Difference between revisions of "Beal's theorem"

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The 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>
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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>
  
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'''Proof''': The claim follows from <math>(a^{\check{m}} + ib^{\check{n}})^2 (a^{\check{m}} - ib^{\check{n}})^2 = (pq)^{2k} = (a^m + b^n - 2b^{\check{n}}(b^{\check{n}} - ia^{\check{m}}))(a^m + b^n - 2b^{\check{n}}(b^{\check{n}} + ia^{\check{m}})) = c^{2k} > 1</math> where <math>p \in \mathbb{P}</math> and <math>q \in \mathbb{N}^{*}</math> imply <math>d^2(pq)^k = 4b^n</math> for <math>d \in \mathbb{N}^{*}</math> and thus <math>p \mid b.\square</math>
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'''Proof''': For <math>p \in {}^{\omega}\mathbb{P}</math> and <math>r, s \in {}^{\omega}\mathbb{Q}</math>, the preceding theorem leads to the complete nontrivial representations for <math>c^k > 1</math> as <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> hence imply <math>p \mid</math> gcd<math>(a, b, c)</math> as well as the claim despite of certain (non-) rational <math>r</math> and <math>s</math> (continuity!).<math>\square</math>
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'''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>
  
 
== See also ==
 
== See also ==

Revision as of 03:46, 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: For [math]\displaystyle{ p \in {}^{\omega}\mathbb{P} }[/math] and [math]\displaystyle{ r, s \in {}^{\omega}\mathbb{Q} }[/math], the preceding theorem leads to the complete nontrivial representations for [math]\displaystyle{ c^k \gt 1 }[/math] as [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] hence imply [math]\displaystyle{ p \mid }[/math] gcd[math]\displaystyle{ (a, b, c) }[/math] as well as the claim despite of certain (non-) rational [math]\displaystyle{ r }[/math] and [math]\displaystyle{ s }[/math] (continuity!).[math]\displaystyle{ \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]

See also

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