The Fermat's Last Theorem Project #74
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This is extremely cool, good luck with the project! |
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385 (1637-2022) years later, Fermat's proof of his last theorem has finally been see "https://www.mdpi.com/2227-7390/10/23/4471", that is " On the Nature of Some Euler’s Double Equations Equivalent to Fermat’s Last Theorem" |
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sounds really cool |
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Thanks for the positive comments! Andrea -- I'll believe your low-level approach if and only if you can formalise it in Lean. I have been sent many many low-level approaches in my life and none of them stand up to the kind of scrutiny which Lean gives. Until then, I think I'll stick with my belief that the only way known to humanity to prove FLT is via elliptic curves and modular forms. |
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Dear Prof. Kevin Buzzard, See Best regards Prof. Andrea Ossicini |
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Hi Andrea, Against my better judgment I attempted to understand both proofs, and I believe I found the flaw. Theorem 3 of the paper says: "Fermat’s Last Theorem is true if and only if a solution is not possible in integers of Equation (8) with the Q non-zero integer", where equation (8) is equivalent to Essentially, I believe that equation (8) has no solutions regardless of the truth of FLT, and so a proof that (8) has no solutions does nothing to show FLT is true. (This is basically denying a simple proof of Theorem 3, although we can prove it by using Fermat-Wiles to prove FLT and then note that both sides of the iff are true....) The issue is that if we suppose FLT to be false, then we have a V=T=P solution to the first equation of 2.10, but the second equation has |
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Hola , buenos días . no soy matemático de profesión , pero me encantan las matemáticas , siempre me gustaron , y en especial Fermat . Yo creo que es en lo que se basó Fermat para enunciar su teorema , lo registré ante notario y lo publiqué en mi periódico local " La Voz de Galicia " como " Fermat y las ternas Pitagóricas ". Es una visión que nadie la ha visto , es un ejemplo sencillo y maravilloso cómo decía Fermat . No hablo inglés , y no encuentro un matemático que tenga el nivel para verlo . Por eso recurro a vosotros Kbuzard , . Creo que os gustará y lo podréis comprender . LAS TERNAS PITAGÓRICAS El teorema fue conjeturado por Pierre de Fermat y demostrado por Wiles,(ayudado por Richard Taylor), en 1995 . Ciertamente este magistrado de Profesión y matemático de corazón contribuyó y mucho de una forma provocadora y simpática al desarrollo de las matemáticas de nuestro tiempo. Cómo pudo deducir Fermat en aquellos tiempos cuando no existían calculadoras, y menos ordenadores, tuvo que ser a base de pluma, papel y lo más importante intuición matemática. Uno piensa que éste hombre tendría a buen seguro un “punto de apoyo en que basarse, y seguramente a mi entender, pienso que fue el teorema de Pitágoras, y por ende las ternas pitagóricas. Cuando observamos que en el triángulo rectángulo 3² + 4² = 5² , vemos que hay números enteros , muchos , diríamos que infinitos , que cumplen esta condición , por citar unos pocos : (3,4,5)-(5,12,13)- (7,24,25)-(8-15-17)-(9-40-41)-(11,60,61)- (12,35,37)-(13,84,35)…etc… . Esto se cumple para infinitos trios de números para potencias igual a “2” ( superficies),pero no se cumple para potencias superiores a “2” , ( lease 3-4-5 ..) . En todas las “ ternas Pitagóricas “, vemos que están formadas por un número par y dos impares . Vamos con unos ejemplos sencillos : 3²+ 4² = 5² / 9 + 16 = 25 Ahora dividiremos entre “2” los números (3/2)² + ( 4/2)² = ( 5/2)² / 2,25 + 4 = 6,25 (3/4)²+ (2/4)² = (5/4)² / 0,5625+ 1 =1,5625 Observamos que hay un ”maridaje” perfecto entre “dos impares” y un “par”. Vayamos ahora con los cubos 3³+ 4³ + 5³ = 6³ / 27 + 64 + 125 = 216 Tenemos miles de ejemplos para los “cubos”, ( también hay ejemplos para potencias superiores con cuatro componentes , pues a partir del “cubo” , 3 , van a verse todas afectadas por las propiedades de los volúmenes , que echaron abajo el teorema de Euler , y que para mí , Fermat lo “vio” ) Observemos ahora el “ maridaje”que se produce en los cubos 3³ + 4³ +5³ = 6 3 / (3/2) 3 + ( 4/2) 3 + (5/2) 3 =( 6/2)³ (1,5)³ + 2³ + (2,5)³ = 27 3,375 + 8 + 15,625 = 27 Dividiendo nuevamente ( 0,750)³ + 1 3 + (1,25)³ = ( 1,5)³ 0,421875 + 1 + 1,953125 = 3,375 Dividiendo nuevamente entre 2 ( 0,375 )³ + ( 0,5)³+ ( 0,625)³ = ( 0,750)³ 0,052734375 +0,125 + 0,244140625 = 0,421875 Resumiendo : Pierre de Fermat , se dio cuenta de que son necesarias en la suma de dos potencias superiores a “2” ( 3,4..) , estarán formadas al menos por cuatro componentes dos de ellos procedentes de números “ impares “ y los otros dos de números “pares , estando condicionados los volúmenes por el triángulo de Pitágoras , que obliga necesariamente a los cuatro componentes mínimos en la ecuación Vemos que en la igualdad de ( 2682440) 4 + (15365639) 4 + ( 18796760) 4 = = ( 20615673) 4 Par( 2-4-8) + impar(1) + par(2-4-8) = impar ( 3) +- par +-impar +- par = +-impar +-impar +-par +- impar = +-par |
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Hola amigos , seguro que ya se habréis comprobado , pero en la suma de potencias superiores a "2" , estarán formadas por al menos dos números impares y dos números pares , por ejemplo en la de : Clemente García Estévez |
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Dear Mario Carneiro,
I'm sorry, but I don't agree. You evidently know little about the
techniques of classical Diophantine analysis.
I am sending you as an attachment the certainty that my proof is correct.
In my paper, have you seen who the Academic Editor is who authorized the
publication?
Sincerely
Andrea Ossicini
Il giorno ven 3 mag 2024 alle ore 21:14 Mario Carneiro <
***@***.***> ha scritto:
… Hi Andrea,
Against my better judgment I attempted to understand both proofs, and I
believe I found the flaw. Theorem 3 of the paper says: "Fermat’s Last
Theorem is true if and only if a solution is not possible in integers of
Equation (8) with the Q non-zero integer", where equation (8) is equivalent
to $$X_1^nV^2+Y_1^2T^2=Z_1^2P^2$$, $$V^2-T^2=Z_1^2Q^2$$ (which is 2.10 in
the second paper). The proof of this theorem is not given in the first
paper as far as I can tell (it is stated in section 4 with a promised proof
in section 6 but there is no proof in this section but rather a bunch of
unrelated algebraic manipulations followed by an apparent application of
the theorem in the last sentence - "The double equations of Euler are
discordant forms and so the F.L.T. turns out to be true"). There is
something close to a proof of it in the second paper, but the proof
actually shows the opposite: if the FLT is false then this diophantine
system has no solutions with V=T=P and Q nonzero.
Essentially, I believe that equation (8) has no solutions regardless of
the truth of FLT, and so a proof that (8) has no solutions does nothing to
show FLT is true. (This is basically denying a simple proof of Theorem 3,
although we can prove it by using Fermat-Wiles to prove FLT and then note
that both sides of the iff are true....) The issue is that if we suppose
FLT to be false, then we have a V=T=P solution to the first equation of
2.10, but the second equation has $$V^2-T^2=Z_1^2Q^2$$ which has no
solutions with $$Q\ne 0$$ since the LHS is zero and the RHS is positive.
So a counterexample to FLT does not give us an integer solution to 2.10 and
hence equation (8), and even if .
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I'm sorry, I should not have continued this discussion on a public post about an unrelated matter. If you are interested in continuing this you can email me (see @digama0 ) or DM me on https://leanprover.zulipchat.com/ . |
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You formalise it in Lean, I'll believe it. |
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Hi friends, |
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the final proof is here: https://www.youtube.com/watch?v=OLsRS--dAfQ |
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You formalise it in Lean, I'll believe it. |
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Pls check the following code for my step sum definition:
import Mathlib.Data.Real.Basic
import Mathlib.Algebra.Group.Defs
import Mathlib.Algebra.BigOperators
namespace SumOperator
…-- Define the basic index, scale factor (rho), step, and the sum
operation
def scaleFactor (ρ : ℝ) : ℕ → ℝ
| n => n * ρ
def step (ρ : ℝ) (idx : ℕ) : ℝ :=
scaleFactor ρ idx
def generalizedSum (f : ℝ → ℝ) (ρ : ℝ) (LL UL : ℝ) : ℝ :=
let n_steps := Nat.floor ((UL - LL) / ρ)
let idx_seq := List.range (n_steps + 1)
List.sum (idx_seq.map (λ i => f (LL + step ρ i)))
-- Some examples of applying the generalized sum
def exampleFunction1 (x : ℝ) : ℝ := x^2 + 2*x + 1
def exampleFunction2 (x : ℝ) : ℝ := sin x
#eval generalizedSum exampleFunction1 (1.0 / 2) 0 2 -- Summation with
step = 1/2
#eval generalizedSum exampleFunction2 (π / 4) 0 π -- Summation with
step = π/4
end SumOperator
Il 2024-08-06 13:04 Kevin Buzzard ha scritto:
You formalise it in Lean, I'll believe it.
--
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New link to the video here (thanks to some hacker my old YT channel was closed...): Fermat the Last Theorem proof via New Sum properties https://www.youtube.com/watch?v=2GzU8rALZh0 Thanks for checking and sharing. It already has a doi and it is well known I'm on from long time ago... |
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Dear Sir/Madam, I am sending you my article as my friends have suggested. I will now present my friend's speech from the IAEA: Fermat's Theorem - A Proof by Fermat Himself The Russian nuclear physicist Grigoriy Leonidovich Dedenko has reconstructed the original reasoning of Pierre Fermat, which led Fermat to conclude that the sum of two identical natural powers of rational numbers, raised to an exponent greater than two, is not representable. This is known as Fermat's Last Theorem. As you may know, in 1637, Fermat wrote a note in the margins of his copy of Diophantus's "Arithmetic" stating his discovery and adding, "I have discovered a truly marvelous proof, but this margin is too narrow to contain it." According to G.L. Dedenko, Fermat analyzed power differences using a method that was novel at the time: decomposing these differences into a sum of pairwise products, later known as the "Newton binomial". Fermat discovered that the coefficients in this expansion satisfied simple conditions equivalent to a logarithmic equation (a concept still developing in the mid-17th century) for the degree of the sum (or difference). This equation has only two solutions: the numbers one and two. Thus, the margins of the book were indeed too narrow to contain the complete proof. Fermat's proof would have required the introduction and development of new concepts, such as expansion with combinatorial coefficients (Newton's binomial) and logarithms. It remains unclear whether Fermat ever wrote down his detailed reasoning, and if so, whether this record survives in an unexpected archive. Historians of natural history are encouraged to search anew. Sincerely, Grigoriy Dedenko The final version №25 is available at this link: |
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Dear colleagues,
https://osf.io/preprints/osf/jbdas
"The original proof of Pierre Fermat's famous theorem, reconstructed after
35 years of work.
17th-century elementary mathematics. Verified in a simplified version using
Coq and in full using other neural networks.
вт, 23 июл. 2024 г. в 09:46, Complicate-Modulus ***@***.***>:
… Hi friends,
My final Wonderful and Simple FLT Proof hereafter is based on the
Telescoping Sum property. So any student knowing Sum, Limit and Calculus
can follow it, really understanding why Fermat & Wiles are right, still if
they have no ide aof what a Ring is. Representing Powers as Sum of Integers
like A^n=\sum_{1}^{A}(X^n-(X-1)^n) one can start to work with any power
problem (Beal too). But I finally discover that we don't know all the
properties for the Summand Operator (we know how to cut, and how to shift
the limits but no more !), then I first develop and prove 2 new rules:
Scaling and Stretching. Scalig is a new property allow us to mover Step=1/K
instead of mute integer index i just. It shows that just a Power of an
Integer is K invariantive, it means it not depeds on how fine 1/K is in the
Sum we use to represent it: so integer step, rational step 1/K or pushing
the Sum to the limit for k to infility, so step dx the result of the Sum is
always f.ex. A^3. Then I was able to represent Fermat The Last as 2 sums
from 1 to A one for sure representing A^n, the other one C^n-B^n, in where
A,B,C since limits of a finite integer Sum are for sure Integers. So the
final proof comes once one understand that beeing a power of an integer
(f.ex. a cube) means it must be a K invariantive structure, so do not
depends on its numerical value. The first simple prooffor the case n=2
prove Pitagora is right, the following one for n=3 and n=5 shows that
Fermat-Wiles are right since working with C^3-B^3 (or at the 5th power)
like it is a genuine power, means a K invariantive Summation, produce the
absurdum we are looking for: rising the "neutral" K we choose, f.ex.
A^{3m}/A^{3m}, the result for C^3-B^3*(A^{3m}/A^{3m}) once rewriteen in sum
taking the factor A^3m into the upper limit as A^m multiplier, and the
1/A^3m factor as divisor of any term of the Sum, produce a K varinative
result: rising m, the resulg go closer and closer to a diferent integer
from C^n-B^n we start from. Once one understand all this new Complicate
Modulus Algebra (a two hand clock modulat math) one can understand (I hope)
the final more wonderful 1 line proof . For what Beal theorem proof
follows. Here the link:
https://www.researchgate.net/publication/382328938_A_new_useful_denition_for_the_Sum_Operator
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The Fermat's Last Theorem Project
Kevin Buzzard discusses the project to prove Fermat's Last Theorem in Lean.
Introduction
Fermat's Last Theorem (FLT) is the claim that some abstract equation has no solutions in positive integers.
Th
https://leanprover-community.github.io/blog/posts/FLT-announcement/
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