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February 19th, 2016, 11:38 PM   #1
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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.
complicatemodulus is offline  
 
February 26th, 2016, 05:39 AM   #2
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From: Barto PA

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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.
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February 26th, 2016, 07:11 AM   #3
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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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