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June 27th, 2016, 12:39 AM   #1
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Fermat the Last (I hope)

Here the last turn for FLT:
(for clarify what Step Sum, step $1/K$ are and works, pls check my several post on)

Giving for true under Fermat's conditions (for $n>2$, $A<B<C \in N$):

$\displaystyle A^n = C^n - B^n $

we know that, under the same conditions:

(1) $\displaystyle (A-1/K)^n = (C-1/K)^n - B^n $

will be true if $K\to\infty$.

Turning the (1) in Step Sum, Step $1/K$ (so we are now able to work in Q) we can re-write the (1) as:

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

lowering the index in the right hand:

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

$\displaystyle =\sum_{1/K}^{C-B-1/K} \left(3x^2/K- 3x/K^2+1/K^3+6Bx/K-3B/K^2+3B^2/K \right). $

Considering the right hand only, pulling out what we immediately recognize as terms that build a cube $(3x^2/K-3x/K^2+1/K^3)$, a square $3B(2x/K-1/K^2)$ and a simple term non x dependent we have:

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

Now subtracting the littlest cube (right hand) from the bigger one (left hand) :

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

Again lowering the limits (without changing the result): $(C-B)$ to $1/K$ imply we remove also: $(C-B-1/K)$ from the upper limit, and we must add it to the index dependent terms, so we have

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

Again grouping the known Cube, the known Square, and non index dependent terms:

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

$\displaystyle +\sum_{1/K}^{A+B-C} 6Cx/K-9C/K^2- 6Bx/K+9B/K^2-6x/K^2+6/K^3-6BC/K+3B^2/K+3C^2/K = $

$\displaystyle =3B (C-B-1/K)^2 + 3B^2/K (C-B-1/K)^2 $

Considering just the last part of the left hand and grouping what known:

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

$\displaystyle +\sum_{1/K}^{A+B-C}\left(-6BC/K+3B^2/K+3C^2/K-6C/K^2+6B/K^2+3/K^3 \right) $

Solving term by term to be clear:

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

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

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

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

$\displaystyle \sum_{1/K}^{A+B-C}\left(-6BC/K+3B^2/K+3C^2/K-6C/K^2+6B/K^2+3/K^3 \right) = $

$\displaystyle \left(-6BC/K+3B^2/K+3C^2/K-6C/K^2+6B/K^2+3/K^3 \right) * (A+B-C) =$

$\displaystyle =(- 6ABC/K+3AB^2/K+3AC^2/K-9CB^2/K+9BC^2/K+3B^3/K-3C^3/K+ $

$\displaystyle + 6AB/K^2-6AC/K^2-12BC/K^2+6B^2/K^2+6C^2/K^2+3A/K^2+3B/K^2-3C/K^2) $

On the right hand we will have:

$ 3B (C-B-1/K)^2 + 3B^2/K (C-B-1/K)^2 = 3B^2C^2/K-6CB^3/K+3BC^2-6CB^2+3B^4/K+3B^3-6CB^2/K^2-6BC/K+6B^3/K^2+6B^2/K+3B^2/K^3+3B/K^2 $


Reordering :

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


$-3C(A+B-C)^2+3B(A+B-C)^2+3/K(A+B-C)^2-(-6ABC/K+3AB^2/K+3AC^2/K+9BC^2/K-9CB^2/K+3B^3/K-3C^3/K+6AB/K^2-6AC/K^2-12BC/K^2+6B^2/K^2+6C^2/K^2+3A/K^2+3B/K^2-3C/K^2)$


$= -3C(A+B-C)^2 + 3B(A+B-C)^2 + 3/K(A+B-C)^2 + 6ABC/K+3AB^2/K+3AC^2/K+3CB^2/K-3BC^2/K+3B^3/K-3C^3/K+6AB/K^2-6AC/K^2-12BC/K^2+6B^2/K^2+6C^2/K^2+3A/K^3+3B/K^3-3C/K^3+$
$+3B^2C^2/K-6CB^3/K+3BC^2-6CB^2+3B^4/K+3B^3-6CB^2/K^2-6BC/K+6B^3/K^2+6B^2/K+3B^2/K^3+3B/K^2 $

that if $K\to\infty$ becomes (it remember Newton's infinite descent):

$\displaystyle (A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+9BC^2+3B^3-3C^3 $

$\displaystyle (A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+9BC^2-3A^3 $

The right hand we know is NOT the right tri-nomial develop, so this is the absurdum we are looking for.

The reason why it will works for $n=2$ and not for $n>2$ can be summarized as: it's due to mixed terms of higher degree (from 2).

And a short cut will be: first derivate = curve.

I know one or more bugs can affect this "process" I won't call "proof", so I'll wait for an expert check... (sub judice)

It was a very hard work for me (8 years long)... I hope it's close to the full stop (for many reasons).

Thanks
Stefano
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June 27th, 2016, 10:45 AM   #2
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Sorry, still 2 computation bugs, but result don't change in the behaviour....

I'll check better asap...

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Stefano
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June 27th, 2016, 09:34 PM   #3
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Here the last version with fixed develope bugs:

On the right hand we will have (here first error fixed):

$ 3B (C-B-1/K)^2 + 3B^2/K (C-B-1/K) = 3BC^2-3CB^2-6BC/K+3B^2/K+3B/K^2 $


Reordering (here the second error fixed):

$ (A+B-C)^3 = $

$-3C(A+B-C)^2+3B(A+B-C)^2+3/K(A+B-C)^2-(-6ABC/K+3AB^2/K+3AC^2/K+9BC^2/K-9CB^2/K+
3B^3/K-3C^3/K+6AB/K^2-6AC/K^2-12BC/K^2+6B^2/K^2+6C^2/K^2+3A/K^2+3B/K^2-3C/K^2)+
3BC^2-3CB^2-6BC/K+3B^2/K+3B/K^2$


$= -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3+6ABC/K+12CB^2/K-3AB^2/K-
3AC^2/K-9BC^2/K+3C^3/K-6B^3/K+6AB/K-6AC/K-12BC/K+3A^2/K+3C^2/K+9B^2/K+6AC/K^2+
12BC/K^2-6AB/K^2-6C^2/K^2-9B^2/K^2+3C/K^2-3A/K^2 $


that if $K\to\infty$ becomes:

$ (A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3 $

So it's clear that the right hand is NOT the right tri-nomial develop, so this is the absurdum we are looking for.


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Ciao
Stefano
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June 29th, 2016, 11:17 PM   #4
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...I hope someone kindly answer...

The point is if it's possible to works with the limits for $K\to\infty$ in this way or not, so if the turn it's true, or false...

I hope it's true since moving parts from left to right or viceversa we respect the inequality, just respecting the sing rule, and for the limits since we have just Sums nothing has to change...

I already did another trip, just doble longer, where will be clear that assuming $A^=C^n-B^n$ for true in the integers there is no continuty if we assume, and computate in this way from right (+1/K) or from the left side (-1/K), since the two limits, once reduced in this way are different.

But I still be prepared to the worst...

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Ciao
Stefano
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July 7th, 2016, 12:36 AM   #5
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OK, I've a nice discussion with italian friends... (that just erase the topics ...)

It's clear that:

$ (A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3 $

is NOT the right tri-nomial develop, but it has, unfotunately, some integer solution... so what we have to do is to compare this result to the result coming from the limit from the right side so of:

(3) $\displaystyle (A+1/K)^n= (C+1/K)^n-B^n$

At the moment it (once reduced with sum and pushed to the limit) give to me another result, so seems the two limits are different and this is the right absurdum since we know they must be the same on a continuos function...

I will check the computation to avoid errors, and I've also to check witch of the possible cooling, or warming, is the shortest one to show the absurdum.

...it's a long way to tipperary...it's a long way to go...

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Ciao
Stefano
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July 7th, 2016, 02:10 AM   #6
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I check:

(3) $\displaystyle (A-1/K)^n= (C)^n-(B-1/K)^n$

and sems to finish at:

$\displaystyle (A+B-C)^3 = 3BC^2-6CB^2+3B^3 $

So if all computation is right I'm alowd to say that this prove that the initial equation must be wrong ?

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Ciao
Stefano
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