Catalan's theorem
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]