Euler's and Fermat's last theorems, the Simpsons and CDC6600
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- Опубліковано 23 бер 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 :)
Joke Time:
Q: What do you get if you pour root beer into a square glass?
A: Beer.
Unless your root beer was imaginary, in which case any beer around you will mysteriously disappear.
how about a cube glass?
Root Beer --> Cube => Square Beer
Unfortunately, since squares are 2-dimensional, square beer would be FLAT :(
@@dlevi67 this sentence looks like something from discworld
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 :)
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!!??
to much math
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.
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!
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.
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!
That's correct. In fact, most proofs of A^4+B^4=C^4 has no solutions argue via Pythagorean triples :)
I would have expected this. It has for me the same „feeling“ as 2 p^2/q^2
By definition, they are not P Triples. Much more fun to make them from Fibonacci like series.
Jeffrey Bernath wow
Good to know. Might take note of this for an algorithm
Can some of them be squares?
Euler's Conjecture: Exists
CDC6600: I'm about to end this man's whole career
You got it wrong... Euler's Conjecture says "Doesn't exist."
M F Hasler don’t know if ur joking but nice
"Maybe not to us mere mortals, but the demigod, Euler was pretty convinced."
I disagree on demigod, he, in fact is a GOD and possibly the GOAT
@@aradhya_purohit Computer say no! 😉
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.
Exactly :)
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.
Thank you so much for your explanation!
(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.
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.
Larger than 2? I also can't find a solution for A^0+B^0=C^0.
I can but the comment section is too short to fit my proof
hahahaha :'D
it has only 1 single solution where A = B = C = 0, because
undefined + undefined = undefined :)
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.
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 :)
Mathologer and 3B1B have been coordinating their upload schedule for the past few weeks. It's great!
I am SO grateful you were able to put this video together! I absolutely adore the more complicated bits, thank you so much!
It sucks hard that Euler's conjecture turned out to be wrong. I like that conjecture.
Yes, very sad, but most of it could still be true :)
Yes indeed such a beautiful structure and a very elegant generalization. Then maths showed us the middle finger.
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) 😌.
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:) ❤️❤️
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
i have found a elegant proof for the reimann hypothesis but it is too long to put it into a youtube comment
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*
That hypothetical reimann hypothesis is probably quite uninteresting compared to the famous Riemann hypothesis...
@@pluto8404 The Read more button is fake!
@@alexandermizzi1095 Exactly
Me 2
Could you imagine if that proof was published in the Simpson's?
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 ua-cam.com/video/J65GNFfL94c/v-deo.html
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.
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.
Indeed 🎯.
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.
Haha nice one, though it should be squares ;) Funny metrics they would be using with cubes.
Some times I find the math with 3 dimensions to be laborious, I've never thinked of beings perceiving 4D! O.O
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.
Busted_Bullseye It is a new level of nightmares! XD
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...
A new Mathologer video on a Saturday is always amazing.
In my opinion this is, by far, your best video. Wow! what a ride!
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!
IndiBrony a hello to another math pony lover
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)
Love your videos Mathologer! keep 'em coming!
Thank you so much for making this proof accessible. These videos are great.
I don't have what it takes but I will keep watching after the warning anyways!
same haha
i hope one day we can have a collaboration video of Mathloger and 3Blue1Brown. That would be something really special ;)
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
Really interesting video, thank you for the content. Watching from Spain and enjoying it, congratulations!
I have a really great comment in mind about this video but its far too long to contain here.
JGLP haha. Great comment.
... the internet is to small for your comment. Hihi.
You dared me to forget... I actually forgot until i rewatched this video....
Wonderful video, as ever.
Thanks for your good work.
Thank you for such a nice and informative and elegant proof.
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.
Yep, luckily not where I actually do the proof. Actually a great one to pinpoint who is really paying close attention to detail :)
His shirt: "I took the RHOMBUS," can also be read as "I took the WRONG BUS." This is too much, man!
Congratulations you got the pun
I was waiting for this video for a long time!!
Consider the t-shirt as an explanation for why you had to wait so long :)
This is beautiful. I really enjoyed this video.
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
Unfortunately, any decent pocket calculator would display the two values as ( LHS ) 2.1367508E9 and ( RHS ) 2.1367506E9 ... so clearly they would differ. :-\
I will decompose the RSA of any complexity into multipliers. Fast and not expensive.
Good job! You make those proofs look easy.
Beautiful proof by contradiction! Followed all of it and it makes perfect sense! Thanks!
Where do you get all those amazing T-shirts? :P
Really all over the planet and quite a few I make myself :)
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)
man, your shirt choices are on point.
14:39 - I love it, lol. Direct and to the point, no need for a lengthy paper.
Now I know how to say "Euler" after years of mispronouncing it. I've thought it's "Eu" in "Euclid".
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.
Yep, that's it :)
You assume that M is odd, it seems to me. Couldn't K either be odd or even?
It doesn't matter the status of K because the power is even and 2*K is always even
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
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.
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.
Wow that was a wild ride!! Glad I had my seat belt on :)
I have a marvelous proof that information is infinitly compressible, but it is too big to fit in this comments section...
What a gem :)
😂😂. Got me....
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.
I was bored today. Now I'm thrilled again.
Awesome video! Thank you.
9:07 9 Digits, not 8. You missed the 5.
I believe he was talking about the limitations of older calculators, namely 8 digit display width.
But those digits being nine seems like something special..
@@rucker69 My 1970s-vintage TI-58C could show 12 digits, and actually calculated internally to a bit more than that.
@@rucker69 My 1970s-vintage TI-58C could show 12 digits, and actually calculated internally to a bit more than that.
Oh hi Shalltear
23:53 - Can someone explain why the 4th power that's multiplied by 2 must be odd? He never explained this.
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)
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 :)
Maybe do a livestream of 7-8 hours to do the whole proof. It would be glorious! Regards from Greece. Love your videos
And I'd be dead :)
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.
16:36 - need to add 1 BROWN!
Azazeo Ainamart
Too bad there's no brown heart emoji. 💙💙💙
I don't get what you mean...
steamroller82
As in 3Blue1Brown?
🤎
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.
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.
But he was doing a greater disservice to himself by constraining his thoughts to desire integral or rational solutions - let alone transcendental.
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!
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!
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.
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 :)
Luckily one of those self-correcting mistakes :)
Really great video
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.
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.
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 :)
Yo my dude, I have that book too! What is it you're trying to figure out? How Fermat did what for n=2?
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.
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?
*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.
Sees "1 or 0", starts thinking of applications in cryptography
Mod 9 is my go-to for checking arithmetic.
9:12
Hey, I also found the '5' after that orange-marked place are very similar
9:10 Looks like it's the first 9 digits, actually...
Look again
He squared it wrong
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?
Oh wait nvm I'm stupid
yes, you can
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 :)
but x and y should be integers here. your analogy would mean that you need a pythagorean triplet where two members are perfect squares
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.
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.
I really like proving this by infinite descent. It just feels really cool.
"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)
9:05, the first *nine* digits, yes?
Well, you know what they say: mathematicians cannot count :)
You probably put that in there just to see who is paying attention. ;-)
yes but 8 is a significant number in digital calculating
first 10 if you round
witch you should not do at that point because the numbers are displayd for more digids
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.
I don't begin to have what it takes, but I love this guy's personality and voice. Maybe something will click...
One day someone will read this comment.
That comment won't stop me because I can't read!
@@endermage77 ........................
Sorry, was just skipping through the comments, did you write something...
5:24 Mathologer proves there is no God.
:)
Take that, theists!
Thank y0u for unveiling difficult mathematics.
Finally I understand how computer proofs work. Thanks!
(0^n)+(0^n) = (0^n)
Solved XD
:)
but if we allow 0 then we can't say for sure that 0^0 + 0^0= 0^0 :(
Brian Cotuinho alright (0^n + 0^n = 0^n : nEZ, n>0)
Except that it isn't. 0^n = 1 for all n≠0, so you're saying that 1+1=1
Oh lol, I was reading n^0 rather than 0^n :X
Forst. This means I'm early
Well yours and three blue one brown are certainly two most amazing UA-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.
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.
the studio light on the right side (your left) is way too bright this time.
but i love your math explanations and simpsons references.
i had no time to watch the video, so i just liked for the t-shirt
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.
Can u please make a video on justification of a line of 0 thickness/width or a mathematical line segment like in the kakeya needle problem!!!!
This video was also awesome as always!!!!!
It seems that De Guas theorem from your Pythogorean video should provide counter examples since areas squared should be intergers to the 4th power.
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.
Yaar isne toh UA-cam me aag laga Di
🔥🔥🔥🔥
Odd to see a Hindi speaker
Where can I get that t-shirt!!!!..it's so cool ! ..
Great video by the way, kudos from all your fans here in India
3BLUE1BROWN + MATHOLOGER?!
YAAS
Really good!
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
In the Simpons example the 9th digit is also the same. Not just the first 8.
I have found a truly marvelous margin of this, which this proof is too contain to narrow.
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.
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.
@Mathologer 16:15 You got up to 2 decimal places. Pretty nice. 👍🏻
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.
About the proof Fermat was talking about in the margin, I was wondering if it could be some kind of recursive algorithm (by composing functions on themselves).
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
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.