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Theorems of the month

Three-Cube Theorem

By Fermat’s little theorem, [math]\displaystyle{ k \in {}^{\omega }{\mathbb{Z}} }[/math] is sum of three cubes if and only if

[math]\displaystyle{ k=(n - a)^3 + n^3 + (n + b)^3 = 3n^3 - a^3 + b^3+ 3c \ne \pm 4\mod 9 }[/math]

and [math]\displaystyle{ a, b, c, d, m, n \in {}^{\omega }{\mathbb{Z}} }[/math] implies both [math]\displaystyle{ (a^2 + b^2)n - (a - b)n^2 = c =: dn }[/math] and [math]\displaystyle{ m^2 = n^2 - 4(b^2 - bn + d) }[/math] for [math]\displaystyle{ 2a_{1,2} = n \pm m.\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{ m := 2n - 1 }[/math], it can be stated for the exact standard measure [math]\displaystyle{ \mu }[/math], where [math]\displaystyle{ \mu }[/math] for [math]\displaystyle{ n = 2 }[/math] needs to be replaced by the Euclidean path length [math]\displaystyle{ L }[/math], that:

[math]\displaystyle{ \hat{m} < r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) < m. }[/math]

Proof:

Since the underlying extremal problem has its maximum for rectangles with the side lengths [math]\displaystyle{ s }[/math] and [math]\displaystyle{ s + 2d0 }[/math], min [math]\displaystyle{ r = s/(3s - 2d0) \le r \le }[/math] max [math]\displaystyle{ r = (3s - 2d0)/s }[/math] holds. The proof for [math]\displaystyle{ n > 2 }[/math] is analogous.[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics