Difference between revisions of "Catalan's theorem"

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It holds that <math>\{(m, n, x, y) \in {}^{\omega}\mathbb{N}_{\ge 2}^4 : 1 + x^m = y^n\} = \{(3, 2, 2, 3)\}</math>.
 
It holds that <math>\{(m, n, x, y) \in {}^{\omega}\mathbb{N}_{\ge 2}^4 : 1 + x^m = y^n\} = \{(3, 2, 2, 3)\}</math>.
  
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'''Indirect proof:''' [[Beal's theorem]] implies <math>m = 3</math> and <math>n = 2</math>, since vice versa for <math>\acute{x}, y \in {}^{\omega}2\mathbb{N}</math> the contradictions <math>1 + x^2 \equiv 2 \equiv y^n \equiv 0</math>, <math>1 + 4x^m \equiv 5 \equiv y^n \equiv 0</math> and <math>1 + 16x^m \equiv 5 \equiv y^n \equiv 0</math> mod <math>8</math> arise. For <math>\acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4}</math>, the contradiction <math>1 + 2^mx^m \equiv (1 + 2^{\acute{m}}s)^2 \equiv 1 + 2^ms + 4^{\acute{m}}s^2 \equiv y^2</math> mod <math>4^{\acute{m}}</math> arises. Thus <math>1 + {\hat{x}}^3 = (\hat{y}^2 + 1)^2</math>, <math>2{\hat{x}}^3 = s^3(s^3 + 1)</math> and <math>y^3 - z^3 = z^3 - 1</math> for <math>y, z \in {}^{\omega}\mathbb{N}^{*}</math> are left and yield the solution specified.<math>\square</math>
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'''Indirect proof:''' [[Beal's theorem]] gives min<math>(m, n) = 2</math>.  Squaring shows <math>1 + 4{\check{x}}^2 = (1 + \acute{y})^n</math> and <math>(1 + 2^rz)^n \equiv 1 + 2^r{\check{x}}^2 \equiv 1</math> mod <math>2^{\overset{\scriptsize{\grave{}}}{r}}</math> for every <math>r \in {}^{\omega}\mathbb{N}_{\ge 3}</math>, and <math>1 + x^2 \equiv 2 \ne 0 \equiv 2^n{\check{y}}^n</math> mod <math>8</math> where <math>\acute{y} = 4z</math> and <math>n</math> is odd. Odd <math>m</math> and <math>s \in {}^{\omega}\mathbb{N}^{*}</math> imply <math>x^m = \acute{y}\overset{\scriptsize{\grave{}}}{y}</math> where <math>s^m \ne \acute{y} \in {}^{\omega}2\mathbb{N}^{*}</math> such that <math>x^m = 8\check{s}\acute{s} = 8.\square</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 01:44, 20 August 2024

It holds that [math]\displaystyle{ \{(m, n, x, y) \in {}^{\omega}\mathbb{N}_{\ge 2}^4 : 1 + x^m = y^n\} = \{(3, 2, 2, 3)\} }[/math].

Indirect proof: Beal's theorem gives min[math]\displaystyle{ (m, n) = 2 }[/math]. Squaring shows [math]\displaystyle{ 1 + 4{\check{x}}^2 = (1 + \acute{y})^n }[/math] and [math]\displaystyle{ (1 + 2^rz)^n \equiv 1 + 2^r{\check{x}}^2 \equiv 1 }[/math] mod [math]\displaystyle{ 2^{\overset{\scriptsize{\grave{}}}{r}} }[/math] for every [math]\displaystyle{ r \in {}^{\omega}\mathbb{N}_{\ge 3} }[/math], and [math]\displaystyle{ 1 + x^2 \equiv 2 \ne 0 \equiv 2^n{\check{y}}^n }[/math] mod [math]\displaystyle{ 8 }[/math] where [math]\displaystyle{ \acute{y} = 4z }[/math] and [math]\displaystyle{ n }[/math] is odd. Odd [math]\displaystyle{ m }[/math] and [math]\displaystyle{ s \in {}^{\omega}\mathbb{N}^{*} }[/math] imply [math]\displaystyle{ x^m = \acute{y}\overset{\scriptsize{\grave{}}}{y} }[/math] where [math]\displaystyle{ s^m \ne \acute{y} \in {}^{\omega}2\mathbb{N}^{*} }[/math] such that [math]\displaystyle{ x^m = 8\check{s}\acute{s} = 8.\square }[/math]

See also

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