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Topic: my Second proof of FLT, a pure geometry proof-- hopefully
Replies: 8   Last Post: Oct 2, 2017 12:37 AM

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Registered: 3/31/08
my Second proof of FLT, a pure geometry proof-- hopefully
Posted: Oct 1, 2017 2:04 AM
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Alright, I need to start over, for I ran into too many obstacles at the end there, trying to span the gap of exp3 into exp4 and higher. My troubles were all about the illusionary 4th dimension and higher. If FLT were just exp3, I would have no problems with a pure geometry proof. But since I have to deal with exp4 and beyond, there was no natural bridge once I proved for exp3, no natural spans to do exp4, 5th, 6th etc.

But, what keeps me interested, is that FLT is the test case theorem, in how to cover-up or conquer dimensions. What I mean is the world has only 3 proper dimensions, and 4th and higher are just human imagination gone awry. There are no ghosts, no witches no fire breathing dragon and there is no 4th and higher dimension. So, what theorem of mathematics is going to teaching us how to handle that fantasy-- and I believe, quite strongly that FLT is that theorem that teaches us how to handle and cope with all cases in math where higher dimensions become a problem.

Once we prove no solutions to A^n + B^n = C^n when n=3, how do we thence use that to say -- so, no solutions to exp4 and beyond?

The idea I was going on, was that say exp5 is (AA)A^3 + (BB)B^3 = (CC)C^3 and that somehow, I can whisk away the AA, the BB, the CC as numerical scalars or parameters, leaving the A^3, B^3, C^3 as the determining geometry. And since there is never a solution to A^3 + B^3 = C^3, no matter what scalars you tack onto that primary vector equation of geometry, no matter what scalars, if the vector primary equation has no solution, then the scalar supplement has no solution.

So, that issue is far far more important than even FLT proof. For that issue is run into almost every day in mathematics, where we are faced with equations of exp4 and larger.

Now in starting all over again with a Geometry proof of FLT, a question arises, for which my memory is at a loss at the moment.

2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 125
6^3 = 216
7^3 = 343
8^3 = 512
9^3 = 729
10^3 = 1000
11^3 = 1331
12^3 = 1728

And here I have a question, a question that is important on whether a Sequence proof is viable, or not viable.

Notice at 5^3 + 6^3 = 7^3 -2 that it is a near miss of just 2

then at

9^3 + 10^3 = 12^3 +1 is a near miss of just 1

The question is, are those near misses unique, or is there another near miss of just 2 down the pike, and another near miss of just 1 down the pike. If memory holds, I believe there is another near miss of just 1 not far down the pike. So if true, means, a recurring near miss of just 1 and recurring near miss of just 2 happens often for exp3. And what that signifies, is that I cannot use a Sequence argument in the proof.

Now, what I am going to retain and carry forward in this new proof attempt, is the idea of notching the numbers and seeing if a notched number can be a X^3 number.

Now earlier, I tried notching just the successor cubes of x^3, (x+1)^3 where I take the successor and subtract the former, such as 64-27 = 37, then ask is there an X^3 that is equal to 37.

But looking above at the near miss of 1 generated at adding 9^3 + 10^3 an missing 11^3 in the act, suggests that I need to expand the subtractions, not just to successors but in that case x^3, (x+1)^3, (x+2)^3, (x+3)^3

So in my first go around, I was not general enough. To be more general, I have to notch not just with successors, but notch a given arbitrary cube, all the way down to 1^3. Now this may seem to put a frightening huge bizarre complication over the proof, but we must realize, that such complication is just a temporary setback, for we notice as the numbers get larger, the NOTCHED CANDIDATES worth consideration get smaller. For example 8^3 is out of the picture by the time we reach 12^3.

And, here again, if there is a second near miss of just 1, will it look like the 9,10,12 triple or will it be a triple of x, x+1, x+2.

So, BASIC OUTLINE OF LOGIC OF PROOF, of course geometry proof, so keep a CUBE in mind, and what we want to do is Find out if a Notched Cube can equal a full cube. What is meant by a Notched Cube?

A notched cube is any cube that has a cube missing out of it. If we have a cube that was 3 by 3 by 3 = 27 and we cut out of that cube another cube of 2 by 2 by 2 = 8 we have remaining a Notched Cube of 19. We just took a cube of 27 and cut out of it another cube of 8 and what remains is a notched cube of 19.

So, the logic is clear, easy and straightforward.

We list all possible Cubes, such as the above which I started, beginning with 1^3

Now, we make a new list of all Possible Notched Cubes.

Finally, we ask in that list of all Possible Notched Cubes, do any of those Notched Cubes equal to a normal cube of m^3? If one exists, then FLT has a solution in exp3, if none exists, then no solutions in exp3.

Alright, I am safe there.

And, I even had a proof for exponent 3 in the idea that the Algebra, or Arithmetic of Notched Cubes is such as to say no solutions.

Successor Notched Cubes (x+1)^3 - x^3 is equal to 3x^2 + 3x +1. So, we ask the important question is there a "m" such that m^3 = 3x^2 + 3x +1. And by Arithmetic we can say no, it is impossible for such m to be a counting number.

Now we can do the same argument with staggered Notched Cubes.

But what is that Arithmetic that disallows "m"

Perhaps that arithmetic also bridges over to exp4 and higher, and so I need no Cover Rule.


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