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Topic: my second proof of FLT-- purely geometry with concept of three-faced Cube
Replies: 13   Last Post: Oct 1, 2017 1:00 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
my second proof of FLT-- purely geometry with concept of three-faced Cube
Posted: Sep 28, 2017 9:08 AM

Newsgroups: sci.math
Date: Thu, 7 Sep 2017 22:36:55 -0700 (PDT)

Subject: A direct, simple Geometry proof of FLT Re: The Irish say-- if it
works, don't fix it Re: A new attempt at FLT, not using the Generalized FLT
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Fri, 08 Sep 2017 05:36:55 +0000

A direct, simple Geometry proof of FLT Re: The Irish say-- if it works, don't fix it Re: A new attempt at FLT, not using the Generalized FLT

Geometry proof of FLT

In Generalized FLT we prove it with condensed rectangles

And we can carry that over into 3rd dimension with a shared face instead of a shared side in 2nd dimension.

So, in 3rd dimension a rectangular solids A and B

3
|         1
|  __2/     this is A a rectangular box 3 by 2 by 1

3         3
|         /
|  __2/     this is B a rectangular box 3 by 2 by 3

Now, joining them on their shared face we form a new third box of 3by2by4

We note the volume of A plus B equals volume of C. This would be Generalized FLT

Now, what would regular FLT be, for exponent 3?

Obviously it would be a cube, and so, instantly, automatically we recognize no cube A plus cube B is ever going to form a new cube C, at best they will form a notched box like this:

_________
|            |______
|            |         |
|            |         |
--------------

So, what FLT was asking for, was something truly impossible, was to add one cube to another cube and expect a new cube to form. If the cube A and B were identical, then the C would be a rectangular box. But if A and B differed, the result was never a new cube but-- not even a rectangular box but a odd shaped figure of two cubes joined.

Now, exponent 4 and larger are treated the same way, and we see their impossibility.

So, yes, here is a proof of FLT, pure geometry, that needs no Generalized FLT.

Here is a proof I can gladly go into any High School where all the students can absorb and understand. There is no high wire walking circus act of logic in this proof. In fact, the Ancient Greeks could have proven FLT.

AP

Newsgroups: sci.math
Date: Thu, 7 Sep 2017 22:51:28 -0700 (PDT)

Subject: Simple Geometry proof of Fermat's Last Theorem, FLT, that does not
rely upon Generalized FLT proof
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Fri, 08 Sep 2017 05:51:28 +0000

Simple Geometry proof of Fermat's Last Theorem, FLT, that does not rely upon Generalized FLT proof

Now in a prior post to this one below, I commented that why climb Everest if you immediately turn around and climb back down. Why not spend some time at the summit, look around, experience the moment, the situation. Thus, I had proven Generalized FLT and which thence gave me FLT, but was not happy with that deliverence, for it seemed contorted in logic. So, I looked around, and fell into this super easy proof.

Newsgroups: sci.math
Date: Thu, 7 Sep 2017 22:36:55 -0700 (PDT)

Subject: A direct, simple Geometry proof of FLT Re: The Irish say-- if it
works, don't fix it Re: A new attempt at FLT, not using the Generalized FLT
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Fri, 08 Sep 2017 05:36:55 +0000

A direct, simple Geometry proof of FLT Re: The Irish say-- if it works, don't fix it Re: A new attempt at FLT, not using the Generalized FLT

Geometry proof of FLT

In Generalized FLT we prove it with condensed rectangles

And we can carry that over into 3rd dimension with a shared face instead of a shared side in 2nd dimension.

So, in 3rd dimension a rectangular solids A and B

3
|         1
|  __2/     this is A a rectangular box 3 by 2 by 1

3         3
|         /
|  __2/     this is B a rectangular box 3 by 2 by 3

Now, joining them on their shared face we form a new third box of 3by2by4

We note the volume of A plus B equals volume of C. This would be Generalized FLT

Now, what would regular FLT be, for exponent 3?

Obviously it would be a cube, and so, instantly, automatically we recognize no cube A plus cube B is ever going to form a new cube C, at best they will form a notched box like this:

_________
|            |______
|            |         |
|            |         |
--------------

So, what FLT was asking for, was something truly impossible, was to add one cube to another cube and expect a new cube to form. If the cube A and B were identical, then the C would be a rectangular box. But if A and B differed, the result was never a new cube but-- not even a rectangular box but a odd shaped figure of two cubes joined.

Now, exponent 4 and larger are treated the same way, and we see their impossibility.

So, yes, here is a proof of FLT, pure geometry, that needs no Generalized FLT.

Here is a proof I can gladly go into any High School where all the students can absorb and understand. There is no high wire walking circus act of logic in this proof. In fact, the Ancient Greeks could have proven FLT.

AP

Newsgroups: sci.math
Date: Fri, 8 Sep 2017 04:13:15 -0700 (PDT)

Subject: Re: Simple Geometry proof of Fermat's Last Theorem, FLT, that does
not rely upon Generalized FLT proof
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Fri, 08 Sep 2017 11:13:16 +0000

Alright, let us start all over again on Fermat's Last Theorem and let us go way way back in history to before the Pythagorean theorem to solve FLT. Let us pretend we do not even know the Pythagorean theorem, and only concerned about FLT.

So, in this dawn of mathematics, before the Pythagorean theorem, let us prove it, only without the slightest notion of the right triangle involved in A^2 + B^2 = C^2

So we know what a square is, 4 equal sides, 4 vertices.

Now, let us define a new sort of Square, call it the Perpendicular-Square which consists of a shape that looks like an L only the sides are equal. So now, the number 3 forms a Perpendicular-Square, so does 5, so does 7 so does 9. An even number cannot form a Perpendicular-Square because, obviously, one side has more than the other.

Now the Perpendicular Square is important in proving that a Square A and Square B when added forms a new square C. Here, the idea is that you have a given Square A, and you have a Perpendicular Square B, such that you take the B, and surround the A by B forming a new square C.

For example, we have the square 4^2 which is 4x4 which is 16

Now, if we surround the one side by 4 more unit squares and then on top, 4 more unit squares and then add a vertex square, 4+4+1 = 9 unit squares we have a new square.

Think of the Perpendicular Square shaped like this
====
||
||

And think of the normal square as this

_____
|      |
|      |
-----

And, what you have done is thus, taken the Perpendicular Square and covered the normal square producing a new square

In essence, you have taken apart a square and formed a new figure that will facilitate the production of another square.

3^2 + 4^2 equals a new square 5^2

By disassembling 9 into a L shaped Perpendicular Square we had a A + B = C.

Now, that idea transfered to FLT. Instead of Perpendicular Square we need a 3 faced figure, and let me call it a three faced figure. We do not know if we can get a cube to be a three faced figure.

So now we go through all the cubes of math

1^3
2^3   and 8-1 = 7
3^3   and 27-8 = 19
4^3   and 64-27 = 37

Now all of these are going to be odd numbers this three faced figure.

Think of the cube as a cube house, and think of the three faced figure as sliding up against the cube house forming a larger cube house.

So, here, we ask, do those odd numbers of Three Faced Cubes come directly from a number N^3?

Is there a N^3 = 7? equal to 19? equal to 37? equal to 61? equal to 91? equal to 127?

If not, why not?

If so, then FLT has a solution in A^3 + B^3 = C^3

AP

Newsgroups: sci.math
Date: Fri, 8 Sep 2017 04:18:17 -0700 (PDT)

Subject: exp higher than 3 Re: Simple Geometry proof of Fermat's Last Theorem,
FLT, that does not rely upon Generalized FLT proof
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Fri, 08 Sep 2017 11:18:17 +0000

exp higher than 3 Re: Simple Geometry proof of Fermat's Last Theorem, FLT, that does not rely upon Generalized FLT proof

How exactly do we treat 4 and higher exponents. The case of 3 exponent is simple straightforward, but since there is no 4th dimension or higher how do we treat A^4 + B^4 = C^4

Well, here I do what I ended up with doing some years back. I say A^4 is the same as A*A^3, just a multiplication of a cube in 3rd dimension which is a rectangular solid. So if A is 2 then A^4 is 2 cubes of 8 cubic volume. If it were 7 exponent A^7 and A was 2 it would be 16 cubes of 8 cubic volume which actually does form a overall large cube.

So, here we see that since exp3 never has any solution, that higher exponents are just multiples of exp3 and they have no solution.

AP

Newsgroups: sci.math
Date: Fri, 8 Sep 2017 14:24:21 -0700 (PDT)

Subject: new concepts of Perpendicular-Square, and the Three Faced Cube Re:
Simple Geometry proof of Fermat's Last Theorem, FLT
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Fri, 08 Sep 2017 21:24:21 +0000

new concepts of Perpendicular-Square, and the Three Faced Cube Re: Simple Geometry proof of Fermat's Last Theorem, FLT

Alright, a proof that two squares can add up to a third new square is all in this chart using the Perpendicular-Square

1^2
2^2   and 4-1 = 3
3^2   and 9-4 = 5
4^2   and 16-9 = 7
5^2   and 25-16 = 9
6^2   and 36-25 = 11
7^2   and 49-36 = 13
8^2   and 64-49 = 15
9^2   and 81-64 = 17
10^2   and 100-81 = 19
11^2   and 121-100 = 21
12^2   and 144-121 = 23
13^2   and 169-144 = 25
14^2   and 196-169 = 27

Do you see why there is a A +B = C in that the Perpendicular-Square 9 exists two-ways, just as 25 perpendicular-square exists.

Now we shift to the Three Face Cubes and see if these objects are repeated

1^3
2^3   and 8-1 = 7
3^3   and 27-8 = 19
4^3   and 64-27 = 37
5^3   and 125-64 = 61
6^3   and 216-125 = 91
7^3   and 343-216 = 127
8^3   and 512-343 = 169
9^3   and 729-512 = 217
10^3   and 1000-729 = 271

Do you see where there is no duplication of a cube and a Three Faced Cube? In Squares, they followed a pattern of the next one was 2 units larger 3, 5, 7, 9, 11, ... In Three Faced Cubes they are spread out too much and no pattern apparent 7, 19, 37, 61, 91, 127, . . .

So, a proof of FLT is simply find what is preventing the Three Faced Cube from equalling a cube in terms of volume.

Now, some are going to ask, AP, why did you invent these two new figures-- perpendicular-square and the Three Faced Cube, and the answer is quite clear-- instead of working antimony of Pythagorean theorem against FLT, it is best to work in concordance, work in harmony, so that one guides the other. And the way to get concordance is to apply a similar concept to both.

AP