Difference between revisions of "Beal's theorem"
Borishaase (talk | contribs) (Beal's theorem) |
Borishaase (talk | contribs) (Beal's theorem) |
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| − | Equation <math>a^m + b^n = c^k</math> where <math>a, b, c \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies gcd<math>(a, b, c) > 1.</math> | + | Equation <math>a^m + b^n = c^k</math> where <math>a, b, c, d, e, s \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies <math>d :=</math> gcd<math>(a, b, c) > 1.</math> |
| − | ''' | + | '''Indirect proof''': Let <math>(dc)^n - (db)^n = (da)^m</math> the necessary ansatz to cancel <math>d</math> after computing modulo <math>d</math>. Then transforming back forces <math>c^n - b^n = e^m (\cdot d^{sm})</math> to yield <math>b = 1</math> and <math>c = d</math>. With respect to the second general form <math>1 + d^n = e^m</math>, the claim follows by contradiction.<math>\square</math> |
'''Conclusion''': The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd<math>(a, b, c) > 1</math> that no <math>n \in {}^{\omega}\mathbb{N}_{\ge 3}</math> satisfies <math>a^n + b^n = c^n</math> for arbitrary <math>a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square</math> | '''Conclusion''': The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd<math>(a, b, c) > 1</math> that no <math>n \in {}^{\omega}\mathbb{N}_{\ge 3}</math> satisfies <math>a^n + b^n = c^n</math> for arbitrary <math>a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square</math> | ||
Revision as of 18:50, 4 August 2024
Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [math]\displaystyle{ a, b, c, d, e, s \in \mathbb{N}^{*} }[/math] and [math]\displaystyle{ k, m, n \in \mathbb{N}_{\ge 3} }[/math] implies [math]\displaystyle{ d := }[/math] gcd[math]\displaystyle{ (a, b, c) \gt 1. }[/math]
Indirect proof: Let [math]\displaystyle{ (dc)^n - (db)^n = (da)^m }[/math] the necessary ansatz to cancel [math]\displaystyle{ d }[/math] after computing modulo [math]\displaystyle{ d }[/math]. Then transforming back forces [math]\displaystyle{ c^n - b^n = e^m (\cdot d^{sm}) }[/math] to yield [math]\displaystyle{ b = 1 }[/math] and [math]\displaystyle{ c = d }[/math]. With respect to the second general form [math]\displaystyle{ 1 + d^n = e^m }[/math], the claim follows by contradiction.[math]\displaystyle{ \square }[/math]
Conclusion: The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] that no [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 3} }[/math] satisfies [math]\displaystyle{ a^n + b^n = c^n }[/math] for arbitrary [math]\displaystyle{ a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square }[/math]