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Theorems of the month
Gelfond-Schneider theorem
It holds [math]\displaystyle{ a^b \in {}^{\omega} \mathbb{T}_\mathbb{C} }[/math] where [math]\displaystyle{ a, c \in {}^{\omega} \mathbb{A}_\mathbb{C}^{*} \setminus \{1\}, Q := {}^{\omega} \mathbb{R} \setminus {}^{\omega} \mathbb{T}_\mathbb{R} }[/math] and [math]\displaystyle{ b, \varepsilon \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus Q }[/math].
Proof:
Where [math]\displaystyle{ b \in Q }[/math] puts the minimal polynomial [math]\displaystyle{ p(a^b) = p(c^q) = 0 }[/math], assuming [math]\displaystyle{ a^b = c^{q+\varepsilon} }[/math] for maximum [math]\displaystyle{ q \in Q_{>0} }[/math] leads to the contradiction [math]\displaystyle{ 0 = (p(a^b) - p(c^q)) / (a^b - c^q) = p^\prime(a^b) = p^\prime(c^q) \ne 0.\square }[/math]
Three-Cube Theorem
Three-cube theorem: It holds [math]\displaystyle{ S := \{n \in \mathbb{Z} : n \ne \pm 4\mod 9\} = \{n \in \mathbb{Z} : n = a^3 + b^3 + c^3 + 3(a + b)c(a - b + c) = (a + c)^3 + (b - c)^3 + c^3\} \subset a^3 + b^3 + c^3 + 6{\mathbb{Z}} }[/math], since independent mathematical induction by equitable variables [math]\displaystyle{ a, b, c \in {\mathbb{Z}} }[/math] first shows [math]\displaystyle{ \{0, \pm 1, \pm 2, \pm 3\} \subset S }[/math], and then the claim.[math]\displaystyle{ \square }[/math]
Fickett's Theorem
For any relative positions of two overlapping congruent rectangular [math]\displaystyle{ n }[/math]-prisms [math]\displaystyle{ Q }[/math] and [math]\displaystyle{ R }[/math] with [math]\displaystyle{ n \in {}^{\omega }\mathbb{N}_{\ge 2} }[/math] and [math]\displaystyle{ \grave{m} := \hat{n} }[/math], the exact standard measure [math]\displaystyle{ \mu }[/math] implies, where [math]\displaystyle{ \mu }[/math] for [math]\displaystyle{ n = 2 }[/math] is the Euclidean path length [math]\displaystyle{ A }[/math]:
Proof:
The underlying extremal problem has its maximum for rectangles with side lengths [math]\displaystyle{ s }[/math] and [math]\displaystyle{ s + \hat{\iota} }[/math]. Putting [math]\displaystyle{ q := 3 - \hat{\iota}\tilde{s} }[/math] implies min [math]\displaystyle{ r = \tilde{q} \le r \le }[/math] max [math]\displaystyle{ r = q }[/math]. The proof for [math]\displaystyle{ n > 2 }[/math] works analogously.[math]\displaystyle{ \square }[/math]