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

[@Mathologer]

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]

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...]

to much math

[@kennethgee2004]

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]

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]

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.

[@edensaquaponics1941]

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]

Exactly :)

[@edensaquaponics1941]

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]

Thank you so much for your explanation!

[@txikitofandango]

(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]

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.

[@jeffreybernath6627]

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]

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

[@robertgumpi7235]

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

[@christopherellis2663]

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

[@sldecka]

Jeffrey Bernath wow

[@ThePharphis]

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

[@johnchessant3012]

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

[@SuperDreamliner787]

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]

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]

Indeed 🎯.

[@trained-wreckscience-strug8932]

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

[@thecwd8919]

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

[@aradhya_purohit]

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

[@RobBCactive]

@@aradhya_purohit Computer say no! 😉

[@gabest4]

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

[@darealpoopster]

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

[@ghrissiabdeltif5397]

hahahaha :'D

[@NizarHaddad]

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

[@lovaaaa2451]

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]

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 :)

[@krillbilly1435]

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

[@M-F-H]

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

[@skylermagnificent5422]

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

[@IndiBrony]

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!

[@lucifersdevilishdetails.]

IndiBrony a hello to another math pony lover

[@Craznar]

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

[@dlevi67]

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

[@kmlo3784]

how about a cube glass?

[@AgentFriday]

Root Beer --> Cube => Square Beer

[@AgentFriday]

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

[@f.jideament]

@@dlevi67 this sentence looks like something from discworld

[@alejandrapaz2864]

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

[@pluto8404]

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]

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

[@alexandermizzi1095]

@@pluto8404 The Read more button is fake!

[@TonyStark-kh4eb]

@@alexandermizzi1095 Exactly

[@idon.t2156]

Me 2

[@B3Band]

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]

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

[@FernandoBiaziNascimento]

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

[@islandfireballkill]

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]

Busted_Bullseye It is a new level of nightmares! XD

[@Arikayx13]

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...

[@anselmschueler]

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

[@Mathologer]

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 kzbin.info/www/bejne/gGeYeIF8m7FsasU

[@bailey125]

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.

[@Tehom1]

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]

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

[@MrRyanroberson1]

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)

[@kirstenwilliams6056]

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

[@johnsnow5305]

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

[@jeromelee428]

same haha

[@twistedgwazi5727]

A new Mathologer video on a Saturday is always amazing.

[@lawrencedoliveiro9104]

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]

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.

[@pronoy91]

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

[@adamsvoboda7717]

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]

Yep, that's it :)

[@anselmschueler]

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

[@davidrheault7896]

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

[@bluestrawberry679]

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]

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.

[@jack002tuber]

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.

[@masked_mizuki]

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

[@robertgumpi7235]

JGLP haha. Great comment.

[@robertgumpi7235]

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

[@morphx666]

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

[@TrimutiusToo]

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

[@dontask23]

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

[@Mathologer]

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

[@hernanipereira]

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

[@podemosurss8316]

19:14 Very easy: If we take X^(4n) + Y^(4n) = Z^(4n) we can rearrange this to be (X^n)^4 + (Y^n)^4 = (Z^n)^4, for which no solutions exist.

[@LivingVacuum]

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

[@rohitkumar-rq6qh]

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

[@GumbyTheGreen1]

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

[@harold3802]

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)

[@logicalfundy]

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

[@n-wordjim1724]

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

[@rucker69]

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

[@sillybears4673]

But those digits being nine seems like something special..

[@lawrencedoliveiro9104]

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

[@lawrencedoliveiro9104]

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

[@justarandomjojofan6674]

Oh hi Shalltear

[@Ricocossa1]

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

[@TheToric]

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

[@artofgameplaying]

What a gem :)

[@cinvest2411]

😂😂. Got me....

[@elevated__arts]

Love your videos Mathologer! keep 'em coming!

[@aakash_kul]

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

[@temra7063]

Congratulations you got the pun

[@antonionanni6683]

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!

[@eliascaeiro5439]

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]

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

[@andyoncam1]

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.

[@KateTheSleepyTeacher]

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 :)

[@brocpage4204]

man, your shirt choices are on point.

[@geertcremers1588]

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.

[@DitDede]

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.

[@thanosAIAS]

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

[@Mathologer]

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

[@hansalexander905]

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

[@BruceLCM]

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

[@uchihamadara6024]

Proof of exercise at 11:20: Given that n is an odd number, n may be expressed as 2k+1 (k is an integer). and 2k+1 to an even power (let's choose 2 as our base case) is equal to: (2k+1)^2 = 4k^2 + 4k + 1, which leaves a remainder of 1 when divided by 4. Note that an odd number to any power is also an odd number. This can be proven by induction but I think it's obvious enough. So therefore any even power (2k+1)^2m can be expressed as: ((2k+1)^m)^2 where (2k+1)^m is odd, we may write (2k + 1)^m = 2j + 1 where j is another integer. And we've already shown that an odd number of this form to an even power has remainder 1. QED

[@AzazeoAinamart]

16:36 - need to add 1 BROWN!

[@denelson83]

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

[@chessandmathguy]

I don't get what you mean...

[@denelson83]

steamroller82 As in 3Blue1Brown?

[@shinymoonlightteaches7179]

🤎

[@shitzoalc6v]

About the proof using divisibility by 3 at 12:10. In essence: Divisibility by any number and non-divisibility by a prime number will be inherited through exponentiation. And a little more detailed: If a number n is divisible by m any power of n will be divisible by m as well. This can be seen when you write n=m*x with some integer x, so n^a=(m*x)^a=m^a*x^a which is divisible by m again (using any integer exponent a). And also if n is not divisible by a prime m no power of n will be divisible by m. This can be seen when we take the prime factorisation of n, which doesn't contain m and take it to any power a, the prime factorisation will repeat itself a times but will still not contain m obviously. (It doesn't work for non prime m btw, for example 6 isn't divisible by 4 but 36=6^2 is.) Both the bases on the left hand side (3987 and 4365) are divisible by 3 so the entire left hand side of the equation is divisible by 3. The base on the right hand side (4472) is not divisible by 3 so the entire right hand side isn't divisible by three, so the equation can't be correct.

[@S.R.400]

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]

Luckily one of those self-correcting mistakes :)

[@TheYourbox]

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

[@Luigicat11]

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

[@colinwang3774]

Look again

[@colinwang3774]

He squared it wrong

[@MrRyanroberson1]

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

[@VerSalieri]

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.

[@randomguy8461]

11:05 Let's choose to write the odd integer n as either (x + 1) or (x + 3) where x is the largest possible integer multiple of 4, less than n. If this is the case, n^2 can be expressed as either (x + 1)^2 or (x + 3)^2, or (x^2 + 2x + 1) or (x^2 + 6x + 9). Because x is defined to be a multiple of 4, all monomials mentioned in the line above must also be a multiple of 4, leaving a remainder of 1 in the first case, and a remainder of 9 - 2(4) = also 1 in the second case, meaning n^2 must have a remainder of 1 when divided by 4. This means n^2 can be written as (z +1), where z is some multiple of 4. When n^2 is raised to any power, it will result in n^(some even number). When (z+1) is raised to any power, the only term in the resulting polynomial that doesn't contain a z will be the last one, which will always be 1. Therefore, n^(an even number) when divided by 4 will always have a remainder of 1, as long as n is an odd integer.

[@MrConverse]

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

[@Mathologer]

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

[@MrConverse]

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

[@rubenjanssen1672]

yes but 8 is a significant number in digital calculating

[@earthbjornnahkaimurrao9542]

first 10 if you round

[@rubenjanssen1672]

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

[@chessandmathguy]

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

[@Suimobile]

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]

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]

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]

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]

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]

*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.

[@robertgumpi7235]

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

[@mk-allard3788]

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]

Oh wait nvm I'm stupid

[@SilisAlin]

yes, you can

[@Mathologer]

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]

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]

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.

[@TheLuckySpades]

This is beautiful. I really enjoyed this video.

[@gregbernstein7524]

5:24 Mathologer proves there is no God.

[@Mathologer]

:)

[@robo3007]

Take that, theists!

[@earthbind83]

Good job! You make those proofs look easy.

[@erikperik1671]

"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)

[@overlordprincekhan]

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

[@PinochleIsALie]

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

[@pianoingels7128]

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

[@ichthysking863]

Forst. This means I'm early

[@MrRyanroberson1]

19:16 it can be shown for all x^(4a)+y^(4b)=z^(2c), emphasis on the 2. remember that x4+y4=/=z2, so for all x,y,z are themselves nth powers, not even necessarily respectively, there are no solutions.

[@numero7mojeangering]

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

[@Mathologer]

:)

[@brcoutme]

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

[@NgenDoesGaming]

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

[@oisyn]

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

[@oisyn]

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

[@TheOnlyRizzy]

11:28: odd^even will always have a remainder of 1. Why? Let's represent odd as even+1. So we have (even+1)^even. All even numbers are divisible by 2, so let's factor 2 out of both even numbers to get: (2n+1)^2m (n and m can be any integers ) Using exponent rules we can rewrite this as: [(2n+1)²]^m Now if we expand the part inside the square brackets we get: [4n² + 4n +1]^m No matter what integers we assign to n and m, 4n² and 4m will always be evenly divisible by 4, so all we're left with is 1, which will be the remainder.

[@TheOnlyRizzy]

I guess what this does is prove that odd^even is equivalent to saying n must be evenly divisible by 4 in (n+1)^m.

[@muditgupta628]

One day someone will read this comment.

[@endermage77]

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

[@arnavanand8037]

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

[@alanhilder1883]

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

[@PC_Simo]

I already read it.

[@tensevo]

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.

[@moskthinks9801]

12:06 We can prove the second equation wrong also by divisibility by 3. Suppose that every number (even or odd, doesn't matter) has a remainder of 0, 1, or 2 upon dividing by 3. We can use some basic modulo arithmetic to not care about (3n+remainder) values but just use the remainders directly. 0^(even power) is just 0. ((3n)^(2k) = (3^2k)(n^2k)=3(stuff)) 1^(even power) is just 1. 2^(even power) = 4^(power) which is the same as 1^(power) which is just 1. We can conclude that if a number is divisible by 3, then when raised to an even power is 0. We can conclude that if a number is not divisible by 3, then when raised to an even power is 1. Back to the Simpson's second equation: 3987^12+4365^12=4472^12 Use the divisibility rule for 3 on each number: 3987: 3+9+8+7=27=9x3. Divisible By 3. 4365: 4+3+6+5=18=6x3. Divisible By 3. 4472: 4+4+7+2=17. Not divisible by 3. So we have on the left side: 0+0=0. We have on the right side: 1. If the equation were true, then 0=1, but it doesn't equal. Thus, the equation is not true. Q.E.D.

[@moskthinks9801]

To justify what I did even more, see this. The case for when the remainder is 0 is easy to explain. (3n)^(2k) = (3^2k)(n^2k)=3(stuff) That has also a remainder of 0. For the case for when the remainder is either 1 or 2, binomial theorem might help us. Let's raise 3n+1 or 3n+2 to the power of 2k. Case 1: (3n+1)^2k=((3n+1)^2)^k=(9n^2+6n+1)^k=(3(3n^2+2n)+1)^k=(3m+1)^k Case 2: (3n+2)^2k=(9n^2+12n+3+1)^k=(3(3n^2+4n+1)+1)^k=(3m+1)^k (Note the m in case 1 is not the same as in case 2, but used to show the connection that leads to the same remainder) Case 1 & 2 (Continued): By the use of Binomial Theorem, we can expand the power as (3m+1)^k=(3m)^k+k*(3m)^(k-1)+...+(k!/(n!(k-n)!))(3m)^(k-n)+...+3km+1=3(junk)+1 Which is clearly the same as 1 mod 3. Q.E.D.

[@mohammedhubail1607]

9:50 If u checks divisiblity by 3 you 'll notice that both squares are divisible by 3 and the l.h.s. isn't

[@TheJohnblyth]

Finally I understand how computer proofs work. Thanks!

[@megalul4141]

11:27 proof: 3 is congruent to (-1)[mod4] We suppose that n is a even integer Therefore n= 2k where k is an integer 3^n= 3^(2k) Therefore 3^n is congruent to (-1)^n[mod4] which is congruent to (-1)^(2k)[mod4] congruent to ((-1)^2)^k[mod4] congruent to 1^k[mod4] Congruent to 1[mod4]

[@unoriginalusernameno999]

Is this how they check? List numbers from 1 to some huge positive integer "n" Then declare 3 variable - x,y,z. List all the permutations of x,y,z out of those "n" integers. Substitute each into the equation x^k + y^k = z^k Then we do this... If z^k/(x^k+y^k)=1 and (z^k)%(x^k+y^k)=0 then we find an example for k=!2 disproving Fermats last theorem...You need supercomputers for that...Hasn't anyone assigned this task to a supercomputer yet?

[@dimosthenisvallis3555]

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

[@Mathologer]

And I'd be dead :)

[@xenxander]

9:10 "the only coincide with the first eight digits..' pause, counts.. then sees a '5' next to them... seems they coincide the first nine digits even if you round.

[@michaelbyrd1674]

the 6th equation at 20:23 reads (u^2-y)(u^2-Y)=4v^4 but should read (u^2-y)(u^2+y)=4v^4. I am still trying to figure out the res.

[@michaelbyrd1674]

I don't understand how (2r^4)(2s^4)=4v^4 leads to r^4 +s^4=u^2

[@MeisterTentaLP]

If you have x^n while x is odd and n is even, you can write n as 2m. Thus, x^n = (x^m)^2. If you divide this by 4, you get (x^m)^2 / 2^2 = ((x^m)/2)^2. Since x is odd, x^m is odd and thus (x^m)/2 = o + 0.5 (o is an integer). (o+0.5)^2 = o^2 + o + 0.25, so the result has a part after the comma equal to 0.25 which corresponds to a remainder of 1.

[@jaideepmishra6961]

Yaar isne toh KZbin me aag laga Di 🔥🔥🔥🔥

[@shambosaha9727]

Odd to see a Hindi speaker

[@dominiorrr6510]

9:05 They actually coincide in the first 9 digits. 5 outside the orange rectangle is uncounted for some reason in this video (mistake?)

[@MasDingos]

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

[@itays7774]

12:03 (I'm a bit late for that, but I'll do it anyway) There is a property of numbers, of the sum of all the digits of a number is divisible by 3 iff the number itself is divisible by 3 ( I don't know how to prove it formally, so take my word for it or check it out for yourself). 3+9+8+7=27 which is divisible by 3, and so is 4+3+5+6=18, this divisibility fact does not change for multiplying the number of applying powers to it. Therefore, the number on the left, 4472, should also be divisible by 3, but summing 7+4+4+2 gives 17, which is clearly not divisible by 3, making 4472 not divisible by 3, and so we reached a contradiction. Any questions?

[@traewatkins931]

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]

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.

[@PC_Simo]

Well, any multiple-of-4-th powers are, themselves, 4th powers, expressible like this: X^(4n) = (X^n)^4. Therefore, it follows that: A^4 + B^4 ≠ C^4 implies that: (A^n)^4 + (B^n)^4 ≠ (C^n)^4 A^(4n) + B^(4n) ≠ C^(4n). *Q.E.D. 𑀩*

[@YoniMek]

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.

[@tobiasactually]

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

[@Mathologer]

:)

[@dozenazer1811]

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.

[@JohnDoe-jy7sv]

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

[@kayak8700]

6:50, for n=2 this can be done in with one variable, n^2=n^2, just disproved the euler conjecture.

[@DrZip]

@24:18 No, it doesn't necessarily mean it's 8 times an odd power, it could be 2^(3+4n), times an odd power though, right? Regardless, the proof still works though - mostly? The contradiction still works @29:22. But the equation @30:10 would be r^4 + s^4 = 2^(4n) * u^2, which for n >= 1, would just mean the new solution set gets smaller faster and thus should be found even sooner... But, it's asserted and not proven here; the smallest non-trivial solution could still be some crazily big ugly natural numbers. Haven't looked deeper into other proofs, presumably they're more air-tight, but that's entirely possible in this situation (like what the CDC6600 found @14:38).

[@elraviv]

16:11 I only looked at the last digit 7^4 is 1 plus 2^4 which is 6 give us *7* , while 5^x is *5*

[@elamvaluthis7268]

Thank y0u for unveiling difficult mathematics.

[@PC_Simo]

9:11 They actually coincide, in the 1st *_9_* digits; the 9th digit being 5, in both numbers. 🙂