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> | + | For <math>1 + x^m = y^n</math> where <math>(m, n, x, y) \in {}^{\omega}\mathbb{N}_{\ge 2}^4</math>, only the solution <math>(3, 2, 2, 3)</math> exists. |
| − | '''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</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 | + | '''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</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> |
== See also == | == See also == | ||
Revision as of 01:45, 11 August 2024
For [math]\displaystyle{ 1 + x^m = y^n }[/math] where [math]\displaystyle{ (m, n, x, y) \in {}^{\omega}\mathbb{N}_{\ge 2}^4 }[/math], only the solution [math]\displaystyle{ (3, 2, 2, 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 }[/math], [math]\displaystyle{ 2{\hat{x}}^3 = s^3(s^3 + 1) }[/math] and [math]\displaystyle{ y^3 - z^3 = z^3 - 1 }[/math] for [math]\displaystyle{ y, z \in {}^{\omega}\mathbb{N}^{*} }[/math] are left and yield the solution specified.[math]\displaystyle{ \square }[/math]