Beal's theorem
Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [math]\displaystyle{ a, b, c, d, e, s \in \mathbb{N}^{*} }[/math] and [math]\displaystyle{ k, m, n \in \mathbb{N}_{\ge 3} }[/math] implies [math]\displaystyle{ d := }[/math] gcd[math]\displaystyle{ (a, b, c) \gt 1. }[/math]
Indirect proof: Let [math]\displaystyle{ (dc)^n - (db)^n = (da)^m }[/math] the necessary ansatz to cancel [math]\displaystyle{ d }[/math] after computing modulo [math]\displaystyle{ d }[/math]. Then transforming back forces [math]\displaystyle{ c^n - b^n = e^m (\cdot d^{sm}) }[/math] to yield [math]\displaystyle{ b = 1 }[/math] and [math]\displaystyle{ c = d }[/math]. With respect to the second general form [math]\displaystyle{ 1 + d^n = e^m }[/math], the claim follows by contradiction.[math]\displaystyle{ \square }[/math]
Conclusion: The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] that no [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 3} }[/math] satisfies [math]\displaystyle{ a^n + b^n = c^n }[/math] for arbitrary [math]\displaystyle{ a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square }[/math]