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May 23rd, 2017, 01:39 AM   #1
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Why I claim I solve Beal problem

The trick solve several Power of Integers Problem is:

- the fact that a Power of an integer can be represented by the area below the derivate of $Y=X^n$,

- that such area can be squared in the Integer (or Rational) as a Sum of Integer Gnomons $M_n= (X^n-(X-1)^n)$, or Rational: $M_{n,k}$

For so $M_n$ can be called the Integer Derivate, and $M_{n,k}$ the Rational one.

- Since any n-th problem power be "Linearized" as an area below a Linear Integer Derivate of the type $Y'=2X-1$ in case the Power $n=Even$ or $Y'= 2mX-m$, in case the Power is $m^n$ with $n=Odd$

- Since an equality of the type $A^x+B^y=C^z$ involves just Power of integers

- then it is equal to find the 3 possible trapezoidal area (the ones below their linear Integer derivate)

- It means that since we are talking of the possible ways to combine the integers gnomons of A,B,C, there must be a Common Base Factor to let us put the Gnomons one after another, or one over another.

- this Comomn Base can be just 1, or a Common Factor

under this conditions is clear that few combination of such trapezoides can be found:

1: The power $A^x$, represented as a trapezoides on a cartesian plane fit exactly the first part of $C^z$, and $B^y$ fits the rest respecting the rule that Exceeding Area is equal to the Missing One.

Here an old graph as example that needs to be adjusted (some error in the lables) and to better show the common base of the Rectangular Gnomons:



2: $A^x$ and $B^y$ has different trapezoidal shapes but the Exceeding part of one balance the missing one of the other, and vice versa (some possible subcase is possible here, and I've already posted the pictures)

3: $A^x$, $B^y$, $C^z$ share the same Base (picture already posted here)

This proof depends on the (trivial) fact that the unit $1$ can be here represented as a unitary area $1x1$ that cannot be divided (just transformed in shape, in case, so scaled) and on the simple equality $1=1$

I know A. Grothendieck works on this concept (pushing it as his limit in the most general case), but I'm not able to put the hands on all his work...

Thanks
ciao
Stefano

Last edited by skipjack; May 23rd, 2017 at 10:53 PM.
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May 23rd, 2017, 06:17 AM   #2
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I hope this graph will be more clear:



- All this are what I call Step Sum, since the index X moves step 3,6,9...

- In the first most simple case $A^x$ is equal to the first part of $C^z$,

- while $B^y$ has to be re-arranged, but using the same base is used by $A$ and $C$.

- $B^y$ area must be putted after the area of $A^x$,

- So there must be a continuity on the First Upper Limit, and the Second Lower Limit,

So the Step Sum for $B^y$ must be re-arranged in a Sum of Gnomons that starts from the next Step here is $A^2+A$ and the only condition we have is to respect the fact that Missing Area has to be equal to the Exceeding one.

- Is not hard to discover that this case is just a multiple of the identity: $3^2 = 1+2^3$

The other case are little more complicate, but no soo much...
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May 23rd, 2017, 09:52 PM   #3
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What I hope will be clear is that there is a simple Math Law to find a solution:

- the adding Power (area), so a number of gnomons must be shifted and redistributed in height (I call it "Shift and Balancing") usign JUST the know parameter fixed in the problem.

The Balancing imply there is a redistribution of the gnomons or a Rotation of their Top that at the end lives the area unchanged and equal to the one is required in the problem.

So now we know the missed "Balancing Equation" (or set of equations in case more terms) let us computate the solution or let us say "no way to have a solution", since the shift we ask (like Fermat the Last for n>2) imply an impossible Balancing of the Integer / Rational Gnomons.

Is also clear that below the solution there is an Integer / Rational Differential Equation that fix the ratio between Shift and Rotation (of the top of the gnomons) that is possible or impossible depending on the condition we fix...

Is anybody out there?

Ciao
Stefano

Last edited by skipjack; May 23rd, 2017 at 10:54 PM.
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May 24th, 2017, 04:02 AM   #4
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I'm so glad you have reached a conclusion in your works, I feel the same way with my Prime number formula. You don't realise how much of your unconscious mind is working on it until you have finished doing it.

I haven't actually read any of the posts on this thread here, but I will look into Beal problem and then read this.
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May 24th, 2017, 05:06 AM   #5
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So is the Beal problem the Beal Conjecture? That has 1 million dollars for who ever solves it.
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May 24th, 2017, 05:28 AM   #6
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Thanks !

Still if I'm still working on because till is not on ArXiv... is just waste paper (and unfortunately ...time..).

To avoid you loss of time I've a .pdf still available for few days on my web site since the final version will be for publishing as a kindle (for pinuts cost).

Maruelli-The-two-Hand-Clock-1of2.pdf

It is the Vol 1/2 just where you can read the fist (simple) algebra (I invented for myself)

Here you can find an Old Draft for the Vol.2 on what I'm working on.

Maruelli-The-two-Hand-Clock-Vol.2-draft.pdf


Searching here for Beal you will find the same examples and concernings but nothing of complete and error free.

Working in the "free" time or in office require me to put on LaTex / Pictures, than return on, several times, to clean errors.

Finally all is at your own risk since nobody read/correct it... and I'm not a professor.

Thanks
Ciao
Stefano
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May 24th, 2017, 05:32 AM   #7
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Quote:
Originally Posted by HawkI View Post
So is the Beal problem the Beal Conjecture? That has 1 million dollars for who ever solves it.
I suspect the prize is just a joke and in any case the conditions are "pubblication", and nobody wanna take care of my "work" (from several years ago)...
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May 24th, 2017, 05:34 AM   #8
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So are you saying you have proved it or disproved it?
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May 24th, 2017, 05:54 AM   #9
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Quote:
Originally Posted by HawkI View Post
So are you saying you have proved it or disproved it?
Yes I (hope) I prove solutions require a common shared factor, but seems "solutions" are also just multiple of trivial (Powers) case like:

Case 1) $3^2 = 1+2^3$

Case 2a) $ 1+1 = 2 $

etc...

Here the correct picture of the case 1:



And the equality comes from the re-arrangement of the 2th Stem Sum (step=3 here, pls refer to the picture if not clear) into:

$\displaystyle A^3=\sum_{x=A}^{A^2} 2x-A $

$\displaystyle C^3=\sum_{x=A=C}^{A^3=C^3} 2x-C (or A)$

Than:

$\displaystyle B^3=\sum_{x=B}^{B^2} 2x-B$

has to be shifted to:

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

than re-arranged copling gnomons as:

$\displaystyle B^3=C^5-A^3 = \sum_{x=A^2+A}^{C^3} 2x-C$

You can play just with known constant (A and C here), Linear Terms, and Upper and Lower Limits of the Step Sums, remembering B^y has to follow A^y without leaving holes in the base of C^z (and cannot exceed this base).

Other little more complicate case require just some more concerning.

Last edited by complicatemodulus; May 24th, 2017 at 06:11 AM.
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May 24th, 2017, 08:03 AM   #10
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Quote:
Originally Posted by complicatemodulus View Post

And the equality comes from the re-arrangement of the 2th Stem Sum (step=3 here, pls refer to the picture if not clear) into:
Pls read as:

And the equality comes from the re-arrangement of the 2th STEP Sum (step=3 here, pls refer to the picture if not clear) into:...


Quote:
Originally Posted by complicatemodulus View Post

than re-arranged copling gnomons as:
Pls read as:

Than re-arranged coupling gnomons (2 by 2) as:...

Following the simple rule that the Exceeding Area has to be equal to the missing one so in this case the number of gnomons must be the same and there must be a balancing point in the middle of the Remaining area that has to be equal to $B^3$
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