Matematické Fórum

check_drummer · 25. 07. 2020 08:44

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

vlado_bb · 25. 07. 2020 12:30

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.

check_drummer · 25. 07. 2020 18:24

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

Dacu · 26. 07. 2020 08:16

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

check_drummer · 26. 07. 2020 22:58

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

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

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

check_drummer · 27. 07. 2020 20:45

↑ 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?

Dacu · 28. 07. 2020 14:03

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

check_drummer

↑ 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.

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

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

check_drummer · 02. 08. 2020 09:30

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

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

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

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

check_drummer · 05. 08. 2020 20:56

↑ 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.

Matematické Fórum

#26 25. 07. 2020 08:44

Re: Fermatova Velká Věta

#27 25. 07. 2020 12:30

Re: Fermatova Velká Věta

#28 25. 07. 2020 18:24

Re: Fermatova Velká Věta

#29 26. 07. 2020 08:16

Re: Fermatova Velká Věta

#30 26. 07. 2020 22:58

Re: Fermatova Velká Věta

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

Re: Fermatova Velká Věta

check_drummer napsal(a):

#32 27. 07. 2020 20:45

Re: Fermatova Velká Věta

#33 28. 07. 2020 14:03

Re: Fermatova Velká Věta

check_drummer napsal(a):

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

Re: Fermatova Velká Věta

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

Re: Fermatova Velká Věta

check_drummer napsal(a):

#36 02. 08. 2020 09:30

Re: Fermatova Velká Věta

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

Re: Fermatova Velká Věta

check_drummer napsal(a):

#38 05. 08. 2020 20:56

Re: Fermatova Velká Věta

Zápatí