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June 15th, 2016, 03:25 AM   #21
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"I wonder you keep talking- no one marks you."

William Shakespeare, "Much Ado about Nothing".
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June 15th, 2016, 04:51 AM   #22
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'And, when you want something, all the universe conspires in helping you to achieve it.'

Paulo Coelho

....FLT a part...

Stefano ;-P
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June 16th, 2016, 05:10 AM   #23
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Any void turn push me to study and search more, and more...

Unfortunately each turn becomes time by time more tricky.

The problem is to find out a trick where the property of the Powers (non K dependence), under FLT equation fail.

I've another trick here that use all what shown till now, but, i hope, will lead to a correct final answer. The problem is that it's soo long in computation that it's very hard to state, before start if I'm on the same ring... or I found the right exit...

The start it's the same:

A Power of an integer is squarable indipendently by $K$, with $1<K<\infty$, and also for some $K>1$ at condition $K|A$:

So we can write ( remembering $x=X/K$ ):

$\displaystyle A^2 = \sum_{1}^{A} 2X-1 = \sum_{1/K}^{A} 2x/K-1/K^2 = \lim_{K\to\infty} \sum_{1/K}^{A} 2x/K-1/K^2 = \int_{0}^{A} 2x dx= A^2 $

If we can find, starting from $A^n=C^n-B^n$, an equation in A,B,C,K where we show the K dependence (with K under correct conditions) we have the proof of why Fermat is right.

so another turn can be, starting from the same equation of the previous trick:


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

but turning it in Step Sum:

$\displaystyle \sum_{1}^{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 $

and subtracting term by term, as possible, the right hand from the left hand we arrive to a monstre equation K dependent in the form...

$\displaystyle (A+B-C)^n = f(K)$

But before bore you with the long computation, the question is if someone can immediately see, before calculation that we will be, again, in a round trip (or in a yellow submarine) ?

After long computation (to be cecked) the solution looks like:

$(A+B-C)^3 =3B^2C^2/K-6CB^3/K-12ABC+3BA^2+6AB^2+6AC^2+9BC^2-3CA^2-9CB^2+3B^4/K+3B^3-3C^3-6ABC/K+3BC^2/K-3AB^2/K-3AC^2/K-3CB^2/K+3C^3/K-3B^3/K+3A^2/K+3B^2/K+3C^2/K+3BC^2/K^2-12CB^2/K^2+6AB/K-6AC/K-6BC/K+9B^3/K^2+6AC/K^2+12BC/K^2-6AB/K^2-6B^2/K^2-6C^2/K^2-6BC/K^3+9B^2/K^3+3C/K^3-3A/K^3-3B/K^3+3B/K^4$

Thanks
ciao
Stefano

Last edited by complicatemodulus; June 16th, 2016 at 05:15 AM.
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June 21st, 2016, 07:03 AM   #24
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This means that pulling $K\to\infty$, we have to return to the known Fermat equation, but we have:

$\displaystyle (A+B-C)^3 =3B^2C^2/K-6CB^3/K12ABC+3BA^2+6AB^2+6AC^2+9BC^2-3CA^2-9CB^2+3B^4/K+3B^3-3C^3-6ABC/K+3BC^2/K-3AB^2/K-3AC^2/K-3CB^2/K+3C^3/K-3B^3/K+3A^2/K+3B^2/K+3C^2/K+3BC^2/K^2-12CB^2/K^2+6AB/K-6AC/K-6BC/K+9B^3/K^2+6AC/K^2+12BC/K^2-6AB/K^2-6B^2/K^2-6C^2/K^2-6BC/K^3+9B^2/K^3+3C/K^3-3A/K^3-3B/K^3+3B/K^4$

that becomes:

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

or:

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

or:

$\displaystyle A^3-3AB^2+6ABC-3AC^2-2B^3-6BC^2+2C^3+6CB^2=0$

or:

$C_1= -A/2 +B$

$C_2= A+B$

$C_3= A+B$

So when all will be well checkd we can give, or not, an answer on FLT end...

Still like a check on it...

Thanks
Ciao
Stefano

Last edited by complicatemodulus; June 21st, 2016 at 07:06 AM.
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June 23rd, 2016, 09:02 AM   #25
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OK this is the dirty job PART 1:

From:

(1) $\displaystyle A^3=C^3-B^3$ given for true,

"Cooling" both terms we know:

(2) $\displaystyle (A-1/K)^3 \neq (C-1/K)^3-B^3$

But it return true to the (1), if we push $K\to\infty$

So we can make any kind of correct develope/reduction on the (2) than, at the end, pushing $K\to\infty$ we have to return to the (1) or to another TRUE condition. (this is the point, pls check...)

But we will se we will arrive to an absurdum.

The (2) In Step Sum: we left the equal sign, just remembering at the end K has to go infinity:

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

or:

$\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 right hand limits:

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

Solving the inthernal term and pulling out the known cube:

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

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

Subtracting the cube from the left hand will rest:

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

Again lowering the limits in the left hand:

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

$\displaystyle \sum_{1/K}^{A+B-C}(6Cx/K-6BC/K-6Bx/K+3B^2/K+3C^2/K+3x^2/K+9B/K^2-9C/K^2-9x/K^2+7/K^3) = $

Again extracting the right cube:

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

So we retun to:

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

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

...second part will follow...

Thanks
ciao
Stefano
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June 23rd, 2016, 09:51 PM   #26
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Just working on the right hand:

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

Grouping and extracting the known (2x-1) squares, and non index dependente terms:

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

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

or...

$\displaystyle =6BCA^2/K+18ACB^2/K-18ABC^2/K-3A^2B^2/K-3A^2C^2/K-18B^2C^2/K+6AC^3/K+12BC^3/K+12CB^3/K-6AB^3/K-12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2-3B^4/K-3C^4/K+6B^3-3C^3+3CB^2/K+6AB-6AC-6BC-3B^3/K+3A^2+3B^2+3C^2+24ABC/K^2+6CA^2/K^2+18CB^2/K^2-6BA^2/K^2-12AB^2/K^2-12AC^2/K^2-18BC^2/K^2-6BC/K+6C^3/K^2-6B^3/K^2+6B^2/K-3B^2/K^2+6AC/K^3+6BC/K^3-6AB/K^3-3A^2/K^3-3B^2/K^3-3C^2/K^3+3B/K^2$

and now knowing that for $K\to\infty$ all K franctions will vanish... and if the initial statment was true than this develope must be true again.

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

or:

$\displaystyle A^3-3A^2-3AB^2+6ABC-6AB-3AC^2+6AC-5B^3-3B^2-9BC^2+6BC+2C^3-3C^2+12CB^2=0$

(sorry I've no time for better checking before posting...)


That means if FLT equation has a result for n=3, than we found a new trinomial develope....

This seems an absurdum, so Fermat is Right.



The main point now is: check, and check again if no bugs here...

The same process can be made for any following $n$ and we are sure that the mixed product terms coming from the second degree (and following) terms will always create a non K dependent terms that will not vanish.

Thanks
ciao
Stefano
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June 24th, 2016, 12:53 AM   #27
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i open this i think you talk about fft but flt never listen
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June 24th, 2016, 02:35 AM   #28
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Fore sure it's a booring think, but also my fault... I've no time (/friends/professors) to push a fully perfect version, so I left till now some bugs in the develope ...I'm trying to self check...

I found 2 and in my last revision of this morning the end is:

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

So the "concept" for what FLT equation fails from $n>3$ seems right... but...

Still waiting for a check on the concept, so if I'm aloud to make that algebra "reduction", than again push $K\to\infty$, or if this trick is wrong by concept...

Thanks
ciao
Stefano
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June 24th, 2016, 03:14 AM   #29
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Quote:
Originally Posted by MMath View Post
I open this I think you talk about fft (signal processing), but FLT never listened ?

Nice, thanks, it means all the rest it's true...

Fermat Last Theorem... shorten as FLT in lot of place...

Thanks
Ciao
Stefano
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