
the sheer dazzling beauty of this Third proof of FLT Re: Three different proofs of Fermat's Last Theorem
Posted:
Oct 2, 2017 10:37 PM


On Monday, October 2, 2017 at 9:47:20 AM UTC5, Archimedes Plutonium wrote: (snipped) > 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 > >
Now, the beauty of this third proof of FLT is its sheer simplicity. Say you want to think there is a solution in exp4, where A^4 + B^4 = C^4. Say you contemplate that, then the solution has to by Arithmetic law of exponents , the solution is cascaded down to where the C = 1 and the A and B are fractions between 0 and 1. Fine so far, but, why cannot that be true, why cannot there be a solution in exponent 4? Because there already is a solution of those very same numbers in exponent 2 of some specific righttriangle with that solution set of A,B,C, and thus you have a A^2 = A^4. Same goes for exp5 or any exponent you wish to dream of.
What makes this proof of FLT work, and work so well, work so charmingly beautiful, is that any solutions of exponents 3 or larger, are already spoken for by exponent 2. What those solutions of C=1 and then A and B decimal fractions between 0 and 1, where A and B are complements of each other well, this is Trigonometry of a right triangle complementary duals of angles. If one angle is 30, then the dual angle is 60 degrees. If one angle A is 15 degrees, the dual angle B is 75 degrees.
So, mathematics cannot afford to duplicate a solution of A+B=C in a higher exponent from exp2 since exp2 covers each and every possible A+B=C.
Now, would I have ever discovered that proof, without going through and proving the Generalized FLT first, then the Geometry proof of FLT in notched cubes? No, I doubt it. I doubt I would have seen all the disparate elements of Arithmetic barriers.
And, if I had discovered this Third proof first, I likely would never have discovered the preceding two proofs.
AP

