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Topic: Three different proofs of Fermat's Last Theorem(1) Generalized FLT
(2) Geometry using notched cubes (3) Duality of Numbers to Angles

Replies: 5   Last Post: Oct 4, 2017 12:38 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
FLT turns out to be addition of angles in Trigonometry Re: the sheer
dazzling beauty of this Third proof of FLT

Posted: Oct 4, 2017 12:36 AM

On Tuesday, October 3, 2017 at 6:43:53 PM UTC-5, Archimedes Plutonium wrote:
> On Tuesday, October 3, 2017 at 4:21:17 AM UTC-5, Archimedes Plutonium wrote:
> > Yes, what a beautiful proof that was, the third proof of FLT, simply stated, no solution exists for exp3 or higher, because any solution there would cascade down to a x,y,1 solution such as 3,4,5 is .6, .8, 1 and so all possible solutions are covered only by exp2. Otherwise you would have A^2 = A^3 or A to the whatever exponent.
> >

And in trigonometry the .6 and .8 would correspond to the Pythagorean theorem where the angles involved are sin37 degrees and sin53 degrees totalling to 90 degrees or, 1.

> > Now that proof does the Array honor, because the proof statement is one paragraph and equal to or smaller than the Theorem statement.
> >
> >

>
> Alright, so, let us try a test run sample of the meaning of this proof of FLT. We have a good sample in exp3 of 9^3 + 10^3 = 12^3 + 1 where it misses by 1.
>
> So let us ignore the 1 and then go ahead with pretending that it is a solution.
>
> So we have .75^3 + .833..^3 = 1^3
> that is .4218... + .5787... =1
>

And here, we want to know the angles involved so we took the square root of .42.. and the square root of .57... and got .649 and .760 respectively, and so what angles are those?

Turns out they are sin40 degrees and sin50 degrees respectively

> Now, how does the proof work?
>
> Well, it says that if any solution, any at all exists is exp3 or higher, any solution.
>
> By arithmetic of law of exponents, that solution can be reduced to a form of x,y,1 where x and y are between 0 and 1 such as our .42... and .57...
>
> So, what is wrong with that solution? What is impossible about it?
>
> It is impossible because we already have that solution in exp2 of Pythagorean theorem
>
> For if we take the square root of .42... and .57... we have .649 and .76 and 1 and converting that in exp2 to be 649, 760, 1000 (close, but of course off by a little bit).
>
> So, the trouble with ever having a solution in exponents higher than 2, is that there already exists a solution in 2 and thus math would be having a A^2 = A^3 and that only occurs when A= 0 or 1.
>

So I am pretty sure that I would never have found this proof, without going through the prior other two proofs, seeing what Arithmetic Barrier is placed on these triples for solutions in exp3 and higher. Only by doing the two proofs first, can one see the Arithmetic barrier placed on "no solutions".

No sequence type of argument is going to be a proof, because the "missing by 1" is a recurring event in exp3, as seen by <9,10,12> then by <64, 94, 103> near misses by 1, repeats itself over in more near misses on down the road. This tells you the reason no solution is a "fine, refined barrier" that you can come so close, so often, yet always miss being a solution.

And in my proof of FLT by Notched Cubes, told me that it was a A^2 cannot equal a A^3 (except 0 and 1) and it was my Generalized FLT proof, that told me the cascade downward is a barrier. So putting those two together-- the lights came on, that any solution in exp3 or higher is a duplication of exp2 solutions. Thus, --- A^2 = A^3 impossible.

The angles in right triangles are finely tuned angles so that the A and B in A+B = 90degrees is a finely tuned angle A and a finely tuned angle B.

And in the proof of FLT, one so often visualizes that as a ho hum ho addition of A and B, nothing special about addition. But, as the proof shows, that addition of A^n + B^n = C^n is a very very special type of addition for it is a addition of Complementary Pairs, dual pairs of numbers that add up to 1, for they are angles that add up to 90 degrees.

AP