August 10th, 2014, 10:15 PM  #131 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
An elegant proof is for sure more direct. Algebraic division can solve, but is still not clear to 99% of readers... Once again: is just question to linearize and transfrom in 2 triangles: One of area: A^2 The other of area: B^2 = C^2 A^2 If you correctly linearize the problem you'll see that the 2 triangle are not similar for the conditions A,B,C integers and coprimes, hence the proof. But the most elegant point is that from n>=3 the derivate is a curve, for so no way to square in the integers... or (for n=3): can be rewritten for sure for sorry is trivial now... BUT IF WE WRITE LIKE FERMAT is asking himself : is possible that what rest is again a (same) power of another integer ? But to do that without make mistake we can just write: and so: This is for sure always true. BUT if we wanna rest in the integers we can just correctly rewrite (and ask): (1) I have to remember once again that this is a play between sum and integrals. now we well know that the one become the other passing to the limit. Probably this is also well clear to Fermat, that also very well know ALSO the properties of the numbers so that is also possible to have again a power using a sum like this: A^3 = 1+7+19+... 3x^23x+1 or the most general case a STEP SUM like this: A^3 = 1/k + 3x^2/k 3x/k^2+1/k^3 if we jump in x with step 1/k, so x1=1/k, x2= 2/k etc.. SO the (1) IS NO LONGER A CONTINUE FUNCTION SINCE YOU'VE A GAP IN CASE YOU LOOK TO THE AREA BELLOW THE DERIVATE (that viceversa is continue) FROM "A" AND "B+1". Since we knoe the properties of Sum and Integrals, this is enought to say; NO SOLUTION, but to prove we can still ask (and correctly write a sum step 1 in the integers, or a step sum that is more precise, as precise as we want (1/k), till the limit k>infinite) So can be possible to find a solution in the integers for n>=3 ? But this GEAR can't work, or has no solution if A,B,C are integers and coprimes since under a continue derivate curve y'= 3x^2 (in this case) we can continue to make littlest and littlest columns, (or triangles) but we can't find the solution since the gear teeth of C^nB^n are not similar to the one of A^n and C^n (or the two area coverted in two triangleas, are not similar): (I remember I use a mine special "Step sum", if you don't like pls trasnfrom in a normal sum with changeing the variable in the sum so have the sum from 1 to.... and not from 1/k step 1/k as I wrote here when you see lower limit 1/k) So rising K in N we found no solution since A,B,C coprimes... so If you've to understood what I mean for step sum we cannot complete (or square in "N" or "Q") the area bellow the curve till we go to the limit or we write: so we have to accept 2 integers and one (A in this case) NON integer. This is the Dedekind section definition for an irrational. Ciao Stefano here what we can do with a linear derivate: and what we cannot (if we won't use the limit so we rest in N or in Q): Last edited by complicatemodulus; August 10th, 2014 at 10:52 PM. 
October 8th, 2014, 12:00 PM  #132 
Senior Member Joined: May 2013 Posts: 118 Thanks: 10  PART THREE Once more,all numbers involved are natural What we already know $\displaystyle X^N+Y^N=Z^N$ (1), N┼XYZ, X=NA+a, Y=NB+b, Z=NC+c (a,b,c are the remainders of divisions X/N, Y/N and Z/N respectively).Of course N┼abc a+b=c (2), $\displaystyle N^2aYbX$, $\displaystyle N^2aZcX$, $\displaystyle N^2bZcY$ (relationships (3)) Sentences used: Sentence I:$\displaystyle N^mDE \rightarrow N^{m+1}D^NE^N$ and $\displaystyle N^mD+E \rightarrow N^{m+1}D^N+E^N$ Sentence II: $\displaystyle Na_1a_2…a_m\rightarrow Na_1$ or $\displaystyle Na_2$ or …or $\displaystyle Na_m$ Sentence III: If $\displaystyle N^kK=N^mM+N^pP$ with k>m, k>p and N┼MP, then m=p. The proof is rather simple, assuming e.g. m<p and dividing the relationship by $\displaystyle N^m$. Corrolary I. If If $\displaystyle N^maYbX$ and $\displaystyle N^maZcX$ then $\displaystyle N^mX+YZ$. The proof is quite simple Actually,if two of the following 4 conditions are true, so are the other two: 1st: $\displaystyle N^maYbX$, 2nd: $\displaystyle N^maZcX$ 3rd: $\displaystyle N^mbZcY$, 4th $\displaystyle N^mX+YZ$. Theorem I If $\displaystyle N^maYbX$ and $\displaystyle N^maZcX$ then $\displaystyle N^{m+1}a^N+b^Nc^N$ Proof: Sentence I : $\displaystyle N^maYbX\rightarrow N^{m+1}a^NY^Nb^NX^N$ and likewise: $\displaystyle N^{m+1}a^NZ^Nc^NX^N$ Substracting and as $\displaystyle X^N+Y^N=Z^N$: $\displaystyle N^{m+1}a^NX^N+b^NX^N c^NX^N \rightarrow\ (sentence II) N^{m+1}a^N+b^N c^N$ Actually,if two of the following 4 conditions are true, so are the other two: 1st: $\displaystyle N^maYbX$, 2nd: $\displaystyle N^maZcX$ 3rd: $\displaystyle N^mbZcY$, 4th $\displaystyle N^{m+1}a^N+b^Nc^N$. Theorem II If $\displaystyle N^maYbX$ and $\displaystyle N^maZcX$, then $\displaystyle N^{m+1}a^{N1}b^{N1}$, $\displaystyle N^{m+1}a^{N1}c^{N1}$, $\displaystyle N^{m+1}c^{N1}b^{N1}$ Proof: First a memo: by notating $\displaystyle D^EF$ we mean $\displaystyle D^EF$ but $\displaystyle D^{E+1}\dagger F$ . So let $\displaystyle N^qa^{N1}b^{N1}=N^qQ$ , $\displaystyle N^rc^{N1}b^{N1}=N^rR$ and $\displaystyle N^ua^{N1}c^{N1}=N^uU$ (of course N┼QRU).We know q,r and u are natural numbers (Fermat’s little theorem).Assuming they are all smaller than m+1,we’ll end to absurdity. By theorem I $\displaystyle N^{m+1}a^N+b^Nc^N=aa^{N1}+b^Ncc^{N1} = a(N^qQ+b^{N1})+b^Nc(N^rR+b^{N1}) = N^qQaN^rRc+b^{N1}(a+bc) = N^qQaN^rRc$ (4). Since N┼Qa and N┼Rc, by sentence III:q=r .Likewise q=r=u and (4) turns to $\displaystyle N^{m+1} N^q(aQcR)\rightarrow N^{m+1q} aQcR$ Likewise $\displaystyle N^{m+1q} cUbR$ . Substracting these last two and because a+b=c: $\displaystyle N^{m+1q} cUaQ+aR$ (5) But $\displaystyle a^{N1}b^{N1}=c^{N1}b^{N1}+a^{N1}c^{N1}\leftrightarrow N^qQ=N^qR+N^qU\rightarrow Q=R+U$ which turns (5) to $\displaystyle N^{m+1q} cUaU\rightarrow (sentence II) N^{m+1q} ca=b$, absurd Therefore q=r=u≥m+1. Hence: $\displaystyle N^{m+1}a^{N1}b^{N1}$, $\displaystyle N^{m+1}a^{N1}c^{N1}$ , $\displaystyle N^{m+1}c^{N1}b^{N1}$ Theorem III: If $\displaystyle N^maYbX$ (6) and $\displaystyle N^maZcX$, then $\displaystyle N^mX^{N1}Y^{N1}$, $\displaystyle N^mX^{N1}Z^{N1}$ , $\displaystyle N^mZ^{N1}Y^{N1}$ Proof: $\displaystyle N^maYbX\rightarrow (sentence\ I) N^{m+1}a^NY^Nb^NX^N$ (7) and (theorem II) $\displaystyle N^{m+1}a^{N1}b^{N1}\rightarrow N^{m+1}a^Nbab^N = N^{m+1}K$ and so by (7): $\displaystyle N^{m+1}(N^{m+1}K+ab^N)Y^N/bb^NX^N\rightarrow N^{m+1} ab^NY^Nb^{N+1}X^N\rightarrow (sentence\ II) N^{m+1} aY^NbX^N =$ (by (6)) $\displaystyle (N^mL+bX)Y^{N1}bX^N\rightarrow N^m bXY^{N1}bX^N\rightarrow (sentence\ II) N^m Y^{N1}X^{N1}$ Likewise: $\displaystyle N^mX^{N1}Z^{N1}$ , $\displaystyle N^mZ^{N1}Y^{N1}$ END OF PART THREE 
January 15th, 2015, 09:21 AM  #133 
Senior Member Joined: May 2013 Posts: 118 Thanks: 10 
While I’m banging my head on the wall to help ideas come about the first case of FLT, I thought we might say some words about the second case, when NXYZ in $\displaystyle X^N+Y^N=Z^N$ In the text, symbol ┼ means “does not divide” and rel.means relationship Sentences to use: Sentence I: If N is a prime and Nab, then Na or Nb. It holds for more than two factors a and b Sentence II: With E,D and F any natural numbers E┼DF & $\displaystyle (E,D^6+F^6)=1$, then $\displaystyle E ┼ D^{6Μ5}D^{6Μ6}F+D^{6Μ11}F^6D^{6Μ12}F^7+…+D^7F^{6Μ12}D^6F^{6Μ12}+DF^{6Μ6}F^{6Μ5}$ The proof lies in #125 post of this thread (theorem II there) So,assume NZ with no loss of generality.Of course N doesn’t divide X or Y, or X,Y,Z wouldn’t be pairwise prime Then it is known that $\displaystyle N^2X^{N1}1$, $\displaystyle N^2Y^{N1}1$, $\displaystyle N^2X+Y$ (1), If N=6q+1 the first two of the rel.above lead to: $\displaystyle N^2(X^4+X^2Y^2+Y^4)(X^{N6}X^{N7}Y+X^{N12}Y^6X^{Ν13}Y^7+…+X^7Y^{Ν13}X^6Y^{Ν12}+XY^{N7}Y^{N6)}$ (2) For details on how this long rel.is achived,see #125 post again On the other hand,due to (1),it’s elementary to show: $\displaystyle NX^K+Y^K$ (3) for every odd K If (3) held for some even K,we would have: $\displaystyle NX^K+Y^K =Nm$ and as K+1 is odd, $\displaystyle NX^{K+1}+Y^{K+1}=XX^K+Y^{K+1}= X(NmY^K)+Y^{K+1} \rightarrow N XY^K+Y^{K+1} = Y^k(YX)$ and since N┼Y we imply $\displaystyle N\dagger Y^K$ ,therefore sentence I commands: N YX absurd, because then,due to (1) NY & NX So $\displaystyle N\dagger X^6+Y^6$ and as N is a prime $\displaystyle (N,X^6+Y^6)=1$ Given that and since N6=6q5, after sentence II we conclude: $\displaystyle N┼ X^{N6}X^{N7}Y+X^{N12}Y^6X^{Ν13}Y^7+…+X^7Y^{Ν13}X^6Y^{Ν12}+XY^{N7}Y^{N6}$ Then (2) and sentence I result to: $\displaystyle N^2 X^4+X^2Y^2+Y^4 = (X^2+XY+Y^2)(X^2XY+Y^2) \rightarrow (sentence I) N X^2+XY+Y^2$ or $\displaystyle NX^2XY+Y^2$ So if $\displaystyle NX^2+XY+Y^2 = (X+Y)^2XY$ and due to (1):NXY → NY or NX, absurd On the other hand if $\displaystyle NX^2XY+Y^2 = (X+Y)^23XY$ and again N3XY and as N≠3 sentence I leads us to the same absurdity Conclusion: All 6q+1 primes satisfy 2nd case of FLT,as they do with the 1st one Or I'm wrong.In which case I'll change my name to crankshaft 
January 20th, 2015, 12:49 AM  #134 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
Sorry for borken link in the previous post (probably nobody care of it...): Here one of the way to transform powers of integers in trapezoid, and what happen in FLT trick (I still believe in graph visualization to show immediately what we are doing, still if it's a very hard work...): Sorry there is some erro on the lower picture: the height of the bigger trapezoid is of course C (not A that is the height of the littlest one ). Also the eight of the new trapezoid Taonc must be expressed in term of C and Tan(Gamma) where Gamma is the angle.... Is easy to have the abscissa X, or the eight X of the new trapezoid Taonc, with a simple identity: Exceding area over X = Missing Upper Area between 0 and X This two areas are not proportional for FLT request, so no way to tessellate with a common non infimus triangle. And this is true for any n>=3 All the 2th degree equation that solve X (just parameters will change) gives back an irrational, so Tb cannot be an integer area, so no way for B to be an integer. Nothing change if we decide to strech all the trapezoides in the abscissa x multiply it by an integer K... (more concerning are necessary). Last edited by complicatemodulus; January 20th, 2015 at 01:00 AM. 
March 7th, 2015, 10:14 PM  #135 
Senior Member Joined: May 2013 Posts: 118 Thanks: 10 
For a number of reasons this thread has run its course.We'll share some new thoughts we have in a very short while

March 8th, 2015, 10:37 AM  #136 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
Where is the error now: I fond that to respect the rule of the sum that gives powers: (example for n=3) Fermat ask if: (1) so he ask if: (2) But we know that : and that for (my ;P) rule of the sum we can lower both the limits of the second sum without changeing the result from A+1 to 1 with this trick: (3) So since the (3) is for sure right the (2) will be wrong or: while Fermat ask if (applying on the (2) the same rule as in the (3)): Cvd... since A<B. It works for n=2 since the term of the sum is linear so it is possible to arrange the trick to be a square... Not if the complicate modulus has others non linear terms This is what I mean when I say: derivate = curve imply Fermat is right... (but I for sure made some mistake....) 

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