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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. Tags end, flt, story Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post V3Slayer New Users 1 February 11th, 2016 03:09 PM kingkos New Users 0 January 23rd, 2013 01:10 PM AnnieJo Algebra 1 July 30th, 2010 12:35 PM rody406 Calculus 0 December 31st, 1969 04:00 PM

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