
Three different proofs of Fermat's Last Theorem(1) Generalized FLT (2) Geometry using notched cubes (3) Duality of Numbers to Angles
Posted:
Oct 2, 2017 10:47 AM


ARRAY PROOFS:
(1) First Proof
Generalized FLT:: Find for GeneralizedFLT a A+B=C conglomerated exponents {1, 8, 16, 27, 32, 64, 81, 125, 216, 243, 256, . .}. Define Condensed Rectangle as rectangles not with one side as 1.
Impossible for unnotched rectangle to be equal to notched rectangle
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Hence, a solution of C in A+B=C must have one side of A,B,C the same side, in order to stack or unstack.
First a Lemma is needed to navigate from Generalized FLT to FLT.
LEMMA:: Any solution found in the Generalized FLT, cannot have a cascade downward, nor a cascade upward of new solutions.
Lemma Proof:: Given factor k when we have A^x + B^y = C^z that never is it true that (A*k)^x + (B*k)^y = (C*k)^z. Why, because Arithmetic exponent law does not allow such. For example:: 2^3 + 2^3 = 2^4 common factor of 2
k=2 thence (2*2)^3 + (2*2)^3 = (2*2)^4 gives 4^3 +4^3 = 4^4
That is 64 + 64 = 256 and clearly false
FLT PROOF:: When we have a solution of Generalized FLT, the A,B,C share a common factor in A^x + B^y = C^z. If we had it where x=y=z for 3 or higher, then by Arithmetic, we have a cascade downwards of at least one new solution and a infinite cascade upwards of new solutions. The Lemma finds this impossible, hence never is the case of x=y=z.
QED
(2) Second Proof
Theorem Statement geometry proof of Fermat's Last Theorem using Notched Cubes:: Given A,B,C, n as Counting numbers, 1,2,3,4,?.. A^n + B^n = C^n where n is 3 or larger, there are no solutions to be found. The pure geometry proof is based on cubes of geometry and notched cubes, where we subtract the volume of one cube from another cube. And in the proof we define all further exponents higher than 3 as extended cubes which are rectangular solids, so that A^5, for example is the scalar A*A multiplied by the cube A^3. This is a geometry proof for it delves into the features, characteristics and traits of the cube. And the proof logic rests on the idea that if no Notched Cube is equal to a Perfect Cube itself, exists, then, hence no A,B,C exists to deliver a solution for A^n + B^n = C^n. For exponent 4 and higher, we generalize notched cubes to notched rectangular solids.
Proof Statement::
Successor Notched Cubes 1^3 2^3 and 81 = 7 3^3 and 278 = 19 4^3 and 6427 = 37 5^3 and 12564 = 61 6^3 and 216125 = 91 7^3 and 343216 = 127 8^3 and 512343 = 169 9^3 and 729512 = 217 10^3 and 1000729 = 271 11^3 and 13311000 = 331 12^3 and 17281331 = 397
Some Examples of Staggered Notched Cubes
3^3 and 271 = 26 4^3 and 648 = 56 5^3 and 12527 = 98 6^3 and 21664 = 152 7^3 and 343125 = 218 8^3 and 512216 = 296 9^3 and 729343 = 386 10^3 and 1000512 = 488 11^3 and 1331729 = 602 12^3 and 17281000 = 728
We conglomerate all notched Cubes into a sequence such as 7, 19, 26, 37, 56, 61, ?..
And ask, is there a Perfect Cube that can equal the volume of any of these numbers of the list of all Notched Cubes?
If we examine the Algebra of a Notched Cube we see it follows along these lines;; (x+1)^3  x^3 = 3x^2 +3x +1
So, in the geometry proof, we ask for a m^3 = 3x^2 + 3x +1 where m,x are counting numbers. And that is a impossible Arithmetic of Law of Exponents.
Now, for exp4 and higher, this arithmetic impossibility is met with again in that the notched figure is a exponent of 1 short of the figure. Hence, never any solutions to A^n + B^n = C^n. QED
Now, the great lesson of FLT. The great lesson it teaches us, is that dimensions beyond 3rd are nonexistent. There is no 4th dimension or 5th or higher cube. Cubes stop at 3rd dimension. So that A^4 or A^5 are A*A^3 and A*A*A^3. Where we take the extra A's as scalars of a Perfect Cube A^3.
So that A^4 is a Rectangular Solid and that A^5 is another Rectangular Solid. And we can have Notched Rectangular Solids or UnNotched Rectangular Solids. This is the great lesson learned from FLT. The proof was difficult. But the application of FLT, that is important, is that mathematics must now always treat 4th dimension and higher, as merely extensions of 3rd dimension. So if presented with a A^6 + B^6 = C^6, we realize those are j*A^3 + k*B^3 = h*C^3.
(3) Third Proof
Now a new Third Proof of FLT, using angles, trigonometry to prove FLT Re: Two proofs of FLT
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On Monday, October 2, 2017 at 1:34:04 AM UTC5, Archimedes Plutonium wrote: > Now here is a interesting question. Is the Arithmetic Impossible Barrier in the Generalized FLT > FLT is it the same as the Arithmetic Impossible Barrier in the Geometry proof of FLT? > > If you recall, the Lemma in Generalized FLT to FLT says the solutions of A^n + B^n= C^n cannot have a cascade down nor a cascade upward of solutions for which a A^n + B^n = C^n requires and so the Arithmetic Impossible Barrier is no cascade of solutions. > > Now, compare that Arithmetic Impossible Barrier in the Geometry FLT proof. There the barrier was the fact that exp3 needs a m^3 = 3x^2 + 3x + 1 where m and x are counting numbers. And that barrier is that no m^3 is possible due to Arithmetic. > > So, is the cascade barrier the same as the exponent lowered by 1 for m^n barrier? Both are Arithmetic barriers, but how closely related are they? >
Yes, those two Arithmetic Impossible Barriers are related. For one can say that the Cascade down and up of solution sets, is the inverse of, m^3 =/ m^2, or, more general m^n =/ m^(n1).
A cascade up and down of solution sets is inverse to m^n =/ m^(n1)
Now, this revelation allows me to make a totally new proof of FLT, because once a person knows a proof of something, he/she can tinker with the proof details to engineer a new proof, a new proof that perhaps is more aerodynamic or more easy or more intuitive.
So, knowing what forces there to not be any solutions in exponent 3 or higher is able to let me design a new proof centered on that arithmetic barrier.
Now, none of these three proofs of FLT are pure algebra without any geometry nor pure geometry without any algebra. It appears that in FLT proofs, you are faced with some geometry and some algebra mixed together in order to make the proof work.
Remember, in the Generalized FLT I had to muster the Condensed Rectangle, and thus, not a purely Algebraic proof, and in the geometry proof of FLT using notched cubes, I had to summon some arithmetic, for I admit my geometry proof of FLT is not purely geometry for I was forced to inject the Arithmetic Impossible barrier of m^n=/m^(n1).
What this THIRD proof of FLT is, is using both Arithmetic Barriers of the previous other two proofs of FLT.
It is a short and very sweet proof.
THIRD PROOF OF FLT, and wondering if there is no pure algebra or pure geometry proof ever.
Algebra allows us to have a cascade up and cascade down of solutions, given any one solution set. So, if we have for exp2 the solution set 3,4,5 then also, the set .6, .8, 1 is a solution. Now, if exp3 or higher has a solution, then we have a cascade downward of that solution so that the solution is x,y,1 where x and y are between 0 and 1. Savvy so far?
Now, we do not care what the exponent is, whether 3,4,5, 6, etc etc, or some huge ugly large number. Any solution is reverted to x,y,1 because of what? Because Arithmetic of exponents allows it.
Now here we inject a theorem of Angles. I am guessing it is a theorem but not sure and not sure of the name of it. Let me just call it the Complimentary Angle theorem. For the idea is that if we take 90 degrees to be 1, then the angle .36= .6^2 and the angle .64 =.8^2 are complimentary angles adding up to be 90 degrees as 1 in a right triangle. This is just standard data of trigonometry, know one vertex angle is x then the other of the right triangle is 1x or 90  x. In a sense, the Pythagorean Theorem is angle compliments adding up to 1 = 90 degrees. For, we can convert all Pythagorean triples to be angles, instead of numbers. So here we have the Duality of Numbers with Angles.
Now we apply angles to FLT.
Numbers are dual to angles, the 3,4,5 right triangle is about a 53 degree angle compliment or dual to a 37 degree angle, as that of .8 and .6 respectively.
So we have Complimentarity of angles and numbers in the Pythagorean theorem, a duality that adds to compliments equaling 1 by addition.
Now consider a solution to A^n + B^n = C^n where n is 3 or larger. Consider any such solution, then by Arithmetic we can divide C into A,B,C and have A and B fractions between 0 and 1 and C = 1. This means we have another compliment dual angles, to an already existing compliment dual in exponent 2. Suppose for example the exp3 was this other solution, then you have A^2 + B^2 = 1 = E^3 + F^3 and thus, A^2 = E^3 and B^2 = F^3. Impossible. QED
AP

