Catalan's theorem

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{ \overset{\scriptsize{\grave{}}}{r}, s, t \in {}^{\omega}\mathbb{N}^{*} }[/math]. The Beal's theorem shows min[math]\displaystyle{ (m, n) = 2 }[/math]. Putting [math]\displaystyle{ \acute{n} \in {}^{\omega}2\mathbb{N}^{*} }[/math] implies [math]\displaystyle{ 1 + x^2 \equiv 2 \ne 0 \equiv 2^n{\check{y}}^n }[/math] mod [math]\displaystyle{ 8 }[/math] and [math]\displaystyle{ 1 + 4{\check{x}}^2 = (1 + 2{\check{y}})^n }[/math] or [math]\displaystyle{ 4{\check{x}}^2 = 2\check{y}((1 + 2\check{y})^{\acute{n}} + ... + 1) }[/math] where [math]\displaystyle{ \check{y} = \hat{z} \in {}^{\omega}\mathbb{N}^{*} }[/math] and [math]\displaystyle{ 1 + 4{\check{x}}^2 \equiv r \ne -1 \equiv (1+ 4z)^n }[/math], since gcd[math]\displaystyle{ (x, 2 + 4z) \nmid (1 + 4z) }[/math] leads to [math]\displaystyle{ 2{\check{x}}^2 \equiv r \ne -1 }[/math] mod [math]\displaystyle{ (2 + 4z) }[/math] due to [math]\displaystyle{ {\check{x}}^2 \gt z }[/math]. Putting [math]\displaystyle{ \acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4} }[/math] implies [math]\displaystyle{ 1 + x^m \ne y^2 }[/math] due to [math]\displaystyle{ x^m \ne s^m(2 + s^m) }[/math] and [math]\displaystyle{ t^m \ne 2 + s^m }[/math]. Thus [math]\displaystyle{ 1 + {8\check{x}}^3 = (1 + 4\check{y}^2)^2 }[/math], [math]\displaystyle{ 2{\check{x}}^3 = s^3(1 + s^3) }[/math] and [math]\displaystyle{ s^3 - t^3 = t^3 - 1 }[/math] are left and yield the solution specified.[math]\displaystyle{ \square }[/math]

See also

• List of mathematical symbols
• Catalan's conjecture