Beal's theorem

The 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: The claim follows from [math]\displaystyle{ (a^{\check{m}} + ib^{\check{n}})^2 (a^{\check{m}} - ib^{\check{n}})^2 = (pq)^{2k} = (a^m + b^n - 2b^{\check{n}}(b^{\check{n}} - ia^{\check{m}}))(a^m + b^n - 2b^{\check{n}}(b^{\check{n}} + ia^{\check{m}})) = c^{2k} \gt 1 }[/math] where [math]\displaystyle{ p \in \mathbb{P} }[/math] and [math]\displaystyle{ q \in \mathbb{N}^{*} }[/math] imply [math]\displaystyle{ d^2(pq)^k = 4b^n }[/math] for [math]\displaystyle{ d \in \mathbb{N}^{*} }[/math] and thus [math]\displaystyle{ p \mid b.\square }[/math]