Euler's and Fermat's last theorems, the Simpsons and CDC6600

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  • เผยแพร่เมื่อ 23 มี.ค. 2018
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    This video is about Fermat's last theorem and Euler's conjecture, a vast but not very well-known generalisation of this super theorem. Featuring guest appearances by Homer Simpson and the legendary supercomputer CDC6600. The video splits into a fairly easygoing first part and a hardcore second part which is dedicated to presenting my take on the simplest proof of the simplest case of Fermat's last theorem: A^4 +B^4=C^4 has no solution in positive integers A, B, C.
    The proof in question is taken from the book Lectures on elliptic curves by J.W.S. Cassels (pages 55 and 56). Here is a scan of the relevant bits: www.qedcat.com/misc/cassels_pr... This writeup of the proof actually contains a few little typos, can you find them? In the video I attribute the proof to John Cassels the author of this book because I've never seen it anywhere else. It's certainly not Fermat's proof as one may be led to believe reading Cassel's writeup of this proof.
    The Wiki page on Euler's conjecture contains a good summary of the known results and a good list of references: en.wikipedia.org/wiki/Euler%2...
    There is one aspect of this conjecture that I did not go into in the video. The conjecture says that for n greater than 2 at least n nth positive integer powers are necessary to make another nth integer power. On the other hand, it is not known whether for every such n there is an example of n nth powers summing to another nth power. In fact, even my example of a sum of five 5th powers summing to another 5th power in the video was not known to Euler. Anyway, the wiki page also has a summary of what's known in this respect.
    Today's t-shirt should be easy to find, just google what it says on the t-shirt.
    Thank you very much to Danil for his continuing Russian translation support, Marty for his very thorough nitpicking of the script and all this help with getting the explanations just right and Michael for his help with filming and editing.
    Enjoy!
    Typo:
    (Someone who's really paying attention :) Great video as usual (even though I already knew the proof). There's a small mistake at 17:48, on line 5 it should be (u^2-Y)(u^2+Y)=4v^2 instead of (u^2-Y)(u^2-Y)=4v^2.
    (M) Yep, luckily not where I actually do the proof. Actually a great one to pinpoint who is really paying close attention to detail :)

ความคิดเห็น • 942

  • @Mathologer
    @Mathologer  6 ปีที่แล้ว +622

    I really struggled with Mathologerizing that proof at the end. Started working on this video sometime last year but then gave up on it. Pretty happy that it's finally done :)

    • @soumyadipsarkar7242
      @soumyadipsarkar7242 6 ปีที่แล้ว +6

      Mathologer ,,u did not use the word " Fermat's descent method",,that's kinda cool that u have used computer programming logic,,,anyways,,can u please make some videos on Fermat's theorem for higher powers!!??

    • @CardDeclined...
      @CardDeclined... 6 ปีที่แล้ว +1

      to much math

    • @kennethgee2004
      @kennethgee2004 6 ปีที่แล้ว +2

      I am sorry you lost me with the powers of four and splitting of the 2's This makes no sense to me. the power 4 is itself 2^2 but that does not make it clear that all n^4 must be divisible by 4.

    • @FernandoBiaziNascimento
      @FernandoBiaziNascimento 6 ปีที่แล้ว +3

      This is only true for even n, but not only n^4, it is for any even power of n, being n^2, n^4, n^6, n^8, ...
      1) Let n be even
      2) so whe can have an integer m that satisfies n = 2m;
      3) n^2 = (2m)^2 = (2^2) * (m^2) = 4(m^2)
      4) n^(2k) = ((2m)^2)^k = ((2^2) * (m^2))^k = 4^k * m^(2k)
      So it will be divisible by 4.
      Hmm, that makes me think that {(2m)^p | p >= 2} will ever be divisible by 4 also for odd p!
      I think it may be better to go have a break before thinking on this!!!! XD
      EDIT: After a brief break, taking from step 3) above and let "a" be a positive integer:
      4a) n^(2+a) = (2m)^(2+a) = (2m)^2 * (2m)^a = (2^2)*(m^2)*(2^a)*(m^a) = 4*(2^a)*(m^(2+a))
      So n^p with even n will be divisible by 4 for any p >=2, being n^2, n^3, n^4, n^5, ...
      Andy Arteaga (below): Nice explanation of the (odd)^(even), I was thinking that should be something like that and would think in that after another break. XD
      Thumbs up!

    • @andresxj1
      @andresxj1 6 ปีที่แล้ว +5

      Kenneth Gee If the number is even then we can write it in the form of *2n* , and if we have an even power then we can write it as *2k* . So we have *(2n)^(2k)* and that's the same as *[(2n)^2]^k* *=* *[(2^2)^k]·[(n^2)^k]* *=* *(4^k)·(n^2k)* so we can see the number will be divisible by 4.

  • @jeffreybernath6627
    @jeffreybernath6627 6 ปีที่แล้ว +473

    I am just now realizing that if A^2+B^2=C^2 has solutions, and A^4+B^4=C^4 has no solutions, that means there is no Pythagorean triple where A, B, & C are all squares. Out of all of the infinite Pythagorean triples out there, none are made up of square numbers. That's nuts!

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +198

      That's correct. In fact, most proofs of A^4+B^4=C^4 has no solutions argue via Pythagorean triples :)

    • @robertgumpi7235
      @robertgumpi7235 6 ปีที่แล้ว +12

      I would have expected this. It has for me the same „feeling“ as 2 p^2/q^2

    • @christopherellis2663
      @christopherellis2663 6 ปีที่แล้ว +5

      By definition, they are not P Triples. Much more fun to make them from Fibonacci like series.

    • @sldecka
      @sldecka 6 ปีที่แล้ว +4

      Jeffrey Bernath wow

    • @ThePharphis
      @ThePharphis 6 ปีที่แล้ว +4

      Good to know. Might take note of this for an algorithm
      Can some of them be squares?

  • @thecwd8919
    @thecwd8919 3 ปีที่แล้ว +201

    "Maybe not to us mere mortals, but the demigod, Euler was pretty convinced."

    • @aradhya_purohit
      @aradhya_purohit 3 ปีที่แล้ว +9

      I disagree on demigod, he, in fact is a GOD and possibly the GOAT

    • @RobBCactive
      @RobBCactive 2 ปีที่แล้ว +3

      @@aradhya_purohit Computer say no! 😉

  • @Craznar
    @Craznar 6 ปีที่แล้ว +574

    Joke Time:
    Q: What do you get if you pour root beer into a square glass?
    A: Beer.

    • @dlevi67
      @dlevi67 6 ปีที่แล้ว +80

      Unless your root beer was imaginary, in which case any beer around you will mysteriously disappear.

    • @kmlo3784
      @kmlo3784 5 ปีที่แล้ว +2

      how about a cube glass?

    • @AgentFriday
      @AgentFriday 5 ปีที่แล้ว +13

      Root Beer --> Cube => Square Beer

    • @AgentFriday
      @AgentFriday 5 ปีที่แล้ว +33

      Unfortunately, since squares are 2-dimensional, square beer would be FLAT :(

    • @f.jideament
      @f.jideament 5 ปีที่แล้ว +9

      @@dlevi67 this sentence looks like something from discworld

  • @johnchessant3012
    @johnchessant3012 6 ปีที่แล้ว +31

    Mathologer and 3B1B have been coordinating their upload schedule for the past few weeks. It's great!

  • @krillbilly1435
    @krillbilly1435 4 ปีที่แล้ว +136

    Euler's Conjecture: Exists
    CDC6600: I'm about to end this man's whole career

    • @M-F-H
      @M-F-H 4 ปีที่แล้ว +3

      You got it wrong... Euler's Conjecture says "Doesn't exist."

    • @skylermagnificent5422
      @skylermagnificent5422 3 ปีที่แล้ว +1

      M F Hasler don’t know if ur joking but nice

  • @alejandrapaz2864
    @alejandrapaz2864 6 ปีที่แล้ว +291

    i have found a elegant proof for the reimann hypothesis but it is too long to put it into a youtube comment

    • @pluto8404
      @pluto8404 6 ปีที่แล้ว +114

      Its not too difficult. I developed a super simple method in my undergrad that i can write in less than 3 sentences.
      Basically all you have to do is...
      *Read more*

    • @mfhasler
      @mfhasler 5 ปีที่แล้ว +20

      That hypothetical reimann hypothesis is probably quite uninteresting compared to the famous Riemann hypothesis...

    • @alexandermizzi1095
      @alexandermizzi1095 4 ปีที่แล้ว +5

      @@pluto8404 The Read more button is fake!

    • @TonyStark-kh4eb
      @TonyStark-kh4eb 3 ปีที่แล้ว +9

      @@alexandermizzi1095 Exactly

    • @idon.t2156
      @idon.t2156 3 ปีที่แล้ว +2

      Me 2

  • @gabest4
    @gabest4 6 ปีที่แล้ว +420

    Larger than 2? I also can't find a solution for A^0+B^0=C^0.

    • @darealpoopster
      @darealpoopster 6 ปีที่แล้ว +402

      I can but the comment section is too short to fit my proof

    • @ghrissiabdeltif5397
      @ghrissiabdeltif5397 6 ปีที่แล้ว +12

      hahahaha :'D

    • @NizarHaddad
      @NizarHaddad 6 ปีที่แล้ว +66

      it has only 1 single solution where A = B = C = 0, because
      undefined + undefined = undefined :)

    • @lovaaaa2451
      @lovaaaa2451 6 ปีที่แล้ว +42

      0^0=1 and is not undefined. Besides this I cannot see a justification for why you could claim equivalence of two undefined notions, in fact if you could set up for instance 1/0=1/0 and claim that this is defined then you can get infinity=-infinity => infinity+1=-infinity+1 => 2infinity+1=1 => infinity=1 or whatever other equality you desire, so this statement is contradictory.

    • @NizarHaddad
      @NizarHaddad 6 ปีที่แล้ว +61

      when striking 0^0 with limits and functions in real analysis, you always end up with 0^0 = 1
      but when going to complex functions & analysis, limits vary much around that point, therefore we concluded that it is undefined.
      and about my statement "undefined + undefined = undefined" it is just for kidding :)

  • @edensaquaponics1941
    @edensaquaponics1941 6 ปีที่แล้ว +88

    11:30 proof of (odd^even) mod 4 = 1
    An odd number can always be expressed as 2n+1, therefore odd² = (2n+1)² = 2n×2n + 2n×1 + 1×2n + 1×1 = 4n² + 4n + 1
    Q.E.D.

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +22

      Exactly :)

    • @edensaquaponics1941
      @edensaquaponics1941 6 ปีที่แล้ว +32

      Mitchel Paulin, so I forgot to include that:
      Any even number can be expressed as 2n, and therefore x²ⁿ = (xⁿ)²
      Therefore the "special" case for odd^2 extends to odd^even.

    • @ryanchild292
      @ryanchild292 6 ปีที่แล้ว +3

      Thank you so much for your explanation!

    • @txikitofandango
      @txikitofandango 6 ปีที่แล้ว +7

      (2k + 1)^(2m) = (4k^2 + 4k + 1)^m. For any power m you would raise this to, you would have a bunch of coefficients that are multiples of 4 (because you're multiplying some combination of 4 and 1)... plus 1^m at the end. That's why you always get a remainder of 1.

    • @ryanprov
      @ryanprov 5 ปีที่แล้ว +2

      You can get an even simpler proof if you remember that xy mod n = (x mod n)(y mod n) mod n -- the same is true for addition, and this means you can mod at any time in the middle to make things easier without changing the result, just make sure to mod at the end (this is because the integers modulo n form a ring over + and *). Then, any odd number is either 1 or 3 mod 4, and 1*1 mod 4=1*3 mod 4=3*3 mod 4=1. So any odd number squared is 1 mod 4, and 1 to any power is still 1.

  • @trained-wreckscience-strug8932
    @trained-wreckscience-strug8932 6 ปีที่แล้ว +10

    I am SO grateful you were able to put this video together! I absolutely adore the more complicated bits, thank you so much!

  • @TomatoBreadOrgasm
    @TomatoBreadOrgasm 6 ปีที่แล้ว +182

    It sucks hard that Euler's conjecture turned out to be wrong. I like that conjecture.

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +84

      Yes, very sad, but most of it could still be true :)

    • @GrandMoffTarkinsTeaDispenser
      @GrandMoffTarkinsTeaDispenser 6 ปีที่แล้ว +40

      Yes indeed such a beautiful structure and a very elegant generalization. Then maths showed us the middle finger.

    • @PC_Simo
      @PC_Simo ปีที่แล้ว +3

      I agree. I like that conjecture, as well. But, at least, it turned out to be wrong in the positive way (that there *_WERE_* solutions to equations that there wasn’t supposed to be; rather, than the other way around) 😌.

    • @jixpuzzle
      @jixpuzzle ปีที่แล้ว +2

      No point in getting sad about it. That's the reality. Reality is not obligated to be beautiful, nice and symmetrical! Ans that's the beauty of reality:) ❤️❤️

    • @tiberiusmagnificuscaeser4929
      @tiberiusmagnificuscaeser4929 8 หลายเดือนก่อน +2

      I wonder if there is some kind of structure underlying such equations, but it’s more complicated than just “if n is greater than k, no solution exists.” Maybe something like “if n is greater than f(k), no solution exists” and f(k) happens to equal k for k=2 and 3

  • @anselmschueler
    @anselmschueler 6 ปีที่แล้ว +102

    Could you imagine if that proof was published in the Simpson's?

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +35

      They actually did publish one original proof in Futurama (same crowd as the Simpsons). I did a video about this Futurama theorem very early on th-cam.com/video/J65GNFfL94c/w-d-xo.html

  • @B3Band
    @B3Band 6 ปีที่แล้ว +124

    Kids living in a 4D world memorize the 3-4-5-6 right tetrahedron, or whatever it would be called since it obviously wouldn't be a tetrahedron.

    • @cryme5
      @cryme5 6 ปีที่แล้ว +2

      Haha nice one, though it should be squares ;) Funny metrics they would be using with cubes.

    • @FernandoBiaziNascimento
      @FernandoBiaziNascimento 6 ปีที่แล้ว +11

      Some times I find the math with 3 dimensions to be laborious, I've never thinked of beings perceiving 4D! O.O

    • @islandfireballkill
      @islandfireballkill 6 ปีที่แล้ว +9

      Fernando Biazi Nascimento Solving rigid body equilibrium problems sounds like a pain. 4 force vectors and 6 moments gives you 10 total equations. I don't even want to imagine the labouriousness in quadruple intergration for centers of mass.

    • @FernandoBiaziNascimento
      @FernandoBiaziNascimento 6 ปีที่แล้ว +1

      Busted_Bullseye It is a new level of nightmares! XD

    • @Arikayx13
      @Arikayx13 6 ปีที่แล้ว +2

      It makes you wonder what a 4D kid would even perceive as math thanks to the 4D connections in their brains. Like we've used numbers to condense what amounts to adding up tick marks both for symbolic and systematic ease.
      Might a 4D brain perhaps have numbers that encode more information or be able to recall large amounts of facts on each number that makes a connection seem trivial that is high level to use so you start zoning out and thinking about 4D brains...

  • @SuperDreamliner787
    @SuperDreamliner787 5 ปีที่แล้ว +17

    I finally got my divisibility by 3 prove of the equation at 11:42 going. At first I thought I need to do an analysis with the possible remainders like in the video. Since I could not draw any conclusion from that, I needed to find another way. Fortunately I remembered this trick by using the digit sum. Since the digit sum of 3987 is 27 and therefore divisible by 3, 3987^12 must also be divisible by 3. 4365^12 yields the same result. So on the left side, there is remainder 0. On the right side however, the digit sum of 4472 (which is 17) does not divide 3, so 4472^12 also does not. Therefore the right side has a remainder, which is not equal to 0. So the remainder-equation reads: 0 does not equal 0, which completes the prove. q.e.d.

    • @TruthNerds
      @TruthNerds 5 ปีที่แล้ว

      Good thinking! I, on the other hand, am a lazy slob, and just entered the numbers into ghci[1]:
      Prelude> 3987^12 + 4365^12
      63976656349698612616236230953154487896987106
      Prelude> 4472^12
      63976656348486725806862358322168575784124416
      Prelude> 3987^12 + 4365^12 - 4472^12
      1211886809373872630985912112862690
      bc[2] also works, with the same syntax, and is preinstalled on many Unix(oid) systems. :o)
      [1] The Glasgow Haskell Compiler's interactive environment.
      [2] The "basic calculator", actually standardized by POSIX as I now learned, so it should be present in all POSIX-conforming systems.

    • @PC_Simo
      @PC_Simo ปีที่แล้ว

      Indeed 🎯.

  • @IndiBrony
    @IndiBrony 6 ปีที่แล้ว +6

    I'm not a Mathematician - I finished school with a B (should have been higher but I was a lazy kid) - and never pursued maths into college and university. As such, I only occasionally catch on to the smaller concepts, but the way these videos are presented and broken down is fascinating to listen to and watch, and make the whole process of understanding just that much easier to someone who has little to no concept of things outside of the basic stuff we learned at school.
    Keep up the good work and, who knows, one day I may come to realise I understand a bit more about maths than I thought!

  • @johnsnow5305
    @johnsnow5305 6 ปีที่แล้ว +22

    I don't have what it takes but I will keep watching after the warning anyways!

  • @masked_mizuki
    @masked_mizuki 6 ปีที่แล้ว +170

    I have a really great comment in mind about this video but its far too long to contain here.

    • @robertgumpi7235
      @robertgumpi7235 6 ปีที่แล้ว

      JGLP haha. Great comment.

    • @robertgumpi7235
      @robertgumpi7235 6 ปีที่แล้ว +1

      ... the internet is to small for your comment. Hihi.

  • @TrimutiusToo
    @TrimutiusToo 4 ปีที่แล้ว +8

    You dared me to forget... I actually forgot until i rewatched this video....

  • @twistedgwazi5727
    @twistedgwazi5727 6 ปีที่แล้ว

    A new Mathologer video on a Saturday is always amazing.

  • @morphx666
    @morphx666 6 ปีที่แล้ว

    In my opinion this is, by far, your best video. Wow! what a ride!

  • @hernanipereira
    @hernanipereira 6 ปีที่แล้ว +15

    i hope one day we can have a collaboration video of Mathloger and 3Blue1Brown. That would be something really special ;)

  • @eliascaeiro5439
    @eliascaeiro5439 6 ปีที่แล้ว +12

    Hi Mathloger, great video as usual (even though I already knew the proof). There's a small mistake at 17:48, on line 5 it should be (u^2-Y)(u^+Y)=4v^2 instead of (u^2-Y)(u^2-Y)=4v^2.

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +12

      Yep, luckily not where I actually do the proof. Actually a great one to pinpoint who is really paying close attention to detail :)

  • @Ricocossa1
    @Ricocossa1 6 ปีที่แล้ว +1

    Thank you so much for making this proof accessible. These videos are great.

  • @Dominik-jk4yq
    @Dominik-jk4yq 6 ปีที่แล้ว

    Love your videos Mathologer! keep 'em coming!

  • @dickybannister5192
    @dickybannister5192 3 ปีที่แล้ว +3

    nice video. I was wondering whether you could tackle the easy parts of the Catalan Conjecture (only fully proven in 2002) using similar visuals. some bits make for a great intro to the rules of field extensions (without really having to do too much. I think you have probably done something similar anyway for other stuff)

  • @bailey125
    @bailey125 5 ปีที่แล้ว +11

    6:00
    95,800^4 + 217,519^4 + 414,560^4 = 422,481^4
    27^5 + 84^5 + 110^5 + 133^5 = 144^5
    There are more examples, but these are the simplest ones.

  • @pronoy91
    @pronoy91 6 ปีที่แล้ว +1

    Wow! A lot of knowledge at one place. Made sense of most of it but I must use pen and paper to get the satisfaction. Kudos for putting this together

  • @dontask23
    @dontask23 6 ปีที่แล้ว +1

    I was waiting for this video for a long time!!

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +4

      Consider the t-shirt as an explanation for why you had to wait so long :)

  • @aakash_kul
    @aakash_kul 6 ปีที่แล้ว +7

    His shirt: "I took the RHOMBUS," can also be read as "I took the WRONG BUS." This is too much, man!

    • @temra7063
      @temra7063 3 ปีที่แล้ว

      Congratulations you got the pun

  • @thanosAIAS
    @thanosAIAS 6 ปีที่แล้ว +39

    Where do you get all those amazing T-shirts? :P

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +21

      Really all over the planet and quite a few I make myself :)

  • @robertgumpi7235
    @robertgumpi7235 6 ปีที่แล้ว +1

    Wonderful video, as ever.
    Thanks for your good work.

  • @rohitkumar-rq6qh
    @rohitkumar-rq6qh 6 ปีที่แล้ว

    Thank you for such a nice and informative and elegant proof.

  • @BruceLCM
    @BruceLCM 6 ปีที่แล้ว +5

    Now I know how to say "Euler" after years of mispronouncing it. I've thought it's "Eu" in "Euclid".

  • @Tehom1
    @Tehom1 6 ปีที่แล้ว +12

    16:11 If you could have included 192 one more time, you'd be there:
    167^4 + 192^4 - 46225^2 = 192
    or to put it differently:
    167^4 + 192(192^3 + 1) = 46225^2

    • @irrelevant_noob
      @irrelevant_noob 5 ปีที่แล้ว

      Unfortunately, any decent pocket calculator would display the two values as ( LHS ) 2.1367508E9 and ( RHS ) 2.1367506E9 ... so clearly they would differ. :-\

  • @hansalexander905
    @hansalexander905 6 ปีที่แล้ว

    Really interesting video, thank you for the content. Watching from Spain and enjoying it, congratulations!

  • @earthbind83
    @earthbind83 6 ปีที่แล้ว +1

    Good job! You make those proofs look easy.

  • @n-wordjim1724
    @n-wordjim1724 5 ปีที่แล้ว +50

    9:07 9 Digits, not 8. You missed the 5.

    • @rucker69
      @rucker69 4 ปีที่แล้ว +7

      I believe he was talking about the limitations of older calculators, namely 8 digit display width.

    • @sillybears4673
      @sillybears4673 4 ปีที่แล้ว

      But those digits being nine seems like something special..

    • @lawrencedoliveiro9104
      @lawrencedoliveiro9104 4 ปีที่แล้ว +1

      @@rucker69 My 1970s-vintage TI-58C could show 12 digits, and actually calculated internally to a bit more than that.

    • @lawrencedoliveiro9104
      @lawrencedoliveiro9104 4 ปีที่แล้ว +1

      @@rucker69 My 1970s-vintage TI-58C could show 12 digits, and actually calculated internally to a bit more than that.

    • @justarandomjojofan6674
      @justarandomjojofan6674 4 ปีที่แล้ว +1

      Oh hi Shalltear

  • @LivingVacuum
    @LivingVacuum 6 ปีที่แล้ว +6

    I will decompose the RSA of any complexity into multipliers. Fast and not expensive.

  • @TheLuckySpades
    @TheLuckySpades 6 ปีที่แล้ว

    This is beautiful. I really enjoyed this video.

  • @brocpage4204
    @brocpage4204 6 ปีที่แล้ว +1

    man, your shirt choices are on point.

  • @MrRyanroberson1
    @MrRyanroberson1 6 ปีที่แล้ว +7

    17:11 the third-to-last row seems like a misprint
    y^2=u^4-4v^4
    4v^4=u^4-y^2
    4v^4=(u^2+y)(u^2-y)

  • @TheToric
    @TheToric 5 ปีที่แล้ว +50

    I have a marvelous proof that information is infinitly compressible, but it is too big to fit in this comments section...

  • @chessandmathguy
    @chessandmathguy 6 ปีที่แล้ว

    Beautiful proof by contradiction! Followed all of it and it makes perfect sense! Thanks!

  • @andyoncam1
    @andyoncam1 3 ปีที่แล้ว +1

    The UK writer Simon Singh has not only written a book on Fermat's Last Theorm but also one called 'The Simpsons and Their Mathematical Secrets' in which he explains how the show's writers , mostly ex-mathematicians, sneak maths jokes into many episodes. To quote the book's back cover blurb, '...everything from pi to Mersenne primes, Euler's equation to P vs NP, perfect numbers to narcissistic ones...'. Well worth a read even for non-mathematicians like me.

  • @adamsvoboda7717
    @adamsvoboda7717 6 ปีที่แล้ว +7

    There is my solution, why odd number to the power of even number always gives remainder of one when dividing by four:
    The power is even, so we can write it in the form of 2*K.
    Take the initial number N to the power of K. You will receive a new odd number, let's call it M.
    Case one:
    M mod 4 = 1
    Then M = 4*Z + 1
    Now we take a square of this number: (4*Z + 1)^2 = 16Z^2 + 8Z + 1 = 4(4Z^2 + 4Z) + 1
    The remainder is 1.
    Case two:
    M mod 4 = 3
    Then M = 4*Z + 3
    (4*Z + 3)^2 = 16Z^2 + 24Z + 9 = 4(4Z^2 + 6^Z + 2) + 1
    As we can see, the remainder is again 1.

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +2

      Yep, that's it :)

    • @anselmschueler
      @anselmschueler 6 ปีที่แล้ว

      You assume that M is odd, it seems to me. Couldn't K either be odd or even?

    • @davidrheault7896
      @davidrheault7896 6 ปีที่แล้ว

      It doesn't matter the status of K because the power is even and 2*K is always even

    • @bluestrawberry679
      @bluestrawberry679 6 ปีที่แล้ว

      i think you can do it shorter, by writing any odd number as 2n+1
      if you square this, you get (2n+1)^2=4n^2+4n+1, where the remainder is clearly 1

    • @stro5179
      @stro5179 6 ปีที่แล้ว

      Yes, and there is an even shorter way that involves very few symbols.
      However, you need an extra lemma about how exponentiation changes the remainder after division.

  • @lawrencedoliveiro9104
    @lawrencedoliveiro9104 4 ปีที่แล้ว +8

    16:14 One thing that makes this sort of thing easier nowadays is the existence of interactive languages like Python that have built-in infinite-precision integer arithmetic. No rounding errors if you avoid fractions!

    • @falklumo
      @falklumo 2 หลายเดือนก่อน

      Only if your calc app uses Python. Otherwise, you’ll use your desktop and other computer languages came with their BigMath libraries anyway, like BigDecimal in Java, or likewise for C++. OTOH, Python really is too slow to search by brute force for near misses etc.

  • @TheYourbox
    @TheYourbox 4 ปีที่แล้ว +1

    I was bored today. Now I'm thrilled again.

  • @kirstenwilliams6056
    @kirstenwilliams6056 ปีที่แล้ว

    Wow that was a wild ride!! Glad I had my seat belt on :)

  • @VerSalieri
    @VerSalieri 6 ปีที่แล้ว +3

    Fermat, Euler, and the Simpsons.... what else could I ask for? Thank you very much.
    Btw, one of my professors back in my college years told me that the proof supplied in 93 (or 91, can’t remember) was the result of more than 15 years of work.
    Let x=2k+1, x^2=8k^2+4k+1=4k’+1, i.e. x^2=1 mod4.

  • @AzazeoAinamart
    @AzazeoAinamart 6 ปีที่แล้ว +35

    16:36 - need to add 1 BROWN!

    • @denelson83
      @denelson83 6 ปีที่แล้ว +7

      Azazeo Ainamart
      Too bad there's no brown heart emoji. 💙💙💙

    • @chessandmathguy
      @chessandmathguy 6 ปีที่แล้ว +1

      I don't get what you mean...

    • @denelson83
      @denelson83 6 ปีที่แล้ว +2

      steamroller82
      As in 3Blue1Brown?

    • @shinymoonlightteaches7179
      @shinymoonlightteaches7179 หลายเดือนก่อน

      🤎

  • @DitDede
    @DitDede 6 ปีที่แล้ว +1

    A small typo error in the condensed proof. At the 5th arrow-bulletv line, one of the terms should have+.
    Very nice video, and putting the condensed proof ( as a spoiler ) enables viewers to pause and fill in the gaps on their own.

  • @logicalfundy
    @logicalfundy 6 ปีที่แล้ว +1

    14:39 - I love it, lol. Direct and to the point, no need for a lengthy paper.

  • @GumbyTheGreen1
    @GumbyTheGreen1 5 ปีที่แล้ว +3

    23:53 - Can someone explain why the 4th power that's multiplied by 2 must be odd? He never explained this.

    • @harold3802
      @harold3802 5 ปีที่แล้ว +4

      As the highest common factor is 2, if it were even it would imply the highest common factor was 4 (even numbers are multiples of 2)

  • @S.R.400
    @S.R.400 6 ปีที่แล้ว +5

    I hate to be that guy, you clearly worked very hard on the video, but at 17:09 you wrote on the screen that u^4 - Y^2 = (u^2 - Y)(u^2 - Y). One of those should be a plus. You got it right in the last part of the video though :)

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +3

      Luckily one of those self-correcting mistakes :)

  • @pianoingels7128
    @pianoingels7128 3 ปีที่แล้ว +2

    i had no time to watch the video, so i just liked for the t-shirt

  • @KillianDefaoite
    @KillianDefaoite 6 ปีที่แล้ว

    Awesome video! Thank you.

  • @Suimobile
    @Suimobile 6 ปีที่แล้ว +4

    Actually, you can get in deep water even without higher powers. Quadratic fields are enough of a problem. I'm looking at my copy of the book by David A. Cox "Primes of the Form X^2 + n*Y^2". I would really like to see a proof with visual reasoning for almost anything in that book. I just ran into a trap while trying to explain how Fermat likely proved that for n=2. (He only gets partial credit for a correct answer because he didn't show his work, but we give him that because we don't know anyone else he could have copied from.)
    My mistake was using something true for real quadratic fields while dealing with a complex quadratic field. I was trying to make a clear proof using mappings of point lattices.
    Now, if someone could show a visual proof concerning the complex field extension dealing with sqrt(-163) that would really be something.

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +1

      I find that visual reasoning is very tough to get going with this sort of maths, really struggled a lot more than usual Mathologerising this proof at the end :)

    • @noahtaul
      @noahtaul 6 ปีที่แล้ว

      Yo my dude, I have that book too! What is it you're trying to figure out? How Fermat did what for n=2?

    • @allsortsofinterests1
      @allsortsofinterests1 6 ปีที่แล้ว

      Mathologer Wiles eludes to it in the title: Modular elliptical curves. It's all about frequencies, harmony, and interference patterns expanded to an infinite degree. Some solutions get close, but with more granularity we realize they are not exact. My intuition leads me to believe all math can be visually represented by wave patterns.

    • @Suimobile
      @Suimobile 6 ปีที่แล้ว +1

      noahtaul I was trying to understand what Fermat invented before he got into his later work. Both he and Newton were sure Diophantus was actually using geometric reasoning, though there is nothing explicitly geometric in his work. Both of them were familiar with classical languages and ancient geometry. Fermat was at the begining of modern mathematics and lacked many tools that make things easier today. With that in mind I am amazed he didn't get more wrong. He was wrong about Fermat primes, but he called that a conjecture. He was right about a long list of claims. We know there are many pitfalls in this area that caught bright people centuries later. My challenge was to prove his theorem on the n=2 case without using anything not known at his time, or if that was not possible, to introduce a minimal invention I would be willing to credit him with. For example, geometric reasoning with complex numbers could make life easier, but even Caspar Wessel was in his future. Was that one of his secret weapons?

    • @noahtaul
      @noahtaul 6 ปีที่แล้ว

      *alludes And no, this really has nothing to do with frequencies/harmony/interference patterns. The closest you come is writing a modular form as a function of an exponential variable, because it satisfies f(x+N)=f(x). But beyond this, it's a lot of algebra. Not wave-pattern numerology.

  • @PinochleIsALie
    @PinochleIsALie 6 ปีที่แล้ว +8

    Sees "1 or 0", starts thinking of applications in cryptography

  • @GoGoTwice
    @GoGoTwice 4 ปีที่แล้ว

    Really great video

  • @dimosthenisvallis3555
    @dimosthenisvallis3555 6 ปีที่แล้ว +1

    Maybe do a livestream of 7-8 hours to do the whole proof. It would be glorious! Regards from Greece. Love your videos

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +5

      And I'd be dead :)

  • @Luigicat11
    @Luigicat11 4 ปีที่แล้ว +3

    9:10 Looks like it's the first 9 digits, actually...

  • @mk-allard3788
    @mk-allard3788 6 ปีที่แล้ว +3

    For that equation X^4 + Y^4 =Z^2, can't you just say that it's the same as (X^2)^2+(X^2)^2=Z^2?

    • @mk-allard3788
      @mk-allard3788 6 ปีที่แล้ว

      Oh wait nvm I'm stupid

    • @SilisAlin
      @SilisAlin 6 ปีที่แล้ว

      yes, you can

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +7

      Actually to rewrite the equation like this is the first step in other proofs by contradiction. The second step is to then use the formula that generates all Pythagorean triples and to conclude that we must be able to express the Pythagorean triple X^2, Y^2, Z in terms of this formula :)

    • @anantmishra9813
      @anantmishra9813 6 ปีที่แล้ว +1

      but x and y should be integers here. your analogy would mean that you need a pythagorean triplet where two members are perfect squares

    • @FernandoBiaziNascimento
      @FernandoBiaziNascimento 6 ปีที่แล้ว +1

      I don't think you have been stupid! It seems to me that it makes sense to resolve a simpler problem and then throw the result on the first one to analyse the consequences.
      Yameromn: Agreed, and may be that this could also be used as a statement to proof or discard an hypothesis.

  • @KateTheSleepyTeacher
    @KateTheSleepyTeacher 6 ปีที่แล้ว +2

    That shirt. I love it !!! I love watching math videos. I learn so much. I teach third grade so this is a bit over my students heads but I can learn so much still which is so exciting. If you ever want to make a math collab video I would love to do something with you :)

  • @elamvaluthis7268
    @elamvaluthis7268 4 ปีที่แล้ว

    Thank y0u for unveiling difficult mathematics.

  • @erikperik1671
    @erikperik1671 5 ปีที่แล้ว +9

    "So you think you've got what it takes?" - that's the best way to provoke me!
    (sprichst du eigentlich Deutsch im echten Leben? dein Akzent klingt so)

  • @MrConverse
    @MrConverse 6 ปีที่แล้ว +10

    9:05, the first *nine* digits, yes?

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +15

      Well, you know what they say: mathematicians cannot count :)

    • @MrConverse
      @MrConverse 6 ปีที่แล้ว

      You probably put that in there just to see who is paying attention. ;-)

    • @rubenjanssen1672
      @rubenjanssen1672 6 ปีที่แล้ว +1

      yes but 8 is a significant number in digital calculating

    • @earthbjornnahkaimurrao9542
      @earthbjornnahkaimurrao9542 6 ปีที่แล้ว

      first 10 if you round

    • @rubenjanssen1672
      @rubenjanssen1672 6 ปีที่แล้ว

      witch you should not do at that point because the numbers are displayd for more digids

  • @nooneatall5612
    @nooneatall5612 6 ปีที่แล้ว

    I really like proving this by infinite descent. It just feels really cool.

  • @andie_pants
    @andie_pants 6 ปีที่แล้ว

    I don't begin to have what it takes, but I love this guy's personality and voice. Maybe something will click...

  • @gregbernstein7524
    @gregbernstein7524 6 ปีที่แล้ว +22

    5:24 Mathologer proves there is no God.

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +4

      :)

    • @robo3007
      @robo3007 6 ปีที่แล้ว

      Take that, theists!

  • @ichthysking863
    @ichthysking863 6 ปีที่แล้ว +6

    Forst. This means I'm early

  • @TheJohnblyth
    @TheJohnblyth 6 ปีที่แล้ว

    Finally I understand how computer proofs work. Thanks!

  • @markdstump
    @markdstump 6 ปีที่แล้ว

    Mod 9 is my go-to for checking arithmetic.

  • @numero7mojeangering
    @numero7mojeangering 6 ปีที่แล้ว +13

    (0^n)+(0^n) = (0^n)
    Solved XD

    • @Mathologer
      @Mathologer  6 ปีที่แล้ว +7

      :)

    • @brcoutme
      @brcoutme 6 ปีที่แล้ว +3

      but if we allow 0 then we can't say for sure that 0^0 + 0^0= 0^0 :(

    • @NgenDoesGaming
      @NgenDoesGaming 6 ปีที่แล้ว

      Brian Cotuinho alright (0^n + 0^n = 0^n : nEZ, n>0)

    • @oisyn
      @oisyn 6 ปีที่แล้ว

      Except that it isn't. 0^n = 1 for all n≠0, so you're saying that 1+1=1

    • @oisyn
      @oisyn 6 ปีที่แล้ว

      Oh lol, I was reading n^0 rather than 0^n :X

  • @muditgupta628
    @muditgupta628 4 ปีที่แล้ว +12

    One day someone will read this comment.

    • @endermage77
      @endermage77 4 ปีที่แล้ว +1

      That comment won't stop me because I can't read!

    • @arnavanand8037
      @arnavanand8037 4 ปีที่แล้ว

      @@endermage77 ........................

    • @alanhilder1883
      @alanhilder1883 3 ปีที่แล้ว +1

      Sorry, was just skipping through the comments, did you write something...

  • @jaideepmishra6961
    @jaideepmishra6961 5 ปีที่แล้ว +2

    Yaar isne toh TH-cam me aag laga Di
    🔥🔥🔥🔥

  • @RichardASalisbury1
    @RichardASalisbury1 6 ปีที่แล้ว

    Beautiful job! I could follow it, or almost all of it. Sometime I plan to go through this again, pausing the video so I can verify each of your steps. This works for me because, as a math major decades ago, I lost interest when we got to calculus (because many of the answers are approximations, as epsilon approaches zero and the exact limit is in fact never reached; bah, humbug!; I wanted exact answers, preferably integral or rational). But I love algebra and number theory (and am endlessly fascinated by the primes), so this was right up my alley.

    • @diribigal
      @diribigal 6 ปีที่แล้ว

      The area of a circle of radius 1 unit is exactly pi square units, despite being defined in Calculus as a limit. If your Calculus teacher said "the exact limit is in fact never reached" you were done a disservice.

    • @Llanchlo
      @Llanchlo 6 ปีที่แล้ว

      But he was doing a greater disservice to himself by constraining his thoughts to desire integral or rational solutions - let alone transcendental.

  • @henryg.8762
    @henryg.8762 5 ปีที่แล้ว +1

    3BLUE1BROWN + MATHOLOGER?!
    YAAS

  • @geertcremers1588
    @geertcremers1588 4 ปีที่แล้ว +2

    BBc horizon made a great documentary called "BBC Horizon Fermat's last theorem" about Andrew Wiles and how he got to his proof. I didn't understand a single word of it, but it was impressive to see the whole process and determination.

  • @dozenazer1811
    @dozenazer1811 5 ปีที่แล้ว

    The remainder of odd number to the 4n power is 1 because if you make a multiplication table for base-4 system, it would be that 1x1 is 1 and 3x3 is 21 -> converges into 1 in the end number, which is 9 in decimal.

  • @aee220phmunirabad
    @aee220phmunirabad 5 ปีที่แล้ว

    My mind exploded just now..

  • @overlordprincekhan
    @overlordprincekhan 4 ปีที่แล้ว +2

    9:12
    Hey, I also found the '5' after that orange-marked place are very similar

  • @MasDingos
    @MasDingos 5 ปีที่แล้ว +1

    So, add all digits of the two numbers 3+9+8+7+ 4+3+6+5 (the will exponents change nothing ) = 45, repeat process until single digit remains, 4+5 = 9. If at the end of doing this, the single digit number is 3, 6 or 9, then the entire number is divisible by 3. So left side is divisible by 3. But the right side totals to 8. Ergo, not equal. Also, this entire thing works with 9. Left is divisible by 9, right isn't - not an equality.
    if this helps, the number 111,111,111 must be divisible by 9. And 222,222,222,222,222,102 at a glance is divisible by 3

  • @JohnDoe-jy7sv
    @JohnDoe-jy7sv 6 ปีที่แล้ว

    I have found a truly marvelous margin of this, which this proof is too contain to narrow.

  • @hikarustarr
    @hikarustarr 4 ปีที่แล้ว

    the studio light on the right side (your left) is way too bright this time.
    but i love your math explanations and simpsons references.

  • @MrRyanroberson1
    @MrRyanroberson1 6 ปีที่แล้ว +1

    29:50 it can also be shown that v4=r4+s4=u2, which implies that (u^2+y)(u^2-y)=u^2, which demands that y=0
    this means you don't even need the presumption that this is the simplest solution, simply that it is one, and the presence of a solution can only occur when y=0 in this way, which indeed works as x^4+0=z^2 has infinitely many solutions

  • @tobiasactually
    @tobiasactually 6 ปีที่แล้ว

    Herr Polster, das war jetzt aber heftig. Am Schluss haben Sie zur Recht gewarnt. Danke. Grüsse aus der Schweiz.

  • @farshadrezvan4661
    @farshadrezvan4661 ปีที่แล้ว +1

    Hi Mathologer, I really appreciate all of your works, here in this video I have an unsolved question. Why do you choose to prove the theorem by reduced Z power to 2 instead of Z power to 4? I traced your solution and couldn't find benefits of this reduction. To be clear my question is that why you do not deal with X^4 +Y^4 = Z^4 and just dealing with X^4+Y^4=Z^2, what is its benefits? thank you for all

  • @heymatheists5562
    @heymatheists5562 6 ปีที่แล้ว

    Well yours and three blue one brown are certainly two most amazing TH-cam channel related to mathematics. Sometimes I wonder why you don't collaborate to make a video, you two praise each other quite often and I am waiting to see you guys together.
    Regards
    Vidyanshu Mishra.

  • @peterb9481
    @peterb9481 6 ปีที่แล้ว

    Really good!

  • @tensevo
    @tensevo 3 ปีที่แล้ว +2

    What you are highlighting here is incredibly important in terms of understanding the modern world.
    A problem when calculated to 8 decimal places appears to be correct.
    The same problem computed to 32 or 64 or more decimal places, is no longer correct.
    Many ppl would benefit from taking caution before proclaiming something to be true or not. This is why I love the visual proofs you do, they are more unambiguous since once understood, anyone can prove for themselves, no need to rely on expert validation and verification.

  • @OlafDoschke
    @OlafDoschke 6 ปีที่แล้ว

    The arguing reminds me of the numberphile video about the Catalan Conjecture, where Holly Krieger showed how you can see why x^2-y^3=1 only has one solution, when it is rewritten as the easier to analyze (x+1)(x-1)=y^3. She shows if y is odd this has no solutions by arguing about which qualities the factors need to have.finally arriving at x-1 and x+1both needing to be cubes, but no two cubes have difference 2.
    There seems to be some kind of entanglement in the science communication vloggers sphere.
    But hats off, you did take on the harder to show (Z-Y^2)(Z+Y^2)=X^4 factorization. I like how this idea of turning a sum to a product allows to test some implications.
    Holly also already hinted this is used in the proof of Fermat's last theorem.

  • @MarioWenzel
    @MarioWenzel 6 ปีที่แล้ว

    I like the explanation that for every solution we would have, there is a smaller one, which in turn must have a smaller one, and we can't have infinitely decreasing chains in the natural numbers, better than the computer-program one.
    I usually only use this argument when there is indeed a smallest solution and larger solutions that are homomorphic to the smallest wrt some property in order to ignore the larger ones since we already know the property from the smallest one.

  • @YoniMek
    @YoniMek 6 ปีที่แล้ว +1

    About Euler's conjecture:
    I suspect there exists an infinite number of solutions that disprove it, but they are very sparse and I wonder which statistical rule gives the chance to find a solution in a given volume.
    In other words, as the "log" rule crudely describes the scarcity of primes in 1d, which equivalent rule describes the scarcity of solutions for any given k in the sum of powers.

  • @jack002tuber
    @jack002tuber 3 ปีที่แล้ว

    Most all these videos, I go about 3/4 of the way in, then my eyes roll back in my head and I have to say, ok, yes, I think so, whatever you say, man.

  • @antonionanni6683
    @antonionanni6683 5 ปีที่แล้ว +1

    A little generalization on Odd^Even mod 4 = 1.
    Let B mod A =r and AC +r = B, then, B^n mod A = r^n mod A -- you can use the binomial expansion of (AC + r)^n to see this easily!

  • @7infernalphoenix
    @7infernalphoenix 6 ปีที่แล้ว

    Where can I get that t-shirt!!!!..it's so cool ! ..
    Great video by the way, kudos from all your fans here in India

  • @temoku
    @temoku 2 ปีที่แล้ว

    Wonderful

  • @terryendicott2939
    @terryendicott2939 6 ปีที่แล้ว

    Using the computer search is a cool way of explaining infinite descent.

  • @traewatkins931
    @traewatkins931 5 ปีที่แล้ว

    You made brain all mushy. In all seriousness, I can never understand how mathheads and do this it simply amazes me the intuitive leaps you make ... I would never see some of these steps.

    • @BladeOfLight16
      @BladeOfLight16 5 ปีที่แล้ว

      The steps are not intuitive. He's not doing this on the fly. He's referencing work that's been thoroughly peer reviewed and analyzed, and he's doing that with preparation! It's not like you sit down one day and can do all this. It takes _practice,_ careful analysis, and a lot of beating your head against the wall trying to find the right direction to go. Sometimes you can reuse a technique you've used in other contexts to speed up the process, but you generally can't just up and do all this on a whim.

  • @thejumperkin
    @thejumperkin 6 ปีที่แล้ว +1

    love the shirt