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>. | ||
| − | '''Indirect proof:''' Let <math>s, t \in {}^{\omega}\mathbb{N}^{*}</math>. [[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 \ne y^n | + | '''Indirect proof:''' Let <math>s, t \in {}^{\omega}\mathbb{N}^{*}</math>. [[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 \ne 0 \equiv y^n</math>, <math>1 + 4x^m \equiv 5 \ne 0 \equiv y^n</math> and <math>1 + 16x^m \equiv 1 \ne 0 \equiv y^n</math> mod <math>8</math> arise. From <math>\acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4}</math>, the contradiction <math>1 + {\hat{x}}^m \ne y^2 \equiv (1 + 2^{\acute{m}}s)^2 = 1 + 2^ms + 4^{\acute{m}}s^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>s^3 - t^3 = t^3 - 1</math> are left and yield the solution specified.<math>\square</math> |
== See also == | == See also == | ||
Revision as of 16:22, 11 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: Let [math]\displaystyle{ s, t \in {}^{\omega}\mathbb{N}^{*} }[/math]. Beal's theorem implies [math]\displaystyle{ m = 3 }[/math] and [math]\displaystyle{ n = 2 }[/math], since vice versa for [math]\displaystyle{ \acute{x}, y \in {}^{\omega}2\mathbb{N} }[/math] the contradictions [math]\displaystyle{ 1 + x^2 \equiv 2 \ne 0 \equiv y^n }[/math], [math]\displaystyle{ 1 + 4x^m \equiv 5 \ne 0 \equiv y^n }[/math] and [math]\displaystyle{ 1 + 16x^m \equiv 1 \ne 0 \equiv y^n }[/math] mod [math]\displaystyle{ 8 }[/math] arise. From [math]\displaystyle{ \acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4} }[/math], the contradiction [math]\displaystyle{ 1 + {\hat{x}}^m \ne y^2 \equiv (1 + 2^{\acute{m}}s)^2 = 1 + 2^ms + 4^{\acute{m}}s^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{ s^3 - t^3 = t^3 - 1 }[/math] are left and yield the solution specified.[math]\displaystyle{ \square }[/math]