Homer Simpson vs Pierre de Fermat - Numberphile
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- Опубліковано 8 лют 2025
- Simpsons Book: amzn.to/1fKe4Yo -- Main Fermat Theorem video: • Fermat's Last Theorem ... -- Water Ballons: • Science of Water Ballo...
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Author Simon Singh on Fermat's Last Theorem in popular culture, especially The Simpsons.
Also mentioning Al Jean and David X Cohen.
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Simon might just have the most beautiful handwriting of all the Numberphile guests
egalomon Hes a writer. It makes sense.
"Doing a troll"
I love you Brady.
+Badatstuff I thought he said "being" a troll, haha
The Drüid lol
He definitely said "Being a troll."
1:23 there is also an attempt at breaking topology at the bottom
Sachin Sahay Nice catch!
Sachin Sahay too bad it wasn’t REAL.
thats Poincaré's conjecture, isn't it?
Homerlogy.
it looks like a donut with bites taken out of it
I had my own near miss solution with a^3 + b^3 = c^3 where a=242, b=720, and c=729. So close, yet so far away
starreactor Again, only because of the odd-even properties ;)
Haha you were 1 away from being right with that near miss solution (729^3 - (242^3 + 720^3) = 1)
+Bengt Bengt 1^3 + 1^3 = 1^3 also near-miss tho XD
Pius Pambudi actually not because x, y and z must be different numbers
I've never seen the stipulation that they need to be. It is easily provable that they cannot be - in a^n+b^n=c^n (a, b, c, n all integer; a, b, c >=1; n>2) c obviously needs to be larger than both a and b, since otherwise you would add something positive to a number and it would stay the same. But also a and b have to be different, because if a=b, then a^n+b^n=a^n+a^n=2*a^n=c^n. Taking the n-th root, we's get (n-th root of 2)*a=c, but since both a and c must be integers but the n-th root of 2 isn't even rational, this simply is not possible.
When will we see a video on pancake flipping?
^
3master o
WarpRulez YOU SHOULD KNOW THE ANSWER TO THAT BEING FROM THE FUTURE
I wish that mathematics wasn't taught so dully and uninterestingly in our culture. math is an art and we should be emphasizing the importance of math more than anything
Math isn't an art. That's what makes it so great.
Well, the best way to do it is to do away with the more utilitarian mindset behind education, after all technically most stuff we get taught can be considered to be "useless" and it's not just us liberal arts majors who can be looked down upon as knowing "useless" stuff.
What is useless info anyways? I think how you answer the question can tell about your mentality behind this question, if your meaning of life is making money then only some business majors and most engineering careers are useful studying.
Perhaps because of these two answers?
The one side is very strict and says: "shouldn't treat it as an art at all", lets keep the classes dull.
and the other goes of on an extreme tangent, preferring to attempt to answer broad/philosophical questions, rather than tackle it practically.
I'm not sure it IS about the "importance", but simply, appealing to students. Everything CAN be important, but if it isn't being taught in a way that people will (want to) apply it, it's still not useful.
Personally from highschool, I remember endless equations of things like: solve (x+4)(x-12)=0
While, now, I realize I'd love to solve stuff like that, back then I just couldnt get the hang of it, by ONLY having the dry, stuffy, equation.
What's the purpose? What is this solving, in real life? Can I get a handle on it that takes it OUT of the theoretical and gives me some insight, rather than just having to crunch numbers again and again.
Perhaps ironically, when I see people like Matt Parker, I get excited about it, and want to learn. All it takes, for me, is to have it presented in a fun way that makes me want to come back for more.
Maybe it's the teachers? Most I've had didn't make the lessons very fun or engaging. Some did. The books didn't really help, they had SOME humor and real-life experiences in it, but they were the lame kind. "Alice and Bernard walked up to their friend Charles". Clearly just to describe points A,B,C. More pedantic than appealing.
how is math an art
Says you
Just want to say how thankful I am that he wrote that book. It got me back my long lost and burning passion for maths, and lifted me out of a long and horrible depression. It reminded me that life can be exciting, and actually worth living.
Yeah that's the spirit. Dont let others control your life. You know I had gone through the same circumstance.
Another example would be Rowan Atkinson who studied engineering at university.
+Slim Charles Indeed, and his long-time side-kick Ben Miller - who studied physics.
Is that how he put on shorts without taking off is trousers?
Yes. It's a topological problem.
did anyone notice that the first equation that Homer writes is also a very close approximation of mass of Higgs Boson ? and that episode aired way before we proved the existence of higgs boson
@Xenon Creed it was predicted by Higgs 🤦♂️ the guy the particle is named after.
@@woobilicious. 🤣
you think he wrote a book about the Simpsons, met the writers (of both shows) and did nothing about Futurama!
Stay tuned for more (or buy his book!)
This whole phenomenon of serious mathematics being referenced in light entertainment reminds me of Charles Dodgson's mathematical references in "Alice in Wonderland", although mathematics plays a much more central role in his book than it does in the Simpsons.
Actually, here's a good idea for Numberphile, assuming they haven't already done it: Do an episode on maths in "Alice in Wonderland" (and "Through the Looking Glass")!
Just came back here to visit another Parker Square solution to Fermat's last theorem.
I literally only read the comments to see if someone referenced Parker for that almost-solution for Fermat's theorem.
1782^12+1841^12=1922^12
It actually equals to 1921,999999955...^12 ._.
VEEEEERY close
Yup, not bad.
8730^12 + 7974^12 ≈ 8944^12 is closer, though.
(8944.00000001412-8944)/8944 ≈ 1.58e-12
(1921.999999955-1922)/1922 ≈ -2.34e-11
or better: 48767^4 + 24576^4 ≈ 49535^4 or e.g. 89,970^3+8,999^3 ≈ 90,000^3 and 90,000^3+9001^3 ≈ 90,030^3
My attempt 0000^12+0000^12 =0000^12.... Fricken math.. no idea what I'm doing but that equation checks out as far as I'm concerned.
@Sandra Braithwaite Sure. That was for fun.
The reason these are being dropped into the episodes is not because mathematicians-turned-into-comedy-witers want to make a nod at their alternate occupation. It is because they are very effective comedy writers. They write comedy for ALL intellectual levels.
That's the beauty of the series and the probable reason it has survived as long as it has. Trailer Park Joe laughs at the bodily function jokes, Office Jack reads the humor in popular culture spoofs and Science Peter gets a kick out of why would a 10 year-old have a Kasparov lunch box.
Case in point: The "faux solution" to Fermat's Last Theorem is a very effective bit of comedy (...even if buried almost as an Easter-egg...) because, under the show's logic and being Homer who he "is", had he - by some divine intervention - really turned into a researcher, Dr. Simpson being convinced he reached the actual theorem solution via a false proposition would be the kind of elevated blunder, leading to all sort of problems, he most certainly would be involved in.
To be fair, you have to have a very high IQ to understand the Simpsons. The humour is extremely subtle, and without a solid grasp of mathematics most of the jokes will go over a typical viewer’s head. There’s also Bart’s nihilistic outlook, which is deftly woven into his characterisation- his personal philosophy draws heavily from Narodnaya Volya literature, for instance. The fans understand this stuff; they have the intellectual capacity to truly appreciate the depths of these jokes, to realise that they’re not just funny- they say something deep about LIFE. As a consequence people who dislike the Simpsons truly ARE idiots- of course they wouldn’t appreciate, for instance, the humour in Homer’s existential catchphrase “D’oh,” which itself is a cryptic reference to Turgenev’s Russian epic Fathers and Sons. I’m smirking right now just imagining one of those addlepated simpletons scratching their heads in confusion as Al Jean’s genius wit unfolds itself on their television screens. What fools.. how I pity them. 😂
And yes, by the way, i DO have a Simpsons tattoo. And no, you cannot see it. It’s for the ladies’ eyes only- and even then they have to demonstrate that they’re within 5 IQ points of my own (preferably lower) beforehand. Nothin personnel kid 😎
There are two kinds of people.
Anybody care what this guy thinks?
NO!
To be fair, it could be for both of those reasons, as they are not mutually exclusive.
It's Anatoly Karpov on the lunchbox, not Garry Kasparov.
Amazing video, I love this Simpsons, Math, and Science.
Great video. Thanks for sharing that. I have just bought Simon Singh's book and cannot wait to read it!
If you want the brown paper from this video there is a link in the vid description
(it is just another way to help support our videos - but the best thing you can do is just watch us and share with friends!!!)
My new favourite Numberphile video! Thanks guys.
Man that is a really clever reference. Props to the Simpsons writers.
I'll defiantly be buying this book, sounds like very interesting stuff. Thanks for sharing this Brady.
The real solution is 3987^12 + 4365^12 = 4472^11 X 4472.00000008
No, not quite, it has to be to the 12th each, what you’ve done is reduce it and then add a decimal.
It doesn't really fit the problem.
@@acewmd.Well what he is saying is that if he did put it in the form of x¹², then the decimal part would be smaller and more precise, so he's just compressing it.
Really love these Simon videos. He should come back and do more!
a near miss of the order of 10^30.
i wish i could do that in my physics exams!
great job brady, thanks to you and everyone providing great content on these channels. love your work, keep it up. i feel that you and the people who follow you are the next great generation of educators. far, far, more engaging than the average university professor, you bring us the great communicators of knowledge.
One of the reasons you see stuff like the 'Karpov' joke in The Simpsons is that they are willing let a joke be missed by most viewers. Most TV shows are controlled to the tiniest detail and a reference to something so arcane would be seen as a weakness that needs to be sent around the writers' room again.
The Simpsons *used* to be novel in that they would put tons of mini-gags into the show that wouldn't be noticed at all by 99% of those watching. These days everyone does it of course.
Ah thanks! Feels satisfying finally knowing how it's called. Being a mystery is one, but not knowing what that mystery is called makes it extra difficult.
Fermat's Last Theorem applies for n = 0 (trivially) and n < -2 as well. You can find general solutions for n = -1 and n = -2 near the bottom of Wikipedia's article for Fermat's Last Theorem.
Man I love Numberphile!! You should upload way more!
I have another problem with no solution; how do you get a USB to fit properly on the first try?
+Arun Singhal you can, but It's hard to explain on text xD
+Arun Singhal
the answer lies in elliptic curves
+Arun Singhal I have the solution...but the whole data that can fit on the Internet is too little to fit it inside
the side with the USB symbol will always be the top side.
also i know this is late but it is worth mentioning
[GD] Unearth not if it's sideways
I am currently reading his book on Fermat last theorem that coincidently I both a day before you uploaded the video. Love the channel Brady, thank you! Loving the book also! Highly recommended
I just remember "In This House We Obey The Laws of Thermodynamics" hahaha funny
thanks!
"This is a solution."
"A solution? That's problematic!"
I love mathematicians :')
Absolutely brilliant! I love this channel!
"When you can't get a job in mathematics, you find a job in arts"
Phew.
I like Simon Singh. He's well spoken and has a very cool presentation style. More please!
futurama is an even better example
Loving the new logo Brady! Congrats on another great video, too.
He underestimates modern phone calculators.
True. Most even those simple ones (including mine) use 52bits of precision. 204290^3 + 146996^3 = 227033^3 does the trick, though.
David is a wonderful writer for what was Futurama.
Great show
It's off by about 10^33 :-)
exscape less than one part per billion then
You guys should definitely do a video on the futurama theorem
It's impossible you say? We'll see about that
David X Cohen is the cocreator of Futurama along with Matt Groening, and that series is full with numbers and science references in each episode, you should do a numberphile episode about that, I´m sure most of the people you interview are big fans of it, and they can speak lots of those references that we dont get.
futurama has many more math jokes in it and David X Cohen comments on every single episode on the DVDs. RIP Futurama :(
I'm surprised you haven't had a numberphile about e yet. It's a fascinating number that I'm sure you could delve quite deeply into. Euler's Identity would be another excellent (and related) topic.
I only came because UA-cam recommended it and because I read Homer Simpson....
I loves the article you wrote in the newspaper about this :D it was great
I'm wondering : am I the only one who's bad at math but enjoys numberphile ?
Because of the narrower focus of the show, there are more opportunities for the creative team to play on Futurama, and you see more math references there. The most notable is probably the novel and rigorous proof about the mathematics of body swapping in "The Prisoner of Benda".
Is this the episode where Homer makes the Lazy Man Reclining Toilet Chair?
yep, then accidentally leaves it in the Thomas Edison museum, and Edison gets credited for it.
Edison invented it, Homer just stole the idea
Love this channel.
Saw this joke on the Simpsons yesterday :
√-1 2^3 sum of (Sigma) Pi
And it was delicious
What does it mean ??
D Gurung i ate some pie...and it was delicious
Haha! Thanks
+D Gurung √-1 = imaginary number, denoted by letter 'i' = I
2^3 = 8 = eight = ate
sum = some
of = of
Pi = pie
...
And it was delicious
I assume the words "sum" and "of" are not supposed to be there because that's what the sigma is for. :)
I ate some pie
D Gurung i 8 some pi aka I ate some Pie
Yep. Died this year (for the second time). RIP best show on television.
the Pythagorean theorem, anyone?
***** Men it says for n>2, and is allways proven when x+y>z.
There are lots of (infinitely many) solutions for n = 2. 5^2+12^2=13^2 works. Fermat's Last Theorem states that there are no solutions for n > 2. At 0:55 the n > 2 restriction is listed. I recommend watching the main Fermat video. Link in the description.
ua-cam.com/video/ReOQ300AcSU/v-deo.html
+Zero Ryoko It's a^n+b^n=c^n for positive a, b, and c. 0 doesn't count because it isn't positive (it's neither positive nor negative).
j
we've done a video on that!
rd r r
hardy harr harr, get it?
This guy is doing a talk at my school! Can't wait!
The theorem just skips the easy one. 3^2+4^2= 5^2 => 9 + 16 = 25 => 25=25
The theorem has to has n>2 just to skip this one.
Also the other easy one x^0 + y^0 = Z^0
Sam Crosswell, Nope, 1+1 is not equal to 1. You're probably thinking of 0^n+0^n=0^n.
I also like the transition of the torus into a sphere on the bottom of the blackboard
Hey Bert!, where's Ernie?
I put forward another recommendation to Simon's book, I've read his stuff before and he is an excellent popular maths writer.
If you were really smart, you wouldn't need to be a mathematician to know that Simpsons isn't providing a "solution" to Fermat's Last Theorem, rather he was trying to DISPROVE Fermat's Last Theorem. It seems NPR and Singh have trouble with basic semantic logic.
At 2:25 you can tell that the equation is wrong right away by comparing the sums of the numbers' digits. The numbers on the left side of the equation are such that the sums of their digits are multiples of 3 (3+9+8+7=27 and 4+3+6+5=18), meaning that they are also multiples of 3, and hence so are their 12th powers, but the result is NOT a multiple of three.
this stuff is amazing
So if I'm looking at this correctly Fermat's last theorem could be stated not only with integers but also with rational numbers. if A, B, and C are all rational numbers then there exists an integer by which each could be multiplied to produce an integer, and the LCM of each of these integers (call it "K") could be multiplied by each rational number term in A^n+B^n=C^n and if the original equation works then the final one will because if A^n+B^n=C^n then (KA)^n+(KB)^n=(K^n)*(A^n+B^n)=(K^n)*C^n
The required followup video to this one would probably be too long to post, but I'm pretty sure it would be epic: a similar examination of the Simpson's sister show, Futurama. which had many of the same people on staff, and more nerdcrumbs than I can count.
Saw your article in the Independent the other day, great read!
Yes
thanks
I really enjoy this guy, please more!
In fact, you can use a compressor on your master channel with a high ratio which will keep your levels under the threshold level you set (also on the compressor), and you can crank up the gain (also on the compressor) so that the levels are as loud as desired. Nice and simple auto leveler.
the problem is that your calculator might not have enough precision to calculate it, as it can't be done integerwise. (as x is quite a bit larger than 2^64).
floating point wise, it can be done, but there is not room for a lot of precision. As soon that x^n is calculated, the error caused by loss of information due floating point, might be large enough to annihilate the difference. Thus your equation might in some cases still give 0 as a result.
Brady you really really need to make a video (series) on the math and science of futurama :) dave and the others really packed em in there:)
Note to creators of Numberphile: I think you make some good quality episodes that I enjoy a lot, but watch your audio levels! I noticed it especially on this video jumping all over the place. Whatever your using to edit, just watch the dB meter and try to keep it constant.
these 3 from this guy have been very good!
Try x^24 + y^24, x^48 + y^48, and x^96 + y^96 for even closer near misses to Femat’s last theorem. You have to have a calculator that shows up to 100 digits after the decimal point.
Love The Simpsons math references. Can you do session on the mathematics that show up in Futurama or for other shows where math references are made?
no problem, they did an episode on it a few weeks ago if youd like to know more actually.
The Amazon link is for the UK site... the US Amazon has it slightly cheaper (according to my currency conversion software), but it doesn't release until Oct 29th!
Oh, I remember reading this article in the newspaper!
hey brady, I like the explanation of the other vids at the end. good thought.
It'd be great if you looked at the maths from Futurama, too. Most of that was David X. Cohen and Ken Keeler, the latter creating what is now known as the 'Futurama theorem'.
Because pythagorean theorem only works for the power of 2, fermat was trying to find out why anything beyond 2 wouldnt work or if there were any powers that would work and then he died before it was revealed, thus fermats last theorem is: X^n + Y^n = Z^n with n>2
2 in base 3 is 2 in base ten. Two is an integer in general. Changing bases does not change whether a number is an integer. I don't see how you could think it does.
Well, I am a bit confused then. My interpretation of your original post is "I wonder if there are any solutions to x^n + y^n = z^n, n is an integer > 2, and x, y, and z are positive integers, when written in base twelve". Obviously if it can't be done in base ten, it also can't be done in any other base.
In the first example I used a DM32 calculator and found a difference of 1.21189*10E33 between the numbers on both sides of the equation. The sum on the left side of the equation is 6.39766564969861261…*10E43 while 6.3976653848672580686235…*E43 is the number on the right side of the equation. When you subtract the sum on the left side of the equation from the number on the right side of the equation with a high accuracy calculator you get the difference mentioned above. This is a near miss which is further proof of Fermat’s equation.
Great stuff!
Yes, I was referring to x, y, and z. I didn't know that they couldn't be equal to each other. Thanks for clarifying.
1:18
The reason why the equation doesn't work:
3987 is a multiple of 3, so 3987^12 is a multiple of 3
Also, 4365 is a multiple of 3, so 4365^12 is a multiple of 3
Therefore: 3987^12+4365^12 is a multiple of 3
However, 4472 has a remainder of 2 when divided by 3 . . . so 4472^12 has a remainder of 2^12=4096 (or remainder 1) when divided by 3.
Since a number clearly cannot have a remainder of *both 0 and 1* when divided by 3, the two sides of the equation are unequal.
I also find it very surprising that there hasn't been a video on the sine, cosine, and other trigonometric functions, and their marvelous relationship with i, the imaginary number, and e.
It works out correctly. You have a 3 dimensional shape with all right angles, and you want to find the length between the farthest corners.
Call your dimensions x, y, and z. if you only wanted to find the hypotenuse between x and y you would use x^2 + y^2 = h^2 so to find the hypotenuse between all 3 dimensions you would need to find h^2 + z^2 = H^2. Because x^2 + y^2 = h^2, you can substitute x^2 + y^2 for h^2 and get x^2 + y^2 + z^2 = H^2.
It's a bit difficult to explain without a diagram.
So that's mr Singh! Loved his code book.
Thanks for clearing that up.
Yeah, there is validation to this guy's theory on why David X Cohen drops math based stuff in The Simpsons and even Futurama, he leaves it as little teasers to other math nerds in the world, of course there are others on the team but it was mostly Cohen explaining all of this on DVD audio commentary, there is other stuff he drops in like fake alien languages that they come up with and it leaves a secret message or something that is mostly in the futurama universe though.
I can highly recommend "Trick or Treatment" by Simon Singh and Edzard Ernst. Not much math trough.
If there are solutions to that, I'm guessing they are few and far between, considering that if your cuboid has side lengths of a, b, and c, {a, b, c, d} has to be a Pythagorean quadruple, but furthermore {a, b, e}, {b, c, f} and {a, c, g} have to be Pythagorean triples. And of course they all have to be positive integers.
Love the Mohawk