My Math Forum FLT (end of the story fo all n)

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 February 19th, 2016, 11:38 PM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 FLT (end of the story fo all n) I'm not so happy I've not received any comments on FLT for n=2 and n=3... So I post here what happen for all n just using words: - I already explain why I use X instead of i,m,n or else as index of the Sum: is because I wanna stick what I'm doing on a X,Y plane, and show via Natural or Rational "squaring" or Infimus integration, the property of y'=nx^n curves. - This means that once we prove and understood what happen for n=2 and for n=3 the followings n will NOT require MORE monky computation on higher n (just induction we know on the binomial develope) - Since the same value of X=C will appear (and must be an integer for FLT conditions, but it is not) ALSO in the following derivates of y=x^n - So for the "squarable" property of powers and its derivates I've shown, we can "lower" each odd "n" : n>3, to a multiple (!) of n=3 "equation/term" case. If you don't believe, just derive several times: 5x^4-10x^3+10x^2-5x+1 till discover you arrive to a multiple (!) of 3x^2-3x+1, I already prove doesn't work... Of course the process works in derivative or integration sense, but I think derivative is more easy to be understood. Really disappointed for no feedback on... probably because I left too many bugs so it's very difficult to follow my crazy posts... Never stop studying & working to clean as possible... Thanks ciao Stefano Last edited by skipjack; February 20th, 2016 at 01:52 AM.
 February 26th, 2016, 05:39 AM #2 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 Here is what you wrote in post #1 of your thread _FLT (hope the end of the story...)_: "If we can prove there is no solution in the integers (c1) for A,B,C having a COMMON FACTOR $\displaystyle K\in N$ (so still if we remove the condition c2) we have proven FLT. So be: $\displaystyle A = a*K$ $\displaystyle B = b*K$ $\displaystyle C = c*K$ We now rewrite FLT as: (1) $\displaystyle (a*K)^n <> (c*K)^n-(b*K)^n$ ..." Your basic error was introducing a factor K of A, B, C. K can be removed any time provided your algebra is correct.* {It's a moot point if your algebra is _not_ correct.} * If your algebra is correct, the ratios A/B, B/C, A/C can not change by removing K. Then you are back to square one: having to prove FLT by supposing (A, B, C) = 1; so you may as well have started with that.
 February 26th, 2016, 07:11 AM #3 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Thanks, After many request (from not flexible minds, I respect) I adbandone the 1/K STEP SUM proof (FLT in the Rational) and the revision of that proof in all my old post (that, sorry, several times are just my though I wrote while working... I'm the boss, so don't care of lost hours ;-P), to show that it works for known sum K=1. I kindly ask comments on the n=2 : n=3 solution, because that was several time checked (at my best)... Viceversa for the proof into the rational I'm working to remove bugs I introduce in the post as though while writing, that are already fixed. But I think I've no chance to see Step Sum recognised as usefull, till was not confirmed all works (as I hope) for n=2 and n=3 If so, I hope, I'll gain little respect on this trick and that will be necessary tolet less flexible mind open to understand and accept the Step Sum too... I repat: i can do all without 1/K since is question of "scaling", but all becomes longer and since are simple computation I think is better to digest the Step Sum... Yes you've to digest that at the limit x/K becomes XdX in the integral, but I hope I can better exaplain this point too (with the final help of my wife)... Thanks Ciao Stefano NOTE: The right first step to show "no solution also" for C in Q, is to start with A,B integers and let be/check if there is a solution with a for sure $\displaystyle C\in Q$ and it's power that can be represented via Step Sum... Last edited by complicatemodulus; February 26th, 2016 at 07:52 AM.

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