Difference between revisions of "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 \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies gcd<math>(a, b, c) > 1.</math> | ||
| − | '''Proof''': | + | '''Proof''': Real points between <math>r, s \in {}^{\omega}\mathbb{Q}</math> do not avoid that every nontrivial equation for <math>c^k > 1</math> is given by <math>(a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s)</math> where all relations <math>a^{m-\hat{r}} = b^{n-\hat{s}}</math> then imply <math>p \mid</math> gcd<math>(a, b, c)</math> as well as the claim for <math>p \in {}^{\omega} \mathbb{P}.\square</math> |
'''Conclusion''': The preceding theorem enables an infinite descent because of gcd<math>(a, b, c) > 1</math> such 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 preceding theorem enables an infinite descent because of gcd<math>(a, b, c) > 1</math> such 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> | ||
Latest revision as of 16:59, 1 March 2023
Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [math]\displaystyle{ a, b, c \in \mathbb{N}^{*} }[/math] and [math]\displaystyle{ k, m, n \in \mathbb{N}_{\ge 3} }[/math] implies gcd[math]\displaystyle{ (a, b, c) \gt 1. }[/math]
Proof: Real points between [math]\displaystyle{ r, s \in {}^{\omega}\mathbb{Q} }[/math] do not avoid that every nontrivial equation for [math]\displaystyle{ c^k \gt 1 }[/math] is given by [math]\displaystyle{ (a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s) }[/math] where all relations [math]\displaystyle{ a^{m-\hat{r}} = b^{n-\hat{s}} }[/math] then imply [math]\displaystyle{ p \mid }[/math] gcd[math]\displaystyle{ (a, b, c) }[/math] as well as the claim for [math]\displaystyle{ p \in {}^{\omega} \mathbb{P}.\square }[/math]
Conclusion: The preceding theorem enables an infinite descent because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] such 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]