Beal's theorem
Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [math]\displaystyle{ a, b, c \in \mathbb{N}^{*} }[/math] and [math]\displaystyle{ k, m, n \in \mathbb{N}_{\ge 3} }[/math] implies gcd[math]\displaystyle{ (a, b, c) \gt 1. }[/math]
Proof: For [math]\displaystyle{ p \in {}^{\omega}\mathbb{P} }[/math] and [math]\displaystyle{ r, s \in {}^{\omega}\mathbb{Q} }[/math], the preceding theorem leads to the complete nontrivial representations for [math]\displaystyle{ c^k \gt 1 }[/math] as [math]\displaystyle{ (a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s) }[/math], where all relations [math]\displaystyle{ a^{m-\hat{r}} = b^{n-\hat{s}} }[/math] hence imply [math]\displaystyle{ p \mid }[/math] gcd[math]\displaystyle{ (a, b, c) }[/math] as well as the claim despite of certain (non-) rational [math]\displaystyle{ r }[/math] and [math]\displaystyle{ s }[/math] (continuity!).[math]\displaystyle{ \square }[/math]
Conclusion: The preceding theorem enables an infinite descent because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] such 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]