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April 23rd, 2016, 05:13 AM   #11
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Luckily my last post was blown away for an unknown reason....

The point is show that a proportional condition on $Y$ (here Delta/Delta=1) on a curve $Y=X^2$, that has a first linear derivate can again "generate" a proportional condition on $X$, here $(C-A)/(C-B)= p/q$ with $(p,q)\in N$.

While this cannot happen on $Y=X^n$ wiht $n>2$ because the non linear first derivates define an irrational ratio between the "middle" integer height ($C^n\in N$) and the correspondent irrational X ($C$), that is his n-th root.

Math asap.... but picture already tell the story...

Happy to hear that Mr. Whiles becomes Sir. and 750KUsd reach, while we, humans, are already searching for a more usefull solution to FLT problem ;-P

Last edited by complicatemodulus; April 23rd, 2016 at 05:18 AM.
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May 14th, 2016, 11:39 PM   #12
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I return in math remembering my first post in this topic:

Part 1) FLT rewritten with Sum is:

$\displaystyle A^3 = \sum_{X=B+1}^{C} (3X^2-3X+1)$

or

$\displaystyle A^3 = \sum_{X=1}^{C-B} (3(X+B)^2-3(X+B)+1)$

or

$\displaystyle A^3 = \sum_{X=1}^{C-B} (3X^2+6BX+3B^2-3X-3B +1)$

or

$\displaystyle A^3 = \sum_{X=1}^{C-B} ((3X^2-3X +1) +6BX-3B +3B^2) $

or

$\displaystyle A^3 = (C-B)^3 + 3B \sum_{X=1}^{C-B} (2X-1) +3B^2(C-B)$

or

$\displaystyle A^3 = (C-B)^3 + 3B (C-B)^2 +3B^2(C-B)$

so $(C-B)>=1$ is a factor of A

Part 2) for the same reason/process will be:

$\displaystyle B^3 = (C-A)^3 + 3A (C-A)^2 +3A^2(C-A)$

so $(C-A)>=1$ is a factor of B

This is true for any $n>=2$ since rising n the limits are the same and change just the term $Mn= (X^n-(X-1)^n)$

Part 3) Why for n=2 there is a solution:

Remembering via Step Sum we prove we are able to represent also Rational value/powers, for n=2 we can write:

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

Introducing the Step Sum, keeping: $K=A$ and calling $x=X/A$ I already show nothing change if we will rewrite the (1) as the Step Sum, Step $x_1=1/A$, $x_2= 2/A$; $x_3= 3/A$ etc...:

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

And now we can divide both terms by $A^2$, dividing the upper limit of the Step Sum by $A$:

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

But remembering the is: $ A= \pi_1 *(C-B) $

And solving and grouping the new known square:

$\displaystyle 1= \sum_{x=1/A}^{(C-B)/A)} (2x/A-1/A^2)) + 2B/A *((C-B)/A) $

or:

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

or:

$\displaystyle 1 = 1/{\pi_1}^2 + 2B/((C-B)* \pi_1) $


or:

$\displaystyle {{\pi_1}^2} * (C-B) = (C-B)+2B $

or:

$\displaystyle {\pi_1}^2 = (C+B)/(C-B) $

that fit for many values: f.ex

$\displaystyle {\pi_1}^2 = (5+4)/(5-4) = 9$

$\displaystyle {\pi_1}=A = 3 $

As already written it works for any $A>=3$

infact in case "A" is ODD:

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

in case "A" is EVEN:

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

Part 4) Why Fermat is right for $n=3$

Because doing the same tedious "process" starting from:

$\displaystyle A^3 = \sum_{X=B+1}^{C} (3X^2-3X+1)$

Sorry for this mental torture... I'll jump directly to the reduced Step Sum already divided by $A^3$:

$\displaystyle \sum_{x=1/A}^{1} (3x^2/A-3x/A^2+ 1/A^3) = \sum_{x=1/A}^{(C-B)/A} (3(x+B)^2/A -3(x+B)/A^2+1/A^3))$

so.... remembering the is: $ A= \pi_1 *(C-B) $ or $ B= \pi_2 *(C-A) $

we arrive to the non elegant solution:

$\displaystyle {\pi_1}^3- 3B^2{\pi_1}/(C-B) - 3B-1 =0 $

or (if we start from $B^3=C^3- A^3$) to:

$\displaystyle {\pi_2}^3- 3A^2{\pi_1}/(C-A) - 3A-1 =0 $


that is not elegant AT ALL since rising n we have again to repeat all this torture...

...but I hope here is much easy to "see" what is the dust (what I call the mixed product) that lock the gears rising n...

Waiting for your kindly check... and probably a way to add some shortcut using your more powerfull math...

Thanks
Ciao
Stefano

Last edited by complicatemodulus; May 14th, 2016 at 11:44 PM.
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May 15th, 2016, 02:01 AM   #13
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One minute more on, let me have an hope for the final elegant proof just remembering the binomial develope rule...


Thanks
Ciao
Stefano
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May 15th, 2016, 09:21 AM   #14
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Quote:
Originally Posted by complicatemodulus View Post
FLT n=3 means A divisible by C-B ?
Without touching upon your considerations

A^n=(C-B)[C^(n-1)+....+B^(n-1)]=(C-B){(C-B)^(n-1)+nCB[C^n-3)-...+B^(n-3)]}

C-B and polynomial in brackets are coprime unless C-B is divided by n. The rest of its factors must be to the power n (including n=3) and divide A.
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May 15th, 2016, 08:51 PM   #15
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Quote:
Originally Posted by McPogor View Post
Without touching upon your considerations

A^n=(C-B)[C^(n-1)+....+B^(n-1)]=(C-B){(C-B)^(n-1)+nCB[C^n-3)-...+B^(n-3)]}

C-B and polynomial in brackets are coprime unless C-B is divided by n. The rest of its factors must be to the power n (including n=3) and divide A.
Thanks, that was very clear... but the point is very far from the the topic's title...
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