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 February 16th, 2014, 02:29 AM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 WHAT FERMAT TEACH... I'm still working onto Fermat the Last. Can someone check/correct this: 1) We know we can reduce the FLT on a X,Y plane where we plot a) Y = X^n b) Y' = (n-1) X^(n-1) What Fermat discover is that a) if n=2 Y' =2x is a line, so some tricks can happen in the integers on the derivate Y' : (1) C^2 = A^2 + B^2 for some A,B,C Since in general we can write: $C^n= \sum_{x=1 }^{C} {X^n-(X-1)^n}$ $A^n= \sum_{x=1 }^{A} {X^n-(X-1)^n}$ $B^n= \sum_{x=1 }^{B} {X^n-(X-1)^n}$ so if: $C^n= A^n +B^n$ is true, than is also alway true that: $C^n= \sum_{x=1 }^{C} {X^n-(X-1)^n} = \sum_{x=1 }^{A} {X^n-(X-1)^n} + \sum_{x=1 }^{B} {X^n-(X-1)^n}$ so putting togheter the similar terms: $C^n= \sum_{x=1 }^{C} {X^n-(X-1)^n} = 2* \sum_{x=1 }^{A} {X^n-(X-1)^n} + \sum_{x=A+1 }^{B} {X^n-(X-1)^n}$ $C^n= \sum_{x=1 }^{C} {X^n-(X-1)^n} = 2* \sum_{x=1 }^{B} {X^n-(X-1)^n} - \sum_{x=A+1 }^{B} {X^n-(X-1)^n}$ And this sound me like a: in C the increment in Y at right, must be the same of the decrement in Y at left side. If this is applied on a Y'= line there can be infinite solution since this is always true Y'' =0 I finally discover that for any integer A, there is a couple of B,C (where C=B+1 if A=odd) that satisfy the (1), as probably known. But if i try to apply this rule to a Y' = curve ??? b) Fermat discover that the area bellow the derivate, Y' (from 0 till x=A or else) is the value of the "main" function in A (etc...) c) But he discover ALSO that IF and only IF Y' is a line, the area bellow it, is "squarable" in the integers, or can be cutted in a finite number of triangles. What I think was the the elegant proof in His mind is: Since for n>2 Y' = curve, so NO way to REDUCE the Area in Finite number of Triangles, so no way to have an integer area bellow the curve Y'. This is a clear concept, but why is it correct ? I think we have too confidence with limits, so first think one can say is: ok, is "generally" true, but can be wrong ....since we well know that an infinite serie can have a finite limit, ....so can happen that an area bellow a curve IN A FINITE INTERVALL A-B has a finite INTEGER value ? In my opinion NO, No for two reasons: 1) is proven by Wiles (UNDER Fermat CONDITIONS) and the theory start from Abel : so that the n over K combination needs all the terms: x^(n-1), x^(n-2)---- till +/- 1 ot the power is not complete... 2) What is missed, but I'm probably not able to explain, is the "soul" of the infinitesimal calculus: so that using the infinitesimal rulues we can have an answer to any kind of "wrong" calculation, like the "AREA" bellow a curve (and else...), but we have to be aware of this when we build our problem, as probably he is... So if we HAVE to go infinitesimal, if we are talking of curve, we cannot return integers (ALWAYS ! ? ) What is the soul I try to "argue" is that when we go at n=3 (or plus) we have 3 free axis problem (3x^2, 3x, 1) or more following the Newton's develope rules, but FERMAT IS ALWAYS asking FOR a 2D Solution: a couple of integers ( C; C^n) so a point (X,Y) where both are integers... So NO WAY ! A problem in 3D (ot n-ht Dimensions) cannot have an INTEGER answer in 2 dimension just (in this case for sure...). We can have a solution JUST if we go infinitesimal, so we make "a big mess of all", and we accept a NON finite solution. I'm not able to find a counter example... also because if you find one you disprove that the area bellow a curve can be "squared" by a raw and a compass (or not ?) ...but i know I'm a donky... so I ask you your opinion about... Thanks, Ciao S.
February 16th, 2014, 05:20 AM   #2
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Re: WHAT FERMAT TEACH...

Quote:
 Originally Posted by complicatemodulus (1) C^2 = A^2 + B^2 for some A,B,C
This holds if C is the hypotenuse of a right triangle with legs A and B. Unfortunately, you haven't proved that such a right triangle exists. (It doesn't.)

February 16th, 2014, 06:48 AM   #3
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Re: WHAT FERMAT TEACH...

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by complicatemodulus (1) C^2 = A^2 + B^2 for some A,B,C
This holds if C is the hypotenuse of a right triangle with legs A and B. Unfortunately, you haven't proved that such a right triangle exists. (It doesn't.)
...Keep lefty... you see it at works...

Once again n=2: A = any odd from 3 ; B= (A^2 -1)/2 ; C= (A^2 + 1)/2

A = any even from 2 ; B=((A/2)^2 - 1) ; C=((A/2)^2 + 1)

But this is not the question on the table...That is what happen for n>2 ;-P

I'm still waiting for serious concernings on...

 February 16th, 2014, 10:34 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: WHAT FERMAT TEACH...

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