From: dreier@jaffna.berkeley.edu (Roland Dreier)
Newsgroups: sci.math
Subject: Re: 1 page proof of FLT; this is what Fermat
had in mind when he wrote "margin to small".
Fermat reasoned that the proof was so simple that
others would easily rediscover it.
Date: 13 Aug 93 13:32:17
Organization: U.C. Berkeley Math. Department.
Lines: 12
Message-ID: (DREIER.93Aug13133217@jaffna.berkeley.edu>
References: (CBpno3.JqM@dartvax.dartmouth.edu>

In article (CBpno3.JqM@dartvax.dartmouth.edu>
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes:
For the case n=3 suppose there exists a P-triple
which satisfies FLT. Then there has to exist a smallest
P-triple for n=3. If true implies there exists a number N
which has the property N+N+N=NxNxN=M.

Not that I expect a coherent answer, but let me ask anyway:
why does this follow? In other words, given positive
integers a,b,c with a^3+b^3=c^3, how do I get a positive
integer N with N+N+N=N*N*N ?

--
Roland "Mr. Excitement" Dreier dreier@math.berkeley.edu

Answering Roland Dreier FLT post of 1993, and why 2+2 = 2*2 as AdditiveMultiplicative Identity AMI proves FLT overall 

From: dreier@durban.berkeley.edu  (Roland Dreier) 
Newsgroups: sci.math 
Subject: Re: 1 page proof of FLT 
Date: 18 Aug 93 14:55:02 
Organization: U.C. Berkeley Math. Department. 
Lines: 42 
Message-ID: (DREIER.93Aug18145502@durban.berkeley.edu> 
References: (CBxp0H.817@dartvax.dartmouth.edu> 
(24s7de$cv4@outage.efi.com> 
(CByoqr.7Aw@dartvax.dartmouth.edu> 
(1993Aug18.173020.10123@Princeton.EDU> 

In article (1993Aug18.173020.10123@Princeton.EDU> 
kinyan@fine.princeton.edu (Kin Chung) writes: 
In article (CByoqr.7Aw@dartvax.dartmouth.edu> 
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium) writes: 
LP     Hardy in Math..Apology said words to the effect that the 
LP  understanding of any math proof is like pointing out a peak in the 
LP  fog of a mtn range and you can only point so long and do other 
LP  helps and hope the other person will see it and say Oh yes now I 
LP  see it. But you can not exchange eyeballs. Again I repeat the 
LP  arithmetic equivalent of FLT is that for exp2 there exists a 
LP  number equal under add & multiply i.e. 2+2=2x2=4. Immediately a 
LP  smallest P triple is constructible for exp2 i.e. (3,4,5>. But no 
LP  number exists like 2 for exp3 or higher in order to construct P- 
LP  triples for these higher exp. I am very sorry that I cannot make it 
LP  any clearer than that. Time to take a break and reread Hardy Math 
LP  Apology. 

KC   You also say that a smallest P-triple is constructible for exp2   
KC   immediately from the existence of a number N such that 
KC   N+N=NxN, namely N=2. How do you construct a P-triple given N 
KC   with this property? Please note that I am not asking how you do 
KC    it for exp3, but for exp2. 

Before I continue, let me say that this post does not in any way constitute 
an endorsement of LvP's "proof"; what I am about to explain does not 
extend to exponent 3 in the least. However, things are rather easy for 
exponent two.  (Not to be critical, but you really could have figured this 
out yourself :-) 

So suppose we have an N with 2xN=N+N=NxN. Set a=N+1, b=N+N=NxN. 
Then we get 
                     a^2 = (N+1)^2 = N^2+2xN+1 = 2xN^2+1 
also 
                     b^2 = (N+N)^2 = 4xN^2. 
So 
                     a^2+b^2 = 6xN^2+1. 
Now set c=2xN+1.  Then 
                     c^2 = (2xN+1)^2 = 4xN^2 + 4xN + 1 = 4xN^2 + 2xN^2 + 1       
                      = 6xN^2+1. 
So magically a^2+b^2=c^2, just as desired! ! 

If you can figure out how to do that for exponent 3, make yourself famous. 

Roland 
-- 
Roland "Mr. Excitement" Dreier                 dreier@math.berkeley.edu 

Newsgroups: sci.math
Date: Fri, 8 Mar 2019 21:32:46 -0800 (PST)
Subject: My 1993 basis vector pure algebra proof of Regular FLT Re: AP's 2014
 proof of Generalized FLT Fermat's Last Theorem and proof of Regular FLT as a
 corollary thereof//updated 2019 for more clarity
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sat, 09 Mar 2019 05:32:47 +0000

My 1993 basis vector pure algebra proof of Regular FLT Re: AP's 2014 proof of Generalized FLT Fermat's Last Theorem and proof of Regular FLT as a corollary thereof//updated 2019 for more clarity

Ever since 1993, I have been making proof attempts for a Pure Algebra Proof of FLT. My first attempt was the specialness of 4 with 4 = 2+2 = 2*2, arguing that exp2 has solutions due to the fact of this specialness of 4. Now if another integer had a property of N+N+N = N*N*N = N^3, the argument would go that this special number allows for a solution set in exp3. Similarly for higher exponents. 

To this day I maintain the logic of that is correct and is a Pure Algebra proof of Fermat's Last Theorem. For you can consider that special number as a basis vector of a exponent allowing for a solution set. 

But as the years rolled by, came 2014 and sort of tired of arguing the basis vector proof. I took a plunge and dive at proving Generalized FLT and did so in 2014, and lo and behold to my surprise, a easy proof of Regular FLT comes out as a corollary. Ever since 2014, I sort of bypassed or mildly forgotten the basis vector pure algebra proof. 

But I think I should resurrect it. The foolish math professors of my time just will never understand a basis vector proof-- the simpleton fools cannot even accept the fact they were misguided for 2 thousand years on the ellipse for it is never a conic. So can anyone expect such fools to have a sensible idea of a proof by basis vector? Of course not. 

My 6th published book

World's First Valid Proofs of Fermat's Last Theorem, 1993 & 2014 // Math proof series, book 5 Kindle Edition
by Archimedes Plutonium (Author) (Amazon's Kindle)

Last revision was 29Apr2021. This is AP's 6th published book.

Preface: Truthful proofs of Fermat's Last Theorem// including the fake Euler proof in exp3 and Wiles fake proof.

Recap summary: In 1993 I proved Fermat's Last Theorem with a pure algebra proof, arguing that because of the special number 4 where 2 + 2 = 2^2 = 2*2 = 4 that this special feature of a unique number 4, allows for there to exist solutions to A^2 + B^2 = C^2. That the number 4 is a basis vector allowing more solutions to exist in exponent 2. But since there is no number with N+N+N = N*N*N that exists, there cannot be a solution in exp3 and the same argument for higher exponents. In 2014, I went and proved Generalized FLT by using "condensed rectangles". Once I had proven Generalized, then Regular FLT comes out of that proof as a simple corollary. So I had two proofs of Regular FLT, pure algebra and a corollary from Generalized FLT. Then recently in 2019, I sought to find a pure algebra proof of Generalized FLT, and I believe I accomplished that also by showing solutions to Generalized FLT also come from the special number 4 where 2 + 2 = 2^2 = 2*2 = 4. Amazing how so much math comes from the specialness of 4, where I argue that a Vector Space of multiplication provides the Generalized FLT of A^x + B^y = C^z.

Cover Picture: In my own handwriting, some Generalized Fermat's Last Theorem type of equations.

As for the Euler exponent 3 invalid proof and the Wiles invalid FLT, both are missing a proof of the case of all three A,B,C are evens (see in the text).

Product details
• ASIN ‏ : ‎ B07PQKGW4M
• Publication date ‏ : ‎ March 12, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 1503 KB
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 156 pages
• Best Sellers Rank: #4,327,817 in Kindle Store (See Top 100 in Kindle Store)
◦ #589 in Number Theory (Kindle Store)
◦ #3,085 in Number Theory (Books)

Actually Roland Dreier in August of 1993 had a full very short proof of FLT, but he just did not recognize it. And of course Andrew Wiles never had a proof of FLT other than his con-art chicanery of a mindless wreck with his fake-proof.

Roland's proof is logically this-- every time you can have A^x + B^y = C^z, then you can break that down into being the addition of 2+2=2x2.

The only mistake Roland made, is that he thought he had to show that 2+2= 2x2 had to be installed into A^3 + B^3 = C^3.

No, Roland, you need not have to install 2+2= 2x2 = 4 into exponent 3, but simply just say there is no N+N+N = NxNxN.

So, just a hoopla of logic, stopped Roland from recognizing he had a proof of Fermat's Last Theorem by August of 1993. Allowing for the con-artist Andrew Wiles to permeate his fake FLT.

On Saturday, April 22, 2023 at 11:55:23 PM UTC-5, Archimedes Plutonium wrote:
> From: dre...@jaffna.berkeley.edu (Roland Dreier)
> Newsgroups: sci.math
> Subject: Re: 1 page proof of FLT; this is what Fermat
> had in mind when he wrote "margin to small".
> Fermat reasoned that the proof was so simple that
> others would easily rediscover it.
> Date: 13 Aug 93 13:32:17
> Organization: U.C. Berkeley Math. Department.
> Lines: 12
> Message-ID: (DREIER.93A...@jaffna.berkeley.edu>
> References: (CBpno...@dartvax.dartmouth.edu>
>
> In article (CBpno...@dartvax.dartmouth.edu>
> Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:
> For the case n=3 suppose there exists a P-triple
> which satisfies FLT. Then there has to exist a smallest
> P-triple for n=3. If true implies there exists a number N
> which has the property N+N+N=NxNxN=M.
>
> Not that I expect a coherent answer, but let me ask anyway:
> why does this follow? In other words, given positive
> integers a,b,c with a^3+b^3=c^3, how do I get a positive
> integer N with N+N+N=N*N*N ?
>
> --
> Roland "Mr. Excitement" Dreier dre...@math.berkeley.edu
>
>
> Answering Roland Dreier FLT post of 1993, and why 2+2 = 2*2 as AdditiveMultiplicative Identity AMI proves FLT overall
>
> From: dre...@durban.berkeley.edu (Roland Dreier)
> Newsgroups: sci.math
> Subject: Re: 1 page proof of FLT
> Date: 18 Aug 93 14:55:02
> Organization: U.C. Berkeley Math. Department.
> Lines: 42
> Message-ID: (DREIER.93A...@durban.berkeley.edu>
> References: (CBxp0...@dartvax.dartmouth.edu>
> (24s7de$c...@outage.efi.com>
> (CByoq...@dartvax.dartmouth.edu>
> (1993Aug18.1...@Princeton.EDU>
>
> In article (1993Aug18.1...@Princeton.EDU>
> kin...@fine.princeton.edu (Kin Chung) writes:
> In article (CByoq...@dartvax.dartmouth.edu>
> Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:
> LP Hardy in Math..Apology said words to the effect that the
> LP understanding of any math proof is like pointing out a peak in the
> LP fog of a mtn range and you can only point so long and do other
> LP helps and hope the other person will see it and say Oh yes now I
> LP see it. But you can not exchange eyeballs. Again I repeat the
> LP arithmetic equivalent of FLT is that for exp2 there exists a
> LP number equal under add & multiply i.e. 2+2=2x2=4. Immediately a
> LP smallest P triple is constructible for exp2 i.e. (3,4,5>. But no
> LP number exists like 2 for exp3 or higher in order to construct P-
> LP triples for these higher exp. I am very sorry that I cannot make it
> LP any clearer than that. Time to take a break and reread Hardy Math
> LP Apology.
>
> KC You also say that a smallest P-triple is constructible for exp2
> KC immediately from the existence of a number N such that
> KC N+N=NxN, namely N=2. How do you construct a P-triple given N
> KC with this property? Please note that I am not asking how you do
> KC it for exp3, but for exp2.
>
> Before I continue, let me say that this post does not in any way constitute
> an endorsement of LvP's "proof"; what I am about to explain does not
> extend to exponent 3 in the least. However, things are rather easy for
> exponent two. (Not to be critical, but you really could have figured this
> out yourself :-)
>
> So suppose we have an N with 2xN=N+N=NxN. Set a=N+1, b=N+N=NxN.
> Then we get
> a^2 = (N+1)^2 = N^2+2xN+1 = 2xN^2+1
> also
> b^2 = (N+N)^2 = 4xN^2.
> So
> a^2+b^2 = 6xN^2+1.
> Now set c=2xN+1. Then
> c^2 = (2xN+1)^2 = 4xN^2 + 4xN + 1 = 4xN^2 + 2xN^2 + 1
> = 6xN^2+1.
> So magically a^2+b^2=c^2, just as desired! !
>
> If you can figure out how to do that for exponent 3, make yourself famous..
>
> Roland
> --
> Roland "Mr. Excitement" Dreier dre...@math.berkeley.edu
>
> Newsgroups: sci.math
> Date: Fri, 8 Mar 2019 21:32:46 -0800 (PST)
> Subject: My 1993 basis vector pure algebra proof of Regular FLT Re: AP's 2014
> proof of Generalized FLT Fermat's Last Theorem and proof of Regular FLT as a
> corollary thereof//updated 2019 for more clarity
> From: Archimedes Plutonium <plutonium....@gmail.com>
> Injection-Date: Sat, 09 Mar 2019 05:32:47 +0000
>
> My 1993 basis vector pure algebra proof of Regular FLT Re: AP's 2014 proof of Generalized FLT Fermat's Last Theorem and proof of Regular FLT as a corollary thereof//updated 2019 for more clarity
>
> Ever since 1993, I have been making proof attempts for a Pure Algebra Proof of FLT. My first attempt was the specialness of 4 with 4 = 2+2 = 2*2, arguing that exp2 has solutions due to the fact of this specialness of 4. Now if another integer had a property of N+N+N = N*N*N = N^3, the argument would go that this special number allows for a solution set in exp3. Similarly for higher exponents.
>
> To this day I maintain the logic of that is correct and is a Pure Algebra proof of Fermat's Last Theorem. For you can consider that special number as a basis vector of a exponent allowing for a solution set.
>
> But as the years rolled by, came 2014 and sort of tired of arguing the basis vector proof. I took a plunge and dive at proving Generalized FLT and did so in 2014, and lo and behold to my surprise, a easy proof of Regular FLT comes out as a corollary. Ever since 2014, I sort of bypassed or mildly forgotten the basis vector pure algebra proof.
>
> But I think I should resurrect it. The foolish math professors of my time just will never understand a basis vector proof-- the simpleton fools cannot even accept the fact they were misguided for 2 thousand years on the ellipse for it is never a conic. So can anyone expect such fools to have a sensible idea of a proof by basis vector? Of course not.
>
>
>
> My 6th published book
>
> World's First Valid Proofs of Fermat's Last Theorem, 1993 & 2014 // Math proof series, book 5 Kindle Edition
> by Archimedes Plutonium (Author) (Amazon's Kindle)
>
> Last revision was 29Apr2021. This is AP's 6th published book.
>
> Preface: Truthful proofs of Fermat's Last Theorem// including the fake Euler proof in exp3 and Wiles fake proof.
>
> Recap summary: In 1993 I proved Fermat's Last Theorem with a pure algebra proof, arguing that because of the special number 4 where 2 + 2 = 2^2 = 2*2 = 4 that this special feature of a unique number 4, allows for there to exist solutions to A^2 + B^2 = C^2. That the number 4 is a basis vector allowing more solutions to exist in exponent 2. But since there is no number with N+N+N = N*N*N that exists, there cannot be a solution in exp3 and the same argument for higher exponents. In 2014, I went and proved Generalized FLT by using "condensed rectangles". Once I had proven Generalized, then Regular FLT comes out of that proof as a simple corollary. So I had two proofs of Regular FLT, pure algebra and a corollary from Generalized FLT. Then recently in 2019, I sought to find a pure algebra proof of Generalized FLT, and I believe I accomplished that also by showing solutions to Generalized FLT also come from the special number 4 where 2 + 2 = 2^2 = 2*2 = 4. Amazing how so much math comes from the specialness of 4, where I argue that a Vector Space of multiplication provides the Generalized FLT of A^x + B^y = C^z.
>
> Cover Picture: In my own handwriting, some Generalized Fermat's Last Theorem type of equations.
>
> As for the Euler exponent 3 invalid proof and the Wiles invalid FLT, both are missing a proof of the case of all three A,B,C are evens (see in the text).
>
> Product details
> • ASIN ‏ : ‎ B07PQKGW4M
> • Publication date ‏ : ‎ March 12, 2019
> • Language ‏ : ‎ English
> • File size ‏ : ‎ 1503 KB
> • Text-to-Speech ‏ : ‎ Enabled
> • Screen Reader ‏ : ‎ Supported
> • Enhanced typesetting ‏ : ‎ Enabled
> • X-Ray ‏ : ‎ Not Enabled
> • Word Wise ‏ : ‎ Not Enabled
> • Print length ‏ : ‎ 156 pages
> • Best Sellers Rank: #4,327,817 in Kindle Store (See Top 100 in Kindle Store)
> ◦ #589 in Number Theory (Kindle Store)
> ◦ #3,085 in Number Theory (Books)