Difference between revisions of "Catalan's theorem"

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m (Catalan's theorem)
m (Catalan's theorem)
 
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For <math>1 + x^m = y^n</math> where <math>m, n, x, y \in {}^{\omega}\mathbb{N}_{\ge 2}</math>, only the solution <math>m_0 = 3</math>, <math>n_0 = 2</math>, <math>x_0 = 2</math> and <math>y_0 = 3</math> exists.
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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>.
  
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'''Indirect proof:''' Assuming <math>\acute{x}, y \in {}^{\omega}2\mathbb{N}</math> yields the contradiction <math>1 + x^m \equiv 2 \equiv y^n \equiv 0</math> resp. <math>4</math> mod <math>8</math>. For <math>p \mid y,\ x = \hat{r},\ y = \hat{s} + 1</math> and <math>2 < m \ne n > 2</math>, subsequently <math>(\hat{s} + 1)^n - 1 \equiv \hat{r}^m \equiv 0</math> mod <math>p \in {}^{\omega}\mathbb{P}_{\ge 3}</math> is assumed. Then also <math>(\hat{s} + 1)^{\hat{n}} - 1 = (\hat{s} + 2)^n\ \hat{s}^n \equiv \hat{r}^m \equiv 0</math> mod <math>p</math> and <math>2^{kn}(t^2 + t)^n\ 2^{-gn} \equiv 2^{lm}\hat{r}^m \equiv 0</math> mod <math>2^{\text{max}(kn,\ lm)}</math> hold where <math>t = \check{s}</math> and maximal <math>g \in \mathbb{N}^{*}</math>, from which <math>kn = lm</math> due to <math>(t^2 + t)2^{-gn}, \hat{r}^m \in \mathbb{N} \setminus 2\mathbb{N}</math> and even <math>m = n</math> is implied contradicting the premise. Both <math>(\hat{s} + 1)^3 - 1 = \hat{r}^2</math> are left, which can be refuted by the [[w:Integer factorization|<span class="wikipedia">integer factorisation</span>]] of <math>r</math> and <math>s</math>, and <math>(\hat{s} + 1)^2 - 1 = 4s(s + 1) = \hat{r}^3</math>, which results in <math>r = s = 1</math> and thus leads to 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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