"A Brief History of Fermat's Last Theorem" by Prof. Kenneth Ribet

[@franzkronseder1035]

Another thank-you for the many amateur manageable pieces of insight stories in this talk :)

[@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...., proved by me :) (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. see pdfs at physicsdiscussionforum dot orgProof 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...., proved by me :) (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. see pdfs at physicsdiscussionforum dot org

[@山山-y4q]

🍎 Fermat's Last Theorem n=1×2×3×4×5×6×7×8×… If n is an infinite product of natural numbers, then from the Riemann zeta function, n=√2√π, using the following relationship, 1=(1/1)=(π/π)=1 In other words, when measured in unit numbers, it can be said that 1≡π. Furthermore, π≡π π=π is an identity. Therefore, √2×√π≡√2=√1×√2 is obtained. x^n+y ^n=z^n ⇒ x^√2√π + y^√2√π= z^√2√π ⇒ // n^2= (√2√π )^2, 1≡π, // x^2 + y^2=z^2 As a result, only n=1 and n=2 can have integer points. Fermat's Last Theorem was easily proven!

[@damianojeda938]

does n = 2 work because there are two variables being used in the equation? (x and y). would n = 3 work if there were three variables such as "a^3 + b^3 + c^3 = z^3" and likewise any other value for N where there are N variables being added? and if it's true, does that also mean that in the same way that only n = 2 works for 2 variables, does n = 3 only work for 3 variables and no other number of variables? does n = 7 only work when there are 7 variables?

[@BabacarKa-kx4uy]

Wonderful ! Very good and wise idea , Man !

[@ProfJigsaw]

Amazing!

[@nnsnumbersandnotesunlimite7368]

More on solutions’ enumeration in Diophantine or Fermat like equations : kzbin.info/www/bejne/bKPNoIJqgMSLq6s

[@gilberttheisen9270]

Le 14/8/2023 Consultez d'autres sites où je développe la solution ""FACILE"" du théorème de FERMAT, ne voulant pas me répéter sans cesse de sites en sites.

[@نمرمحمدعلى]

I have real solution

[@odinedin8265]

= THE GREAT! - THE GREATEST!!! Theorem of the 21st century! = !!!!!!!!!!!!!!!!!!!!! "- an equation of the form X**m + Y**n = Z**k , where m != n != k - any integer(unequal "!=") numbers greater than 2 , - INSOLVable! in integers". !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! /- open publication priority of 22/07/2022 / /-Proven by me! minimum-less than 7-10 pp. @!@!@!

[@protocolfree]

a messy huge proof is no proof , and who exactly understands this proof? 30 people around the world? !!!! thnx for tryin to explain it anyways

[@diribigal]

This talk was about the history, not the many refinements since then which simplified various parts, etc. While few have pored over the details of every part, more and more people know broad strokes and details that suffice for some parts as things get simplified.