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1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

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  1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-20 13:38 -0700
    Generalized FLT Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-20 15:15 -0700
      Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-09-20 22:57 +0000
      examine this algebra principle that does not preserve whole-numbership Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-20 20:02 -0700
        a little uneasy about pure Algebra proof, but nothing to worry Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-21 15:25 -0700
          Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-21 18:24 -0700
            Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-21 21:44 -0700
              Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-21 23:05 -0700
                Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-21 23:26 -0700
                  proof principle, in sqrt[A^3 +B^3] = N, A,B impossible to be whole numbers Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-22 00:04 -0700
                    Lemma proof: sqrt[A^3 + B^3] = N where N is a whole number then impossible for A,B to be whole numbers Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-22 00:34 -0700
                      is the Cubic Formula the proof of FLT Re: Lemma proof: sqrt[A^3 + B^3] = N where N is a whole number then impossible for A,B to be whole numbers Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-22 01:13 -0700
                        cheers, the proof at last-- usually I find them during my bath, now I found this one before my bath Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-22 02:08 -0700
                          cheers, the proof at last-- usually I find them during my bath, now I found this one before my bath Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-22 02:36 -0700
                            Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-09-22 17:34 +0000
                Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT qbwrfmlx@gmail.com - 2018-09-22 00:50 -0700
                  Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-22 01:00 -0700
          Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-09-22 02:34 +0000
      Re: Ghost figures of geometry in Generalized FLT Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-20 23:00 -0700
    Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT qbwrfmlx@gmail.com - 2018-09-20 16:49 -0700
      Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-20 19:09 -0700
        Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT eastside.erik@gmail.com - 2018-09-20 21:49 -0700
          nerds that never belonged in sci.math Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-20 22:07 -0700

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#423940 — 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

Alright, now I can do this in complete satisfaction. And, I shall do it with ARRAY PROOFS. What is an Array Proof of Mathematics, you may well ask. An array proof is a proof stripped of all side talk, side chatter. An array proof is where the author expects the reader to be so familiar with the problem at hand, that all that is shown is the proof mechanism. A proof stripped of everything but the fine line necessity.

ARRAY PROOFS OF FLT of AP's 3 proofs:

(1)

FLT pure Algebra proof STATEMENT: whole numbers A,B,C, there never exists A,B,C and z, such that A^z + B^z = C^z where z is a whole number 3 or larger.
PROOF: Given any whole numbers M and N,and x where x is 3 or greater, M^x = N^x-1 has only two solutions 0 and 1. 
QED

EXAMPLE OF THE PROOF TO MAKE CLEAR: Numbers example: 3^2 + 4^2 = 5^2 in exp2. Pick any arbitrary number in exp3, say we pick 5^3, then is there whole numbers A,B such that  A^3 + B^3 = 5^3.  Add 3^2 +4^2 = 5^2 to that of equation A^3 + B^3 = 5^3, and we have A^3 + B^3 + 9 + 16 = 125 + 5^2
that becomes

A^3 + B^3 -100 = 25
that becomes
sqrt[A^3 + B^3 - 100] = 5

Now, apply the Algebra principle M^x = N^x-1 has only two solutions 0 and 1. That entails that A and B are not whole numbers for only 0 and 1 can survive square root of A^3 and B^3. And since C^3 was arbitrary picked applies to all C^3 and since the rule applies to all exponents 3 or larger, thus FLT pure algebra.


(2)

FLT pure GEOMETRY proof STATEMENT: Given a perfect cube C, there __never exists__ two smaller perfect cubes A,B such that A + B = C, and where, we bridge into Analytical Geometry that cube A, cube B, cube C cover all exponents of whole numbers starting with 3.
PROOF: Given any whole numbers M and N,and x where x is 3 or greater, M^x = N^x-1 has only two solutions 0 and 1. 
QED

EXAMPLE OF THE PROOF TO MAKE CLEAR: We pick a arbitrary Cube C and we dissect it by a perpendicular slice along one of the axes of the cube retrieving two rectangular boxes out of the slice.

So we have boxes A' and B' and all we need do is convert those rectangular boxes into two smaller cubes from the C cube. The volume of C cube equals the volume of box A' plus, box B'.

I arbitrary pick the cube 5*5*5 = 5^3 = 125

Now, here again I borrow from exp2

I borrow the fact that exp2, we have 3^2 + 4^2 = 5^2 would end up as a perpendicular slice along axis yielding two rectangles of 5*1.8 + 5*3.2 = 5*5 

C*k  + C*j  = C*C 

square-root[C*k + C*j] = C 
substituting 
square-root[5*1.8 + 5*3.2] = 5 

Let me go back to 
[C*k + C*j] = C*C 

Now if I multiply that exp2 with another C we have 

CCk + CCj = CCC 

And the C is 5 

And the CC1.8 with CC3.2 are two complementary rectangular boxes 

Now substituting in C= 5 

we have 25*1.8 + 25*3.2 = 125 

45 + 80 = 125 

So here, again, like in my pure algebra proof I added a solution set from exp2 onto exp3 and found an impossibility. Here, I find the impossibility to be that converting 45 and 80 to cubes are not whole numbers. There is an equality of the equation that demands one side equals the other side. And there is the fact that exp2 had WHOLE NUMBERS as Solution. But when you mix the square root of a cube A^3 or B^3, you lose on the maintaining of A and B as whole numbers.

Alright, the mechanism that prevents whole number solutions. When the PPT has a whole number solution, then Algebra allows us to multiply every term by C and still be equal. And the solution of whole numbers was BASED in exp2, now we multiply by C every term, and we have elevated into FLT, for we transitioned from PPT with a whole number solution and now is based in FLT. If you have a whole number solution in exp2, you cannot have a whole number solution in exp3, otherwise you have the algebra of M^3 = N^2 and that only exists for 1 and 0. 

The algebra proof was add exp2 to both sides. The geometry proof is multiply every term of a PPT solution by C and then destroy the chance of a solution set in exp3 because you have a solution set in exp2. 

Now a question quickly arises-- for every cube, C*C*C is there a PPT or PT such that the hypotenuse length = CCC? For example I have 44,117,125 and then 5*5*5 = 125. We can skip 2^3, then jump to 3^3 = 27, and I guess 5,12,13 multiply by 27/13 will do. 

And also, for exp4, exp5 and all higher exponents, those are also CUBES, when we do GHOST FIGURE GEOMETRY. So that something like A^4 + B^4 = C^4 becomes where A^4 is A^2*A^2 and that is two wire squares spaced a distance A apart forming a Ghost Cube. All exponents larger than 3 are Ghost Cubes and follow the same proof mechanism.

Summary of Geometry Proof of FLT:: Give me any cube C^3 in A^3 + B^3 = C^3. I immediately look for a C in A^2 + B^2 = C^2 of Pythagorean Theorem. I then see that multiplying the Pythagorean Theorem solution by another C, to elevate it into exp3 should be another solution, and it is, but because it was a whole number solution in exp2, it can not be a whole number solution in exp3. 


Next time I do the Generalized FLT

Very crude dot picture of 5f6, 94TH 
ELECTRON=muon DOT CLOUD of 231Pu 


                ::\ ::|:: /:: 
                 ::\::|::/:: 
                     _ _ 
                    (:Y:) 
                     - - 
                 ::/::|::\:: 
                ::/ ::|:: \:: 
One of those dots is the Milky Way galaxy. And each dot represents another galaxy. 
            . \ .  . | .   /. 
           . . \. . .|. . /. . 
              ..\....|.../... 
               ::\:::|::/:: 
---------------      ------------- 
--------------- (Y) ------------- 
---------------      -------------- 
               ::/:::|::\:: 
              ../....|...\... 
           . . /. . .|. . \. . 
            . / .  . | .   \ . 

  
http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      

Read my recent posts in peace and quiet.

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
Archimedes Plutonium 

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#423949 — Generalized FLT Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

Alright, I need to incorporate all three proofs of FLT, the pure Algebra proof, the pure Geometry proof and then the Generalized FLT that delivers FLT as a corollary side proof.

Now the reason I am spending so much time on this, is because I want to see for myself the LOGICAL Flow where the mechanism that proves FLT in pure algebra is the same mechanism that proves it in pure geometry, but in Generalized FLT we use the rectangles that were used as Rectangular Boxes in pure geometry proof. And the very same algebra mechanism that proves pure algebra, pure geometry, is also used by Generalized FLT.

I want to fully reveal how all three proofs are interconnected. There are threads running through all three proofs, as there should be.

        (3) GENERALIZED FLT proof

What is Generalized FLT? Glad you asked, for it is: 

A^x + B^y = C^z where A,B,C,x,y,z are positive integers with x,y,z greater than 2 

FLT is of course 

A^n + B^n = C^n where A,B,C, n are positive integers and n is greater than 2 

Newsgroups: sci.math 
Date: Thu, 9 Jun 2016 13:38:48 -0700 (PDT) 

Subject: Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed 
From: Archimedes Plutonium <plutonium....@gmail.com> 
Injection-Date: Thu, 09 Jun 2016 20:38:49 +0000 

The history of this proof starts 1993 when I became interested in proving FLT and started with 2+2 = 2x2 = 2^2, where I noted that addition is the same as multiplication for 2 and 4 as a building block to build all solution triples. The reason exp2 has whole number solutions is because of 2 and 4, and exp3 has nothing like a building block, hence, no solutions.

But by 2014, I proved Generalized FLT and saw FLT as a corollary. The trick I used in 2014, was Condensed Rectangles-- rectangles formed from the factors of numbers, so that 8 would be a condensed rectangle of 2*4, or 12 a condensed rectangle of 3*4, or it could be 2*6.
  

PROOFS of Generalized FLT & FERMAT's LAST THEOREM as corollary 

Editing and commentary corner: Once Generalized FLT and FLT are turned into a geometrical perspective-- condensed rectangles, then the proof is as easy as making homemade cherry pie. 

Summary of text: Both proofs of Generalized FLT and FLT are based on a fact of geometry, that you can represent a number with its cofactors as the sides of a rectangle. And to prove either Generalized FLT or FLT is a simple matter of stacking two rectangles that have equal sides, A and B to produce a third rectangle C which has a side equalling the _shared side_ of A and B. 

So now, here, reflect back to the Pure Geometry proof of FLT where we have a Cube and we put it through a Perpendicular Slicer on a axis of the cube and we end up with two smaller rectangular boxes.

But back to Generalized FLT:

We see the relationship between a Theorem and its Corollary very easily from Generalized FLT and FLT, in that we have all these cases to worry about. The history of FLT was a case study, prove it in the case of exponent this and that, but never a proof of all of FLT, and that is because corollaries are never proven as stand alone theorems and must have their theorem (Generalized FLT) proven first. 

________________________ 
DETAILED PROOF OF Generalized FLT 
________________________ 

It is a constructive proof. 

We make the table of all the numbers possible in the Generalized FLT Conjecture as the conglomeration of exponents of 3 or larger as this set: 

{1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .} 

Here we have conglomerated exp3 and exp4 and exp5 etc etc into one set. 

We know Generalized FLT has solutions of A+B=C in that set for here are three examples: 

2^3 + 2^3 = 2^4 with prime divisor 2 
3^3 + 6^3 = 3^5 with prime divisor 3 
7^3 + 7^4 = 14^3 with prime divisor 7 

What we need to prove is that all solutions have a prime divisor in common, ie all three rectangles of A, B, C so that A+B=C, have one shared side equal to one another. 

Definition of Condensed-Rectangle: given any number in the set of conglomerated exponents, we construct rectangles of that number from its unit squares whose sides are cofactors of the number. For instance, rectangle of 216 units as either 12x18 units, or 9x24 units, or 6x36 units or 3x72 units, or 2x108, but never a 1x216 units. We exclude 1 times the number as a condensed rectangle. So a condensed-rectangle is one in which it is composed of cofactors of the number in question, except for 1, and the number itself for 1x216 units is not a condensed-rectangle. 

Now for the constructive proof that Generalized FLT solutions must have a common prime divisor. 

We stack Condensed-Rectangles of the number-space that Generalized FLT conjecture applies: 

Solution Number Space for Generalized FLT now becomes these condensed-rectangles: 

{ 2*4, 2*8, 4*4, 3*9, 2*16, 4*8, 2*32, 4*16, 8*8,. . . }   

We convert each of those numbers into Condensed-Rectangle, except 1 of course, and where many numbers have several condensed rectangles so the Solution Space of Numbers increases by a large amount. If an A and B as condensed-rectangles have the same side such as 3x9 units and 9x24units 
wherein you stack them on their shared side of 9 and which matches another number of its condensed-rectangle such as 9x27 units, then you have a Generalized FLT solution of A+B=C. For if we were to take the 9 by 27 condensed rectangle it decomposes into 3x9 and 9x24. 

Now, the question is, are all A+B equal to a C, form stacked condensed rectangles that share a common side? 

All stackable condensed-rectangles must have one side the same for the two rectangles to stack, in the case above it is the side 9 with its common divisor of the prime 3. 

If any other solution to Generalized FLT had A stacked upon B without a common side between them, then the figure formed cannot be a overall new rectangle but something that looks like this: 

HHHHHH 
HHHHHH 
HHHHHHHHHH 
HHHHHHHHHH 

That is not a condensed rectangle and all the numbers of the Solution Space, except the number 1, are condensed rectangles. That above figure is 6-sided figure. That is a 6-sided figure, yet a rectangle is only a 4 sided figure. That is not a Condensed Rectangle, represented by its cofactors. So that equality can not be achieved by any other stacking than condensed rectangles equalling condensed rectangles. 

Then the question is, can we have an A and B with a shared common side equal to a C that is a condensed rectangle with no common side shared with A and B? 

The question is, how can I be sure that all the A, B, C such that A + B = C have a common prime divisor? 

As if the question is asking whether the Condensed Rectangles covered the question by forbidding any equality unless there is a common shared side in A in B and in C. 

Well, the answer is easily enough covered for the Condensed Rectangles eliminates any possibility of A,B,C where A+B = C and not have a shared side by all three of A, B, C. It does this by the Distributive law of integers. 

All the numbers in the solution space of Generalized FLT have condensed rectangles, except the number 1. 

All the A, B, C are written as condensed rectangles of (s*t) + (r*p) = (u*v). 

For example (9*3) + (9*24) = (9*27) which is 3^3 + 6^3 = 3^5  which is 27 + 216 = 243. 

In order for (s*t) + (r*p) = (u*v) then the s, r, u have to be equal sides. Have to be (9*t) + (9*p) = (9*v) in the example of 27 + 216 = 243. 

Distributive Law (9*3)/9 + (9*24)/9 = (9*27)/9 where 3 + 24 = 27 

So, give me three numbers A, B, C, chosen arbitrarily from the Solution Space of Generalized FLT {1, 8, 16, 27, 32, 64, 81, 125, 128, 216,  243, 256, . .} except of course 1, and can those three chosen arbitrarily A, B, C obey A + B = C? Only if they obey the Distributive Law can the A + B equal to the C when we write the A, B, C as condensed rectangles. 

QED 


_________________________________________ 
Detailed Proof of FLT using condensed-rectangles 
_________________________________________ 

It is a construction proof method for we show that it is impossible to construct A+B = C inside of a specific exponent. 

Fermat's Last Theorem FLT conjecture says there are no solutions to the equation a^y + b^y = c^y where a,b,c,y are positive integers and y is greater than 2. 

The number Space that governs FLT is this: 

exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .} 

exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .} 

exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .} 

exp6 ..... 
. 
. 
. 
. 

So in FLT we ask whether there are any triples, A,B,C in any one of those _specific exponents_ such that A+B=C. In FLT, our solution space is only one particular exponent such as 3 or 4, or 5 to hunt down and find a A,B,C to satisfy A+B=C. 

In the proof we use Condensed-Rectangles which is defined as a rectangle composed of unit squares of the cofactors of a number, except for 1 x number itself. So the number 27 in exp3 has Condensed Rectangles of 3x9 only. The number 125 in exp3 has condensed rectangles of 5x25 only, and the number 81 in exp4 has condensed rectangles of 3x27 and 9x9. 

Now in the proof of Generalized FLT solutions of A + B = C require a common divisor. 

2^3 + 2^3 = 2^4 with prime divisor 2 
3^3 + 6^3 = 3^5 with prime divisor 3 
7^3 + 7^4 = 14^3 with prime divisor 7 

Here in FLT, we need not even worry about exponents because all solutions to Generalized FLT encompass exponents so that if these were FLT solutions: 

2^? + 2^? = 2^? with prime divisor 2 
3^? + 6^? = 3^? with prime divisor 3 
7^? + 7^? = 14^? with prime divisor 7 

So in that construction of a solution to FLT there is a common divisor and so now we divide the equation by the common divisor, and we get this: 

1^y + 1^y = 1^y 
1^z + 2^z = 1^z 
1^w + 1^w = 2^w 

And those are impossible constructions. So the proof of FLT requires Generalized FLT be proven first in order to display that a Generalized FLT proof cannot co-exist unless FLT has no solutions. 

Stated another way:: should FLT have a whole number set solution of A,B,C, it contradicts Generalized FLT for you can reduce the FLT to that of something as 1^3 + 1^3 = 1^3

QED 

Now in upcoming posts I want to reduce Generalized FLT into a ARRAY proof and I want to comment on the similarities between pure algebra proof, pure geometry proof and generalized FLT proof.


Very crude dot picture of 5f6, 94TH 
ELECTRON DOT CLOUD of 231Pu 


                ::\ ::|:: /:: 
                 ::\::|::/:: 
                     _ _ 
                    (:Y:) 
                     - - 
                 ::/::|::\:: 
                ::/ ::|:: \:: 

        One of those dots is the Milky Way galaxy. And 
each dot represents another galaxy. 


            . \ .  . | .   /. 
           . . \. . .|. . /. . 
              ..\....|.../... 
               ::\:::|::/:: 
---------------      ------------- 
--------------- (Y) ------------- 
---------------      -------------- 
               ::/:::|::\:: 
              ../....|...\... 
           . . /. . .|. . \. . 
            . / .  . | .   \ . 

  
http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
Archimedes Plutonium

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#423950 — Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test

Math Failure Archimedes Plutonium <plutonium.archimedes@gmail.com> fails:

>AP writes: say there Moroney how is your Gay Bed and Breakfast in Boston 
>going for you? Is there showering included in the price? And is it 
>wheelchair accessible

What's this, Plutonium? You are looking for a gay bed and breakfast in
Boston? One with "group" showers and is wheelchair accessible for your
crippled mind?  When are you visiting Boston? I don't know anything that
can help you, as I told you that's not for me, but Boston does have a
large gay community so you should be very happy there. Enjoy your visit!

Oh, here's a spare wheelchair for your crippled mind.

🧠
♿

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#423959 — examine this algebra principle that does not preserve whole-numbership Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

Review Commentary:

In Pure Algebra proof of FLT-- we take a known solution set A,B,C of A^2 + B^2 = C^2 such as 3,4,5

9 + 16 = C*C

We take an arbitrary number in C^3 Space such as for example 5*5*5 = 5^3 = 125

We ask, why is there no A^3 and B^3 that equals 125 ?

A^3 + B^3 = C*C*C = 125

We add 9+16 to both sides of equation

(A^3 + B^3) + 9 + 16 = 125 + 25

subtract 125 from both sides

(A^3 + B^3) -100 =  25

take the square root of both sides

sqrt[(A^3 + B^3) -100] = 5

Inspect the one side sqrt[(A^3 + B^3) -100]
one possible solution is A^3 = 25 while B^3 =100 meaning A= 2.924... B = 4.641...
And any and all combinations of A, B in exp3 are never whole numbers because 3,4,5 P Triples were whole numbers.

In Pure Geometry, staying with 3,4,5 as solutions in exp2, we pick an arbitrary exp3 in FLT number as 125= cube 5^3. We perpendicular Slice into the cube along an axis and produce two rectangular boxes of C*C*k and C*C*j where k+j = 5. We go back to exp2 with its C equaling the C in CCC. We do a perpendicular slice into the square 5*5 and we produce the two rectangles 5*1.8 and 5*3.2

So we have for exp2 in pure Geometry

C*k + C*j = C*C
that is a solution in Whole Numbers in exp2
We multiply every term by C, catapulting the exp2 equation to be a exp3 equation
C*C*k + C*C*j = C*C*C

replacing the C=5, k = 1.8, j= 3.2 we have

25*1.8 + 25*3.2 = 125

Here we have A^3 = 45 and B^3 = 80, thus A = 3.556.... and B = 4.308....
Explanation: the square of numbers in exp2 as a solution of whole numbers is not preserved by the cube root, for it turns those whole number solutions into fractions.

Very crude dot picture of 5f6, 94TH 
ELECTRON=muon DOT CLOUD of 231Pu 


                ::\ ::|:: /:: 
                 ::\::|::/:: 
                     _ _ 
                    (:Y:) 
                     - - 
                 ::/::|::\:: 
                ::/ ::|:: \:: 
One of those dots is the Milky Way galaxy. And each dot represents another galaxy. 
            . \ .  . | .   /. 
           . . \. . .|. . /. . 
              ..\....|.../... 
               ::\:::|::/:: 
---------------      ------------- 
--------------- (Y) ------------- 
---------------      -------------- 
               ::/:::|::\:: 
              ../....|...\... 
           . . /. . .|. . \. . 
            . / .  . | .   \ . 

  
http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      

Read my recent posts in peace and quiet.

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
Archimedes Plutonium

[toc] | [prev] | [next] | [standalone]

#424001 — a little uneasy about pure Algebra proof, but nothing to worry Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

On Thursday, September 20, 2018 at 10:03:03 PM UTC-5, Archimedes Plutonium wrote:
> Review Commentary:
> 

I am a bit uneasy about this proof, in comparison to the pure Geometry proof where we multiply all terms in the equation to get the set up of a sqrtA^3 cannot be a whole number. In the multiply proof, the sqrtA^3 seems more clear.


> In Pure Algebra proof of FLT-- we take a known solution set A,B,C of A^2 + B^2 = C^2 such as 3,4,5
> 
> 9 + 16 = C*C
> 
> We take an arbitrary number in C^3 Space such as for example 5*5*5 = 5^3 = 125
> 
> We ask, why is there no A^3 and B^3 that equals 125 ?
> 
> A^3 + B^3 = C*C*C = 125
> 
> We add 9+16 to both sides of equation
> 
> (A^3 + B^3) + 9 + 16 = 125 + 25
> 
> subtract 125 from both sides
> 
> (A^3 + B^3) -100 =  25
> 
> take the square root of both sides
> 
> sqrt[(A^3 + B^3) -100] = 5
> 
> Inspect the one side sqrt[(A^3 + B^3) -100]
> one possible solution is A^3 = 25 while B^3 =100 meaning A= 2.924... B = 4.641...
> And any and all combinations of A, B in exp3 are never whole numbers because 3,4,5 P Triples were whole numbers.
> 

In the inspection stage of the proof, it is clear that I have a range if numbers to inspect, because we have to get a sqrt25 on the left side. This is why it is important to select a known PPT that will make the range as small as possible, for if I had selected 11,60,61 I would have a mountain of whole numbers to have to eliminate to say that A and B are fractions.

By choosing 3,4,5, the range of values of whole numbers for both A and B can only be 1,2,3,4,5 for if A=1, then the best B can be is 5, if A is 2 the best B can be is 5 again, if A is 3, the best B can be is either 4 or 5, if A is 4 then best B can be is 4

That means the A and B cannot be whole numbers.

So, is there any guidance on what PPT to pick given any arbitrary picked exp3 number? Suppose we selected 1000 as our exp3 number. Then what PPT numbers in exp2 would we use in order to narrow the range of whole number solutions? Would it be the P Triple of 6, 8, 10 giving us 36, 64 and 100, so that we subtract 1000 from both sides and end up with A^3 + B^3 - 900 = 100 causing our range of whole number solutions to be 1 to 10. Well that is no big hardship going through ten numbers. In this manner, we see that 3,4,5 can be used in all arbitrary picked numbers of exp3

> In Pure Geometry, staying with 3,4,5 as solutions in exp2, we pick an arbitrary exp3 in FLT number as 125= cube 5^3. We perpendicular Slice into the cube along an axis and produce two rectangular boxes of C*C*k and C*C*j where k+j = 5. We go back to exp2 with its C equaling the C in CCC. We do a perpendicular slice into the square 5*5 and we produce the two rectangles 5*1.8 and 5*3.2
> 
> So we have for exp2 in pure Geometry
> 
> C*k + C*j = C*C
> that is a solution in Whole Numbers in exp2
> We multiply every term by C, catapulting the exp2 equation to be a exp3 equation
> C*C*k + C*C*j = C*C*C
> 
> replacing the C=5, k = 1.8, j= 3.2 we have
> 
> 25*1.8 + 25*3.2 = 125
> 
> Here we have A^3 = 45 and B^3 = 80, thus A = 3.556.... and B = 4.308....
> Explanation: the square of numbers in exp2 as a solution of whole numbers is not preserved by the cube root, for it turns those whole number solutions into fractions.
> 
> Very crude dot picture of 5f6, 94TH 
> ELECTRON=muon DOT CLOUD of 231Pu 
> 
> 
>                 ::\ ::|:: /:: 
>                  ::\::|::/:: 
>                      _ _ 
>                     (:Y:) 
>                      - - 
>                  ::/::|::\:: 
>                 ::/ ::|:: \:: 
> One of those dots is the Milky Way galaxy. And each dot represents another galaxy. 
>             . \ .  . | .   /. 
>            . . \. . .|. . /. . 
>               ..\....|.../... 
>                ::\:::|::/:: 
> ---------------      ------------- 
> --------------- (Y) ------------- 
> ---------------      -------------- 
>                ::/:::|::\:: 
>               ../....|...\... 
>            . . /. . .|. . \. . 
>             . / .  . | .   \ . 
> 
>   
> http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 
> 
> I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      
> 
> Read my recent posts in peace and quiet.
> 
> https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
> Archimedes Plutonium

[toc] | [prev] | [next] | [standalone]

#424004 — Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

The title spells it all out, and what my chore is, is express that idea as best as possible. For example 8 is 2^3, but if you wanted to take the sqrt of 8, it is never a whole number. So, in all FLT proofs, that idea is somewhere within the text of the proof. It is even in the text of Generalized FLT when proving the corollary of FLT-- and it appears in the reduction of the "claimed solution triple" as being divided by a prime factor, for example 2^3 + 2^3 = 2^4, so when k= 1/2 we have 1^3 + 1^3 = 1^4-- because you supposed a solution set is the supposing there was a whole number in doing the cube root of 2^4.

The Governing-Principle of FLT, is the fact that sqrt of a cube is never a whole number except 0 and 1. The sqrt of a fifth power, or the cube root of a 4th power, the 5th root of a exp6, in all cases, if you have a whole number in an exponent power, it no longer remains a whole number if you take the root of it in a smaller power, coprime.

Does Math need a proof of this fact itself? The fact that integer M, if M^k is whole number where k is integer greater than 2, then M^k to the root of k-1 is never a whole number. That by taking a smaller root of a whole number, you destroy the whole number solution. Does math need a proof of that principle? I think someone long time ago proved this principle. Was it Pierre Fermat in 1637? And when he penned his FLT, was he thinking of sqrtA^3 is never a whole number? I would guess yes, for certainly, anyone who studies mathematics long enough realizes that sqrt of A^3 whole number is never going to be a whole number itself unless 0 or 1.

On Friday, September 21, 2018 at 5:25:09 PM UTC-5, Archimedes Plutonium wrote:
> On Thursday, September 20, 2018 at 10:03:03 PM UTC-5, Archimedes Plutonium wrote:
> > Review Commentary:
> > 
> 
> I am a bit uneasy about this proof, in comparison to the pure Geometry proof where we multiply all terms in the equation to get the set up of a sqrtA^3 cannot be a whole number. In the multiply proof, the sqrtA^3 seems more clear.
> 
> 
> > In Pure Algebra proof of FLT-- we take a known solution set A,B,C of A^2 + B^2 = C^2 such as 3,4,5
> > 
> > 9 + 16 = C*C
> > 
> > We take an arbitrary number in C^3 Space such as for example 5*5*5 = 5^3 = 125
> > 
> > We ask, why is there no A^3 and B^3 that equals 125 ?
> > 
> > A^3 + B^3 = C*C*C = 125
> > 
> > We add 9+16 to both sides of equation
> > 
> > (A^3 + B^3) + 9 + 16 = 125 + 25
> > 
> > subtract 125 from both sides
> > 
> > (A^3 + B^3) -100 =  25
> > 
> > take the square root of both sides
> > 
> > sqrt[(A^3 + B^3) -100] = 5
> > 
> > Inspect the one side sqrt[(A^3 + B^3) -100]
> > one possible solution is A^3 = 25 while B^3 =100 meaning A= 2.924... B = 4.641...
> > And any and all combinations of A, B in exp3 are never whole numbers because 3,4,5 P Triples were whole numbers.
> > 
> 
> In the inspection stage of the proof, it is clear that I have a range if numbers to inspect, because we have to get a sqrt25 on the left side. This is why it is important to select a known PPT that will make the range as small as possible, for if I had selected 11,60,61 I would have a mountain of whole numbers to have to eliminate to say that A and B are fractions.
> 
> By choosing 3,4,5, the range of values of whole numbers for both A and B can only be 1,2,3,4,5 for if A=1, then the best B can be is 5, if A is 2 the best B can be is 5 again, if A is 3, the best B can be is either 4 or 5, if A is 4 then best B can be is 4
> 
> That means the A and B cannot be whole numbers.
> 
> So, is there any guidance on what PPT to pick given any arbitrary picked exp3 number? Suppose we selected 1000 as our exp3 number. Then what PPT numbers in exp2 would we use in order to narrow the range of whole number solutions? Would it be the P Triple of 6, 8, 10 giving us 36, 64 and 100, so that we subtract 1000 from both sides and end up with A^3 + B^3 - 900 = 100 causing our range of whole number solutions to be 1 to 10. Well that is no big hardship going through ten numbers. In this manner, we see that 3,4,5 can be used in all arbitrary picked numbers of exp3
> 
> > In Pure Geometry, staying with 3,4,5 as solutions in exp2, we pick an arbitrary exp3 in FLT number as 125= cube 5^3. We perpendicular Slice into the cube along an axis and produce two rectangular boxes of C*C*k and C*C*j where k+j = 5. We go back to exp2 with its C equaling the C in CCC. We do a perpendicular slice into the square 5*5 and we produce the two rectangles 5*1.8 and 5*3.2
> > 
> > So we have for exp2 in pure Geometry
> > 
> > C*k + C*j = C*C
> > that is a solution in Whole Numbers in exp2
> > We multiply every term by C, catapulting the exp2 equation to be a exp3 equation
> > C*C*k + C*C*j = C*C*C
> > 
> > replacing the C=5, k = 1.8, j= 3.2 we have
> > 
> > 25*1.8 + 25*3.2 = 125
> > 
> > Here we have A^3 = 45 and B^3 = 80, thus A = 3.556.... and B = 4.308....
> > Explanation: the square of numbers in exp2 as a solution of whole numbers is not preserved by the cube root, for it turns those whole number solutions into fractions.
> > 
> > Very crude dot picture of 5f6, 94TH 
> > ELECTRON=muon DOT CLOUD of 231Pu 
> > 
> > 
> >                 ::\ ::|:: /:: 
> >                  ::\::|::/:: 
> >                      _ _ 
> >                     (:Y:) 
> >                      - - 
> >                  ::/::|::\:: 
> >                 ::/ ::|:: \:: 
> > One of those dots is the Milky Way galaxy. And each dot represents another galaxy. 
> >             . \ .  . | .   /. 
> >            . . \. . .|. . /. . 
> >               ..\....|.../... 
> >                ::\:::|::/:: 
> > ---------------      ------------- 
> > --------------- (Y) ------------- 
> > ---------------      -------------- 
> >                ::/:::|::\:: 
> >               ../....|...\... 
> >            . . /. . .|. . \. . 
> >             . / .  . | .   \ . 
> > 
> >   
> > http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 
> > 
> > I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      
> > 
> > Read my recent posts in peace and quiet.
> > 
> > https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
> > Archimedes Plutonium

[toc] | [prev] | [next] | [standalone]

#424017 — Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

It would be instructive to do one's own proof that only the Whole Numbers of 0 and 1 can solve for sqrtA^3 when A is a whole number.

AP


On Friday, September 21, 2018 at 8:24:32 PM UTC-5, Archimedes Plutonium wrote:
> The title spells it all out, and what my chore is, is express that idea as best as possible. For example 8 is 2^3, but if you wanted to take the sqrt of 8, it is never a whole number. So, in all FLT proofs, that idea is somewhere within the text of the proof. It is even in the text of Generalized FLT when proving the corollary of FLT-- and it appears in the reduction of the "claimed solution triple" as being divided by a prime factor, for example 2^3 + 2^3 = 2^4, so when k= 1/2 we have 1^3 + 1^3 = 1^4-- because you supposed a solution set is the supposing there was a whole number in doing the cube root of 2^4.
> 
> The Governing-Principle of FLT, is the fact that sqrt of a cube is never a whole number except 0 and 1. The sqrt of a fifth power, or the cube root of a 4th power, the 5th root of a exp6, in all cases, if you have a whole number in an exponent power, it no longer remains a whole number if you take the root of it in a smaller power, coprime.
> 
> Does Math need a proof of this fact itself? The fact that integer M, if M^k is whole number where k is integer greater than 2, then M^k to the root of k-1 is never a whole number. That by taking a smaller root of a whole number, you destroy the whole number solution. Does math need a proof of that principle? I think someone long time ago proved this principle. Was it Pierre Fermat in 1637? And when he penned his FLT, was he thinking of sqrtA^3 is never a whole number? I would guess yes, for certainly, anyone who studies mathematics long enough realizes that sqrt of A^3 whole number is never going to be a whole number itself unless 0 or 1.
> 
> On Friday, September 21, 2018 at 5:25:09 PM UTC-5, Archimedes Plutonium wrote:
> > On Thursday, September 20, 2018 at 10:03:03 PM UTC-5, Archimedes Plutonium wrote:
> > > Review Commentary:
> > > 
> > 
> > I am a bit uneasy about this proof, in comparison to the pure Geometry proof where we multiply all terms in the equation to get the set up of a sqrtA^3 cannot be a whole number. In the multiply proof, the sqrtA^3 seems more clear.
> > 
> > 
> > > In Pure Algebra proof of FLT-- we take a known solution set A,B,C of A^2 + B^2 = C^2 such as 3,4,5
> > > 
> > > 9 + 16 = C*C
> > > 
> > > We take an arbitrary number in C^3 Space such as for example 5*5*5 = 5^3 = 125
> > > 
> > > We ask, why is there no A^3 and B^3 that equals 125 ?
> > > 
> > > A^3 + B^3 = C*C*C = 125
> > > 
> > > We add 9+16 to both sides of equation
> > > 
> > > (A^3 + B^3) + 9 + 16 = 125 + 25
> > > 
> > > subtract 125 from both sides
> > > 
> > > (A^3 + B^3) -100 =  25
> > > 
> > > take the square root of both sides
> > > 
> > > sqrt[(A^3 + B^3) -100] = 5
> > > 
> > > Inspect the one side sqrt[(A^3 + B^3) -100]
> > > one possible solution is A^3 = 25 while B^3 =100 meaning A= 2.924... B = 4.641...
> > > And any and all combinations of A, B in exp3 are never whole numbers because 3,4,5 P Triples were whole numbers.
> > > 
> > 
> > In the inspection stage of the proof, it is clear that I have a range if numbers to inspect, because we have to get a sqrt25 on the left side. This is why it is important to select a known PPT that will make the range as small as possible, for if I had selected 11,60,61 I would have a mountain of whole numbers to have to eliminate to say that A and B are fractions.
> > 
> > By choosing 3,4,5, the range of values of whole numbers for both A and B can only be 1,2,3,4,5 for if A=1, then the best B can be is 5, if A is 2 the best B can be is 5 again, if A is 3, the best B can be is either 4 or 5, if A is 4 then best B can be is 4
> > 
> > That means the A and B cannot be whole numbers.
> > 
> > So, is there any guidance on what PPT to pick given any arbitrary picked exp3 number? Suppose we selected 1000 as our exp3 number. Then what PPT numbers in exp2 would we use in order to narrow the range of whole number solutions? Would it be the P Triple of 6, 8, 10 giving us 36, 64 and 100, so that we subtract 1000 from both sides and end up with A^3 + B^3 - 900 = 100 causing our range of whole number solutions to be 1 to 10. Well that is no big hardship going through ten numbers. In this manner, we see that 3,4,5 can be used in all arbitrary picked numbers of exp3
> > 
> > > In Pure Geometry, staying with 3,4,5 as solutions in exp2, we pick an arbitrary exp3 in FLT number as 125= cube 5^3. We perpendicular Slice into the cube along an axis and produce two rectangular boxes of C*C*k and C*C*j where k+j = 5. We go back to exp2 with its C equaling the C in CCC. We do a perpendicular slice into the square 5*5 and we produce the two rectangles 5*1.8 and 5*3.2
> > > 
> > > So we have for exp2 in pure Geometry
> > > 
> > > C*k + C*j = C*C
> > > that is a solution in Whole Numbers in exp2
> > > We multiply every term by C, catapulting the exp2 equation to be a exp3 equation
> > > C*C*k + C*C*j = C*C*C
> > > 
> > > replacing the C=5, k = 1.8, j= 3.2 we have
> > > 
> > > 25*1.8 + 25*3.2 = 125
> > > 
> > > Here we have A^3 = 45 and B^3 = 80, thus A = 3.556.... and B = 4.308....
> > > Explanation: the square of numbers in exp2 as a solution of whole numbers is not preserved by the cube root, for it turns those whole number solutions into fractions.
> > > 
> > > Very crude dot picture of 5f6, 94TH 
> > > ELECTRON=muon DOT CLOUD of 231Pu 
> > > 
> > > 
> > >                 ::\ ::|:: /:: 
> > >                  ::\::|::/:: 
> > >                      _ _ 
> > >                     (:Y:) 
> > >                      - - 
> > >                  ::/::|::\:: 
> > >                 ::/ ::|:: \:: 
> > > One of those dots is the Milky Way galaxy. And each dot represents another galaxy. 
> > >             . \ .  . | .   /. 
> > >            . . \. . .|. . /. . 
> > >               ..\....|.../... 
> > >                ::\:::|::/:: 
> > > ---------------      ------------- 
> > > --------------- (Y) ------------- 
> > > ---------------      -------------- 
> > >                ::/:::|::\:: 
> > >               ../....|...\... 
> > >            . . /. . .|. . \. . 
> > >             . / .  . | .   \ . 
> > > 
> > >   
> > > http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 
> > > 
> > > I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      
> > > 
> > > Read my recent posts in peace and quiet.
> > > 
> > > https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
> > > Archimedes Plutonium

[toc] | [prev] | [next] | [standalone]

#424019 — Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

On Friday, September 21, 2018 at 11:44:17 PM UTC-5, Archimedes Plutonium wrote:
> It would be instructive to do one's own proof that only the Whole Numbers of 0 and 1 can solve for sqrtA^3 when A is a whole number.
> 

Alright, this is a nice bedtime proof for math students in college. Here is the challenge.

You are given a Whole Number M greater than 1 and you are cubing M, so as M^3 = M*M*M

Now, you are asked to take the square root of M^3 and further asked if this square root can be a Whole Number solution? Remember, M is greater than 1.

So, how does a College student prove that sqrtM^3 is never a whole number solution.

The proof the student gives, must be capable of proving cuberootM^4 never has a whole number solution. The sqrtM^5 has no whole number solution. The 5throotM^6 has no whole number solution.

So, was Pierre Fermat in 1637 aware of this fact of mathematics? Perhaps Fermat proved it. But such facts are so woven into mathematics that no-one actually feels compelled to prove it-- for they think the proof should be "proof: obvious"

Now I suspect textbooks do not give this proof, and I suspect no website discusses the issue.

So if you are wanting a bedtime math proof, here is a nice one for the weekend bedtime.

Let me get the student started, for I have not worked out the proof myself. 

We know 1 and 0 can solve sqrtA^3 where A is a whole number for consider

sqrt 0*0*0 is sqrt0 and the sqrt of 0 is 0

sqrt 1*1*1 is sqrt1 and the sqrt of 1 is 1

sqrt2*2*2 is sqrt8 and that is not a whole number

So how can we prove that every whole number beyond 2 ends up with sqrtA^3 is not a whole number?

How about this for a proof mechanism, that the square-root is asking for a B*B = A*A*A where both B and A are whole numbers larger than 1. For example B*B = 2*2*2, would be B*B = 8 and we want a whole number for B such that it is B*B = 8. Well 3 comes close in 3*3 = 9, it still misses.

So the proof of the statement sqrtA^3 has only 0 and 1 as whole number solutions. The proof involves the mechanics of a B*B = A*A*A where both B and A are whole numbers larger than 1. If A was 2 then B=3 misses by only 1 as solution.

So, do we have some bright College math students to grind out a proof? Do we have a learned math historian to tell us when in math history such an idea was proven? And did Fermat prove this, but I think the Ancient Greeks would have proven it.

Now, as you examine this exercise, think of FLT itself, because proving the above is a mirror reflection of FLT in that no whole numbers can fit A^3 + B^3 = C^3.

So as you are proving no sqrtA^3 is a whole number, in one sense, you are proving Fermat's Last Theorem itself.

AP

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#424020 — Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

On Saturday, September 22, 2018 at 1:06:05 AM UTC-5, Archimedes Plutonium wrote:
> On Friday, September 21, 2018 at 11:44:17 PM UTC-5, Archimedes Plutonium wrote:
> > It would be instructive to do one's own proof that only the Whole Numbers of 0 and 1 can solve for sqrtA^3 when A is a whole number.
> > 
> 
> Alright, this is a nice bedtime proof for math students in college. Here is the challenge.
> 
> You are given a Whole Number M greater than 1 and you are cubing M, so as M^3 = M*M*M
> 
> Now, you are asked to take the square root of M^3 and further asked if this square root can be a Whole Number solution? Remember, M is greater than 1.
> 
> So, how does a College student prove that sqrtM^3 is never a whole number solution.
> 
> The proof the student gives, must be capable of proving cuberootM^4 never has a whole number solution. The sqrtM^5 has no whole number solution. The 5throotM^6 has no whole number solution.
> 
> So, was Pierre Fermat in 1637 aware of this fact of mathematics? Perhaps Fermat proved it. But such facts are so woven into mathematics that no-one actually feels compelled to prove it-- for they think the proof should be "proof: obvious"
> 
> Now I suspect textbooks do not give this proof, and I suspect no website discusses the issue.
> 
> So if you are wanting a bedtime math proof, here is a nice one for the weekend bedtime.
> 
> Let me get the student started, for I have not worked out the proof myself. 
> 
> We know 1 and 0 can solve sqrtA^3 where A is a whole number for consider
> 
> sqrt 0*0*0 is sqrt0 and the sqrt of 0 is 0
> 
> sqrt 1*1*1 is sqrt1 and the sqrt of 1 is 1
> 
> sqrt2*2*2 is sqrt8 and that is not a whole number
> 
> So how can we prove that every whole number beyond 2 ends up with sqrtA^3 is not a whole number?
> 
> How about this for a proof mechanism, that the square-root is asking for a B*B = A*A*A where both B and A are whole numbers larger than 1. For example B*B = 2*2*2, would be B*B = 8 and we want a whole number for B such that it is B*B = 8. Well 3 comes close in 3*3 = 9, it still misses.
> 
> So the proof of the statement sqrtA^3 has only 0 and 1 as whole number solutions. The proof involves the mechanics of a B*B = A*A*A where both B and A are whole numbers larger than 1. If A was 2 then B=3 misses by only 1 as solution.
> 
> So, do we have some bright College math students to grind out a proof? Do we have a learned math historian to tell us when in math history such an idea was proven? And did Fermat prove this, but I think the Ancient Greeks would have proven it.
> 
> Now, as you examine this exercise, think of FLT itself, because proving the above is a mirror reflection of FLT in that no whole numbers can fit A^3 + B^3 = C^3.
> 
> So as you are proving no sqrtA^3 is a whole number, in one sense, you are proving Fermat's Last Theorem itself.
> 
> 

Alright, well, if you are a college student, this is a nice opportunity of learning a Counterexample proof.

A=100, then A^3 = 10^6 and thus sqrt is 1000

So, well, I need something more, such as adding of a term

sqrt[A^3 + B^3 -100] = 5

And no, this is not a College Level proof exercise.

It is still a mirror reflection of FLT

But, can a proof be a proof if every, every sample has to be checked through a range of integer values, no matter how small the range can be, for in the above, 1,2,3,4 cannot solve that.

So is a proof, a proof if you always need to check a range of values.

AP

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#424022 — proof principle, in sqrt[A^3 +B^3] = N, A,B impossible to be whole numbers Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

So let us say I picked the arbitrary C^3 value of 100 which is 100*100*100 in A^3 + B^3 = 1,000,000

Now I am looking for a exp2 triple that will reduce that god awful large number 10^6 so I need not check many numbers.

I pick the triple 3,4,5 and multiply by 201 for 603, 804, 1005

Giving me A^3 + B^3 + 603^2 + 804^2 = 1,000,000 + 1,010,025

sqrt[A^3 + B^3 + 10,025] = 1005

No, not looking good. What I need is a small tiny range.

Ah, bad night in bean counting. The QBWR bean counting rubbed off on me from yesterday. And, will I ever teach them to look bright and positive, rather than always a deadbeat.

So, let me just multiply by 200

A^3 + B^3 +600^2 + 800^2 = 1,000,000 + 1,000,000
A^3 + B^3 = 1,000,000
sqrt[A^3 + B^3] = 1000

No, I have not made the range any better for in this case I need to check all the integers from 1 to 10

Now I pause and reflect, for I think I can always manage to have the left side be that of sqrt[A^3 + B^3]

And with my previous painful lesson that sqrtA^3 does have whole number solutions, perhaps it is the combination of sqrt[A^3 + B^3] that destroys whole numbers.

So, let me regroup myself for tomorrow to drill down on sqrt[A^3 + B^3] = N where N is a whole number that impossible for A,B to be whole numbers.

AP

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#424023 — Lemma proof: sqrt[A^3 + B^3] = N where N is a whole number then impossible for A,B to be whole numbers

Alright, tomorrow I have my work cut out for me. I must prove that sqrt[A^3 + B^3] = N where N is a whole number, that it is impossible for A,B to be whole numbers.

The only difference between proving the above, and proving the bare naked FLT of A^3 + B^3 = C^3 is that I have a lot more items to work with. I have a postive whole number on the right side of the equation and I have a entire square root over the leftside of the equation. The more tools you have going into a proof, the better are the chances of proving it. So in bare naked FLT they had no square root, no positive number constant.


On Saturday, September 22, 2018 at 2:04:54 AM UTC-5, Archimedes Plutonium wrote:
> So let us say I picked the arbitrary C^3 value of 100 which is 100*100*100 in A^3 + B^3 = 1,000,000
> 
> Now I am looking for a exp2 triple that will reduce that god awful large number 10^6 so I need not check many numbers.
> 
> I pick the triple 3,4,5 and multiply by 201 for 603, 804, 1005
> 
> Giving me A^3 + B^3 + 603^2 + 804^2 = 1,000,000 + 1,010,025
> 
> sqrt[A^3 + B^3 + 10,025] = 1005
> 
> No, not looking good. What I need is a small tiny range.
> 
> Ah, bad night in bean counting. The QBWR bean counting rubbed off on me from yesterday. And, will I ever teach them to look bright and positive, rather than always a deadbeat.
> 
> So, let me just multiply by 200
> 
> A^3 + B^3 +600^2 + 800^2 = 1,000,000 + 1,000,000
> A^3 + B^3 = 1,000,000
> sqrt[A^3 + B^3] = 1000
> 
> No, I have not made the range any better for in this case I need to check all the integers from 1 to 10
> 
> Now I pause and reflect, for I think I can always manage to have the left side be that of sqrt[A^3 + B^3]
> 
> And with my previous painful lesson that sqrtA^3 does have whole number solutions, perhaps it is the combination of sqrt[A^3 + B^3] that destroys whole numbers.
> 
> So, let me regroup myself for tomorrow to drill down on sqrt[A^3 + B^3] = N where N is a whole number that impossible for A,B to be whole numbers.
> 

If I can prove the above statement-- sqrt[A^3 + B^3] = N where N is a whole number that impossible for A,B to be whole numbers

It is equivalent to FLT, for it is called a Lemma proof in mathematics, a helper proof. For once I proved the above, it is easy to see that FLT is a version of the Lemma.

Now Pierre Fermat in 1637 did not prove sqrt[A^3 + B^3] = N where N is a whole number that impossible for A,B to be whole numbers, so we need not involve math historians.

AP

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#424027 — is the Cubic Formula the proof of FLT Re: Lemma proof: sqrt[A^3 + B^3] = N where N is a whole number then impossible for A,B to be whole numbers

On Saturday, September 22, 2018 at 2:35:05 AM UTC-5, Archimedes Plutonium wrote:
> Alright, tomorrow I have my work cut out for me. I must prove that sqrt[A^3 + B^3] = N where N is a whole number, that it is impossible for A,B to be whole numbers.
> 
> The only difference between proving the above, and proving the bare naked FLT of A^3 + B^3 = C^3 is that I have a lot more items to work with. I have a postive whole number on the right side of the equation and I have a entire square root over the leftside of the equation. The more tools you have going into a proof, the better are the chances of proving it. So in bare naked FLT they had no square root, no positive number constant.
> 

Alright, I think I may have a lead in to this problem already. 

Remember that Quadratic Equation we all learned in High School and thereafter never needed it for anything dealing with science, only a game toy of mathematics.

Most of us memorized it

x = (- b +- sqrt[b^2 - 4ac ]) / 2a

Well, most of us never even looked at the cubic equation

source: wikipedia

C = cuberoot[delta1 +- sqrt[delta1^2 - 4delta0^3] / 2


So, what I am thinking, so I can have something to think about in sleep for this problem. What I am thinking is the reason that exp2 has all these solutions is because there is a sqrt of a b^2 in the discriminant

But now the cubic equation, and can we call the A^3 + B^3 = C^3 a cubic equation that obeys this Cubic Formula, I think we can.

If you look at that Cubic Formula you see you have a cube root of a square root, and you will never have a whole number survive in that machinery and come out as a whole number solution.

So, let me sleep on that, and maybe I can find a more simple solution than the Cubic Formula.

If the Cubic Formula tells us it is impossible to have whole number solutions, then, FLT was really, after all, just an acknowledgement of the Cubic Formula as the proof.

AP

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#424028 — cheers, the proof at last-- usually I find them during my bath, now I found this one before my bath

Alright, finally, after all the tossing and turning, I have the FLT proof of Pure Geometry, and of Pure Algebra.

It involves the Pythagorean Triples, in fact the P triples are the crux and backbone of the proof.

What was troubling me, was the addition use of P triples into FLT.

So as I was doing say pick 125 out of exp3 Space and find a A^3  +B^3 = 125, trouble was I was adding a P triple as in A^3 + B^3 + 9+16 = 125 +25
reducing

A^3 + B^3 -100 = 25

Then taking square root of both sides and then becoming stuck, stuck stuck.

And instead of just saying addition of a P-triple just does not work, I wanted to believe it worked.

Whereas in my Pure Geometry proof, I multiplied.

So the same arbitrary C^3 from exp3 is say 125. Then we go to a Pythagorean triple in exp2. We look carefully for a P triple that is conducive with 125, for I do not want to select 11,60,61 when I can select 3,4,5.

So now I find a solution set of where I cut the 5 by 5 square so as to have two squares add up to 25, which we know of course is 3by3 and 4by4. but we represent that 3by3 as C*1.8 and the 4by4 as C*3.2, and the 25 as C*C

Now we have in exp2 this

C*1.8 + C*3.2 = C*C

In FLT we have
A^3 + B^3 = 125

The C*C is 25 and I carefully selected it because we can by Algebra multiply every term by C=5 and retain equality.

So we have

C*C*1.8  + C*C*3.2 = C*C*C
that represents

45  + 80 = 125

Every FLT arbitrary picked number in a exponent space, ours was exp3 and 125, every such number is converted to a Pythagorean Triple in the manner above. Every single one and that is easy because of the Algebra Principle of multiply by k to retain equality, and have your CCC be exactly formed from the picked P-triple.

The only last part of the proof is how do I say, that 45 and 80 do not make for a Whole Number Cube or whatever exponent of FLT. How do I say they are not whole numbers.

I thought it was the different exponents, but no, it was something carried over from the definition of P-triples-- the A and B are coprime, as that 3 and 4 are coprime. So that 45 and 80 cannot be whole numbers once you take the cube root. Same idea for every other number in every exponent of Fermat's Last Theorem.

So I must full use the P-triples to conquer FLT, and only the multiplication of P-Triples works.

Thank goodness.

AP

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#424031 — cheers, the proof at last-- usually I find them during my bath, now I found this one before my bath

Now, how is being coprime a hindrance or deterrent as a whole number? Well the A and B are linked as complementary slices for 1.8 + 3.2 = 5  so one affects the other, if one is not a whole number the other cannot be whole, both are either fractions or both are whole numbers.

Now what does the coprime have to do with why they cannot be whole numbers. Well 29 is prime and the cube root must be fractional, and 27 is not prime and has the possibility of a whole number solution.

If A and B are connected as complementary solutions by PT with the condition of Coprime means all numbers in FLT are forced to be fraction solutions never whole number.

Coprime is the internal mechanism of FLT, and the use of solutions from P-triples is the backbone the framing structure of FLT.

AP

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#424042 — Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test

Math Failure Archimedes Plutonium <plutonium.archimedes@gmail.com> fails:

>Now, how is being coprime a hindrance or deterrent as a whole number? Well
[snip more babbling]

Archie, why are you even wasting your time with FLT, especially with n=3
and n=4?  Fermat himself proved it for n=4, and Euler proved it for n=3 in
1770 (with a mistake which he later corrected). Other specific values for
n also proven, including all nonprime n.

Of course Wiles proved it for all integers almost 25 years ago, in 1994.
So proving something that already has a proof is rather silly, isn't it?

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#424024 — Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

On Friday, September 21, 2018 at 11:06:05 PM UTC-7, Archimedes Plutonium wrote:
> Alright, this is a nice bedtime proof for math students in college. Here is the challenge.
> 
> You are given a Whole Number M greater than 1 and you are cubing M, so as M^3 = M*M*M
> 
> Now, you are asked to take the square root of M^3 and further asked if this square root can be a Whole Number solution? Remember, M is greater than 1.
> 
> So, how does a College student prove that sqrtM^3 is never a whole number solution.

Let M equal 4

Cube M getting 64

Take the square root of 64 getting 8

Hummm... I'm having a very difficult time completing the next
step of the proof that 8 is not a whole number solution.

Can you help with a hint of what that next step might be?

After that you can in a bright positive way decide what to
do with all your posts in the last couple of days that say
only 0 and 1 have the property that the square root of the
cube is an integer.

Thank you
Bean Counter

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#424025 — Re: Proof Principle involved in all FLT proofs-- if A^3 is a whole number then sqrtA^3 is never whole Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

On Saturday, September 22, 2018 at 2:50:22 AM UTC-5, qbwr...@gmail.com wrote:
> On Friday, September 21, 2018 at 11:06:05 PM UTC-7, Archimedes Plutonium wrote:
> > Alright, this is a nice bedtime proof for math students in college. Here is the challenge.
> > 
> > You are given a Whole Number M greater than 1 and you are cubing M, so as M^3 = M*M*M
> > 
> > Now, you are asked to take the square root of M^3 and further asked if this square root can be a Whole Number solution? Remember, M is greater than 1.
> > 
> > So, how does a College student prove that sqrtM^3 is never a whole number solution.
> 
> Let M equal 4
> 
> Cube M getting 64
> 
> Take the square root of 64 getting 8
> 
> Hummm... I'm having a very difficult time completing the next
> step of the proof that 8 is not a whole number solution.
> 
> Can you help with a hint of what that next step might be?
> 
> After that you can in a bright positive way decide what to
> do with all your posts in the last couple of days that say
> only 0 and 1 have the property that the square root of the
> cube is an integer.
> 
> Thank you
> Bean Counter

Good, you are improving your bean counting in leaps and bounds.

But I am more concerned over your downbeat and negative posts, than any math content.

What I need you to work on is a more optimistic and upbeat tone to your posts

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#424008 — Re: Archimedes "Total Failure" Plutonium flunked the Math Test of a lifetime-generation test

Math Failure Archimedes Plutonium <plutonium.archimedes@gmail.com> fails:

>On Friday, September 21, 2018 at 7:09:43 PM UTC-5, Bill Smith wrote:
>> Shut up idiot.

>AP writes:: yes sir, it takes that discerning eye from AIOE to recognize MM
>is a babbling idiot. MM needs to pay more attention to his Gay Bed 
>&Breakfast in Boston. Say there, MM, how many feet away is your B&B from 
>the front door of Harvard and the front door of MIT?

Definitely we need to discuss your sudden projections all the time.
So I take it your own gayness must be bothering you for you to project
it onto me? It looks like you now want to start your own gay B&B. Do you
want to start it on your own property, or near a South Dakota university
or are you planning to move to Massachusetts so you can run your gay Bed
and Breakfast right outside MIT and Harvard? Good choice since many 
college students are gay, so you should have lots of customers (and 
boyfriends!) if you did that.

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#423965 — Re: Ghost figures of geometry in Generalized FLT Re: 1.1- Reviewing AP's 3 proofs of FLT (1) pure algebra (2) pure geometry (3) Generalized FLT

On Thursday, September 20, 2018 at 5:15:59 PM UTC-5, Archimedes Plutonium wrote:
> Alright, I need to incorporate all three proofs of FLT, the pure Algebra proof, the pure Geometry proof and then the Generalized FLT that delivers FLT as a corollary side proof.
> 

Now let me review the Generalized FLT that gives a proof of FLT itself as a side-shoot corollary. I am looking for similarities between Pure Algebra and Pure Geometry proofs of FLT with Generalized.


> Now the reason I am spending so much time on this, is because I want to see for myself the LOGICAL Flow where the mechanism that proves FLT in pure algebra is the same mechanism that proves it in pure geometry, but in Generalized FLT we use the rectangles that were used as Rectangular Boxes in pure geometry proof. And the very same algebra mechanism that proves pure algebra, pure geometry, is also used by Generalized FLT.
> 
> I want to fully reveal how all three proofs are interconnected. There are threads running through all three proofs, as there should be.
> 
>         (3) GENERALIZED FLT proof
> 
> What is Generalized FLT? Glad you asked, for it is: 
> 
> A^x + B^y = C^z where A,B,C,x,y,z are positive integers with x,y,z greater than 2 
> 
> FLT is of course 
> 
> A^n + B^n = C^n where A,B,C, n are positive integers and n is greater than 2 
> 
> Newsgroups: sci.math 
> Date: Thu, 9 Jun 2016 13:38:48 -0700 (PDT) 
> 
> Subject: Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed 
> From: Archimedes Plutonium <plutonium....@gmail.com> 
> Injection-Date: Thu, 09 Jun 2016 20:38:49 +0000 
> 
> The history of this proof starts 1993 when I became interested in proving FLT and started with 2+2 = 2x2 = 2^2, where I noted that addition is the same as multiplication for 2 and 4 as a building block to build all solution triples. The reason exp2 has whole number solutions is because of 2 and 4, and exp3 has nothing like a building block, hence, no solutions.
> 
> But by 2014, I proved Generalized FLT and saw FLT as a corollary. The trick I used in 2014, was Condensed Rectangles-- rectangles formed from the factors of numbers, so that 8 would be a condensed rectangle of 2*4, or 12 a condensed rectangle of 3*4, or it could be 2*6.
>   
> 
> PROOFS of Generalized FLT & FERMAT's LAST THEOREM as corollary 
> 
> Editing and commentary corner: Once Generalized FLT and FLT are turned into a geometrical perspective-- condensed rectangles, then the proof is as easy as making homemade cherry pie. 
> 
> Summary of text: Both proofs of Generalized FLT and FLT are based on a fact of geometry, that you can represent a number with its cofactors as the sides of a rectangle. And to prove either Generalized FLT or FLT is a simple matter of stacking two rectangles that have equal sides, A and B to produce a third rectangle C which has a side equalling the _shared side_ of A and B. 
> 
> So now, here, reflect back to the Pure Geometry proof of FLT where we have a Cube and we put it through a Perpendicular Slicer on a axis of the cube and we end up with two smaller rectangular boxes.
> 
> But back to Generalized FLT:
> 
> We see the relationship between a Theorem and its Corollary very easily from Generalized FLT and FLT, in that we have all these cases to worry about. The history of FLT was a case study, prove it in the case of exponent this and that, but never a proof of all of FLT, and that is because corollaries are never proven as stand alone theorems and must have their theorem (Generalized FLT) proven first. 
> 
> ________________________ 
> DETAILED PROOF OF Generalized FLT 
> ________________________ 
> 
> It is a constructive proof. 
> 
> We make the table of all the numbers possible in the Generalized FLT Conjecture as the conglomeration of exponents of 3 or larger as this set: 
> 
> {1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .} 
> 
> Here we have conglomerated exp3 and exp4 and exp5 etc etc into one set. 
> 
> We know Generalized FLT has solutions of A+B=C in that set for here are three examples: 
> 
> 2^3 + 2^3 = 2^4 with prime divisor 2 
> 3^3 + 6^3 = 3^5 with prime divisor 3 
> 7^3 + 7^4 = 14^3 with prime divisor 7 
> 

And so, right here, let us remake those three examples of Generalized FLT, remake them as Ghost Cubes or Ghost Rectangular Boxes.

We have 2^3 + 2^3 = 2^4 and we need do nothing about 2^3 because that is 3rd D as we normally know it. However, 2^4, what is that all about?

Well, 3rd Dimension is the last and final dimension so anything beyond has to relate to the three known dimensions and so what 2^4 is, is a Ghost Cube of that of 2^2*2^2, or we can make a Ghost Rectangular Box. Now, make a square out of a wire, I used a paper clip, two of them and bent them to resemble each other as a square. Then I placed them at a distance equal to the side of the square, opposite one another. So I have two squares as wires and places opposite at a distance of the side of square. This is a Ghost Cube and we can treat it as any full cube. However for 2^4 we need a Ghost Rectangular Box that is 2by2by4 in order to have 2by2by2 add 2by2by2 equal two cubes add to a box.

Now we have 3^3 + 6^3 = 3^5 and our only concern is 3^5, what geometrical object is that? Well, 3^5 cannot be a Ghost Cube of side 3 for we already have 3^3 and so, what can we interpret 3^5 to be? Well taking a clue from 3^3 + 6^3 as a coefficient of 9 we can see that 3^5 is 3^3(3^2) and is a square of 9 that is elongated by a distance of 27 as being a Rectangular Box of 3by3by27 and is a Ghost Rectangular Box. Then, 3^3 is seen as a cube that is 3by3by3 and then 6^3 is 3by3by24. So, we have two Ghost Rectangular Boxes and one cube.


Finally 7^3 + 7^4 = 14^3  we have a cube of 7by7by7 with a  Ghost rectangular box 7by7by49 and a Ghost rectangular box of 7by7by56

So the Generalized FLT puts prominently in place the need to interpret exponents as either a ghost cube or ghost rectangular box. 

This fits nicely in the Pure Geometry proof where we took Perpendicular to Axis Slices of a starting figure to obtain a ghost rectangular box.

> What we need to prove is that all solutions have a prime divisor in common, ie all three rectangles of A, B, C so that A+B=C, have one shared side equal to one another. 
> 
> Definition of Condensed-Rectangle: given any number in the set of conglomerated exponents, we construct rectangles of that number from its unit squares whose sides are cofactors of the number. For instance, rectangle of 216 units as either 12x18 units, or 9x24 units, or 6x36 units or 3x72 units, or 2x108, but never a 1x216 units. We exclude 1 times the number as a condensed rectangle. So a condensed-rectangle is one in which it is composed of cofactors of the number in question, except for 1, and the number itself for 1x216 units is not a condensed-rectangle. 
> 
> Now for the constructive proof that Generalized FLT solutions must have a common prime divisor. 
> 
> We stack Condensed-Rectangles of the number-space that Generalized FLT conjecture applies: 
> 
> Solution Number Space for Generalized FLT now becomes these condensed-rectangles: 
> 
> { 2*4, 2*8, 4*4, 3*9, 2*16, 4*8, 2*32, 4*16, 8*8,. . . }   
> 
> We convert each of those numbers into Condensed-Rectangle, except 1 of course, and where many numbers have several condensed rectangles so the Solution Space of Numbers increases by a large amount. If an A and B as condensed-rectangles have the same side such as 3x9 units and 9x24units 
> wherein you stack them on their shared side of 9 and which matches another number of its condensed-rectangle such as 9x27 units, then you have a Generalized FLT solution of A+B=C. For if we were to take the 9 by 27 condensed rectangle it decomposes into 3x9 and 9x24. 
> 
> Now, the question is, are all A+B equal to a C, form stacked condensed rectangles that share a common side? 
> 
> All stackable condensed-rectangles must have one side the same for the two rectangles to stack, in the case above it is the side 9 with its common divisor of the prime 3. 
> 
> If any other solution to Generalized FLT had A stacked upon B without a common side between them, then the figure formed cannot be a overall new rectangle but something that looks like this: 
> 
> HHHHHH 
> HHHHHH 
> HHHHHHHHHH 
> HHHHHHHHHH 
> 
> That is not a condensed rectangle and all the numbers of the Solution Space, except the number 1, are condensed rectangles. That above figure is 6-sided figure. That is a 6-sided figure, yet a rectangle is only a 4 sided figure. That is not a Condensed Rectangle, represented by its cofactors. So that equality can not be achieved by any other stacking than condensed rectangles equalling condensed rectangles. 
> 
> Then the question is, can we have an A and B with a shared common side equal to a C that is a condensed rectangle with no common side shared with A and B? 
> 
> The question is, how can I be sure that all the A, B, C such that A + B = C have a common prime divisor? 
> 
> As if the question is asking whether the Condensed Rectangles covered the question by forbidding any equality unless there is a common shared side in A in B and in C. 
> 
> Well, the answer is easily enough covered for the Condensed Rectangles eliminates any possibility of A,B,C where A+B = C and not have a shared side by all three of A, B, C. It does this by the Distributive law of integers. 
> 
> All the numbers in the solution space of Generalized FLT have condensed rectangles, except the number 1. 
> 
> All the A, B, C are written as condensed rectangles of (s*t) + (r*p) = (u*v). 
> 
> For example (9*3) + (9*24) = (9*27) which is 3^3 + 6^3 = 3^5  which is 27 + 216 = 243. 
> 
> In order for (s*t) + (r*p) = (u*v) then the s, r, u have to be equal sides. Have to be (9*t) + (9*p) = (9*v) in the example of 27 + 216 = 243. 
> 
> Distributive Law (9*3)/9 + (9*24)/9 = (9*27)/9 where 3 + 24 = 27 
> 
> So, give me three numbers A, B, C, chosen arbitrarily from the Solution Space of Generalized FLT {1, 8, 16, 27, 32, 64, 81, 125, 128, 216,  243, 256, . .} except of course 1, and can those three chosen arbitrarily A, B, C obey A + B = C? Only if they obey the Distributive Law can the A + B equal to the C when we write the A, B, C as condensed rectangles. 
> 
> QED 
> 
> 
> _________________________________________ 
> Detailed Proof of FLT using condensed-rectangles 
> _________________________________________ 
> 
> It is a construction proof method for we show that it is impossible to construct A+B = C inside of a specific exponent. 
> 
> Fermat's Last Theorem FLT conjecture says there are no solutions to the equation a^y + b^y = c^y where a,b,c,y are positive integers and y is greater than 2. 
> 
> The number Space that governs FLT is this: 
> 
> exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .} 
> 
> exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .} 
> 
> exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .} 
> 
> exp6 ..... 
> . 
> . 
> . 
> . 
> 
> So in FLT we ask whether there are any triples, A,B,C in any one of those _specific exponents_ such that A+B=C. In FLT, our solution space is only one particular exponent such as 3 or 4, or 5 to hunt down and find a A,B,C to satisfy A+B=C. 
> 
> In the proof we use Condensed-Rectangles which is defined as a rectangle composed of unit squares of the cofactors of a number, except for 1 x number itself. So the number 27 in exp3 has Condensed Rectangles of 3x9 only. The number 125 in exp3 has condensed rectangles of 5x25 only, and the number 81 in exp4 has condensed rectangles of 3x27 and 9x9. 
> 
> Now in the proof of Generalized FLT solutions of A + B = C require a common divisor. 
> 
> 2^3 + 2^3 = 2^4 with prime divisor 2 
> 3^3 + 6^3 = 3^5 with prime divisor 3 
> 7^3 + 7^4 = 14^3 with prime divisor 7 
> 
> Here in FLT, we need not even worry about exponents because all solutions to Generalized FLT encompass exponents so that if these were FLT solutions: 
> 
> 2^? + 2^? = 2^? with prime divisor 2 
> 3^? + 6^? = 3^? with prime divisor 3 
> 7^? + 7^? = 14^? with prime divisor 7 
> 
> So in that construction of a solution to FLT there is a common divisor and so now we divide the equation by the common divisor, and we get this: 
> 
> 1^y + 1^y = 1^y 
> 1^z + 2^z = 1^z 
> 1^w + 1^w = 2^w 
> 
> And those are impossible constructions. So the proof of FLT requires Generalized FLT be proven first in order to display that a Generalized FLT proof cannot co-exist unless FLT has no solutions. 
> 
> Stated another way:: should FLT have a whole number set solution of A,B,C, it contradicts Generalized FLT for you can reduce the FLT to that of something as 1^3 + 1^3 = 1^3
> 
> QED 
> 
> Now in upcoming posts I want to reduce Generalized FLT into a ARRAY proof and I want to comment on the similarities between pure algebra proof, pure geometry proof and generalized FLT proof.
> 
> 
> Very crude dot picture of 5f6, 94TH 
> ELECTRON DOT CLOUD of 231Pu 
> 
> 
>                 ::\ ::|:: /:: 
>                  ::\::|::/:: 
>                      _ _ 
>                     (:Y:) 
>                      - - 
>                  ::/::|::\:: 
>                 ::/ ::|:: \:: 
> 
>         One of those dots is the Milky Way galaxy. And 
> each dot represents another galaxy. 
> 
> 
>             . \ .  . | .   /. 
>            . . \. . .|. . /. . 
>               ..\....|.../... 
>                ::\:::|::/:: 
> ---------------      ------------- 
> --------------- (Y) ------------- 
> ---------------      -------------- 
>                ::/:::|::\:: 
>               ../....|...\... 
>            . . /. . .|. . \. . 
>             . / .  . | .   \ . 
> 
>   
> http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 
> 
> I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      
> 
> https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
> Archimedes Plutonium

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#423955

On Thursday, September 20, 2018 at 1:38:14 PM UTC-7, Archimedes Plutonium wrote:
> Alright, now I can do this in complete satisfaction.
> 
> FLT pure Algebra proof STATEMENT: whole numbers A,B,C, there never exists A,B,C and z, such that A^z + B^z = C^z where z is a whole number 3 or larger.
> PROOF: Given any whole numbers M and N,and x where x is 3 or greater, M^x = N^x-1 has only two solutions 0 and 1. 
> QED

If your proof is "the Algebra principle M^x = N^x-1 has only two solutions 0 and 1"
then you have to prove your Algebra principle only has those two solutions.

Precisely correctly exactly in detail what is your algebraic proof that
there can be no A^3+B^3=C^3 for integers A.B.C not equal to zero?

Saying "It can't because it can't. QED" is not a convincing argument for anyone.
You are supposed to demonstrate WHY it can't, not just repeat what you are to prove.
Saying "A^3=C^3*j and B^3=C^3*k" is not a convincing argument for anyone.
Repeatedly making mistakes is not a convincing argument for anyone.
Saying M^2 cannot equal N^3 unless M,N are either 0 or 1 is simply false

   8^2 = 4^3
   27^2 = 9^3
   64^2 = 16^3
   ...
unless you mean something else that you haven't been able to explain.

If you have a proof then put it on the table with each step clearly
logically obviously correctly depending on the previous steps.

You don't need to spiral off into higher dimensions or talk about other
kinds of proofs or proofs of other things, just put your correct detailed
convincing algebraic proof for A^3+B^3=C^3 has no solutions for nonzero A,B,C
on the table.

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