A Simpler Proof of Fermat's Last Theorem:

[primework123]

A Simpler Proof of Fermat's Last Theorem:

Keywords: Forms and the abc Conjecture (please visit link: https://www.physforum.com/index.php?showtopic=120595)

The theorem states there are no integral (all positive integers) solution to Fermat's equation:

1. x^n + y^n = z^n where n > 2 and x * y * z ≠ 0.

We assume there exists an integral solution to equation one.

Equation one implies 2. x^n = z^n − y^n = ( z^(n/2) + y^(n/2) ) ( z^(n/2) − y^(n/2) ).

We define a rational number, ɛ, such that 0 < ɛ < n < ∞ (Do you see a possible consequence of the sound abc Conjecture here?) so that:

3. x^n = x^( n/2 + ɛ/2 ) * x^( n/2 − ɛ/2) ) = ( z^(n/2) − y^(n/2) ) ( z^(n/2) − y^(n/2) ).

We have 4. x^( n/2 + ɛ/2) = z^(n/2) + y^(n/2) and

5. x^( n/2 − ɛ/2) = z^(n/2) − y^(n/2) since equations four and five are of the simple

forms: a * b = c + d and a / b = c − d, respectively. Verify this result until you're satisfied...

Now from equations four and five, we have

6. x^( n/2 + ɛ/2 ) / x^( n/2 − ɛ/2) = x^ɛ = ( z^(n/2) + y^(n/2) ) / (z^(n/2) −y^(n/2) ).

After applying some algebraic manipulation to equation six, we have

7. z^n = y^n * ( ( x^ɛ + 1 ) / (x^ɛ − 1) )^2.

Now, combining equations one and seven to eliminate z^n , we have

8. y = (1/4)^(1/n) * (x^(n − ɛ ) )^(1/n) * (x^ɛ − 1)^(2/n)
= x * (1/4)^(1/n) * ( (x^ɛ − 1 )^2 / x^ɛ )^(1/n)
= (1/4)^(1/n) * x * ( (x^ɛ − 1 )^2 / x^ɛ )^(1/n).

And since the term, (1/4)^(1/n), is irrational for all n > 2, the product of the far right side of equation eight is not a positive integer. This is a clear contradiction since we assume y is a positive integer.

Thus, the Fermat's Last Theorem is true! (May the great Fermat rest in peace...)

P.S. Hmm. For n = 2, ɛ = 1, and x = 9, we verify equations eight and seven:

For equation eight we have y = 9 * (1/4) ^(1/2) * ( (9 −1)^2 / 9)^(1/2) = 9 * 1/2 * (64 / 9)^(1/2) = 9 * 1/2 * 8/3 = 72/6 = 12 or y =12.

And we have 9^2 + 12^2 = 81 + 144 = 225 = z^2 which implies z = 15. Okay!

Equation seven states z^2 = 12^2 ( (9+1)/(9-1) )^2 = 12^2 * (10/8)^2.
Therefore, z = 12 * 10/8 = 3 * 5 = 15. Okay!

P.S. Rational numbers are dense in real numbers, and there are infinitely many Pythagorean triples. Can you find them all?

David Cole
aka primework123
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!

[Guest]

A Simpler Proof of Fermat's Last Theorem:

Keywords: Forms and the abc Conjecture (please visit link: https://www.physforum.com/index.php?showtopic=120595)

The theorem states there are no integral (all positive integers) solution to Fermat's equation:

1. [tex]x^{n}[/tex] + [tex]y^{n}[/tex] = [tex]z^{n}[/tex] where n > 2 and x * y * z ≠ 0.

We assume there exists an integral solution to equation one.

Equation one implies 2. [tex]x^{n}[/tex] = [tex]z^{n}[/tex] − [tex]y^{n}[/tex] = ( [tex]z^{n/2}[/tex] + [tex]y^{n/2}[/tex] ) ( [tex]z^{n/2}[/tex] − [tex]y^{n/2}[/tex] ).

We define a rational number, ɛ, such that 0 < ɛ < n < [tex]\infty[/tex] (Do you see a possible consequence of the sound abc Conjecture here?) so that:

3. [tex]x^{n}[/tex] = [tex]x^{n/2 + ɛ/2}[/tex] * [tex]x^{n/2 - ɛ/2}[/tex] = ( [tex]z^{n/2}[/tex] + [tex]y^{n/2}[/tex] ) ( [tex]z^{n/2}[/tex] - [tex]y^{n/2}[/tex] ).

We have 4. [tex]x^{n/2 + ɛ/2}[/tex] = [tex]z^{n/2}[/tex] + [tex]y^{n/2}[/tex] and

5. [tex]x^{n/2 - ɛ/2}[/tex] = [tex]z^{n/2}[/tex] - [tex]y^{n/2}[/tex] since equations four and five are of the simple

forms: a * b = c + d and a / b = c − d, respectively. Verify this result until you're satisfied...

Now from equations four and five, we have

6. [tex]x^{n/2 + ɛ/2}[/tex] / [tex]x^{n/2 - ɛ/2}[/tex] = [tex]x^{ɛ}[/tex] = [tex](z^{n/2}[/tex] + [tex]y^{n/2})[/tex] / [tex](z^{n/2}[/tex] - [tex]y^{n/2})[/tex].

After applying some algebraic manipulation to equation six, we have

7. [tex]z^{n}[/tex] = [tex]y^{n}[/tex] * (( [tex]x^{ɛ}[/tex] + 1 ) / ([tex]x^{ɛ}[/tex] − 1))^2.

Now, combining equations one and seven to eliminate the term, [tex]z^{n}[/tex] , we have

8. y = [tex](1/4)^{1/n}[/tex] * [tex]x^{(n − ɛ)^{1/n} }[/tex] * [tex](x^{ɛ} - 1)^{2/n}[/tex]
= x * [tex](1/4)^{1/n}[/tex] * [tex](x^{ɛ} - 1)^{2/n}[/tex] / [tex](x^{ɛ})^{1/n}[/tex].


And since the term, [tex](1/4)^{1/n}[/tex], is irrational for all n > 2, the product of the far right side of equation eight is not a positive integer. This is a clear contradiction since we assume y is a positive integer.

Thus, the Fermat's Last Theorem is true! (May the great Fermat rest in peace...)

P.S. Hmm. For n = 2, ɛ = 1, and x = 9, we verify equations eight and seven:

For equation eight we have y = 9 * (1/4) ^(1/2)* (9 −1)^2 / 9)^(1/2) = 9 * 1/2 * (64 / 9)^(1/2) = 9 * 1/2 * 8/3 = 72/6 = 12 or y =12.

And we have 9^2 + 12^2 = 81 + 144 = 225 = z^2 which implies z = 15. Okay!

Equation seven states z^2 = 12^2 ( (9+1)/(9-1) )^2 = 12^2 * (10/8)^2.
Therefore, z = 12 * 10/8 = 3 * 5 = 15. Okay!

P.S. Rational numbers are dense in real numbers, and there are infinitely many Pythagorean triples. Can you find them all?

David Cole
aka primework123
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!

[Guest]

Reference Link: 'Smooth solutions to the abc equation: the xyz Conjecture' by Prof. JCL and Prof. KSound,
https://www.researchgate.net/publication/45885748_Smooth_solutions_to_the_abc_equation_the_xyz_Conjecture.

[Guest]

Proof of Fermat's Last Theorem Updates are at the following link:

http://math.stackexchange.com/questions/1993460/prove-a-statement-about-a-conditional-diophantine-equation.

[Guest]

More Updates on Problem Solving (Math Conjectures, etc) are found at the following link:

https://plus.google.com/collection/43KtnB.

[Guest]

More Updates on Problem Solving (Math Conjectures, etc) are found at the following link:

https://plus.google.com/collection/43KtnB.

Please recall a previous reference link for updates or new developments on a simpler proof of Fermat's Last Theorem:

http://math.stackexchange.com/questions/1993460/prove-a-statement-about-a-conditional-diophantine-equation.

[Guest]

FYI: 'Discover Fermat's Proof of His Last Theorem (FLT)',

https://plus.google.com/communities/103144691065830451539

[Guest]

Proof of Fermat's Last Theorem Updates are at the following link:

http://math.stackexchange.com/questions/1993460/prove-a-statement-about-a-conditional-diophantine-equation.

The algebraic curves have solutions galore. Seek them and you shall have glory! Amen![/quote]

Study the works of Emmy Noether (https://en.wikipedia.org/wiki/Emmy_Noether), the great master of Algebra:

"On Emmy Noether and Her Algebraic Works"

[Guest]

Proof of Fermat's Last Theorem Updates are at the following link:

http://math.stackexchange.com/questions/1993460/prove-a-statement-about-a-conditional-diophantine-equation.

Study the works of Emmy Noether (https://en.wikipedia.org/wiki/Emmy_Noether), the great master of Algebra:

"On Emmy Noether and Her Algebraic Works"

Excellent Reference Textbook on Algebraic Geometry:

'Algebraic Geometry' by Prof. J.S. Milne (see attached file for details).

[Guest]

Sir, kindly go through the attached file.
Thank you.
Prof. R. C. Ray
My observation
We can derive, e=ln((z^(n/2)+y^(n/2))/(z^(n/2)-y^(n/2) ))/ln(x). Example: Let, z=5, y=2, x=4, n=4. Then, e=ln(29/21)/ln(4).
e is irrational, (1/4)1/4 is irrational. It is difficult to understand to take x=4 is irrational.
I have got two (2) methods, shown in the attachment.

[Guest]

Sir, kindly go through the attached file.
Thank you.
Prof. R. C. Ray
My observation
We can derive, e=ln((z^(n/2)+y^(n/2))/(z^(n/2)-y^(n/2) ))/ln(x). Example: Let, z=5, y=2, x=4, n=4. Then, e=ln(29/21)/ln(4).
e is irrational, (1/4)1/4 is irrational. It is difficult to understand to take x=4 is irrational.
I have got two (2) methods, shown in the attachment.

You can start a new topic to discuss your point more effectively with a wider audience. Good Luck!

[Guest]

Sir, kindly go through the attached file.
Thank you.
Prof. R. C. Ray
My observation
We can derive, e=ln((z^(n/2)+y^(n/2))/(z^(n/2)-y^(n/2) ))/ln(x). Example: Let, z=5, y=2, x=4, n=4. Then, e=ln(29/21)/ln(4).
e is irrational, (1/4)1/4 is irrational. It is difficult to understand to take x=4 is irrational.
I have got two (2) methods, shown in the attachment.

WARNING! Is that attached file ('book.pdf') safe to download? Can someone scan that file for any viruses?

It is better to be safe than sorry...

[Guest]

Sir, kindly go through the attached file.
Thank you.
Prof. R. C. Ray
My observation
We can derive, e=ln((z^(n/2)+y^(n/2))/(z^(n/2)-y^(n/2) ))/ln(x). Example: Let, z=5, y=2, x=4, n=4. Then, e=ln(29/21)/ln(4).
e is irrational, (1/4)1/4 is irrational. It is difficult to understand to take x=4 is irrational.
I have got two (2) methods, shown in the attachment. I am eagerly waiting for your opinion. Thank you.

[Guest]

Hello,
I would like mathematicians to stop telling the world that Fermat was bluffing, saying he had proof of theorem.

Such simple proof is here
https://www.henrykdot.com
Fermat may lack space because at that time Newton's binomial and Pascal's triangle were not yet known.
Mathematicians are trying to ignore this proof of theorem.
I share Pelerman's view that correct proof does not need anyone's approval.


Regards
Henryk Zajdel, Katowice, Poland

[Guest]

"A mathematical theory is not to be considered complete until you have made it so clear that you can explain it to the first man whom you meet on the street."

"Besides, it is an error to believe that rigor is the enemy of simplicity. On the contrary, we find it confirmed by numerous examples that the rigorous method is at the same time the simpler and the more easily comprehended. The very effort for rigor forces us to find out simpler methods of proof."

— David Hilbert, a great scientist.

"Everything should be made as simple as possible, but not simpler."

-- Albert Einstein, a great scientist.


FYI: 'Why are people still searching for elementary proof of Fermat's Last Theorem?',

https://math.stackexchange.com/questions/3037280/why-still-people-are-searching-for-elementary-proof-of-fermats-last-theorem.

[Guest]

FYI: 'Why are people still searching for elementary proof of Fermat's Last Theorem?',

Because Wiles built his prestige by undermining Fermat's authority.
Wiles obsessed with winning the prize he said that Fermat lied that he had proof.
Of course, Fermat could have proof because having proof for N = 3 and N = 4 is sufficient.
The proof for N = 3 is the same as for all odd N.
Wiles not only came up with pathetic proof but he lied himself claiming
that Fermat lied because he could not have proof.
The search for simple proof is to protect Fermat's good name and to show Wiles's bad intentions!

[Guest]

FYI: 'Why are people still searching for elementary proof of Fermat's Last Theorem?',

Because Wiles built his prestige by undermining Fermat's authority.
Wiles obsessed with winning the prize he said that Fermat lied that he had proof.
Of course, Fermat could have proof because having proof for N = 3 and N = 4 is sufficient.
The proof for N = 3 is the same as for all odd N.
Wiles not only came up with pathetic proof but he lied himself claiming
that Fermat lied because he could not have proof.
The search for simple proof is to protect Fermat's good name and to show Wiles's bad intentions!

We know Fermat's Last Theorem is correct! That's all that matters.

[Guest]

FYI: 'Why are people still searching for elementary proof of Fermat's Last Theorem?',

Because Wiles built his prestige by undermining Fermat's authority.
Wiles obsessed with winning the prize he said that Fermat lied that he had proof.
Of course, Fermat could have proof because having proof for N = 3 and N = 4 is sufficient.
The proof for N = 3 is the same as for all odd N.
Wiles not only came up with pathetic proof but he lied himself claiming
that Fermat lied because he could not have proof.
The search for simple proof is to protect Fermat's good name and to show Wiles's bad intentions!

We know Fermat's Last Theorem is correct! That's all that matters.

According to the link: "Prove a statement about a conditional Diophantine equation."

https://math.stackexchange.com/questions/1993460/prove-a-statement-about-a-conditional-diophantine-equation,

there are two derived crucial equations,

1. [tex]l−k=λ[/tex]

and

2.[tex]\frac{l^{m-1}−k^{m}}{l−k} = l^{m-1}+l^{m−2}⋅k+⋅⋅⋅+l⋅k^{m−2}+k^{m−1} = \frac{4^{m}}{λ}[/tex]

for some positive rational number, λ, such that [tex]0<λ<4^{m}[/tex].

Equation two contains an irreducible polynomial over the field of rational numbers when [tex]m > 2[/tex].

That result confirms the truth of Fermat's Last Theorem (FLT).

Relevant Reference Link:

https://en.wikipedia.org/wiki/Irreducible_polynomial.

[Guest]

(Update): 2.[tex]\frac{l^{m}−k^{m}}{l−k} = l^{m-1}+l^{m−2}⋅k+⋅⋅⋅+l⋅k^{m−2}+k^{m−1} = \frac{4^{m}}{λ}[/tex]

[Guest]

(Update): 2.[tex]\frac{l^{m}−k^{m}}{l−k} = l^{m-1}+l^{m−2}⋅k+⋅⋅⋅+l⋅k^{m−2}+k^{m−1} = \frac{4^{m}}{λ}[/tex]

That equation is unsolvable over the set of positive rational numbers.