Difference between revisions of "Catalan's theorem"

From MWiki
Jump to: navigation, search
m (Catalan's theorem)
m (Catalan's theorem)
 
(6 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
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>.
  
'''Indirect proof:''' Let <math>\overset{\scriptsize{\grave{}}}{r}, s, t \in {}^{\omega}\mathbb{N}^{*}</math>. [[Beal's theorem]] shows min<math>(m, n) = 2</math>. Putting <math>\acute{n} \in {}^{\omega}2\mathbb{N}^{*}</math> implies <math>1 + x^2 \equiv 2 \ne 0 \equiv 2^n{\check{y}}^n</math> mod <math>8</math> and <math>1 + 4{\check{x}}^2 = (1 + 2{\check{y}})^n</math> or <math>4{\check{x}}^2 = 2\check{y}((1 + 2\check{y})^{\acute{n}} + ... + 1)</math> where <math>\check{y} = \hat{z} \in {}^{\omega}\mathbb{N}^{*}</math> and <math>1 + 4{\check{x}}^2 \equiv r \ne -1 \equiv (1+ 4z)^n</math>, since gcd<math>(x, 2 + 4z) \nmid (1 + 4z)</math> leads to <math>2{\check{x}}^2 \equiv r \ne -1</math> mod <math>(2 + 4z)</math><ref name="Ribenboim">[[w:Harald Scheid|<span class="wikipedia">Scheid, Harald</span>]]: ''Zahlentheorie'' : 1st Ed.; 1991; Bibliographisches Institut; Mannheim; ISBN 3411148411, p. 197.</ref> due to <math>{\check{x}}^2 > z</math>. Putting <math>\acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4}</math> implies <math>1 + x^m \ne y^2</math> due to <math>x^m \ne s^m(2 + s^m)</math> and <math>t^m \ne 2 + s^m</math>. Thus <math>1 + {8\check{x}}^3 = (1 + 4\check{y}^2)^2</math>, <math>2{\check{x}}^3 = s^3(1 + s^3)</math> and <math>s^3 - t^3 = t^3 - 1</math> are left and yield the solution specified.<math>\square</math>
+
'''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 ==
Line 7: Line 7:
 
* [[w:Catalan's_conjecture|<span class="wikipedia">Catalan's conjecture</span>]]
 
* [[w:Catalan's_conjecture|<span class="wikipedia">Catalan's conjecture</span>]]
 
[[Category:Areas of mathematics]]
 
[[Category:Areas of mathematics]]
 
== Reference ==
 
<references />
 
  
 
[[de:Satz von Catalan]]
 
[[de:Satz von Catalan]]

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