What was Fermat’s “Marvelous" Proof? | Infinite Series 

"Typo" at 1:10! The prime may divide *at least one* (not exactly one) of the two integers. (Thanks to some of you for spotting this!)

Technically Fermat was correct, a proof was too large to fit in the margin of the book.

Have you heard the Last Theorem of Lord Fermat, the Wise? Of course not, is not a theorem that a mathematician would prove...

A drunk friend once told me: "If Φ was really called 'phee', then π would be called 'pee' "

I know exactly what Fermat was thinking, but the explanation is too long for a RU-vid comment.

Don't usually comment on videos, but I just want to say: While the transition between hosts was (very) rough, I've really enjoyed the last few videos. I'm glad you've been able to find that sweet spot balancing rigor, entertainment, and video length. Thanks for all you do. =)

Proof of Fermat's Last Theorem for Village Idiots (works for the case of n=2 as well) To show: c^n a^n + b^n for all natural numbers, a,b,c,n, n >1 c = a + b c^n = (a + b)^n = [a^n + b^n] + f(a,b,n) Binomial Expansion c^n = [a^n + b^n] iff f(a,b,n) = 0 f(a,b,n) 0 c^n [a^n + b^n] QED n=2 "rectangular coordinates" c^2 = a^2 + b^2 + 2ab Note that 2ab = 4[(1/2)ab] represents the areas of four right triangles) "radial coordinates" Lete p:= pi, n= 2 multiply by pi pc^2 = pa^2 + pb^2 + p2ab Note that pc^2, pa^2, and pb^2 represent areas of circles, wile p2ab = a(2pb) is the product of a radius (a) and a circumference (2pb). This proof also works for multi-nomial functions. Note: every number is prime relative to its own base: a = a(a/a) = a(1_a) a + a = 2a (Godbach's Conjecture (now Theorem.... :) (Wiles' proof) used modular functions defined on the upper half of the complex plane. Trying to equate the two models is trying to square the circle. c = a + ib c* - a - ib cc* = a^2 + b^2 #^2 But #^2 = [cc*] +[2ab] = [a^2 + b^2] + [2ab] so complex numbers are irrelevant. Note: there are no positive numbers: - c = a-b, b>a iff b-c = a, a + 0 = a, a-a=0, a+a =2a Every number is prime relative to its own base: n = n(n/n), n + n = 2n (Goldbach) 1^2 1 (Russell's Paradox) In particular the group operation of multiplication requires the existence of both elements as a precondition, meaning there is no such multiplication as a group operation) (Clifford Algebras are much ado about nothing) Remember, you read it here first) There is much more to this story, but I don't have the spacetime to write it here.

I'm sad that this channel has stopped posting new videos, but I always wish you guys good luck and happiness.

"...if and only if your ring has a very special property." The one part of this video I actually understood ;-)

That was one of the best videos going off on a tangent out of nowhere and then actually answering the question in the titel while really showing the connection with the seemingly unrelated tangent

Eight years as a mathematician and I thought phi was one of the Greek letters we all pronounced the same way. But now here you are with "fee"

A lot of books stated the theorem: if p divides ab, then p divides either a or b.

Credit were credit is due! This was huge improvement with relation to the last video presented by the same host! It is clear, it presents an interesting topic without occulting it with metaphors and comparisons! I think you listened very effectively to the public criticism and corrected your course! Congratulations and keep up the great work!

Maybe there was no marvellous proof & Fermat was just being mischievous.

6:30: as soon as those factors popped up, I expected a look up at a second camera, a fist raised in frustrated rage, and “Phiiiiiiii!!!” (Okay, less “expected,” more “chuckled at the mental image of,” but, y’know.)

Algebraic number theory is like pure satisfaction... I'd love to see more abstract algebra on the channel!

This video kicked my butt. I didn't know about Definition B and UFDs. Thank you!

I think that this video was a missed opportunity. The main problem in solving equations like the x^n+y^n=z^n is that they contain both multiplication (the n-th power) and addition. Adding a root of unity to the mix helps a lot because then we can transform the equation to be multiplicative in nature. If we know that primes=irreducibles (which is the main subject of the video), then we can use results about unique factorization to attack this problem. This idea gives a good motivation for why we need this UFD definition, while the video mainly gives the definitions and a claim that it is somehow related to Fermat's last theorem. In particular, for x^2+y^2=z^2, every student should learn that sum of squares should almost always lead to the decomposition (x+iy)(x-iy)=z^2. Then, it is a simple exercise to use the unique factorization property in the Gaussian integers Z[i] in order to find all Pythagorean triples.

So coooool I'm learning Abstract Algebra and so excited to master all this subject

The video is nice but it contains an error: A ring being a UFD is not equivalent to atoms (= irreducibles) being prime. This is only true if the ring is atomic, i.e. if every non-zero non unit is the product of atoms. To be more specific: R is a UFD (R is atomic and every atom is prime), for every domain R.

Hey Gabe, don't hold yourself back in the answers. Most of us indeed don't fully understand when people talk advanced maths, doesn't mean we're not interested in hearing what the discussion is about :-) - IMHO it's rather motivating to see how much is left for us to discover!

1 is the unique product of no primes, it’s like the multiplicative zero. Oh, and I f you pronounce the golden ratio “fee” then you have to pronounce the circumference diameter ratio “pee”. And I think I like that.

Wow, I never thought field extensions would come into play with primality. Thank you.

Awesome episode on Number Theory !!! More ! MORE !!!

Fermat's last theorem is a fact. Fermat's did not claim that there are no whole solutions to equation. Fermat's claim that are no solutions to the equation in whole numbers.

As someone who really struggled with abstract algebra. I really wish this video came out 2 years ago. Would have helped a whole lot. Just giving that little nudge in the right direction, that one of the main motivations for rings is that, a ring is the least amount of information needed to talk about primality, would have been incredibly helpful.

Really enjoyed this episode! Great explanations. Keep up the good work :)

Can you do a video about p-adic numbers?

I can't remember where I read that Fermat had some difficulty with the (relatively new) exponentiation notation. He may have been talking about: n^x + n^y = n^z, which is easily shown to have no integer solutions when n > 2.

Here's a puzzle: If you scramble a rubik's cube with any arbitrary algorithm (a random set of moves). Knowing only the result of the scramble. How often do you have to do the same scramble to return to the orignal state?

How we miss this channel.

OK, I didn't understand a lot of things you said, but at least I learned in which space Fermat's theorem was being analyzed.

Or Fermat was the biggest troll in history of humanity

Any episodes on Heegner numbers and UFDs coming up? Because please and thank you!

"if you are looking for a buzzword", nice humor there

I support PBS Space Time, but all the work PBS does here is excellent.

If you want to learn more about rings I recommend "Elements of Number Theory" by John Stillwell It's a textbook I used when I took a number theory course in college.

In oreder for the multiplication operator to exist, both its elements must exist. Russell's Paradox: 1^2 1 # = 2 = 1+1 (first order) Then #^2 = (1 + 1)^2 = [1^2 + 1^2] + [2(1)(1)] = 4(1^2) (second order - via Binomial Expansion) where the first term is existence and the second is interaction (multiiplication, entanglement, entropy) Note that existence and interaction are not 4D (1,1,1,1) which diagonal is 4 elements without multiplication. Every number is prime relative to its own base. n = n(n/n) = n(1_n) Goldbach's Theorem: every even number is the sum of two primes: n + n = 2n n is odd. Godel's characterization of wff's in his meta-language only uses odd numbers (products of primes). Therefore, the sums of odd numbers (even numbers) cannot be represented by hhis wff's. So it is just Goedel's meta-language that is incomplete, not positive real numbers. Together with Fermat's Last Theorem (applied to multinomials of arbitray powers), the arithmetic system is complete and consistent for positive real numbers. There are no negative numbers: -c = a - b, b > a b - c = a, a + 0 = a, a - a = 0.. If there are no negative numbers, there are no square roots of negative numbers. Proof of Fermat's Theorem for Village Idiots (n>2) c = a + b c^n = a^n + b^n +f(a,b,n) (Binomial Expansion) c^n = a^n + b^n iff f(a,b,n) = 0 f(a,b,n)0 c^n a^n + b^n QED Also valid for n > 1 c^2 = [a^2 + b^2] + [2ab]] 2ab < >0 c^2 a^2 + b^2 QED (Pythagoras was wrong; use your imagination) Check out my pdfs in physicsdiscussionforum "dot" org.

Ring thoery was one of my favourite topics at university. It's such a shame that interesting maths like this isn't really presented at schools.

I miss this channel...

:o i haven't been watching for a while, and now there's new people! Including the old spacetime host! :O

hahaha this video ending is amazing XD replayed that more than 5 times haha

I find Gabe much more organic and enjoyable than the other presenter. Any one else agree?

Proof of def A if and only if def B: Let P = XY. Then P divides XY, so that P divides either X or Y. If P = XY divides X, say XYZ = X, then we are led to YZ = 1 and thus Y = 1 or Y = -1. Now let P divide XY. Suppose P does not divide Y. Suppose we know that GCD (P, Y) = 1. Then GCD (XP, XY) = X GCD (P, Y) = X. Since P divides both XP and XY, P must divide X. So all we need is GCD (P, Y) = 1. Clearly GCD (P, Y) must divide P, say Z GCD (P, Y) = P. Since GCD (P, Y) is positive we are led to either Z = 1 or GCD (P, Y) = 1. Assuming Z = 1, we are led to GCD (P, Y) = P which cannot happen since P does not divide Y.

Wonderful presentation

Please do a video on an introduction and overview to ring theory! My apologies if you already have.

for any non americans confused by the term “foil” its an acranym for first outside inside last used to remember the distributive rule. americans use the anidote to refer to the action, where the phrase foiling a product comes from or foiled when done in past tense, this is the same as saying I added the numbers as per the law of addivtivity, or in more common terms it is analagious to saying i used long division to divide two numbers rather then simply saying i divided. This is just the weird nomenclature americans use to remember it in high school and the years after learning the anidote the students are quickly taught the more generalized rule for larger problems

1:10 nope, the prime p may divide both factors. The world 'exactly' shouldnt be there

The definitions of irreducible and prime elements should include the proviso that the element isn't a unit. Otherwise, 1 is irreducible and prime in Z. Elements in a UFD are not necessarily "uniquely" the product of irreducible elements. For example: in Z, 6 = (2)(3) = (-2)(-3). The definition of a UFD is: each non-unit of the UFD can be factored into irreducible elements and for any two factorizations of each non-unit of the UFD into irreducible elements, there is a 1-1 correspondence between the factors so that each factor of one the factorizations corresponds to an associate of itself (that is, that factor times a unit).

There is a possibility of a very large reduction of the general proof of the Fermat theorem. This proof can only be obtained by analyzing the Pythagorean theorem.

Thanks for the reply and the great video as usual. You nailed that last name Gabe!

Maybe I need more coffee. I got lost at 2:52. It seemed like she was making a statement about integers, said it was NOT true, but then said it WAS true for INTEGERS.

This episode is SO GOOD!

Great video!

One question about UTF: Let's say that a,b,c,d are unique primes. Why is ab≠cd at all times? This was obvious to me throughout university studies, but not anymore.

There's a small mistake in the video. The video states that a (commutative, unital) ring is a UFD iff primality is equivalent to irreducibility. But this only gives you uniqueness of factorization if it exists, but you may not have existence of a factorization into irreducibles. So you need some other condition like that the ring needs to be Noetherian or whatever. This is obviously unimportant to viewers who don't know any ring theory, but just in case it confuses any students studying this who come across this video.

This fairly trivially extends into more than two factors (with integers). No matter how you factorize a number n, a multiple of a prime p, p must exist in at least one of those factors because it is a priori a prime factor of n. Any factor of a number that is itself a prime is a prime factor of that number, again by definition.

Define π Sum check solution. Sum(n) = 1/2 n (n+1) Sum(1) = 1/2 n × 1 × (1+1) Sum(1) = 1/2 × 1 × 2 Sum(1) = 1

Wait, so at 5:40 does this mean 3 is not a unit in Z but it is a unit in Q? For the number u=3 in Q, there is a v=1/3 also in Q, so that uv=1 Are all rationals considered units under their own set? Or what am I missing?

Please name the axioms "cheese" and "coco puffs" next time.

What a great presentation! Here is my theory. We know Fermat's note to himself is all the evidence there is, and he never returned to this problem again. This is behavior (not revisiting a "Marvelous" discovery) is associated with realising we are wrong, and is the single most ignored piece of evidence in FLT. I seem to be the only person who has dared to conjecture that the next time Fermat took a look at his "Marvelous" proof he realised it was a trivial proof to an even more trivial problem and never bothered with it again. Following discovery and publication of his private notes, mathematicians did not see the ambiguity of FLT as written, and misunderstood it as a far more complex problem. However, this assumption ignores the case of a "Marvelous" solution to a trivial problem, so trivial in fact, that it really wasn't that "Marvelous" after all. Fermat realised he was wrong (and he was also sometimes wrong without realising it!). All very human. The "trivial" FLT for which the solution is likewise trivial, is to assume the Pythagorean (h^2 = x^2 + y^2) as a given. The "trivial" proof is then one of substitution, requiring about 6 steps. How embarrassing!!! But it is still too long to squeeze into the available space of his margins. However after his death, the mathematicians assumed a non-trivial FLT, and set out to prove the Pythagorean a different way (for integers) rather than assume it. In doing so, it fails to take advantage of the particular case in order to prove the general case, but also sets out to prove the general case as if the Pythagorean (reals) was outside the scope of the Diophantine (integers). I fail to see the need to do this. Its quite easy to show graphically by setting h=1 for all N>2 that a right triangle cannot be formed and the Pythagorean no longer applies and this is true for all reals, not just integers. Only for N=2 is it possible to form a right triangle, and for greater or lesser values of N the right angle increases or decreases. The ration (h=1:x:y) then holds for any integer or real (h,x,y). The proposition that Fermat would simply walk away from such a claim as he made without realising he was wrong, has far less credibility than the proposition that he wittingly abandoned a "Marvelous" discovery 'just like that'. I, for one, am not convinced one iota that mathematicians ever understood the severe triviality of FLT, but embarked a different problem altogether. That is my position, and it is over to Andrew Wiles et. al. to show why my understanding and solution for FLT - as a trivial solution to a trivial problem - is wrong (my solution it is certainly not wrong). Emperor Fermat has no clothes!!!!

The general consensus about Fermat is that he probably had a few special cases and thought he could easily generalize it

Thanks for the great video! Fermat never claimed to anyone that he had a proof. He wrote in the margin of his own book. This was not discovered until after he died. So now he's "the greatest troll ever." Just remember folks, some future archaeologist, digging some landfill, may read some stuff you wrote and threw away.

His proof was almost certainly mistaken.

All Fermat did was construct an exponent table and calculate the differences between different power tables ^2 nets you an ‘even’ difference of odd values, making a nice a^2+b^2=c^2 equivalence with 3, 4, and 5, among other sets X^3 differences get more complicated with a total sum of 6•[0,x)+1, but it is still reasonable to work with, it even has an equivalence set directly based on the Pythagorean Triples, it’s just not a triple set From there on up, though exponential equivalences do persist to the infinite, perfect triplets all but stop existing and get replaced by more complex sums Put simply, the differences between subsequent powers of bases get exponentially more complicated as the power variable increases, preventing triples from being constructed past ^2 And that’s the _condensed_ version, it’s no wonder Fermat didn’t publish his proof, he probably had a textbook worth of numerical dynamics regarding this and other exponential concepts dancing in his dome when he passed

Fermat's Last Theorem Simplified Proof a^n +b^n =c^n odd+even=odd a^3+b^3=c^3 let c=2x+1 8x^3+12x^2+6x+1=c^3 let b=2x b^3+12x^2+6x+1=c^3 a^3=12x^2+6x+1 Only valid solution to a is when x=0 for all c=2x+1,2x+3,2x+5,... n=3,4,5,... No whole number solution for a,b,c when n>2

@4 :12 b has to belong to Q* otherwise if b =0 you just have Q

"Which some people would call FOIL?" Why the question mark? Is FOIL a bad way to teach binomial distribution?

@7:15 check out the references below for why 2 is irreducible- which reference?

You all have it wrong, fermat was looking at squares, I know the secret what fermat was talking about. When I have time to write it up.

I really want to know fermats proof even if it's wrong

7:29 definition A is incorrect over the integers, as it would include 1 as a prime.

Fermat videos will never not be uninteresting

"Behave a lot like integers" is not exactly how I would informally describe a ring.

Im quite sure that since the proof for every power, that is a multiple of 2, is quite easy Fermat thought he could generalize this for every number

all PBS-ers has to move their hands

A unique factorization domain (UFD) is not uniquely defined by primes being the same as irreducibles. Consider the ring Z + Q[x]. Not a UFD, but all irreducibles are prime.

Z[sqrt(5)] includes a/2+b×sqrt(5)/2 where a and b are both odd, as well as when they're both even. So it includes Φ and 1/Φ, which are units. So 2×Φ×2/Φ is not essentially different from 2×2. This is true for all sqrt(p) where p≡1 mod 4. If p=-3, you get the Eisenstein integers. Z[sqrt(5)] is a UFD, but Z[sqrt(-5)] is not.

I have a small quarry with that definition of prime. Assume we are in a integral domain, so that xy=0 implies x=0 or y=0. Clearly, 0|0, since we can take 0=0*1. But even more, assume that 0|x. Then for some y, we have x=0*y=0 and with that we prove that 0|x iff x=0. Now let's check the given definition for prime when p=0: If 0|ab, then ab=0. But we are in a domain, so a=0 or b=0, but that means that 0|a or 0|b, so 0 is prime. In particular, if we are talking about integers, that definition says that 0 is a prime integer, which is untrue (in fact, it is simple to show that 0 is not irreducible, with the given definition). The proper definition would be that p is a prime iff p is not 0 or a unit, and p|ab implies p|a or p|b.

I think we are pretty sure that his mistake was that he didn't know that the factorization in terms of complex numbers isn't unique. If you assume it is, the proof is less than half a page... I say proof, I mean "supposed proof" cause it is obviously incorrect.

9:00 "According to abstract ring theory, irreducible and prime are equivalent concepts if and only if your ring is a UFD." Well, except for some pathological cases. See ramanujan.math.trinity.edu/rdaileda/teach/m4363s07/non_ufd.pdf for an example of a domain satisfying prime = irreducible, but not unique factorisation.

At the core is prime. Surrounding the core is irreducibility. And surrounding this is unique factorization. Then standard factorization comes next. This is the justifcation I needed to try and pry math away from the intangible and the abstract. Thank you.

I have to admit, I don't understand why fractions are introduced into prime number theory.

You made me lol at the end of the video, Gabe, haha.

Was the music the FTL soundtrack?

I do love abstract algebra!

oh no this is bringing back memories from my algebra class

Super nice video.

For those noticing that Definition B should be 'at least' instead of 'exactly': it's the same in this case, you could use 'exactly' without loss of generality because of associativity, you can always rearrange the product to mean exactly one and the definition holds true for any product, no matter how you associate. It's not that the observation is wrong, is just that you can use the definition to mean exactly the same, and keep it compact and elegant at the same time.

Great vid!!! Will comment more later, thanks.

What about complex numbers with this regards to these definition

Actually I have a copy of the proof of Fermat himself regarding his last theorem.

Wrong statement at 9:13. Irreducibles are primes if (but not only if) the ring is a UFD. But an IP ring gotta be atomic (aka any number has at least one factorization to irreducibles).

Fermat's Last Theorem: 홀수 솟수 p에 대하여 x^p+y^p=z^p을 만족하는 자연수 x, y, z는 존재하지 않는다 My Marvelous Proof) x, y, z가 존재한다면 Fermat's Little Theorem에 의해, vw(v+w+2pk)F(v, w, p, k) = p^(p-1) k^p 으로 변형. 그 해는 (x, y, z) = (v+pk, w+pk, v+w+pk) 꼴이다. 그런데 자연수 k= n인 경우 해가 존재한다면 n=1*n 이므로 k=1일 때의 해의 n배의 해를 가져야 하는데... k=1일 때 해가 존재한다면 '홀수=짝수'로 모순. 따라서 해는 존재하지 않음.

You got definition of irr wrong. An element a\in R is called irreducible iff a is not unit nor a 0 and a=pq implies p is a unit or q is a unit.

Gabe: Was just about to call Elon to convince him to patreon a mil, but upon observing how your threads are as flashy as mine....

Isn’t the def of unit number wrong in this video. u is unit number if uv = v right?

During fermat's time the concept of ring was not defined

Pfouah, that's complicated! I feel like it's useful though, maybe someone could do crypto out of the primality/irreducibility in different rings or something?

Very well made video. Keep it up