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October 19th, 2016, 08:18 AM   #11
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...summarising:

FLT can't work in N nor Q, since the $x$ of the solution for FLT equation depends by $2^{(1/n)}$, while the x we can rise Rational step depends by $2^{(1/(n-1))}$

$n=2$ is the only exception.


Last edited by complicatemodulus; October 19th, 2016 at 09:10 AM.
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October 20th, 2016, 08:35 AM   #12
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Here the example for n=2 for who lost my previous post on that:



Solutions comes from:

Re-writing FLT for n=2 with the sum we ask when:

$\displaystyle A^2 = \sum_{X=1}^{A}{(2X-1)} =? \sum_{X=B+1}^{C}{(2X-1)} $


if A is ODD, I found that if we put what well known from the old Greek mathematician, in this equation:


$ B= (A^2-1)/2 $ ; $ C= (A^2+1)/2$


We have immediately:

\[ A^2 = \sum_{x=1}^{A}{(2X-1)} = \sum_{X=B+1}^{C}{(2X-1)} = \sum_{m= {\frac {(A^2-1) }2} +1 }^{\frac {(A^2+1) }2 } {(2X-1)} \]

\[ = \sum_{X= {\frac {(A^2+ 1) }2}}^{\frac {(A^2+1) }2 } {(2X-1)} = \left( {2* {\frac {A^2+1 }2 }} \right)-1 = A^2 \]


So for any A ODD exist a couple B, C as above described, that satisfy the equation.

Or in case A is EVEN we can put:

$B= ((A/2)^2) -1$ ; $C= ((A/2)^2) +1) $


And we come back in the same identity, so also for any A EVEN exist a couple B, C as above described, that satisfy the equation.

Under Fermat's conditions, for any $A\in N$ exist a couple of $B, C \in N$ as above described, that satisfy the:

\[ A^2 = C^2 - B^2 \]


Note: this A,B,C triplets as described, ARE NOT 'ALL' THE POSSIBLE TRIPLETS.
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October 20th, 2016, 11:54 PM   #13
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I hope is now clear that FLT condition (in a well hidden way) state for the derivate $y'=nx^{(n-1)}$:

$Deltax / Deltay \in Q $ = cost.

I hope, I also prove: no solution also for $A,B,C \in Q$ for $n>=3$

(and that it's possible to keep a weak condition: A,B,C pair coprimes, not necessary to be 3 coprimes)

Still if I've no answers on all that, I'm quite confident this "simple" poof is correct.

I'm now living more in pace... free to better prove what already said on Beal...

Thanks to mymathforum for the guest, space ad time !
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November 8th, 2016, 12:19 AM   #14
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I did my best...

Here all my work on Complicate Modulus Algebra, Fermat & Beal:

https://www.academia.edu/29726700/The_Two_Hands_Clock

The arXiv version will folow asap, I hope cleaned enough and of course without Appendix and note for beginners...
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November 9th, 2016, 09:15 AM   #15
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What I don't write but will be clear to who really understand my method

is that for $n>3$ if FLT equation has integers/rational solution for $C$, than any height $C^n/K$, for $K\in N$ taken on $Y=2X^n$, must be squarable in the Rationals once we keep $A, B\in N$, but this cannot be possible because each value coming from putting

$Y= C^n/K$

in

$Y=2X^n$

Will give back a different irrational... so this means that indipendently by the irrational Upper Limit we choose, we can square the area bellow the derivate with the Rational Sum I show...

While I've shown, in general, it's not possible to square an Irrational with a Sum or a Step Sum.

The only condition is the trivial one where the irrational part of the number can be taken out from the number (so from the Sum) as a factor. But in this case we cannot compare Sums with integer/rational index (A, B), with the one multiplied by such irrational...

I confess I really laughing thinking on how many Professors will have to eat a brand new cap.... ;-P

I'm trying to go ArXiv, but they change again the latex formatting standard...

I'll never give up !

Ciao
Stefano

Last edited by complicatemodulus; November 9th, 2016 at 09:19 AM.
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November 10th, 2016, 08:29 AM   #16
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In short words:

Just for $n=2$, Delta Base /Delta Height, for the rational Gnomon rectangular area, lead to Rational results, for all other $n$ this become an irrational so no way to square it using else than infimus step.

Ciao
Stefano

p.s. some error in the previous post, but no time to fix it now.
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