3-More to add to AP's 6th book//World's First Valid Proofs of Fermat's Last Theorem, 1993 & 2014 // Math proof series, book 5 Kindle Edition by Archimedes Plutonium (Author). A scientist, when he does a math proof or a physics theory, none of them

More to add to AP's 6th book//World's First Valid Proofs of Fermat's Last Theorem, 1993 & 2014 // Math proof series, book 5 Kindle Edition by Archimedes Plutonium (Author).

A scientist, when he does a math proof or a physics theory, none of them leave you, none leaves you alone after a while. All of them continually nag you and the nagging never goes away. Such is the case of doing science. And sometimes in this nagging a new twist enters the picture. I have found this to be the case of nearly all my science work. Every time I write something on those discoveries, it is as if a new twist is bursting to come forth.

So on FLT which I proved in early 1990s, as early as 1991, my argument was that of a Basis Vector of Algebra is the reason no exponent 3 or higher has a solution. Of course, there are ample solutions in exponent 2 and more so in exponent 1.

But the new twist that dawned on me, is that a proof of FLT, should involve exp 1 and exp 2 and then exp3 and higher, as a mathematical induction proof.

Maybe we need not start at exp 1, for that is arithmetic A + B = C. Then exp 2 is the Pythagorean Theorem. So we have two starting true cases of the General FLT. For exp 2 we have the basis vector 2+2 = 2x2 =4, where we have a number that is equal under add and multiply. Now for exponent 1 we could say the basis vector is all of Arithmetic. Now for exponent 3, we can have no n+n+n = nxnxn = m, same for higher exponents.

So what I missed in my book was to emphatically suggest that a proof of FLT has to fully incorporate the exponents that do have solutions. Every mathematician before AP , looks at FLT in isolation of exponent 2, and by doing so, cut off their chances of finding a valid proof of FLT. Because the moment your mind asks the question, why no solutions in exp 3 but myriad solutions in exp 2, forces the mind to think that the valid proof has to incorporate in its proof, a mechanism, a mechanism the spans and bridges between exponent 2 and exponent 3, fully incorporate the picture that exp 2 has solutions not exp 3. And that then puts the onus of the mind to look at a Basis Vector where add is the very same as multiply. So that solutions are metaphorically analogous to building concrete block buildings and the concrete blocks are the basis vector.

Every Pythagorean theorem solution in Natural Counting Numbers has its basic building block of 2 and 4, of 2+2= 2x2= 4. You can analyze every P-triple and find it is constructed of 2 and 4. Whereas every exp 3 is wanting a building block for all possible solutions, yet no numbers (not even 0 for the n and m have to be different) have the ability to be n+n+n = nxnxn = m.

So I need to emphatically state in my 6th published book, that a proof of FLT, or even Generalized FLT should look at all exponents and not isolate-out exp2 from the higher exponents.

That is extremely important point of logic, that we tend to shove off to the side and want to focus all our attention on just a part of the puzzle, a part of the problem, separate from the larger problem. We tend to separate, when we should look at the big picture to give us guidance and clues as to the mechanism of the proof.

So, actually, FLT was even absurdly more simple as a math problem and proof than most every other math proof in recorded history. FLT is more simple to prove than even the Pythagorean theorem is to prove. Because this is a proof of FLT. Proof: 2+2= 2x2= 4 allows us to build solutions in exp 2, but there does not exist a n+n+n = nxnxn = m so no solutions ever in exp 3 and the same argument for exp 4 and higher. QED

Totally simple proof is FLT, and if mathematicians had asked, what, ultimately what allows solutions in exp2 and said, well, well, 2+2=2x2 is the building block of all solutions in exp2.

No, my proofs in math and my theories in science and physics will never leave me alone, even if I tried. I can picture myself at my deathbed, and even there, one of my science theories will invade my mind as a die. Such, is the nature of a world of superdeterminism in an Atom Totality.

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)

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

Preface:
Real 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).
Length: 156 pages
File Size: 1503 KB
Print Length: 156 pages
Publication Date: March 12, 2019
Sold by: Amazon Digital Services LLC
Language: English
ASIN: B07PQKGW4M
Text-to-Speech: Enabled 
X-Ray: 
Not Enabled 
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Lending: Enabled
Enhanced Typesetting: Enabled 

Archimedes Plutonium
Jul 7, 2021, 12:01 PM
to sci.math
Now everyone is free to chose who they want to believe, do you want to believe Andrew Wiles with his 100 pages or more of math that is everything including the kitchen sink of mathematics thrown at the Fermat's Last Theorem FLT ? Where most people cannot even understand the 1st page-- what the hell is going on. Or, do you want to chose AP's proof of FLT where he proves it in a sentence that everyone in the entire world, even in Grade School can understand, that 2+2 = 2x2 = 4 gives solutions to Pythagorean theorem and A^2 + B^2 = C^2, but if you want solutions for A^3+B^3 = C^3 or higher, you need a special number of n+n+n = nxnxn = m for n and m in exponent 3, yet there exists no such special numbers n and m to satisfy that, hence, FLT.

So, take your pick, do you believe in B.S. of Wiles with his obnoxious over 100 pages of cluttered together phony baloney mess argument. Wherein Andrew Wiles was so stupid on FLT, he failed to even notice that Euler had **no proof** in exponent 3 of FLT because Euler forgot he had to prove the case of where A,B, C in A^3 +B^3= C^3 were even numbers. Euler forgot he had to prove that; and instead assumed there was no three even Counting numbers were no solution. But Andrew Wiles, the math failure he is, never even noticed that Euler had no proof in exponent 3. So, do you believe in a Andrew Wiles 100 page "hornswaggle mess" of elliptic curve argument. Or do you believe in AP when he says the reason 3^2 + 4^2 = 5^2 is because 2+2 = 2x2 = 4, the only two counting numbers with that feature of addition is the same as multiplication.

Now Andrew Wiles was looking for a proof of FLT in early 1990s, as early as 1993 when AP notified the world public that AP had already proven FLT, for I proved it in 1991, but Andrew Wiles had no proof of FLT, even after 1993..

And there was a exciting exchange of ideas from AP and from Princeton Univ and Berkeley where Roland Dreier gives the SUPPORTING ARGUMENT, that the AP proof of FLT is the world's only valid proof of FLT. Although Roland was not prepared to go that far, it is obvious, these almost 30 years later, that AP had the proof, but Wiles is a con-artist failure of FLT.

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