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#26 25. 07. 2020 08:44

check_drummer
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Re: Fermatova Velká Věta

↑↑ Dacu:
How does from the fact that $x$,$y$ are natural numbers follows that $s$,$t$ are natural numbers?


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#27 25. 07. 2020 12:30

vlado_bb
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Re: Fermatova Velká Věta

Dear colleagues, I certainly do not want to interfere with this discussion, however, Dacu - if you are sure you have an alternative proof of FLT, I strongly recommend to submit it into a suitable journal, in this case probably to the Annals of Mathematics.

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#28 25. 07. 2020 18:24

check_drummer
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Re: Fermatova Velká Věta

It can be useful excercise to find weak spots in presented proofs, so we can continue this discussion.


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#29 26. 07. 2020 08:16

Dacu
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Re: Fermatova Velká Věta

Hello all,

Dear users of the forum, I will soon show that from variant "2) $x=-\bigg (\frac{y^{n-1}}{z^{n-1}}-\cos{C}\bigg )\cdot t$ and $y=\bigg (\frac{x^{n-1}}{z^{n-1}}-\cos{B}\bigg )\cdot t$ where $t\in \mathbb N^*$ with $t=2z^{n-1}\cdot x\cdot y\cdot s$ where $s\in \mathbb N^*$ and $0<x<y<z$ ". results $x=y$.
For personal reasons I have to postpone the next post. Thank you very much for your understanding!

All the best,

Dacu


"Don't worry about your difficulties to math.I assure you that mine are even bigger! ” Albert Einstein

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#30 26. 07. 2020 22:58

check_drummer
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Re: Fermatova Velká Věta

↑ Dacu:
Firstly you have to proove that $t=2z^{n-1}\cdot x\cdot y\cdot s$ and that $s\in \mathbb N^*$.


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#31 27. 07. 2020 07:11 — Editoval Dacu (27. 07. 2020 07:19)

Dacu
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Re: Fermatova Velká Věta

check_drummer napsal(a):

↑ Dacu:
Firstly you have to proove that $t=2z^{n-1}\cdot x\cdot y\cdot s$ and that $s\in \mathbb N^*$.

Hello,

Because in 5) $\bigg (\frac{x^{n-1}}{z^{n-1}}-\cos{B}\bigg )\cdot x+\bigg (\frac{y^{n-1}}{z^{n-1}}-\cos{C}\bigg )\cdot y=0$ we have $0<\frac{x^{n-1}}{z^{n-1}}-\cos{B}=\frac{x^n-y^n-z^{n-2}(x^2-y^2)}{2z^{n-1x}}<1$ and $0<\frac{y^{n-1}}{z^{n-1}}-\cos{C}=\frac{y^n-x^n-z^{n-2}(y^2-x^2)}{2z^{n-1}y}<1$ , then it follows that the equation $\bigg (\frac{x^{n-1}}{z^{n-1}}-\cos{B}\bigg )\cdot x+\bigg (\frac{y^{n-1}}{z^{n-1}}-\cos{C}\bigg )\cdot y=0$ is a Diophantine equation with rational coefficients.
----------------------------------------
Example of Diophantine equation with rational coefficients:

Solve the Diophantine equation in the set of integer numbers $\frac{2}{3}\cdot x+\frac{7}{5}\cdot y=0$.

Solution:

$x=-\frac{7}{5}\cdot t$ and $y=\frac{2}{3}\cdot t$ where$ t=15\cdot s$ and $s\in \mathbb Z$.
I am waiting for the following questions ...
Thank you very much!

All the best,

Dacu


"Don't worry about your difficulties to math.I assure you that mine are even bigger! ” Albert Einstein

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#32 27. 07. 2020 20:45

check_drummer
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Re: Fermatova Velká Věta

↑ Dacu:
Hi,
1) why do both inequalities hold?
$0<\frac{x^{n-1}}{z^{n-1}}-\cos{B}=\frac{x^n-y^n-z^{n-2}(x^2-y^2)}{2z^{n-1x}}<1$
and why it is important that they hold?

2) Why it is important that we got Diophantine equation?


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#33 28. 07. 2020 14:03

Dacu
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Re: Fermatova Velká Věta

check_drummer napsal(a):

↑ Dacu:
Hi,
1) why do both inequalities hold?
$0<\frac{x^{n-1}}{z^{n-1}}-\cos{B}=\frac{x^n-y^n-z^{n-2}(x^2-y^2)}{2z^{n-1x}}<1$
and why it is important that they hold?

2) Why it is important that we got Diophantine equation?

Hello,

1) Correction:

$0<\bigg |\frac{x^{n-1}}{z^{n-1}}-\cos{B}\bigg |=\bigg |\frac{x^n-y^n-z^{n-2}(x^2-y^2)}{2z^{n-1x}}\bigg |<1$ and $0<\bigg |\frac{y^{n-1}}{z^{n-1}}-\cos{C}\bigg|=\bigg |\frac{y^n-x^n-z^{n-2}(y^2-x^2)}{2z^{n-1}y}\bigg |<1$

2) For demonstration .... and in fact $x^n+y^n=z^n$ is a Diophantine equation ...

All the best,

Dacu


"Don't worry about your difficulties to math.I assure you that mine are even bigger! ” Albert Einstein

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#34 28. 07. 2020 20:46 — Editoval check_drummer (28. 07. 2020 20:59)

check_drummer
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Re: Fermatova Velká Věta

↑ Dacu:
So why does the inequalities with absolute values hold?
Edit: I see inequality "<1" - because we are subtracting numbers that are in interval (0;1), but I don't see inequality "<0" ie why those numbers are not equal.


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#35 29. 07. 2020 07:02 — Editoval Dacu (29. 07. 2020 07:13)

Dacu
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Re: Fermatova Velká Věta

check_drummer napsal(a):

↑ Dacu:
So why does the inequalities with absolute values hold?
Edit: I see inequality "<1" - because we are subtracting numbers that are in interval (0;1), but I don't see inequality "<0" ie why those numbers are not equal.

Hello,

I repeat:

"From $\bigg (\frac{x^{n-1}}{z^{n-1}}-\cos{B}\bigg )\cdot x+\bigg (\frac{y^{n-1}}{z^{n-1}}-\cos{C}\bigg )\cdot y=0$ the following possibilities result:

1) $\frac{x^{n-1}}{z^{n-1}}-\cos{B}=0$ and $\frac{y^{n-1}}{z^{n-1}}-\cos{C}=0$.

2) $x=-\bigg (\frac{y^{n-1}}{z^{n-1}}-\cos{C}\bigg )\cdot t$ and $y=\bigg (\frac{x^{n-1}}{z^{n-1}}-\cos{B}\bigg )\cdot t$ where $t\in \mathbb N^*$ with $t=2z^{n-1}\cdot x\cdot y\cdot s$ where $s\in \mathbb N^*$ and $0<x<y<z$.

Which of the two possibilities can exist, 1) or 2)?"
----------------------------------------------------------------

What inequality are you talking about, what equality of numbers are you talking about?Thank you very much!



All the best,

Dacu


"Don't worry about your difficulties to math.I assure you that mine are even bigger! ” Albert Einstein

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#36 02. 08. 2020 09:30

check_drummer
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Re: Fermatova Velká Věta

↑ Dacu:
So - you dont know the answer and this is the question for the whole forum?

Why in 2) this must hold: $t=2z^{n-1}\cdot x\cdot y\cdot s$, $s\in \mathbb N^*$ ?


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#37 04. 08. 2020 06:31 — Editoval Dacu (04. 08. 2020 06:32)

Dacu
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Re: Fermatova Velká Věta

check_drummer napsal(a):

↑ Dacu:
So - you dont know the answer and this is the question for the whole forum?

Why in 2) this must hold: $t=2z^{n-1}\cdot x\cdot y\cdot s$, $s\in \mathbb Z^*$ ?

Hello,

I reformulate variant 2):

2) $x=-\bigg (\frac{y^{n-1}}{z^{n-1}}-\cos{C}\bigg )\cdot t$ and $y=\bigg (\frac{x^{n-1}}{z^{n-1}}-\cos{B}\bigg )\cdot t$ where $t\in \mathbb Z^*$ and $0<x<y<z$.
Let $\frac{y^{n-1}}{z^{n-1}}-\cos{C}=\frac{p}{q}$ and $\frac{x^{n-1}}{z^{n-1}}-\cos{B}=\frac{r}{s}$ two irreducible fractions , then there are the following possibilities:
a) $p<0$ , $r>0$ , $x>0$ and $y>0$ where $t=q\cdot s\cdot u$ with $u\in \mathbb N^*$.
b) $p>0$ , $r<0$ , $x>0$ and $y>0$ where $t=q\cdot s\cdot u$ where $u\in \mathbb Z^{-}$.
Because $0<x<y<z$ and $x$ , $y$ , $z$ must have no common divisors, then it is necessary that $|t|=|q\cdot s\cdot u|=1$ and so $x=y$ which contradicts $x<y$.

All the best,

Dacu


"Don't worry about your difficulties to math.I assure you that mine are even bigger! ” Albert Einstein

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#38 05. 08. 2020 20:56

check_drummer
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Re: Fermatova Velká Věta

↑ Dacu:
So you are trying to prove that variant 2) cannot hold, right? So I have two questions:
I) How do you know that $t$ is whole number and not rational number? (From what fact daes it follow?)
II) Why other variants except a) and b) cannot hold - ie that p,q have the same sign?
III) Why eg in a) equality $t=q\cdot s\cdot u$ holds?
Thank you.


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