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

If Fermat had a little more room in his margin, what proof would he have written there?

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Resources:
Contemporary Abstract Algebra by Joseph Gallian

Standard Definitions in Ring Theory by Keith Conrad

Rings and First Examples (online course by Prof. Matthew Salomone)

Fermat's Enigma by Simon Singh

Who was first to differentiate between prime and irreducible elements? (StackExchange)

Previous Episodes:
What Does It Mean to be a Number?

What are Numbers Made of?

Gabe's references from the comments:

Blog post about the Peano axioms and construction of natural numbers by Robert Low:

Recommended by a viewer for connections to formulation of numbers in computer science:

In 1637, Pierre de Fermat claimed to have the proof to his famous conjecture, but, as the story goes, it was too large to write in the margin of his book. Yet even after Andrew Wiles’s proof more than 300 years later, we’re still left wondering: what proof did Fermat have in mind?

The mystery surrounding Fermat’s last theorem may have to do with the way we understand prime numbers. You all know what prime numbers are. An integer greater than 1 is called prime if it has exactly two factors: 1 and itself. In other words, p is prime if whenever you write p as a product of two integers, then one of those integers turns out to be 1. In fact, this definition works for negative integers, too. We simply incorporate -1. But the prime numbers satisfy another definition that maybe you haven’t thought about: An integer p is prime if, whenever p divides a product of two integers, then p divides at least one of those two integers.

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[pbsinfiniteseries]

"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!)

[BuleriaChk]

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.

[mheermance]

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

[lexolotlgod]

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. =)

[lovaaaa]

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

[DiegoBQZ]

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

[atrumluminarium]

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

[MatthewLeeKnowles]

"...if and only if your ring has a very special property."

The one part of this video I actually understood ;-)

[halbeard]

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

[omargaber]

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

[Ouvii]

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

[SKyrim]

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!

[alexmcgaw]

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"

[LwLiPp]

Perhaps Fermat thought he could generalize his technique of the infinite descent, which can be used to show the non-solvability of the equation x^4+y^4=z^4.

[amirhomayounnejah]

I just want to say congratulations. Your channel is awesome.

[heaslyben]

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

[joshmyer]

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 (Okay, less “expected, ” more “chuckled at the mental image of, ” but, y’know.)

[KcKc-bhlu]

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

[FernandoVinny]

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

[tomrivlin]

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