# A Simpler Proof of Fermat's Last Theorem:

### A Simpler Proof of Fermat's Last Theorem:

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!
primework123

Posts: 22
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### Re: A Simpler Proof of Fermat's Last Theorem:

primework123 wrote: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 < $$\infty$$ (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 the term, $$z^{n}$$ , we have

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

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!

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### Re: A Simpler Proof of Fermat's Last Theorem:

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.
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### Re: A Simpler Proof of Fermat's Last Theorem:

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

https://plus.google.com/collection/43KtnB.
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### Re: A Simpler Proof of Fermat's Last Theorem:

Guest wrote: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.
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### Re: A Simpler Proof of Fermat's Last Theorem:

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

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