Andrew Wiles, you are a con-artist, fraud of mathematics, so dumb in math you could never spot the error in Euler, so stupid in math you could never do a Geometry proof of Calculus Fundamental Theorem Re: Fermat status.

Andrew Wiles and his fake FLT proof, so dumb on FLT he could not even spot Euler's flaw of exp 3 FLT, and so dumb as a mathematician, he never could do a geometry proof of calculus, FTC. Thanks to Roland Dreier of Berkeley back in 1993.

Errors in Wiles's work on Fermat's Last Theorem
1) So dumb in logical reasoning that Andrew never saw the gaping hole in Euler's so called proof of FLT in exponent 3. The hole is that Euler had to prove no A,B,C all even numbers are a solution in A^3+B^3=C^3. One would think that a person wanting to prove all of FLT would have a good enough logical mind to spot a error in a proof of just exponent 3. No, not Andrew Wiles, for he is chasing after fame and fortune but never the truth of mathematics.

2) Andrew was stupid in math, for the most pressing problem of all of mathematics of the 1900s and now 2000s is a geometry proof of Calculus Fundamental Theorem. This was #1 urgent problem in all of mathematics because for the simple reason, it corrects and cleans out the entire house of mathematics in the "giving of a geometry proof of calculus". Reals are not the true numbers of mathematics. Functions and equations have to be turned into polynomials to be valid, just to list two key items that a geometry proof of Calculus would clean up and clean out. But where was Andrew? Of course, chasing after fame and fortune, but not math truth.

3) Andrew was intensely stupid in Logic, for you need logic to prove FLT, but there was Andrew with his Boole logic that 2 OR 1 = 3, that Either..Or...Or.. Both is acceptable, with Andrew accepting the logic that 2 AND 1 is subtraction, that the truth table of AND for Andrew was TFFF when in reality it is TTTF. That Andrew thought the truth table of IF-Then was TFTT when in reality it is TFUU where U means unknown. And where Reductio ad Absurdum, RAA is not a valid proof method for mathematics, yet, stupid Andrew uses RAA for his fakery FLT.

4) Andrew's FLT is so dumb, that he never thought about Shadowy FLT where you get a equation of A^x+B^y= C^z where one of x,y,z is exponent 2 and it turns that these Shadowy FLT follow the same course of geometry-- Condensed Rectangles as does Generalized FLT, except the one exception of 1^3 +2^3 = 3^2. This Shadowy FLT was totally unknown to Andrew Wiles as was a proof of Generalized FLT. So Andrew Wiles is a con-artist and fraud mathematician that has duped the world down his insane path away from the real truth of mathematics.

Andrew Wiles and his fake FLT proof, so dumb on FLT he could not even spot Euler's flaw of exp 3 FLT, and so dumb as a mathematician, he never could do a geometry proof of calculus, FTC.

Thanks to Roland Dreier of Berkeley back in 1993. For I was working on the algebra, but Roland was too fast for me and had it before I did.
Berkeley's Roland Dreier was extremely generous in 1993, and he needed not state that AP had proven FLT, for it is obvious that AP had proven FLT and Roland had given that part of the proof with his (below proof) that Pythagorean Triples are built from 2+2 = 2x2 = 2^2 = 4.


On Friday, December 3, 1993 at 7:36:50 PM UTC-6, Andrew Wiles wrote:
> In view of the speculation on the status of my work on the
> Taniyama-Shimura conjecture and Fermat's Last Theorem I will give a
> brief account of the situation. During the review process a number of
> problems emerged, most of which have been resolved, but one in
> particular I have not yet settled. The key reduction of (most cases
> of ) the Taniyama-Shimura conjecture to the calculation of the Selmer
> group is correct. However the final calculation of a precise upper
> bound for the Selmer group in the semistable case (of the symmetric
> square representation associated to a modular form) is not yet
> complete as it stands. I believe that I will be able to finish this
> in the near future using the ideas explained in my Cambridge
> lectures.
> The fact that a lot of work remains to be done on the
> manuscript makes it still unsuitable for release as a preprint . In
> my course in Princeton beginning in February I will give a full
> account of this work.
>
> Andrew Wiles.

Andrew, your FLT is junk and a sham proof. So dumb on FLT are you, Andrew, that you never spotted the error of Euler in his exponent 3 of FLT, the error that Euler could never prove the case of when all three A,B,C are even, A^3 + B^3 = C^3. You never spotted that error of Euler and yet you are so pompous that you think you found a proof of all of FLT. No, Andrew, actually you are a math failure for you never recognized that the pressing problem in all of mathematics of our generation is to give a Geometry proof of Fundamental Theorem of Calculus (see below at end). Instead, you, Andrew chased after fame and fortune, but never the "truth of mathematics".

5-Andrew Wiles and his fake FLT proof, so dumb on FLT he could not even spot Euler's flaw of exp 3 FLT, and so dumb as a mathematician, he never could do a geometry proof of calculus, FTC.

Archimedes Plutonium
Jul 7, 2021, 11:10:15 PM
to sci.math
For thirty years, 30 years, AP has been at it on Fermat's Last Theorem. It was 1991, that I saw that 2+2=2x2=4 was the heart and crux of the proof of FLT. And it was a hard and bumpy ride in those 30 years, with much fanfare and intrigue. And where the fame and fortune of proving FLT by AP was stolen from him, stolen by Andrew Wiles. But I am not sorry of that stealing because in the meantime, I had far far more important work and discoveries to do, than to claim back my proof and success of FLT. But now, here in 2021, some 30 years later, I am not so generous, not so lenient, and now I want my proof to have its rightful historical place mark. FLT was never proven by Andrew Wiles and his alleged proof is a massive joke. And a measure of how dumb and a joke that Wiles offering was, is easily seen in asking Wiles, how his offering proves that exponent 2 has solutions. Ask Wiles how his technique or mechanism of elliptic curves shows A^2+B^2=C^2 has solutions but not A^3+B^3=C^3 with no solutions. You see, Andrew Wiles has few logical marbles to ever be doing a mathematics proof, let alone FLT. Let alone asking Andrew to do a geometry proof of Fundamental Theorem of Calculus. AP reclaims his "world's first valid proof of Fermat's Last Theorem".

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.

Alright, so in the true proof of FLT via Generalized FLT we accomplish this via Condensed Rectangles and FLT is a corollary of Generalized FLT. Another proof of FLT is directly from the Algebra of 2+2=2x2=2^2=4. Then a direct proof of Generalized FLT is also from 2+2=2x2=2^2=4.

And as for geometry we are left with two geometry figures, the Condensed Rectangles for Generalized and for Shadowy FLT. And Right Triangles for Pythagorean Theorem.

So, a big big question now looms. Since all equations of Form A^x + B^y = C^z except x,y,z = 1 for that is just plain Arithmetic, but for x,y,z = 2 or larger, and A,B,C of = 1 or larger, the big looming question is can we directly link Right Triangles of Pythagorean Theorem with Rectangles of FLT?

We all know that half of a rectangle is a right triangle at diagonal.
__________
|\ |
| \ |
| \ |
|_______\ |

So, can we link up Pythagorean Theorem as right triangles directly with FLT as Condensed Rectangles?

I know of one such link up is Calculus where the Integral is the rectangle while the derivative is the hypotenuse of right triangle. In the specialized function of Y= x the integral is area 1/2 x^2, the area of right triangle, while the derivative is 1 = dy/dx.

So in some sense is the derivative the right triangle hypotenuse and the Condensed Rectangle is the integral.

This would be a beautiful link up of All solutions to equation A^x + B^y = C^z, not Arithmetic but x,y,z = 2 or larger.

And this would make sense in Quantum Mechanics, that numbers 1,2,3,... are quantized and the geometry of calculus is either rectangle or right triangle.

This ties both Pythagorean Theorem together with FLT.

AP
King of Science, especially Physics

Alright, AP is going to see if he can build the whole entire modern day calculus out of just the Pythagorean Theorem and the FLT theorem of Condensed Rectangles, AP's proof that A^x + B^y = C^z for x,y,z = 2 or higher and A,B,C = 1 or higher where A^n +B^n=C^n has only solutions in n=2.

In AP calculus, all the functions have to be Polynomials and all the numbers have to be Decimal Grid Systems starting with 10 Grid. In this manner there is no loss of generality as we say-- the Counting Numbers from 1 to 100 is in fact the 10 Decimal Grid. Where we do this.

1= 0.1
2= 0.2
3= 0.3
..
..
..
98 = 9.8
99= 9.9
100 = 10.0

Similar actions take place on the next Grid System of 100 as 100^2 = 10000 where 1=0.0001, and 10000 = 100.

You see what I am doing here is replacing integers as being representatives of Decimal Grid Numbers in Decimal Grid Systems.

Now in the geometry proof of Fundamental Theorem of Calculus, we need these number systems discrete, with holes and gaps in between one number and the next so that we can divide by 2 the upper side of the Integral rectangle and make a diagonal cut to form a right-triangle that is then lifted up to sit atop the trapezoid and forms the Derivative. This is why the proof of Generalized FLT is Condensed Rectangles, because that is the integral of mathematics. And the proof of Pythagorean theorem is the Right-Triangle.

So, let us get started in showing that Pythagorean Theorem is the Derivative of Mathematics and the FLT is the Integral of Mathematics.

Here I have the first smallest Pythagorean Triples.

3,4,5
5,12,13
7,24,25
8,15,17
9,40,41

And starting with the first one of 3,4,5 would be a right triangle and on the sides of that right triangle would be three squares of 3^2 and 4^2 and 5^2. So in AP calculus, my FLT condensed rectangle is 3*4 = 12 where I cut the 3 into half as 1.5 and lift up a right triangle in that 3 by 4 rectangle.

The area or integral is 3*4=12. Now the perimeter is 3+4+5= 12.

Next is 5,12,13 and that integral area is 5*12 = 60, and perimeter is 30, or 1/2 of 60.

Next is 7,24,25 and that integral area is 7*24 = 168, and the perimeter is 56, or 1/3 of 168.

Next is 8,15,17 and that integral area is 8*15 = 120, and the perimeter is 40, or 1/3 of 120.

Next is 9,40,41 and that integral area is 9*40 = 360, and that perimeter is 90, or 1/4 of 360.

So a pattern is established, with discrete number systems, discrete right triangles, discrete rectangle area, discrete perimeters.

What is hoped for, is that we can build calculus directly from Pythagorean Theorem plus FLT theorem.

See my geometry proof of Fundamental Theorem of Calculus as reference.

World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 19May2021. This is AP's 11th published book of science.
Preface:
Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.

Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?

Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.

Length: 137 pages

Product details
ASIN : B07PQTNHMY
Publication date : March 14, 2019
Language : English
File size : 1307 KB
Text-to-Speech : Enabled
Screen Reader : Supported
Enhanced typesetting : Enabled
X-Ray : Not Enabled
Word Wise : Not Enabled
Print length : 137 pages
Lending : Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
#20 in Calculus (Kindle Store)

y  z
|  /
| /
|/______ x

More people reading and viewing AP's newsgroup than viewing sci.math, sci.physics. So AP has decided to put all NEW WORK, to his newsgroup. And there is little wonder because in AP's newsgroups, there is only solid pure science going on, not a gang of hate spewing misfits blighting the skies.

In sci.math, sci.physics there is only stalking hate spew along with Police Drag Net Spam of no value and other than hate spew there is Police drag net spam day and night.

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of stalkers and spammers, Police Drag Net Spam that floods each and every day, book and solution manual spammers, off-topic-misfits, front-page-hogs, churning imbeciles, stalking mockers, suppression-bullies, and demonizers.  And the taxpayer funded hate spew stalkers who ad hominem you day and night on every one of your posts.

There is no discussion of science in sci.math or sci.physics, just one long line of hate spewing stalkers followed up with Police Drag Net Spam (easy to spot-- very offtopic-- with hate charged content). And countries using sci.physics & sci.math as propaganda platforms, such as tampering in elections with their mind-rot.

Read my recent posts in peace and quiet.
https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe  
Archimedes Plutonium

AP
King of Science, especially Physics

The discovery of calculus by Newton and Leibnitz, more than anything else was started with the polynomial function Y=x^2 where it was noticed that 2x was rate of change and that (1/3)x^3 was area under function graph of x^2.

The discovery of calculus by Newton and Leibniz, more than anything else was started with the polynomial function Y=x where it was noticed that 1 was rate of change as dy/dx and that (1/2)x^2 was area under function graph of Y=x. Of course, the rectangle of Y=x is a square and (1/2)x^2 is area of right triangle inside of square.

So, can I retrieve the entire modern day Calculus by noting that the Pythagorean Theorem is right triangles and FLT solutions are Condensed Rectangles..

It should be there!!!

AP
King of Science, especially Physics

It is of no surprise that the integral as area is the polynomial x^2 for square for every rectangle is composed of a square inside itself.

It is no surprise that the hypotenuse of right triangle is the derivative of calculus.

It is no surprise that a square has right-triangles inside the square.

So a Algebra of all functions as polynomials, where equations as A^x + B^y = C^z where x,y,z are 2 or larger covers all equations of 3 geometrical figures of squares and right-triangles. A geometry figure of something like A^4 such as 3^4 can be written as 9^2 or written as 9 times the square 3x3. The equation A^x + B^y = C^z is comprehensive and encompassing for calculus.

And so we can contemplate that Calculus itself comes directly out of Pythagorean theorem as the hypotenuse of right triangle while the Condensed Rectangles of FLT theorem compose the integral of calculus.

AP
King of Science, especially Physics

So, in my recent foray into Fermat's Last Theorem, especially Generalized FLT, I could not help but notice that when I bring in the Shadowy FLT, those A^x + B^y = C^z with at least one of x,y,z as 2, that they also follow the proof of Condensed Rectangles of Generalized FLT. With the exception of 1^3 + 2^3 = 3^2 (where the 1 can actually be any exponent 2 or larger).

What that tells me is that the geometry of the Calculus is also based on a right triangle being flipped up or down from a rectangle turned into a trapezoid and is the derivative of calculus.

Since the proof of Generalized FLT involves condensed rectangles, means, Calculus comes from both the Pythagorean theorem inside a Generalized FLT theorem. And of course, all of that comes from the specialness of 2 and 4 wherein 2+2=2x2=2^2= 4.

For centuries, it was thought that the FLT was a rather insignificant idea and proof of math. Here, today, we see that FLT gives rise to Calculus, and it is for certain that calculus is the top number one idea in all of mathematics. It is the calculus that is the math of Physics.

AP
King of Science, especially Physics

Alright, this should not be too difficult to prove. What I am proving is that the Pythagorean theorem along with the Generalized FLT theorem proven by AP with condensed rectangles creates the Calculus of mathematics. The right triangles of Pythagorean theorem is the derivative of calculus and the Condensed Rectangles of FLT is the Rectangle Square of integrals of calculus. I have a depiction of the geometry below.

We start with a rectangle for integral in a interval say between 1 and 2 in whole numbers or between 0.1 and 0.2 in 10 Grid System.

_______
|______|
1 2

Now in calculus the above represents the area of a rectangle under a function graph and the above would be a function such as perhaps Y = 3 or Y =4, some straight line function horizontal and parallel to x-axis.

But now we take the midpoint of that rectangle, call it m.

___m___
|______|
1 2

And m would be something like 1.5 if we had 1 and 2 , or it would be 0.15 if we had the interval 0.1 to 0.2.

Notice the midpoint in geometrical-calculus is a number that does not exist in the numbers used, for 1.5 is not a whole number, and 0.15 is not a member of the 10 Grid but a member of the next higher grid system of 100 Grid. So in geometrical-calculus, we always have to borrow from a higher grid to make calculus work. For we need empty space from one point to the next successor point.

You are required to have Discrete Space for Calculus to exist at all.

So, now we have our midpoint of the above rectangle and now what I want to do is outline two squares that exist in the upper portion of the rectangle.

___m___
|___|___|
1 2

And now I want to specify a specific function, the easiest function with a derivative that is equal to 1 and with angle 45degrees. That of course is the function Y = x and is a diagonal inside a square.

The integral of the function Y=x is (1/2)x^2 which makes sense as being the area of a isosceles right triangle. So now let me draw into the above picture the two right triangles in the one square.

___m___
| /_|___|
1 2

Now we lift up the one right triangle on its hinge at m, and place it on the base of the other square, looking like this.

m/__|
/_|___|
1 2

So in this diagram we show the geometrical Derivative of Calculus coming from the geometrical Integral of Calculus.

We have shown how right triangles are embedded inside a rectangle and are hinged at a midpoint of rectangle and when lifted up (or down) create a derivative.

If we started with derivative, we still find the midpoint and hinge the right triangle down to form a rectangle that is the integral, the area of the function graph in that interval.

Now in geometrical calculus, we have to take the derivative and integral in the smallest interval possible and in 10 Grid that would be every 0.1 interval, in 100 Grid that would be 0.01 interval. We cannot do geometrical calculus over a function with a interval larger than the smallest interval for that messes up the midpoint.

As I said earlier, the whole numbers, the Counting numbers from 1 to 100 is a representation of the 10 Grid, where 1 = 0.1 and 10 = 100. In this view, as we do calculus upon the 10 Grid, we can switch our view to saying we did calculus on the Counting numbers from 1 to 100 and thus we can say the Pythagorean theorem is our right triangle we lifted up to form the derivative of Y=x. And the Counting numbers from 1 to 100 is the integral area of the condensed rectangle in Generalized FLT, as in the example above of adding together the two squares.

AP
King of Science, especially Physics

On Saturday, July 17, 2021 at 5:00:57 PM UTC+1, Archimedes Plutonium wrote:

[snip - imagine some monkeys randomly typing stuff]

Andrew Wiles is a genius who cracked Fermat's Last Theorem. The proof has been checked and validated by experts. AP probably can't even understand the reasoning involved in the proof let alone much else.

One day someone is going to have to make a decision: leave AP's comments intact for the benefit of some future internet historian writing about 'quirky online personalities', or to simply delete them all in one go. Would anything of value be lost?

Have a Wonderful Day
QB

Remain Calm and Keep Loving Real Analysis

So, now, the function Y= x is a constant derivative of 1 with dy the same number as dx. And is coming from a 90-45-45 right triangle as the right triangle we hinge up or down on the midpoint. But what about a more complex function of Y = x^2 for the derivative varies from each cell to the next and so the right triangle varies from each cell to the next. Read below as to the structure of the cells.

On Saturday, July 17, 2021 at 11:00:57 AM UTC-5, Archimedes Plutonium wrote:
> Alright, this should not be too difficult to prove. What I am proving is that the Pythagorean theorem along with the Generalized FLT theorem proven by AP with condensed rectangles creates the Calculus of mathematics. The right triangles of Pythagorean theorem is the derivative of calculus and the Condensed Rectangles of FLT is the Rectangle Square of integrals of calculus. I have a depiction of the geometry below.
>
> We start with a rectangle for integral in a interval say between 1 and 2 in whole numbers or between 0.1 and 0.2 in 10 Grid System.
>
> _______
> |______|
> 1 2
>
> Now in calculus the above represents the area of a rectangle under a function graph and the above would be a function such as perhaps Y = 3 or Y =4, some straight line function horizontal and parallel to x-axis.
>
> But now we take the midpoint of that rectangle, call it m.
>
> ___m___
> |______|
> 1 2
>
> And m would be something like 1.5 if we had 1 and 2 , or it would be 0.15 if we had the interval 0.1 to 0.2.
>
> Notice the midpoint in geometrical-calculus is a number that does not exist in the numbers used, for 1.5 is not a whole number, and 0.15 is not a member of the 10 Grid but a member of the next higher grid system of 100 Grid.. So in geometrical-calculus, we always have to borrow from a higher grid to make calculus work. For we need empty space from one point to the next successor point.
>
> You are required to have Discrete Space for Calculus to exist at all.
>
> So, now we have our midpoint of the above rectangle and now what I want to do is outline two squares that exist in the upper portion of the rectangle.
>
> ___m___
> |___|___|
> 1 2
>
> And now I want to specify a specific function, the easiest function with a derivative that is equal to 1 and with angle 45degrees. That of course is the function Y = x and is a diagonal inside a square.
>
> The integral of the function Y=x is (1/2)x^2 which makes sense as being the area of a isosceles right triangle. So now let me draw into the above picture the two right triangles in the one square.
>
>
> ___m___
> | /_|___|
> 1 2
>
> Now we lift up the one right triangle on its hinge at m, and place it on the base of the other square, looking like this.
>
>
> m/__|
> /_|___|
> 1 2
>
> So in this diagram we show the geometrical Derivative of Calculus coming from the geometrical Integral of Calculus.
>
> We have shown how right triangles are embedded inside a rectangle and are hinged at a midpoint of rectangle and when lifted up (or down) create a derivative.
>
> If we started with derivative, we still find the midpoint and hinge the right triangle down to form a rectangle that is the integral, the area of the function graph in that interval.
>
> Now in geometrical calculus, we have to take the derivative and integral in the smallest interval possible and in 10 Grid that would be every 0.1 interval, in 100 Grid that would be 0.01 interval. We cannot do geometrical calculus over a function with a interval larger than the smallest interval for that messes up the midpoint.
>
> As I said earlier, the whole numbers, the Counting numbers from 1 to 100 is a representation of the 10 Grid, where 1 = 0.1 and 10 = 100. In this view, as we do calculus upon the 10 Grid, we can switch our view to saying we did calculus on the Counting numbers from 1 to 100 and thus we can say the Pythagorean theorem is our right triangle we lifted up to form the derivative of Y=x. And the Counting numbers from 1 to 100 is the integral area of the condensed rectangle in Generalized FLT, as in the example above of adding together the two squares.
> AP
> King of Science, especially Physics

On Saturday, July 17, 2021 at 3:07:13 PM UTC-5, Archimedes Plutonium wrote:
> So, now, the function Y= x is a constant derivative of 1 with dy the same number as dx. And is coming from a 90-45-45 right triangle as the right triangle we hinge up or down on the midpoint. But what about a more complex function of Y = x^2 for the derivative varies from each cell to the next and so the right triangle varies from each cell to the next. Read below as to the structure of the cells.

So, now, in the function Y=x^2 we want to examine how all of calculus boils down to having Pythagorean right triangles carved out of a Condensed Rectangle of a Generalized FLT of A^x+B^y= C^z where A,B,C are 1 or higher and x,y,z are 2 or higher. And this proof confirms the fact that just as Pythagorean Theorem is a theorem over discrete space, as well as FLT, that Calculus can only exist in discrete space, not a continuum.

So, we examine the Condensed Rectangle of Y=x^2 in 10 Grid for interval 0..9 to 1.0. This would be similar to interval 9 to 10 for integers. And let me retrace my steps in the earlier post for that was of function Y=x and here I have a more complex function Y=x^2.

We start with a rectangle for integral in a interval between 0.9 and 1.0 in 10 Grid

_______
| |
| |
|______|
0.9 1.0

Now in calculus the above represents the area of a rectangle under a function graph that is the integral of Y= x^2 from 0.9 to 1.0.

But now we take the midpoint of that rectangle, call it m.

___m__
| |
| |
|______|
0.9 1.0
And m is thus 0.95.

And we know that at 0.9 the coordinate point is (0.9, x^2) = (0.9, 0.81) and for 1.0 the coordinates are (1.0, 1.0).

So, what we need is a right triangle carved out of the half square or half rectangle from midpoint m. So that the hypotenuse of right triangle in the leftside of m, the hypotenuse has a length so that it spans to reach 1.0. But we must remember the flat-line that contains m is not at Y= 0.81, so the length is not 0.19.

The integral of this interval is (1/3)x^3 and we find the this area by x=1 gives 0.333 and x=0.9 gives 0.242 and subtracting gives area of rectangle with midpoint m. 0.333-0.242 = 0.091. The width of rectangle is 0.1, the height of the rectangle where m is situated must then be 0.91. The coordinate point of m itself, is thus (0.95, 0.91).

So for Y= x^2, from interval 0.9 to 1.0, I need a right triangle carved out of condensed rectangle to sit atop the trapezoid that takes the y coordinate point from (0.9, 0.81) to that of (1, 1). So I need a hypotenuse of a right triangle with length that spans from 0.81 to that of 1. I already have a base of the trapezoid at elevation of 0.91. So I need a hypotenuse of right triangle with legs of 0.05 and the quantity of 1-0.91 = 0.09. So my right triangle is going to have to be with legs 0.05 and 0.09. This would be a hypotenuse of .05^2 + .09^2 = .0025 + .0081 and taking square root of = 0.102.

Now let us see if everything agrees, agrees together.

Now we spend some time in doing the dy/dx in the Angle Chart.

90    84   78.5    73    68    63    59    55    51    48    45 these are angles made from (0,0)

10/1 10/2 10/3 10/4 10/5 10/6 10/7 10/8 10/9 10/10 slope1.0

90  84   78.5    73    68    63    59    55    51   48    45, slope 1, rt-triangle 10by10

                                                                      /     42, rt-triangle 10by9 blocks

                                                      /            39, rt-triangle 10by8 blocks

                                                     /                    35, rt-triangle 10by7 blocks

                                    /                            31, rt-triangle 10by6 blocks
45degree
                                       /                                    27, rt-triangle 10by5 blocks

                          /                                           22, rt-triangle 10by4 blocks

                        /                                                   17, rt.triangle 10by3 blocks
/
                                                                                11.5, rt-triangle 10by2 blocks
/
                                                                                 6, rt-triangle 10by1 blocks
/
0    .1      .2     .3      .4      .5      .6     .7    .8    .9     1.0

The derivative of Y=x^2 is 2x. The derivative at x=.9 would be thus 1.8

That is nearby to 10by5 which is 27 degrees and its complament is thus 2 and close to 1.8, meaning a angle of about 60 to 63 degrees.

AP
King of Science, especially Physics

Archimedes Plutonium<plutonium.archimedes@gmail.com>
Jul 17, 2021, 6:29 PM
to Plutonium Atom Universe
For me this has solved one of the nagging questions on my mind. Can a derivative slope rise so fast that you need a rectangle to form the right triangle at midpoint. And here the answer is yes, you need a rectangle to form the right triangle at midpoint.

But even a larger nagging question is whether the calculus function rises so rapidly there is no rectangle to hold a right triangle to hing upward and reach the next successor coordinate point of the function graph. Apparently this is true, that at some exponent, the rise is too rapid.

And if true, places a barrier on polynomials and exponents to be calculus. So that algebraically we can compute a derivative of say Y= x^604, but geometrically we run out of room to form a rectangle to form a right triangle to hinge upward.

Is this barrier at exponent 7??? I do not know. I say 7 because I have yet not found a Shadowy FLT for exponent 7. It would be like the barrier of quintic in polynomials.
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Archimedes Plutonium's profile photo
Archimedes Plutonium<plutonium.archimedes@gmail.com>
Jul 17, 2021, 6:53 PM
to Plutonium Atom Universe

Read the below, for this implies a barrier in the heart of Calculus.

If the above is true, would mean that Calculus exists only in a range of polynomial exponents of 1 to 6, ending at 6.

And a nice interpretation of that barrier, is that Physics electromagnetic forces stop at exponent 6, just as the speed of light has a upper limit of speed and no faster speed.

If all of physics is just a interplay of the 6 laws of electricity and magnetism, those 6 laws and their forces would have a upper limit of calculus, because the forces could not generate the force of rate of change or area under function graph.

This would be a beautiful harmonization of physics and math, both lock-step with one another.

So reviewing my recent attempt at finding a Shadowy FLT for 7.

Now looking at Shadowy FLT where one of A,B,C has exponent 2.

2^2 + 2^2 = 2^3
2^2 + 2^5 = 6^2
2^3 + 2^3 = 4^2
3^3 + 3^2 = 6^2
4^2 + 4^2 = 2^5
5^2 + 10^2 = 5^3

I could not find a Shadowy FLT for A,B,C being 7. Sort of like a barrier not just for FLT but more importantly, for Calculus.

y  z
|  /
| /
|/______ x

More people reading and viewing AP's newsgroup than viewing sci.math, sci.physics. So AP has decided to put all NEW WORK, to his newsgroup. And there is little wonder because in AP's newsgroups, there is only solid pure science going on, not a gang of hate spewing misfits blighting the skies.

In sci.math, sci.physics there is only stalking hate spew along with Police Drag Net Spam of no value and other than hate spew there is Police drag net spam day and night.

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of stalkers and spammers, Police Drag Net Spam that floods each and every day, book and solution manual spammers, off-topic-misfits, front-page-hogs, churning imbeciles, stalking mockers, suppression-bullies, and demonizers.  And the taxpayer funded hate spew stalkers who ad hominem you day and night on every one of your posts.

There is no discussion of science in sci.math or sci.physics, just one long line of hate spewing stalkers followed up with Police Drag Net Spam (easy to spot-- very offtopic-- with hate charged content). And countries using sci.physics & sci.math as propaganda platforms, such as tampering in elections with their mind-rot.

Read my recent posts in peace and quiet.
https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe  
Archimedes Plutonium

On Saturday, July 17, 2021 at 8:49:36 PM UTC-5, Archimedes Plutonium wrote:
> Now looking at Shadowy FLT where one of A,B,C has exponent 2.
> 2^2 + 2^2 = 2^3
> 2^2 + 2^5 = 6^2
> 2^3 + 2^3 = 4^2
> 3^3 + 3^2 = 6^2
> 4^2 + 4^2 = 2^5
> 5^2 + 10^2 = 5^3
> I could not find a Shadowy FLT for A,B,C being 7. Sort of like a barrier not just for FLT but more importantly, for Calculus.

I spent some more time today, looking to see if I can muster up a 7 Shadowy FLT, using both 7 and 14, but to no avail. I know that if one exists, it is likely to be a small number, not way out there. I suspect that 7 is going out too far and is linked with the barrier limit of a derivative of calculus, as too steep a rise/run. But, maybe a computer program can find a 7 Shadowy FLT.

Also looking at 1^3 + 2^3 = 3^2 as a Shadowy FLT if I bend the definition around to say that 1 is a common factor in 1,2,3 and that would create a Condensed Rectangle of 1 unit wide and either 2 or 3 units long. This rectangle would be long and thin.

And also, considering a Algebraic proof that 2+2=2x2=2^2 = 4 covers this special FLT of 1^3 + 2^3 = 3^2 in a simple proof of 1+2^3 = (2+1)^2
1+2^3 = 1+ 2(2+2) = 1 + 4 + 4 = 1+2+2+2+2

and

(2+1)
x(2+1) = 4+ 2+2 + 1 = 2+2+2+2 +1

Perhaps the meaning in math of 1^3 + 2^3 = 3^2, is that it is the unit cell of integral before we find the midpoint and split the unit cell into two parts to construct the right triangle to span to the next coordinate point..

AP
King of Science, especially Physics

On Sunday, July 18, 2021 at 2:05:28 AM UTC-5, Archimedes Plutonium wrote:
> On Saturday, July 17, 2021 at 8:49:36 PM UTC-5, Archimedes Plutonium wrote:
> > Now looking at Shadowy FLT where one of A,B,C has exponent 2.
> > 2^2 + 2^2 = 2^3
> > 2^2 + 2^5 = 6^2
> > 2^3 + 2^3 = 4^2
> > 3^3 + 3^2 = 6^2
> > 4^2 + 4^2 = 2^5
> > 5^2 + 10^2 = 5^3
> > I could not find a Shadowy FLT for A,B,C being 7. Sort of like a barrier not just for FLT but more importantly, for Calculus.

I still cannot find a Shadowy FLT involving 7 for A,B,C.

So what I plan to do next is outline this triple 3^3 + 3^2 = 6^2 to see what type of derivative slope it has, whether close to being almost 90 degree slope.

My intuition or hunch is that Calculus as a barrier and that this barrier involves 7. What it means is very important for physics, for it means that the forces of electricity and magnetism end at 7 in A^x + B^y = C^z.

AP
King of Science, especially Physics

My conjecture of a barrier in Calculus due to Shadowy FLT is down the drain.

2^5 + 7^2 = 3^4

I no longer have a Calculus barrier with that counterexample.

However I do have a new conjecture-- every number starting with 1 onward is a number squared in A^x + B^y = C^z wherein that number M as M^2 is a A,B, or C such that M^2 is a solution.

A list of this would be:
1^2 + 2^3 = 3^2
2^2 +2^2 = 2^3
3^2 + 3^3 = 6^2
4^2 +4^2 = 2^5
5^2 +10^2 = 5^3
2^5+ 7^2 = 3^4

So, the conjecture is that every counting number, 1, 2, 3, 4, .... has a square involved in a solution of A^x + B^y = C^z.

This conjecture maybe more difficult to prove than the original FLT or the Generalized FLT.

On Sunday, July 18, 2021 at 2:15:38 PM UTC-5, Archimedes Plutonium wrote:
> On Sunday, July 18, 2021 at 2:05:28 AM UTC-5, Archimedes Plutonium wrote:
> > On Saturday, July 17, 2021 at 8:49:36 PM UTC-5, Archimedes Plutonium wrote:
> > > Now looking at Shadowy FLT where one of A,B,C has exponent 2.
> > > 2^2 + 2^2 = 2^3
> > > 2^2 + 2^5 = 6^2
> > > 2^3 + 2^3 = 4^2
> > > 3^3 + 3^2 = 6^2
> > > 4^2 + 4^2 = 2^5
> > > 5^2 + 10^2 = 5^3
> > > I could not find a Shadowy FLT for A,B,C being 7. Sort of like a barrier not just for FLT but more importantly, for Calculus.
> I still cannot find a Shadowy FLT involving 7 for A,B,C.
>
> So what I plan to do next is outline this triple 3^3 + 3^2 = 6^2 to see what type of derivative slope it has, whether close to being almost 90 degree slope.
>
> My intuition or hunch is that Calculus has a barrier and that this barrier involves 7. What it means is very important for physics, for it means that the forces of electricity and magnetism end at 7 in A^x + B^y = C^z.
> AP
> King of Science, especially Physics

On Monday, July 19, 2021 at 12:42:22 AM UTC-5, Archimedes Plutonium wrote:
> My conjecture of a barrier in Calculus due to Shadowy FLT is down the drain.
>
> 2^5 + 7^2 = 3^4
>
> I no longer have a Calculus barrier with that counterexample.
>
> However I do have a new conjecture-- every number starting with 1 onward is a number squared in A^x + B^y = C^z wherein that number M as M^2 is a A,B, or C such that M^2 is a solution.
>
> A list of this would be:
> 1^2 + 2^3 = 3^2
> 2^2 +2^2 = 2^3
> 3^2 + 3^3 = 6^2
> 4^2 +4^2 = 2^5
> 5^2 +10^2 = 5^3
> 2^5+ 7^2 = 3^4
>
> So, the conjecture is that every counting number, 1, 2, 3, 4, .... has a square involved in a solution of A^x + B^y = C^z.
>

Alright, one of the reasons I did not find 2^5 + 7^2 = 3^4 sooner was that I was looking for a composite of 7. So that this Shadowy FLT with at least one term in exp 2, has no common factor other than if we allowed 1.

So what AP is thinking is that his Shadowy FLT allows for exponent 2, whereas Generalized FLT allows only exponent 3 or higher.

This sounds troubling. And perhaps a contradiction. And perhaps there exists a Generalized FLT of A^x + B^y = C^z where A,B,C is 1 or higher and where x,y,z = 3 or higher, for which the only common factor is 1.

Now an example of Generalized FLT with exp3 or higher is 7^3 + 7^4 = 14^3 and we see the common factor of 7.

Now here is an interesting Shadowy FLT of 3^5 + 11^4 = 122^2. It is not a Generalized FLT because of the one exp2, but then one has to ask the question. Just because you have a exp2, should not hamper the problem such that only a common factor greater than 1 exists.

What I am saying is that, our statement of Generalized FLT is in error. That we should include exp2, so long as not all three of x,y,z are exponent 2. And that our end result is that a solution can exist where the only common factor is 1 unit.

For example: 1^2 + 2^3 = 3^2 and the example of 2^5 + 7^2 = 3^4 where in both cases the only common factor is 1. This would predict or forecast that a Generalized FLT exists where the only common factor of A,B,C is 1. And we just have not yet spotted such a solution equation.

AP
King of Science, especially Physics

On Monday, July 19, 2021 at 12:42:22 AM UTC-5, Archimedes Plutonium wrote:
> My conjecture of a barrier in Calculus due to Shadowy FLT is down the drain.
>
> 2^5 + 7^2 = 3^4
>
> I no longer have a Calculus barrier with that counterexample.
>
> However I do have a new conjecture-- every number starting with 1 onward is a number squared in A^x + B^y = C^z wherein that number M as M^2 is a A,B, or C such that M^2 is a solution.
>
> A list of this would be:
> 1^2 + 2^3 = 3^2
> 2^2 +2^2 = 2^3
> 3^2 + 3^3 = 6^2
> 4^2 +4^2 = 2^5
> 5^2 +10^2 = 5^3
> 2^5+ 7^2 = 3^4
>
> So, the conjecture is that every counting number, 1, 2, 3, 4, .... has a square involved in a solution of A^x + B^y = C^z.
>

Alright, one of the reasons I did not find 2^5 + 7^2 = 3^4 sooner was that I was looking for a composite of 7. So that this Shadowy FLT with at least one term in exp 2, has no common factor other than if we allowed 1.

So what AP is thinking is that his Shadowy FLT allows for exponent 2, whereas Generalized FLT allows only exponent 3 or higher.

This sounds troubling. And perhaps a contradiction. And perhaps there exists a Generalized FLT of A^x + B^y = C^z where A,B,C is 1 or higher and where x,y,z = 3 or higher, for which the only common factor is 1.

Now an example of Generalized FLT with exp3 or higher is 7^3 + 7^4 = 14^3 and we see the common factor of 7.

Now here is an interesting Shadowy FLT of 3^5 + 11^4 = 122^2. It is not a Generalized FLT because of the one exp2, but then one has to ask the question. Just because you have a exp2, should not hamper the problem such that only a common factor greater than 1 exists.

What I am saying is that, our statement of Generalized FLT is in error. That we should include exp2, so long as not all three of x,y,z are exponent 2. And that our end result is that a solution can exist where the only common factor is 1 unit. Or, perhaps go even stronger and call all equations of solutions for A^x + B^y = C^z where x,y,z can all three be 2, call them all as FLT, and that most have a common factor of greater than 1 but many have only a common factor of 1 itself.

For example: 1^2 + 2^3 = 3^2 and the example of 2^5 + 7^2 = 3^4 where in both cases the only common factor is 1. This would predict or forecast that a Generalized FLT exists where the only common factor of A,B,C is 1. And we just have not yet spotted such a solution equation.

So, we have three categories of equations of form A^x + B^y = C^z.
1) Pythagorean theorem
2) Shadowy FLT, at least one exponent 2 but not all three A,B,C
3) Generalized FLT, exp 3 or higher

I am trying to forge those three into being just one, by saying all have common factor of 1, but others have 1 and another number.

Hard for me to believe that Shadowy FLT where one or two of the exponents is 2, the other higher has either a solution with factor 1 or higher, yet every Generalized FLT has solution with only a higher than 1 unit factor solution. All the Pythagorean Theorem solutions are 1 unit factor solutions, so how would a single square cause all solutions be a 1 unit factor or higher solution for Shadowy FLT, but not for Generalized FLT.

On Monday, July 19, 2021 at 3:19:04 AM UTC-5, Archimedes Plutonium wrote:
> On Monday, July 19, 2021 at 12:42:22 AM UTC-5, Archimedes Plutonium wrote:
> > My conjecture of a barrier in Calculus due to Shadowy FLT is down the drain.
> >
> > 2^5 + 7^2 = 3^4
> >
> > I no longer have a Calculus barrier with that counterexample.
> >
> > However I do have a new conjecture-- every number starting with 1 onward is a number squared in A^x + B^y = C^z wherein that number M as M^2 is a A,B, or C such that M^2 is a solution.
> >
> > A list of this would be:
> > 1^2 + 2^3 = 3^2
> > 2^2 +2^2 = 2^3
> > 3^2 + 3^3 = 6^2
> > 4^2 +4^2 = 2^5
> > 5^2 +10^2 = 5^3
> > 2^5+ 7^2 = 3^4
> >
> > So, the conjecture is that every counting number, 1, 2, 3, 4, .... has a square involved in a solution of A^x + B^y = C^z.
> >
>
> Alright, one of the reasons I did not find 2^5 + 7^2 = 3^4 sooner was that I was looking for a composite of 7. So that this Shadowy FLT with at least one term in exp 2, has no common factor other than if we allowed 1.
>
> So what AP is thinking is that his Shadowy FLT allows for exponent 2, whereas Generalized FLT allows only exponent 3 or higher.
>
> This sounds troubling. And perhaps a contradiction. And perhaps there exists a Generalized FLT of A^x + B^y = C^z where A,B,C is 1 or higher and where x,y,z = 3 or higher, for which the only common factor is 1.
>
> Now an example of Generalized FLT with exp3 or higher is 7^3 + 7^4 = 14^3 and we see the common factor of 7.
>
> Now here is an interesting Shadowy FLT of 3^5 + 11^4 = 122^2. It is not a Generalized FLT because of the one exp2, but then one has to ask the question. Just because you have a exp2, should not hamper the problem such that only a common factor greater than 1 exists.
> What I am saying is that, our statement of Generalized FLT is in error. That we should include exp2, so long as not all three of x,y,z are exponent 2. And that our end result is that a solution can exist where the only common factor is 1 unit. Or, perhaps go even stronger and call all equations of solutions for A^x + B^y = C^z where x,y,z can all three be 2, call them all as FLT, and that most have a common factor of greater than 1 but many have only a common factor of 1 itself.
> For example: 1^2 + 2^3 = 3^2 and the example of 2^5 + 7^2 = 3^4 where in both cases the only common factor is 1. This would predict or forecast that a Generalized FLT exists where the only common factor of A,B,C is 1. And we just have not yet spotted such a solution equation.
> So, we have three categories of equations of form A^x + B^y = C^z.
> 1) Pythagorean theorem
> 2) Shadowy FLT, at least one exponent 2 but not all three A,B,C
> 3) Generalized FLT, exp 3 or higher
>
> I am trying to forge those three into being just one, by saying all have common factor of 1, but others have 1 and another number.

On Monday, July 19, 2021 at 7:53:50 AM UTC-5, Archimedes Plutonium wrote:
> Hard for me to believe that Shadowy FLT where one or two of the exponents is 2, the other higher has either a solution with factor 1 or higher, yet every Generalized FLT has solution with only a higher than 1 unit factor solution. All the Pythagorean Theorem solutions are 1 unit factor solutions, so how would a single square cause all solutions be a 1 unit factor or higher solution for Shadowy FLT, but not for Generalized FLT.

Now I do not recall if anyone has proven that in the Pythagorean Theorem A^2 + B^2 = C^2 that the Primitive P Triples cannot be all three A,B,C are even numbers? Is a proof of that statement the simple fact that 3,4,5 is the smallest Primitive P-Triple in integers, and hence all other Primitive P-triples have to follow that pattern of where one of A,B,C is even and the other two odd numbers. Is that a proof that A,B,C cannot all three be even numbers? I believe that is a proof that Pythagorean Primitive P-Triples must have 2 odds and 1 even. For the moment you have the smallest Primitive P-Triple and it is not all three evens for A,B,C, means that you can divide out the evens by 2 and have a smaller Primitive P-Triple.

Of course, in Euler's exp3 FLT fake proof, he cannot have a ready made smallest Primitive Triple Solution and observe that it is all three even numbers or 2 odd with 1 even in the solution A,B,C.
All in Old Math that think Euler's proof of exp3 FLT is sound, point to this idea that if all three A,B,C were even, you can immediately divide out the evens by 2 and obtain a 2 odd with 1 even. But they are gravely logically mistaken for the example of 2^5 + 2^5 = 2^6 so we divide out by 2 to have 1^5 + 1^5 =/= 1^6. What is the logical fallacy that everyone in Old Math believing Euler had no gap in his proof? I do not know if that logical fallacy has a name. A name to refer to the logical mistake of thinking a pattern in exp2 is to hold true in a pattern for a entirely different exponent 3. To think that area of squares pattern holds true in volume of cubes pattern is a Logical Fallacy. So that three cubes of volume in integers of A^3 + B^3 = C^3, does not mean that you can divide out by 2 and still have equality. Yet the logical-igorant in Old Math, seems to believe that Euler had a proof of exp3 by borrowing the pattern in exp2, when obviously you cannot borrow that pattern.

And it is for that reason that Euler could never have a proof of exp3 FLT without a proof that 3 even integers cannot solve for A^3 + B^3= C^3. But try telling that or even convincing the ignorant Andrew Wiles and his con-art fakery of FLT.

So yes, above is the easy proof that Primitive P Triples must be of the pattern 2 odd with 1 even for exponent 2, but obviously none of that transfers to exponent 3, where you need to prove no three evens exist as a solution in A^3 + B^3= C^3. For mathematicians that can calculate are dime a dozen, but mathematicians with a logical mind that do not slip into fallacies, well they are extremely rare.

AP
King of Science, especially Physics

Alright, so, let us examine this situation of where Old Math has 3 groupings for the equation A^x + B^y = C^z.

We have

1) Pythagorean Theorem A^2 + B^2 = C^2 and that is geometrically the forming of a Right Triangle by placing the squares at corners from each other.

2) A^x + B^y = C^z called Generalized FLT where A,B,C = 1 or higher, and x,y,z = 3 or higher and the AP proof of 2014 via Condensed Rectangles show we can adjoin two rectangles with a common side to form the C^z rectangle of common side.

3) Until recently, we have the Shadowy FLT, all A^x + B^y = C^z with A,B,C = 1 or higher and x,y,z 2 or higher. If we said 1 or higher for exponent we would trivially include Arithematic e.g. 2 + 3 = 5 and that would be pointless.

But the question here becomes since Shadowy FLT has equations that have no common factor other than 1. So that questions the truth of Generalized FLT; are there solutions which have 1 as the only common factor.

So looking at some examples of

2^5 + 7^2 = 3^4
1^3 + 2^3 = 3^2
3^2 + 3^3 = 6^2
and
7^3 + 7^4 = 14^3

And as we analyze these such as 7^3 + 7^4 would be two condensed rectangles, one that is side 7 by 49 and one with side 7 by 343. So we stack 7by49 with 7by343 on side 7 and the result is 7 by (49+343=392) and now we have a rectangle of 7 by 392. Now we see if that is the same condensed rectangle of 14^3. And sure enough 7*392 is 2744 as well as 14^3 is 2744.

So the proof of Generalized FLT was always about comparing Condensed Rectangles that HAD to have a common side, all three rectangles having that common side.

But now, we have Shadowy FLT here in 2021, whereas in 2014 we had just Generalized FLT. And Shadowy FLT shows some equations with the only factor in common is the factor of 1.

This of course, begs me to ask the question if Generalized FLT has solutions where the only common factor is 1. Whereas Old Math put Generalized FLT saying the factor in common had to be a prime number, not 1 for 1 is not a prime in Old Math but a units.

So, this is why I am here now, battling whether Generalized FLT has any solutions that have only 1 as a common factor. And I come very close with a example already in that of 2^5 + 7^2 = 3^4 if only there was no exp2.

So, the perplexing question is, does Generalized FLT also have solutions where 1 is the only common factor, or, does Generalized FLT, due to the fact one of the exponents is a 2, somehow avert the chances of a solution with 1 as the only common factor.

So I turn my attention not to volume but to condensed rectangles.

In 2^5 + 7^2 = 3^4 we can have rectangles 1by32 and 1by49 to join to form rectangle 1by81. This was a Shadowy FLT equation. But is there a Generalized FLT equation which can only be assembled with common factor 1. If not, what is the mechanism that a single 2 exponent throws it off and has to be a prime factor {2, 3, 5, 7, 11, ...}.

AP
King of Science, especially Physics

Whenever I am in a tough, tight, perplexed problem like this-- does Generalized FLT have solutions of where 1 is the only common factor, because Shadowy FLT where one or two of the exponents of x,y,z are 2, and the third is 3 or higher, for Shadowy FLT has solutions where 1 is the only common factor, eg, 2^5 + 7^2 = 3^4.

So, which is true, only Shadowy FLT but not Generalized FLT has 1 factor solutions or does both FLT have solutions where 1 is the only common factor.

When stuck with a problem like this, I generally lunge forward with a idea, favoring one answer, and spend the rest of the day pondering that solution..

So, let me lunge forward into the unknown by siding on the side that no Generalized FLT, all exponents 3 or higher has a solution where 1 is the only common factor. Let me side with that, not because that is the history of Old Math siding with that idea, for it is doubtful anyone in Old Math ever realized they had to include the Shadowy FLT in a overall picture of equation A^x + B^y = C^z.

But I side on Shadowy FLT as the only 1 unit factor solutions because Pythagorean Theorem is squares locked together to form right triangles, while Generalized FLT is rectangles joined from sides forming another new rectangle..

So the process of locking together vertex of squares to form in its center a right triangle is a altogether different process of locking together entire sides of rectangles to form a larger rectangle.

So in those two processes, of square + square locked at vertices versus rectangle + rectangle locked together at sides, it would seem to me, the most probably outcome of a Shadowy FLT, a hybrid world of FLT, that the appearance of a square + rectangle, that it would not know where to lock onto a side or a vertex, and that this hybrid process of locking together geometry figures would tend to favor only solutions for Generalized FLT is a locking of side to side, not vertex locking.

So let me mull that idea over and over for the entire rest of the day. I am pretty confident even now, as I write this, that the Generalized FLT has only common factor solutions of {2,3,5,7,...} and because Shadowy FLT has a square involved not a rectangle, it is able to have solutions with no common factor other than 1. This is bolstered from the fact that all Primitive Pythagorean Theorem Triples have only 1 as a common factor.

AP
King of Science, especially Physics

Alright, spent the day thinking about this, and coming closer to a overall proof. I have so far, mini proofs below.

In Pythagorean Theorem A^2 + B^2 = C^2 we have three squares when joined vertex to vertex form a right triangle.

In Generalized FLT A^x + B^y = C^z where x,y,z must all be 3 or higher we have condensed rectangles of A and B when joined at a common side equals the condensed rectangle of C.

In Shadowy FLT, a hybrid of A^x + B^y = C^z, where one or two of the A,B,C is exponent 2 and the other/s are exponent 3 or higher. We can have only a condensed rectangle solution, but with the caveat that some of the condensed rectangles are with a unit 1 length. In Generalized FLT the condensed rectangles joined to sides must be one of {2,3,5,7,....}. In Shadowy FLT, condensed rectangles come from {1,2,3,5,7, ....}.

So, am I closer to a proof of the above? Well it is obvious that since Generalized FLT has no A,B,C with exponent 2, thus, no squares, leaving all A,B,C as condensed rectangles, and that alone would serve as a proof you will not find a unit 1 factor in {1,2,3,5,7, ....} in Generalized FLT.

AP, King of Science

On Tuesday, July 20, 2021 at 1:23:05 AM UTC-5, Archimedes Plutonium wrote:
> Alright, spent the day thinking about this, and coming closer to a overall proof. I have so far, mini proofs below.
>
> In Pythagorean Theorem A^2 + B^2 = C^2 we have three squares when joined vertex to vertex form a right triangle.
>
> In Generalized FLT A^x + B^y = C^z where x,y,z must all be 3 or higher we have condensed rectangles of A and B when joined at a common side equals the condensed rectangle of C.
>
> In Shadowy FLT, a hybrid of A^x + B^y = C^z, where one or two of the A,B,C is exponent 2 and the other/s are exponent 3 or higher. We can have only a condensed rectangle solution, but with the caveat that some of the condensed rectangles are with a unit 1 length. In Generalized FLT the condensed rectangles joined to sides must be one of {2,3,5,7,....}. In Shadowy FLT, condensed rectangles come from {1,2,3,5,7, ....}.
>
> So, am I closer to a proof of the above? Well it is obvious that since Generalized FLT has no A,B,C with exponent 2, thus, no squares, leaving all A,B,C as condensed rectangles, and that alone would serve as a proof you will not find a unit 1 factor in {1,2,3,5,7, ....} in Generalized FLT.
>

So, we can accept the idea that in Shadowy FLT, we can have either a condensed rectangle solution where the rectangles are 1 unit wide and however many units long rectangle, or we can have condensed rectangle solutions whose width is one of the primes {2,3,5,7, ....}.

In Generalized FLT, we can only have condensed rectangle solutions of a prime width.

In Pythagorean Theorem we can only have solutions if the three squares joined at vertices forms a right triangle in center of outlying squares.

But I want to return to this idea of a barrier in Calculus. My earlier attempt of placing a barrier, where the rise is too enormous for the run and thus no geometry for a right triangle hinged atop a trapezoid for the derivative of calculus.

Does Calculus have a internal barrier? The idea is so fascinating, it will not let me go. When I could not immediately find a Shadowy FLT for A^x + B^y = C^z for 7, I thought of a barrier. But later I found 2^5 + 7^2 = 3^4. But there still maybe a internal barrier in Calculus, where you cannot get derivatives for a high rising polynomial function. Of course, you would not ask for a Y=x^500 in 10 Grid for you would have no derivatives. But in 10000 Grid, you have derivatives of x^500. So I am not talking about a Grid encumbrance, but a actual internal barrier in Calculus where derivatives fail. Where you cannot carve out a right triangle from the integral rectangle and place the right triangle atop the trapezoid as derivative reaching for the next successive coordinate point of function graph.

So I plug into my calculator in 100 Grid that of the function Y=x^7, I could do it in 10 Grid but want to be comfortable with lots of room. And I always chose a number for a cell of 1 to 1.01 in 100 Grid or 2 to 2.01 cell. I do not like the numbers between 0 and 1 for they can be deceptive to the mind.

So I calculate 1.01^7 is 1.0721... and see no problems there. But what if I had Y= x^70. Would I run into trouble in 100 Grid? And x^70 for 1.01 is 2.0067... But let me look to see what x^100 is for 1.01. And I get 2.7048.... not in any trouble. But I suppose if I did the cell of 5 to 5.01 that my x^100 would be beyond the 100 mark of 100 Grid.

So is that the take away. That in Calculus in Decimal Grid Systems, the only limitation or barrier is one of the highest number in the grid system such as 10 in 10 Grid, may have no numbers in that Grid System if the exponent is high. For example Y=x^2 so that we have a number for 3.1 that is within 10 Grid, but not for 3.2 which is outside 10 Grid and would have to borrow from 100 Grid to continue beyond 3.1.

So in Grid Systems, I have always believed we can borrow all the way to 10^604, and if that is the case, we never have a Calculus barrier. So maybe that is the answer. Our ability to borrow higher grids eliminates any barrier to calculus.

AP, King of Science

🐒 of Math and 🦍 of Physics Archimedes "AnalButtfuckManure" Plutonium
<plutonium.archimedes@gmail.com> fails at math and science:

> So, we can accept

Your diseased cats also accept that crap?

> the idea that in Shadowy FLT, we can have either a condensed rectan

The letters in "Archimedes Plutonium" can be rearranged to spell "Hi, I
pound male rectums!"

Far easier for a con-artist fraud Andrew Wiles of Math to hire a stalker, than to ever have to confront the gaping holes of his fake and fraud proof Fermat's Last Theorem--- Andrew, do you pay these stalkers???? With your prize money???? For certainly Andrew, you never confront listed mistakes of gaping holes in your con-art proof.

(1) Could not even see Euler had no FLT proof in exponent 3.
(2) Has a wacko understanding of Logic and the true logic connectors where Reductio Ad Absurdum is not a valid proof method of mathematics, even the Intuitionist logic school rejects RAA, but not con-artists of math.
(3) So dumb in math, Andrew Wiles could not be bothered for 5 minutes to place a Kerr lid inside a homemade paper cone and see the Oval is the slant cut, never the Ellipse, and the fool Andrew uses elliptic curves in his nonsense FLT.
(4) So failed in math, Andrew never realized Calculus was geometry, with his "limit analysis" of Fundamental Theorem of Calculus, and so confused in math, that Andrew never realized the onus was upon him to "have" a geometry proof of Fundamental Theorem of Calculus (see below of AP's proofs).

On Wednesday, July 21, 2021 at 1:28:29 AM UTC-5, Michael Moroney wrote:
>"stalking shitwit"
> "AnalButtfuckManure"
> fails at math and science:
> "Hi, I
> pound male rectums!"

Instead of Andrew Wiles discussing why he missed Euler's gaping hole of a proof in exp3 of FLT, or why Andrew never had a geometry proof of Fundamental Theorem of Calculus or why Andrew never had the Logic connectors correct-- to see that Reductio Ad Absurdum is not a valid math proof method. Rather than face questions of his Fermat's Last Theorem, Andrew choses to run and hide from math reality. And I would bet Andrew delights in a foghorn spitting spewing swearing Kibo Parry M to do Andrew's talking.

Andrew, is this all part of the con-art of fake math that dupes the world general public? You failed Mathematics Andrew-- you could not even detect that Euler had a fake proof in exp 3 of FLT, for Euler forgot he had to prove when A,B,C all three are evens A^3+B^3= C^3.

Of course you would miss that gaping hole Andrew because you never had 2 marbles of logic in your entire life in math, for you still believe to this very day that 2 OR 1 = 3 with AND as subtraction, believing that Either..Or..Or..Both is logically sound giving OR truth table as TTTF, when AND truth table is really TTTF, not the Boole hypocrisy of TFFF. No, Andrew, you even failed Logic, not realizing that Reductio Ad Absurdum is not a valid proof method of mathematics, and RAA is your entire fake con-artist Fermat's Last Theorem.

And even worse, Andrew, you were so so dumb in geometry, you could not even see that slant cut in single cone is a Oval, never the ellipse. And a failure like you Andrew thinks he proved Fermat's Last Theorem.

Are you paying Kibo Parry M. to stalk Andrew? Are your prize money's going to fund Kibo to stalk for another 28 years.

Andrew, is it easier to never have to do -- true math and hire and pay a 24-7 stalker rather than engage in the math that you con-arted away with?

Be silent Andrew about a geometry proof of Fundamental Theorem of Calculus, be silent Andrew as your dummy stalker screams and hollers obscenities so you never have to do math when math is needed. Yes, Andrew, Calculus is geometry, yet you never understood that idea, for your "limit analysis" is never a geometry proof, but is a spineless excuse of a fake proof of FTC.

This thread needs to be on the 1st page of a Google Search of "Andrew Wiles" not 5 million hits all saying Andrew Wiles is a math genius, for Andrew is not that at all, but a cowardly con-artist of mathematics, who refuses to ever talk about the gaping holes of his nonsense FLT, or any of the math topics of this post.

11th published book

World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 19May2021. This is AP's 11th published book of science.
Preface:
Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.

Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?

Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.

Length: 137 pages

Product details
ASIN : B07PQTNHMY
Publication date : March 14, 2019
Language : English
File size : 1307 KB
Text-to-Speech : Enabled
Screen Reader : Supported
Enhanced typesetting : Enabled
X-Ray : Not Enabled
Word Wise : Not Enabled
Print length : 137 pages
Lending : Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
#20 in Calculus (Kindle Store)

#8-4, 28th published book

World's First Valid Proof of 4 Color Mapping Problem// Math proof series, book 4 Kindle Edition
by Archimedes Plutonium (Author)

Now in the math literature it is alleged that Appel & Haken proved this conjecture that 4 colors are sufficient to color all planar maps such that no two adjacent countries have the same color. Appel & Haken's fake proof was a computer proof and it is fake because their method is Indirect Nonexistence method. Unfortunately in the time of Appel & Haken few in mathematics had a firm grip on true Logic, where they did not even know that Boole's logic is fakery with his 3 OR 2 = 5 with 3 AND 2 = 1, when even the local village idiot knows that 3 AND 2 = 5 with 3 OR 2 = either 3 or 2 depending on which is subtracted. But the grave error in logic of Appel & Haken is their use of a utterly fake method of proof-- indirect nonexistence (see my textbook on Reductio Ad Absurdum). Wiles with his alleged proof of Fermat's Last Theorem is another indirect nonexistence as well as Hales's fake proof of Kepler Packing is indirect nonexistence.
Appel & Haken were in a time period when computers used in mathematics was a novelty, and instead of focusing on whether their proof was sound, everyone was dazzled not with the logic argument but the fact of using computers to generate a proof. And of course big big money was attached to this event and so, math is stuck with a fake proof of 4-Color-Mapping. And so, AP starting in around 1993, eventually gives the World's first valid proof of 4-Color-Mapping. Sorry, no computer fanfare, but just strict logical and sound argument.

Cover picture: Shows four countries colored yellow, red, green, purple and all four are mutually adjacent. And where the Purple colored country is landlocked, so that if it were considered that a 5th color is needed, that 5th color should be purple, hence, 4 colors are sufficient.
Length: 29 pages

File Size: 1183 KB
Print Length: 29 pages
Publication Date: March 23, 2019
Sold by: Amazon Digital Services LLC
Language: English
ASIN: B07PZ2Y5RV
Text-to-Speech: Enabled 
X-Ray: 
Not Enabled  

Word Wise: Not Enabled
Lending: Enabled
Screen Reader: Supported 
Enhanced Typesetting: Enabled 



#8-5, 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.