Main Page
Welcome to MWiki
Theorems of the month
Leibniz' differentiation rule
For [math]\displaystyle{ f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math] and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \curvearrowright a(x) = a(\curvearrowright x) }[/math] and [math]\displaystyle{ \curvearrowright b(x) = b(\curvearrowright x) }[/math], it holds that
Proof:
Beal's theorem
Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] for [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:
From [math]\displaystyle{ b^n = (c^{k-r} – a^m)(c^r + 1) = c^k – a^m + c^{k-r} – a^mc^r }[/math], it follows that [math]\displaystyle{ a^m =c^{k-\hat{r}} }[/math] and [math]\displaystyle{ r \in {}^{\omega}\mathbb{Q}_{\gt 0} }[/math] by the Gelfond-Schneider theorem, which proves the claim with gcd[math]\displaystyle{ (a,c)\gt 1 }[/math] after exponentiation.[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]