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

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Thank you! Thank Lord GOD!