b = b+a and here this is what you are talking about changing orders . But in that 4 way every term has different sign so they are different but a+b is same as b+a

[@chaddaifouche536]

@@Abhishekkumar-gu5gi We're searching points coordinates so (16,42) and (42,16) are two different points thus here permutations must be taken into account. This wasn't obvious in the video because most examples were small and a was equal to b but rewind and look at 7:04 for 5 = 1²+2², you have only one *pair* of positive numbers whose square add to 5, but you have eight *couples* since (1,2) and (2,1) are different.

[@non-inertialobserver946]

First the Mathvengers: Eulergame, then Numberphile hit pi million subs, and now a new Mathologer video?! Is it Christmas? Oh wait, it is.

[@alperyoloyilmaz5388]

I am really happy to see your videos back. We don't know the value of what we have untill we lose it.

[@NittanyTiger1]

Surprised you didn't call good and bad numbers nice and naughty numbers respectively.

[@Mathologer]

:)

[@shawon265]

Now I can hear him laughing after calling them nice and naughty

[@gcewing]

Number of presents left under tree = number of good children - number of bad children

[@MelindaGreen]

Or impish and admirable

[@arijitkulkarni]

@@gcewing Number of children - Number of bad children maybe ;)

[@AndyGoth111]

Great to see you back!

[@VickyGYT]

A big thank you for acknowledging work by an Indian mathematician! First video doing so. You are a gifted teacher.

[@KresimirYT]

Thank you for all the lovely videos you made this year! I wish to you and Marty all the best in the next year, and I wish myself many more of your great videos... Merry Christmas!

[@TheReligiousAtheists]

My proof that bad odd numbers can't be written as a sum of integer squares: Let our bad odd number be 4n+3, and let's say it's equal to j²+k², for some integers j and k. However, since j² and k² add to an odd number, one of them must be even and the other must be odd. Let's choose j (arbitrarily) to be odd and k to be even. Hence, j=2a+1 and k=2b (for some integers a and b). So 4n+3=(2a+1)²+(2b)². Expanding and rearranging, 4(n-a²-b²-a)=-2 So 2(n-a²-b²-a)=-1. However, this is saying that 2 times an integer is -1, which is something only a drunk man would agree with. Hence, 4n+3 can't be a sum of 2 integer squares. QED.

[@aypfvn]

You can also say that because all squares are congruent to 0 or 1 modulo 4, the sum of two squares must be congruent to 0, 1, or 2 modulo 4. To see that all squares are congruent to 0 or 1 modulo 4, notice that 0^2 = 0, 1^2 = 1, 2^2 = 0, and 3^2 = 1 (mod 4). These are called the quadratic residues modulo 4.

[@TheReligiousAtheists]

@@aypfvn Isn't that the same as what was said in the video?

[@aypfvn]

The Religious Atheists Whoops.. I was reading comments half way through the video and didn't make it to the end yet. Sorry.

[@TheReligiousAtheists]

@@aypfvn No probs

[@l1nk353]

Thanks alot for coming back!!

[@nicklasbrodin6534]

Glad to have you back Mathologer!

[@erutuon]

My mnemonic for this infinite sum now will have to be that it's related to the number of integer points in one quarter of an infinitely large circle. Pi being the area of the whole unit circle. With the Pythagorean theorem, that connects it to the sums of squares. Beautiful proof. I just rewatched this one. You presented it simply enough for a high school math guy like me to understand.

[@kutlokebaikanye6872]

Been following your videos for 4 years and I have never been disappointed. Great stuff mate 👍!

[@aaronleperspicace1704]

Dear Mr Burkard Polster, I sincerely appreciate the fact that you give credit to the right person for the discovery of a mathematical identity/formula (in this case, the Indian mathematician, Mādhava). Earlier also, I learned through you that the fact that e^ix = cos x + i×sin x was known to Roger Cotes before Euler. Cheers and merry Christmas!

[@ceramicsky14]

I love following along and getting that giddy feeling when I can tell where I proof is going. Merry Christmas! Love the Channel since day 1!

[@Mathologer]

That's great. I wish KZbin would give us a list of all subscribers in the order they subscribed. Would be nice to know :)

[@msolec2000]

Glad to see you're back!

[@SoleaGalilei]

Merry Christmas and happy new year! Congratulations on the success of your channel. It's well deserved.

[@geraldillo]

Awesome video once again. I love the way you visualise everything to make the math understandable.

[@hemanthkotagiri8865]

Thank god, my favourite videos are back! Thank you, mr. True Mathologer.

[@MuffinsAPlenty]

Excellent video! Very nice! I've encountered the result about sums of squares for _primes_ before, because that gives us a test for which integer primes remain primes in the Gaussian integers. It's nice to see that there is a much more general version of the result covering _all_ nonzero integers and telling _how many_ ways there are to write as sums of primes. Also, that's a beautiful way to get that infinite series formula :) Merry Christmas!

[@tissuewizardiv5982]

I really appreciated how you saved the explanation for "4 good minus bad" theorem until the end of the video. It cut off almost 9 minutes from the main proof and made it much more manageable. Good job

[@MarcoLiedekerken]

That bell between chapters ..... it gave me goosebumps !

[@slother932]

I love your content and am proud to support your Patreon at the π level!

[@Mathologer]

That's great, thank you very much :)

[@portr002]

The first thing I did this Christmas was watch this whole video! Fantastic as usual thank you!

[@liutiewh]

A great pleasure during Christmas for me. Thank you!

[@RandyLunn]

Beautiful as always! Thank you for an amazing year. Merry Christmas and Best Wishes for the New Year.

[@Mathologer]

:)

[@rc5989]

I enjoyed this holiday gift, from the very best maths channel ever!

[@mayuragarwal9598]

One of the best from mathologer. This proof really blowed up my mind. Amazinnnnggg

[@qingyangzhang887]

Merry Christmas. Your channel is a treasure.

[@maxsch.6555]

I'm glad that you are back :)

[@Socrates...]

YAY! The videos are back!

[@SebastianGodoyMedel]

Just found this video. Amazing way to visualize the proof. Thanks for the hard work!

[@Robinzon__Kruzo]

Thank you for your videos. Happy New Year.

[@chirayu_jain]

Best thing about you, you always tell the real mathematician of the theorem. 😀👍🏻 BTW merry Christmas🥳🥳

[@mariomario-ih6mn]

Hi

[@chirayu_jain]

@@mariomario-ih6mn Hi

[@richardgratton7557]

Really love your channel. The math is great, amazing. But the mathologer is even better!Congratulations on hitting 500 000 subs. Merry Christmas and Happy New year to you. :)

[@pharaohgarmar5611]

Mathologer, you are awesome, and I really look forward to your Mathologerising Gauss’s law of quadratic reciprocity.

[@theglobalgossip1539]

So far the best Christmas present! Thanks "Manta" (Math Santa) and Merry Xmas!

[@Kazetomosuki]

I really enjoyed this present! Dankeschön & Glückwunsch zu mehr als 500.000 !!! 💪 Fröhliche Weihnachten (=

[@enricolucarelli816]

God save internet! I can not even imagine a life no more without having access to these wonderful, interesting, and enlightening videos like the ones you present every now and then! Herzlichen Dank, fröhliche Weihnachten, und bitte weiter so.

[@JCOpUntukIndonesia]

"hardcore mathematics channel"... love it! Merry Christmas professor!

[@TheOneThreeSeven]

I couldn't wait for the video proof of the Christmas thoerem and found something AMAZING on mathoverflow that I wanted to give you a heads up about in case you had not seen it before. I think the famous one sentence proof can be Mathologerized using the incredible geometrical interpretation of the involution in the answer by Moritz Firsching to the MathOverflow question titled "Zagier's one-sentence proof of a theorem of Fermat". I have read many books/papers on polynomial automorphisms and never saw anything like that before in my life. After going through a bunch of proofs I think the one-sentence version is the best hope to do at the level of your videos using that geometry trick to explain where the involution comes from. There is also a wonderful numberphile video on this topic with an outline of the proof.

[@Mathologer]

Very nice. Will ponder this a bit, may well be worth doing a followup video on this insight :)

[@zentecson7415]

I found TheOneThreeSeven!

[@extremeswissgerman2536]

Welcome back!

[@jess_o]

Merry christmas, mathologer!

[@Mihau_desu]

Great video, Love it. I actually found out about those theormes recently when I was trying to solve an olypmpiad problem about sums of two squares. It was pretty difficult problem and I'm really proud I could solve it. Merry Christmas 🎄🎄🎄

[@7eroBubble]

Mathematics is like an intellectual equivalent of Willy Wonka's Everlasting Gobstopper -- it is harder than anything else on Earth, it tastes wonderful and no matter how long you work at consuming it, it never diminishes.

[@petergregory7199]

Not bad Ray, but as a theory it sucks.

[@ejrupp9555]

@@petergregory7199 It's a theory you only get, if you suck at it ... kind'a like a paradox.

[@bruinflight1]

Burkard you are an amazing and wonderful person. Thank you for the great gift of your youtube channel :-)

[@ayrtonsenna9278]

Merry Christmas, this is the best gift ever!!!

[@MelindaGreen]

Frohes neues Jahr und vielen Dank für all die schönen Mathe-Videos!

[@Mathologer]

Fröhliche Weihnachten, hab schon lang nichts mehr von Dir gehört.

[@robertcameron-ellis6518]

Thank you! An excellent Christmas present.

[@mheermance]

Thanks! I always wondered why that formula worked, now I understand the underlying mechanics of it. PS Merry Christmas and Happy New Year.

[@anjanmukherjee7997]

Coolest Christmas present I ever Had .THANK YOU

[@TimMeep]

The fact that such a channel can hit 500K+ subs, and maths videos with millions of views, gives me hope in human kind.... happy to +1 on Patreon

[@Mathologer]

Great :)

[@hakarraji5723]

Bitte mach mehr von deinen Mastervideos. Ich studiere Mathe im ersten Semester und liebe die langen Videos die richtig tief gehen, je tiefer desto besser ;)

[@DrMikeE100]

OMG! You are a fabulously knowledgeable and effective teacher. I am deeply humbled by your excellence. Thanks so much! (Dr. Mike Ecker is a rational skeptic and a Ph.D. mathematician - CUNY 1978, Ph.D. Summa Cum Laude - who researches and writes prolifically. He is also a retired PSU mathematics professor and former computer journalist.)

[@gheffz]

Brilliant! Thank you! Love your approach to explaining things! ( _Merry Christmas! ... let's make 2020 the best year so far!_ )

[@braydentaylor4639]

This comment did not age well

[@gheffz]

@@braydentaylor4639 Nor 2021, but let's be as optimistic as we can be with 2022... it's a decision, a choice... make the best out of whatever happens.

[@braydentaylor4639]

@@gheffz Bro, 2021 was an okay year, eons better than 2020

[@paperEATER101]

2019 hidden in your t shirt ...thanks for this Christmas Present ...Merry Christmas Mathologer and all

[@inyobill]

What a pretty Christmas present. Vielen dank!

[@mi3137]

these are very nicely prepared videos. superb!

[@AnonimityAssured]

Phew! I hope the problem has been solved for good. I was so saddened yesterday.

[@EeshwarBalageethavengateswaran]

Brilliant and Fantastic steps those few from 15:10.. looked like some magic done

[@orstorzsok6708]

Thanks again!!! It was a revelation for me again! I was wondering for a while why pi/4 can be expressible with this inf. series. And thanks for the recommendation of the book of Hilbert and Vossen.

[@ildossi7934]

Thanks and wishing all the best for 2020

[@troemax]

Very nice video, thank you. Und fröhliche Weihnachten!

[@singhania264]

I wonder who the hell can dislike such videos. Love your work though! Your videos are an inspiration to me sir🙏🙏🙏🙏

[@7382121868]

You are one of the Great knowledge provider.... Thanks from India...

[@schnapsdrossel78]

Danke und frohe Weihnachten!!!

[@ernestdecsi5913]

This video is the most beautiful christmas gift. Thanks. - a Slovak pensioner :-)

[@bobengelhardt856]

This other Mathologer video: kzbin.info/www/bejne/r4HPZ2eunsSNkKM shows the hidden circle in Euler's identity Pi^2/6 = Sum(1/n^2). Do all the series that have Pi in their value have hidden circles? It seems that they must, but maths are not always what they seem to be. Then, what of the identities that aren't series, but have Pi in their values? E.g., 1/2! = sqrt(Pi)/2? There are videos showing the derivation of 1/2! using the Gamma function - does the Gamma function have a hidden circle?

[@TylerHNothing]

@13:17 also interesting 1 in 5 dots are green for the row corresponding to 5, the pattern is the same for all the rows, which really makes everything hit home. nice video!!

[@jollyroger9286]

Reminds me of "Pi hiding in Prime numbers" from 3Blue1brown. That was the video that inspired me to do maths. Lovely job, Mathologer

[@Tehom1]

Since 2020/4 = 505 which has prime factors 5 and 101, we just have the power set of {5,101} which is {1,5,101,505}. Since both 5 and 101 are congruent to 1 mod 4, they are both good and so are all their products. You definitely gave us an easy one. So 4 good - 0 bad gives us 16 ways of writing 2020 as the product of squares. To find them, we'll first use the usual trick of dividing out the 4's. We'll find the ways to write 505 as squares and multiply each component by the square root of 4. True confession time: I'm adapting this from work by Dario Alejandro Alpern, whose fsquares program I ported to Gnu GMP. We can find the ways to write 505 by finding the way to write its factors and using the fact that (a^2+b^2) (A^2+B^2) = (aA+bB)^2 + (aB-bA)^2 We'll find the solutions in positive integers, and then convert each such solution (a,b) into 4 solutions, {(a,b)(-a,b)(a,-b)(-a,-b)}. We know that every prime congruent to 1 mod 4 is the sum of two squares. For 5 this is easy: 5 = 1^1 + 2^2 and that's all. For 101 it's not hard either: 101=1^2+10^2, and this confirms that you are pitching us your softest softball. A quick check vs {49, 64, 81} confirms that this is the only way to write 101. Again, this is just the positive/positive solutions. So we have: (1^2+2^2)(1^2+10^2) = (1+20)^2 + (10-2)^2, which gives us: 505 = 21^2 + 8^2 2020 = 2*21^2 + 2*8^2 = 4*441 + 4*64 = 1764 + 256 and consequently 3 other solutions, 2020 = (-42)^2 + (16)^2 = (42)^2+(-16)^2 = (-42)^2+(-16)^2. We also have: (1^2+(-2)^2)(1^2+10^2) = (1-20)^2 + (10+2)^2 = (-19)^2 + 12^2 = 361 + 144 Thus we find 8 ways of writing 2020: 2020 = 42^2+16^2 = -42^2+16^2 = 42^2+-16^2 = -42^2+-16^2 = 19^2+12^2 = -19^2+12^2 = 19^2+-12^2 = -19^2+-12^2 According to the formula, there must be 8 other solutions out there, but I'm not seeing the permutation of these equations that gives them.

[@Tehom1]

Ah, I think I found the other 8. Since in the solutions for 18, we're counting both 3^2+-3^2 and -3^2+3^2 as separate solutions, we seem to be allowing a^2+b^2 and b^2+a^2 as distinct solutions. Thus the other 8 ways to write 2020 are the flips of the ones I gave, starting with 2020 = 16^2+42^2 etc.

[@dlevi67]

@@Tehom1 The other possibility is that since 24^2 + 38^2 = 2020 too, this gives the other 8 with the permutations of signs that you illustrated above? Though of course they don't come from 505... edit: corrected 16 to 24

[@Tehom1]

@@dlevi67 Excellent. I was too focused on 505.

[@adandap]

Tehom Yes, that's right. It's because we're looking for the points on an integer lattice. So (16,42) and (42,16) are different. Took me a while to realise that too.

[@Tehom1]

@@dlevi67 16^2 + 38^2 = 1700

[@willemvandebeek]

Merry Christmas, Burkard and company! :)

[@OhGreatSwami]

Wow. Well done. Engaging to the end.

[@davidbarnett8617]

Glad the issue has been resolved. This is one of the best channels on the internet. I do note the updated 'About' page, removing the GG reference...

[@Mathologer]

Yes, and I guess you also know why. If you want to see what the page looked like in the past, just look it up on the wayback internet archive.

[@ThaJoker128]

Thank god they are back :)

[@hah720]

This video reminded me of Grant Sandersen's (3Blue1Brown) video entitled: "Pi hiding in prime regularities" However I find this video more straightforward. Thanks for the present and Merry Christmas!

[@timh.6872]

I think this one seems more straightforward because it proves less. 3b1b's video only left the "all primes congruent to 3 mod 4 are gaussian primes and all primes congruent to 1 mod 4 can be factored into a pair of conjugate gaussian primes" to be proven. I suspect that the proofs of this video's lemmas Mathologer is trying to adapt are of similar complexity, but they're a bit more integral to the overall proof using this path. Given how jumpy people get around the complex numbers, I suspect it will take a while before we see them.

[@mwalton9526]

@@timh.6872 3b1b's video also doesn't prove that prime factorizations under the gaussian integers are unique.

[@captainsnake8515]

mwalton probably because they aren’t unique... (1-2i)(1+2i)=5 and (-1-2i)(-1+2i)=5

[@mwalton9526]

@@captainsnake8515 unique aside from multiplication by i,-i,-1. (1-2i)*(-1)=-1+2i (1+2i)*(-1)=-1-2i

[@captainsnake8515]

mwalton (1-2i)(1+2i)=(2-i)(2+i)=5 I don’t even know why I’m arguing this point, it’s the worlds dumbest argument ever. I knew what you meant in your first message. I’m just being dumb lol.

[@quantumsigmaqed6312]

So the vids are back

[@nanigopalsaha2408]

Yeaaaaah!

[@RicoCordova]

$1 pledge level? No brainer. I made a comment on a recent video (maybe the most recent) about wanting to see more about how theorems have been used to solve other theorems. I'm not sure if you meant to do that here, but I really like learning about putting together these "looks suspiciously related" theorems to build another result. Thank you very much!

[@hisxmark]

Mathematics never disappoints my sense of wonder.

[@heliy_25]

I am very happy that Your channel continues to work. Your wonderful deeds, o Lord.🙊🙃

[@xugalla3261]

Greeting from China! So nice to see you back~ Kyo

[@xugalla3261]

Waiting for some hardcore video on Abel-Ruffini theorem :D

[@alexandeur]

Welcome back! So happy to see everything back to normal!

[@mwalton9526]

One video is still missing.

[@vytah]

@@mwalton9526 Which one is it? At worst, you might find a copy on Bilibili

[@user-vv6ec8yz9w]

Quadratic reciprocity?! My old professor, way back, wrote a book on number theory, in which he and his coauthor proved the law of quadratic reciprocity several different ways. Then he included a question at the end of the chapter asking the reader to find their own proof! At the time he was teaching analysis but he often slipped in number theory themes in his lectures. I never picked up the book because I wasn't interested in number theory at the time and frankly I was not any good, but I always thought he was special, a genius even, who loved his subject. You remind me of him. Happy New Year!

[@drpope4303]

Thank you for the videos my children love them and I want to wish you a happy holiday

[@anshshah6399]

Absolutely magical 25 mins !! Worth investing !!! :)

[@SaturnCanuck]

Lovely. Merry Chistmas

[@johnchessant3012]

I came across Zagier's one-sentence proof a while ago. I couldn't understand it until I read a longer explanation in "Proofs from the Book". It's really neat!

[@MuffinsAPlenty]

Glad to see things are back to normal. If you ever need anything from me (not sure what that would be, but I thought I would offer), just let me know!

[@Mathologer]

Hmm, would you be interested being part of a proofreading team? If there wasn’t the self-imposed pressure to deliver a video at least once every four weeks, I would keep fiddling with these tricky topics that I am now specialising in much longer. Anyway, those four-week deadlines have the effect that invariably I am running out of time and that the shooting and editing happens last minute and little mistakes can slip in at this stage. Not really a problem but I am thinking of making a preview available to four or five math savvy friends who can have a quick look over the video to see whether they can spot anything. This preview could arrive at some untimely hour and nobody in the group should feel obliged to do anything if the short response time doesn't suit them. As long as I get feedback from one or two in the group each time that should be plenty. If you are interested please just send me an e-mail at [e-mail]? No problem at all this is not for you :) At any rate thanks again for all your help with patiently answering questions over the years.

[@alvarotovar5855]

@@theongrejoy5645 I don't* :)

[@hamiltonianpathondodecahed5236]

Hey I found 2019 in your π-shirt after the 244th digit after the decimal _I am a man of culture_

[@chriszachtian]

Found it myself and had to scroll down VERY far to this spoiler ;-)

[@heliy_25]

Thanks for the video. A Christmas present for you. In binomial decomposition two in power 1 and 2 kernel between units-conditionally 1 (Prime number) and 2 (Prime number). And then the core (sorry slang) - the numbers are not simple. Happy new year and merry Christmas.