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July 4th, 2017, 12:50 AM   #1
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Complicate Modulus Algebra and Class of the Rest

...be prepared....

I discover that also my Complicate Modulus Algebra shows some properties looking at the Congruence Class of certain tricks...

And Fermat the Last is again a good table of work.

Again no limits in going deep...

Last edited by complicatemodulus; July 4th, 2017 at 12:55 AM.
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July 4th, 2017, 01:22 AM   #2
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July 4th, 2017, 10:21 PM   #3
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Ok to let the trick works in the most simple way, so to finally have in the hands the Most Elegant Solution to Fermat's The Last,

I have to generalize my Complicate Modulus Algebra doing this:

Taking as example Fermat's Last Equation rewritten (thanks to the Sum properties) as:

1 $\displaystyle C^n = 2A^n + \Delta$

2 $\displaystyle C^n = 2B^n - \Delta$

Where $\Delta = B^n-A^n$

0) I've to better define some properties of the Rest looking to how it is generated.

For example in a Complicate Modulus Number if we have a Big Rest, it means that it can just comes from a "big enough" Integer Root. But, most clear and useful,

- The $Maximum Rest$ we can have is the difference between Two Following Numbers (here (2*Integer Roots)^n), -1.

- Also we have to define whether we are playing with a Reduced Complicate Modulus Number or a NON Reduced,

- so make a difference if the Integer Root is the Maximum Integer Root we can extract from the Number we are considering, so the Rest is never bigger than The $Maximum Rest$ defined as above.

- and a more general "NON Reduced Complicate Modulus Number" where the Rest is (or can be) bigger as we want, for example because we kept, for some reason, a too little Integer Root.

1) More in general (than shown here in my previous post) we can keep as "modulo" also More Complicated Value,

so not just $(X-(X-1)^n)$

but also other value, for example:

- the difference between: $2X^n$ and $2(X-1)^n$

that is exactly what we need for Fermat's problem.

2) Second:

Positive Rest is just half of the possible representation, so Negative Rest is the other half we need to complete our Tools Box.

So we have a Most General Complicate Modulus Algebra

and we have in the hands now 2 "Two Hand Clocks":

- One Two Hands that works counting Positive Rest, where the Rest Rise (Forward) from $0$ to a $Maximum Rest Value$

- One Two Hands that works with a Bigger integer Root (that exceed the number we are considering) and a Negative Rest, where, vice versa from the first clock, the Rest Falls (Backward) from the $Maximum Rest Value$ to $0$

Fermat's The Last, state that the Number of Steps Forward, that is $\Delta$, has to be equal to the Number of Steps Backward, that is again $\Delta$.

This fix the $Maximum Rest$ that can be, JUST:

$Maximum Rest= 2*\Delta$

The story enter mode deep (differential equations I was looking for long time), and I've no time now, but for those already understood my Complicate Modulus Algebra will not be so hard to realize that the only chance we have to solve is to stick this problem on its analytic extension, so solve this LINEAR Differential Equation where the $Maximum Rest$ lies on:

$y'=2x$

so when, just, $F(x) = X^2$

In all the other case for Bigger $n$ we have in the hands a Too Big $Maximum Rest$, that is defined (sorry no time here but is not so hard to understand) by the way of the first derivate: so $3X^2$ in case we have cube etc...

To be short: in case $n>2$ immediately for $B=A+1$ we have too much division on the Clock than the ones Fermat requires that is: $Maximum Rest= 2*\Delta$.

Here the example of how the Two-Hand-Clocks works in the case $n=2$, where the first derivate is linear and for so where the Rest Rise and Falls in the two Clocks at the same speed, so find after the same time an Equilibrium Point that is exactly at half of the way.

- To find a solution $2A^2$ and $2B^2$ Fermat state they must have the same Class of absolute value of the Rest here 7, 119, 257...

So the Maximum Number of Division of the Clock for the Bigger Number (2B^2) must be exactly $2*\Delta$, independently from who is $n$:



Is not hard to show that $\Delta$, independently by who is $n$ depends on a 2nd degree equation in $A,B,C$ (make the product of the (1)x(2) then solve in $\Delta$).

It will take some months to have the official version... but I won't leave my readers with "just an illusion" (again).

Exactly 10 years... I know it was possible, but I've few hope to really close in this elegant way because I lost in this last year most of the life rest in my body...

"Clean" solution will come asap (So when I'll "be back"). I will also show using my Step Sum Step 1/K how the solution will be found in case $n>2$ going infimus (so with the infinite descent) when the Sum becomes an integral (the process was already shown here).

Last edited by skipjack; July 5th, 2017 at 04:57 AM.
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July 4th, 2017, 11:32 PM   #4
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Sorry time expired while editing...


Ok to let the trick works in the most simple way, so to finally have in the hands the Most Elegant Solution to Fermat's The Last,

I have to generalize my Complicate Modulus Algebra doing this:

Taking as example Fermat's Last Equation rewritten (thanks to the Sum properties) as:

1 $\displaystyle C^n = 2A^n + \Delta$

2 $\displaystyle C^n = 2B^n - \Delta$

Where $\Delta = B^n-A^n$

0) I've to better define some properties of the Rest looking to how it is generate.

For example in a Complicate Modulus Number if we have a Big Rest, it means that it can just comes from a "big enough" Integer Root. But, most clear and useful,

- The $Maximum Rest$ we can have is the difference between Two Following Numbers, here made as $(2*Integer Roots^n), -1$.

- Also we have to define whether we are playing with a Reduced Complicate Modulus Number or a NON Reduced,

- so make a difference if the Integer Root is the Maximum Integer Root we can extract from the Number we are considering, so the Rest is never bigger than The $Maximum Rest$ defined as above.

- and a more general "NON Reduced Complicate Modulus Number" where the Rest is (or can be) bigger as we want, for example because we kept, for some reason, a too little Integer Root.

1) More in general (than shown here in my previous post) we can keep as "modulo" also More Complicated Value,

so not just $(X-(X-1)^n)$

but also other value, for example:

- the difference between: $2X^n$ and $2(X-1)^n$

that is exactly what we need for Fermat's problem.

2) Second:

Positive Rest is just half of the possible representation, so Negative Rest is the other half we need to complete our Tools Box.

So we have a Most General Complicate Modulus Algebra

and we have in the hands now 2 "Two Hand Clocks":

- One Two Hands that works counting Positive Rest, where the Rest Rise (Forward) from $0$ to a $Maximum Rest Value$

- One Two Hands that works with a Bigger integer Root (that exceed the number we are considering) and a Negative Rest, where, vice versa from the first clock, the Rest Falls (Backward) from the $Maximum Rest Value$ to $0$

Fermat's The Last, states that the Number of Steps Forward, that is $\Delta$, has to be equal to the Number of Steps Backward, that is again $\Delta$.

This fix the $Maximum Rest$ that can be, JUST:

$Maximum Rest= 2*\Delta$

The story enter mode deep (differential equations I was looking for long time), and I've no time now, but for those already understood my Complicate Modulus Algebra will not be so hard to realize

that the only chance we have to solve is to stick this problem on its analytic extension, so solve this LINEAR Differential Equation where the $Maximum Rest$ lies on:

$y'=2x$

so when, just, $F(x) = X^2$

In all the other case for Bigger $n$ we have in the hands a Too Big $Maximum Rest$, that is defined (sorry no time here but is not so hard to understand) by the way of the first derivate: so $3X^2$ in case we have cube etc...

To be short: in case $n>2$ immediately for $B=A+1$ we have too much division on the Clock than the ones Fermat requires that is: $Maximum Rest= 2*\Delta$.

Here the example of how the Two-Hand-Clocks works in the case $n=2$, where the first derivate is linear and for so where the Rest Rise and Fells in the two Clocks at the same speed, so find after the same time an Equilibrium Point that is exactly at half of the way.

- To find a solution $2A^2$ and $2B^2$ Fermat state they must have the same Class of absolute value of the Rest here 7, 119, 257...

So the Maximum Number of Division of the Clock for the Bigger Number (2B^2) must be exactly $2*\Delta$, independently from who is $n$:



It's not hard to show that $\Delta$, independently by who is $n$ depends on a 2nd degree equation in $A,B,C$ (make the product of the (1)x(2) then solve in $\Delta$).

It will take some month to have the official version... but I won't left my readers with "just an illusion" (again).

Exactly 10 years... I know it was possible, but I've few hope to really close in this elegant way because I lost in this last year most of the life rest in my body...

"Clean" solution will come asap (So when I'll "be back"). I will also show using my Step Sum Step 1/K how the solution will be found in case $n>2$ going infimus (so with the infinite descent) when the Sum becomes an integral (the process was already shown here).

Last edited by skipjack; July 5th, 2017 at 04:57 AM.
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July 5th, 2017, 08:46 PM   #5
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One more concerning:

Working with Homogeneous Variables (so with the same Complicate Modulus for All the Clocks) helps:

If we, also, write $C^n$ as $2c^n$ we immediately see what the problem is.

Last edited by skipjack; July 5th, 2017 at 08:52 PM.
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July 6th, 2017, 08:11 PM   #6
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July 8th, 2017, 03:47 AM   #7
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Hey CM, had a drink too many?

https://commons.wikimedia.org/wiki/F...CK-N3-0-11.jpg

Last edited by Denis; July 8th, 2017 at 03:51 AM.
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July 8th, 2017, 09:06 PM   #8
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Quote:
Originally Posted by Denis View Post
It was a 9 years Trip. And It looks wondefull and powerfull day by day. Really don't know how many have already understood it...
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July 9th, 2017, 03:04 PM   #9
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What the hell is the meaning of "complicate" here?
You've made 5 consecutive posts in this thread.
Your English is difficult to understand: why don't
you get an English girlfriend to help you post clearly?

I thought it was only me that couldn't "follow" your
posts: but nobody else has commented, so it's not.
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July 10th, 2017, 04:59 AM   #10
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Quote:
Originally Posted by Denis View Post
I thought it was only me that couldn't "follow" your
posts: but nobody else has commented, so it's not.
Well I'm following this ... but for entertainment purposes only.
Thanks from jonah and Denis
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