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Topic:
Summary of the AP proofs of FLT and Beal #1473 Correcting Math
Replies:
2
Last Post:
Jan 28, 2014 8:32 AM




Summary of the AP proofs of FLT and Beal #1473 Correcting Math
Posted:
Jan 27, 2014 6:04 PM


Summary of the AP proofs of FLT and Beal #1473 Correcting Math
Often times a summary of a proof can be as valuable as the proof itself, for it generalizes the problem and shows the bigger picture of what went on.
In that vain, let me summarize the proofs of Fermat's Last Theorem, FLT and Beal conjecture. ______________ Summary of FLT ______________
To prove FLT, all one needs to focus upon is that joining together two condensed rectangles that have a equal side in common such as
2^3 + 2^3 = 2^4 3^3 + 6^3 = 3^5
Where the common side is 2 or 4 for 8 +8= 16 And where the common side is 9 for 27+216 = 243
So the proof revolves around a concept of length confined inside a specific exponent, cannot be made equal to a condensed rectangle because the next number that shares the same length is multiplied out of reach. In the case of 2^3 + 2^3 = 2^4 if we wanted to remain in exp3 we would have to use 4^3 rather than 2^4 and 4^3 is **at minimum 2 times too large**. In the case of 3^3 + 6^3 = 3^5, if we wanted to remain in the same exp3 rather than 3^5, we would have to go to 9^3 and that is **at minimum 3 times too large** for the A+B = C.
So to prove FLT, a new concept arises that to have a A+B=C, where two condensed rectangles having equal sides as A and B and stacked to make a third new condensed rectangle C. That to obtain a C condensed rectangle where the one side matches the common side of A + B, that we need to go 2 units higher up or 3 units higher up at a minimum which thence makes the C area at least 2 times higher than needed. FLT as a theorem says that 2^3 cannot equal 4^3, or, that 3^3 cannot equal 6^3. We we are confined to the same specific exponent, we cannot match a A+B within that exponent for we have to increase the base by at least 2 or at least 3 and the exponent then enlarges the size of the C condensed rectangle that it is not equal to A+B.
Now the summary of the proof of Beal is much easier.
____________________ Summary of Beal Proof ____________________
In Beal proof we know we have solutions. What we do not know is whether all solutions have a prime divisor of its A, B, C in A+B=C. Well, all numbers in the Beal set of numbers are decomposed into condensed rectangles, except for 1 as in 1^3 or 1^4. All numbers except 1 are composite numbers in the Beal set of numbers. That means all numbers except 1 are condensed rectangles in the Beal set of numbers.
In order for an A + B to form a third new condensed rectangle C, the A and B must share a common side, otherwise the stacking of A + B is not a rectangle. The condensed rectangle C must have a equal common side to A and B. Thus, all solutions must have a side that is equal in all three of A, B, C.
Now, someone may say that A and B has no common equal side and stack as in this diagram: HHHHHH HHHHHH HHHHHHHHH HHHHHHHHH
And where a C is of this same diagram.
But I would retort, that the above is not a rectangle and is not a multiplication of a y*z that makes a rectangle and that is begot from cofactors of a composite number. What is the area of the above 6sided figure? And how does it relate to a number such as 27 or 216 or 243 in that of 27+216 =243?
So, what makes Beal proof true, is the fact that all composite numbers are separated into condensed rectangles that stack and if two share a common side they form a third new condensed rectangle, and the only solutions for A+B=C is a prime divisor in all three of A, B, C.
Now I likely will go through several revisions of this summary. 
Recently I reopened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.
https://groups.google.com/forum/?hl=en#!forum/plutoniumatomuniverse
Archimedes Plutonium



