Difference between revisions of "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.
 
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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'''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 -1</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 -1</math> mod <math>p</math> and <math>2^{kn}(t^2 + t)^n\ 2^{-gn} \equiv 2^{lm}r^m \equiv -1</math> mod <math>2^{\text{max}(kn,\ lm)}</math> hold where <math>t = \check{s}</math> and maximal <math>g, k, l \in \mathbb{N}^{*}</math>, from which <math>kn = lm</math> due to <math>(t^2 + t)2^{-gn}, 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, too.<math>\square</math>
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'''Indirect proof:''' [[Beal's theorem]] implies m = 3 and n = 2, 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> und <math>1 + 16x^m \equiv 5 \equiv y^n \equiv 0</math> mod <math>8</math> follow. 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> follows. Thus <math>1 + {\hat{x}}^3 = (\hat{y}^2 + 1)^2 \equiv 1 + 8(x - s) \equiv 1 + 8s \equiv 1</math> mod <math>8</math> is left leading to the solution specified.<math>\square</math>
  
 
== See also ==
 
== See also ==

Revision as of 17:28, 10 August 2024

For [math]\displaystyle{ 1 + x^m = y^n }[/math] where [math]\displaystyle{ m, n, x, y \in {}^{\omega}\mathbb{N}_{\ge 2} }[/math], only the solution [math]\displaystyle{ m_0 = 3 }[/math], [math]\displaystyle{ n_0 = 2 }[/math], [math]\displaystyle{ x_0 = 2 }[/math] and [math]\displaystyle{ y_0 = 3 }[/math] exists.

Indirect proof: Beal's theorem implies m = 3 and n = 2, since vice versa for [math]\displaystyle{ \acute{x}, y \in {}^{\omega}2\mathbb{N} }[/math] the contradictions [math]\displaystyle{ 1 + x^2 \equiv 2 \equiv y^n \equiv 0 }[/math], [math]\displaystyle{ 1 + 4x^m \equiv 5 \equiv y^n \equiv 0 }[/math] und [math]\displaystyle{ 1 + 16x^m \equiv 5 \equiv y^n \equiv 0 }[/math] mod [math]\displaystyle{ 8 }[/math] follow. For [math]\displaystyle{ \acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4} }[/math], the contradiction [math]\displaystyle{ 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]\displaystyle{ 4^{\acute{m}} }[/math] follows. Thus [math]\displaystyle{ 1 + {\hat{x}}^3 = (\hat{y}^2 + 1)^2 \equiv 1 + 8(x - s) \equiv 1 + 8s \equiv 1 }[/math] mod [math]\displaystyle{ 8 }[/math] is left leading to the solution specified.[math]\displaystyle{ \square }[/math]

See also

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