Difference between revisions of "Beal's theorem"

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(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>
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Equation <math>a^m + b^n = c^k</math> where <math>a, b, c, d, e, r, 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>
  
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'''Proof''': For <math>b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr}</math>, the function <math>f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0</math> is continuous in <math>q, r \in {}^{\omega} \mathbb{R}_{>0}</math> and <math>(q_0, r_0) = \left(\check{1}, \check{1}\right)</math> solves the equation. Every further solution in fractions yields after exponentiation gcd<math>(a, c) > 1</math> and thus proves the claim.<math>\square</math>
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'''Indirect proof''': Put necessarily <math>(da)^n + (db)^m = (dc)^n</math> to cancel <math>d</math> after computing mod <math>d</math>. Multiply <math>b^s + e^m = c^s</math> generalised by <math>d^{rm}</math> again to yield <math>b = 1</math>, <math>c = d</math> and <math>s \ne m \notin 2\mathbb{N}</math>. The second general form <math>1 + d^s = e^m</math> shows likewise the claim by contradiction for <math>m \ne s \notin 2\mathbb{N}</math>.<math>\square</math>
  
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'''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>
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'''Conclusion''': Due to <math>m \ne s</math>, 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}^{*}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 07:34, 11 August 2024

Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [math]\displaystyle{ a, b, c, d, e, r, 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: Put necessarily [math]\displaystyle{ (da)^n + (db)^m = (dc)^n }[/math] to cancel [math]\displaystyle{ d }[/math] after computing mod [math]\displaystyle{ d }[/math]. Multiply [math]\displaystyle{ b^s + e^m = c^s }[/math] generalised by [math]\displaystyle{ d^{rm} }[/math] again to yield [math]\displaystyle{ b = 1 }[/math], [math]\displaystyle{ c = d }[/math] and [math]\displaystyle{ s \ne m \notin 2\mathbb{N} }[/math]. The second general form [math]\displaystyle{ 1 + d^s = e^m }[/math] shows likewise the claim by contradiction for [math]\displaystyle{ m \ne s \notin 2\mathbb{N} }[/math].[math]\displaystyle{ \square }[/math]

Conclusion: Due to [math]\displaystyle{ m \ne s }[/math], 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}^{*} }[/math].

See also

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