Please note that Fermat's last theorem was originated from Pythagoras Theorem, where he (Fermat), must had known a very basic and simple trick which is too elementary to prove, what is the trick?,

"A primitive Pythagoras triplets (in co prime integers), are impossible with two sides of a right angle triangle being as powerful numbers"

Powerful number : is an integer which has all of its prime factors exponent are greater than one
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Bassam Karzeddin
·
6 months ago

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We may generalize the exponent to be a real positive algebraic number say (g), the generalization would be as this:

X^g + Y^g = Z^g

have no solution in distinct positive coprime integers, (X < Y < Z), where (g) is greater than two

This has a specific history that was older than accepted proof of FLT
–
Bassam Karzeddin
·
9 months, 3 weeks ago

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I don't know how to generalize Fermat's Last Theorem, but I can give you a link. This paper is Andrew Wiles' original paper on his proof of Fermat's Last Theorem. It is called "Modular elliptic curves and Fermat's Last Theorem".
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Ananth Jayadev
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10 months, 2 weeks ago

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This link is very useful in this regard: http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
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Bassam Karzeddin
·
6 months ago

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@Bassam Karzeddin
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Editt: I mean "A primitive Pythagoras triplets (in co prime integers), are impossible with all sides of a right angle triangle being as powerful numbers", or "A primitive Pythagoras triplets (in co prime integers), are impossible with two sides of a right angle triangle being as powerful numbers of this form (x^n, y^m, z), where (n, m) are positive integers > 1, and (x, y, z) are positive integers"
–
Bassam Karzeddin
·
3 months, 2 weeks ago

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TopNewestPlease note that Fermat's last theorem was originated from Pythagoras Theorem, where he (Fermat), must had known a very basic and simple trick which is too elementary to prove, what is the trick?,

"A primitive Pythagoras triplets (in co prime integers), are impossible with two sides of a right angle triangle being as powerful numbers"

Powerful number : is an integer which has all of its prime factors exponent are greater than one – Bassam Karzeddin · 6 months ago

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We may generalize the exponent to be a real positive algebraic number say (g), the generalization would be as this:

have no solution in distinct positive coprime integers, (X < Y < Z), where (g) is greater than two

This has a specific history that was older than accepted proof of FLT – Bassam Karzeddin · 9 months, 3 weeks ago

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I don't know how to generalize Fermat's Last Theorem, but I can give you a link. This paper is Andrew Wiles' original paper on his proof of Fermat's Last Theorem. It is called "Modular elliptic curves and Fermat's Last Theorem". – Ananth Jayadev · 10 months, 2 weeks ago

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This link is very useful in this regard: http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers – Bassam Karzeddin · 6 months ago

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– Bassam Karzeddin · 3 months, 2 weeks ago

Editt: I mean "A primitive Pythagoras triplets (in co prime integers), are impossible with all sides of a right angle triangle being as powerful numbers", or "A primitive Pythagoras triplets (in co prime integers), are impossible with two sides of a right angle triangle being as powerful numbers of this form (x^n, y^m, z), where (n, m) are positive integers > 1, and (x, y, z) are positive integers"Log in to reply