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= Welcome to MWiki = | = Welcome to MWiki = | ||
== Theorems of the month == | == Theorems of the month == | ||
− | === | + | === Three-Cube Theorem === |
− | + | By Fermat’s little theorem, <math>k \in {}^{\omega }{\mathbb{Z}}</math> is sum of three cubes if and only if | |
− | = | + | <div style="text-align:center;"><math>k=(n - a)^3 + n^3 + (n + b)^3 = 3n^3 - a^3 + b^3+ 3c \ne \pm 4\mod 9</math></div> |
− | + | ||
+ | and <math>a, b, c, d, m, n \in {}^{\omega }{\mathbb{Z}}</math> implies both <math>(a^2 + b^2)n - (a - b)n^2 = c =: dn</math> and <math>m^2 = n^2 - 4(b^2 - bn + d)</math> for <math>2a_{1,2} = n \pm m.\square</math> | ||
− | === | + | === Fickett's Theorem === |
− | For <math> | + | For any relative positions of two overlapping congruent rectangular <math>n</math>-prisms <math>Q</math> and <math>R</math> with <math>n \in {}^{\omega }\mathbb{N}_{\ge 2}</math> and <math>\grave{m} := \hat{n}</math>, the exact standard measure <math>\mu</math> implies, where <math>\mu</math> for <math>n = 2</math> is the Euclidean path length <math>L</math>: |
− | + | <div style="text-align:center;"><math>\tilde{m} < r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) < m.</math></div> | |
==== Proof: ==== | ==== Proof: ==== | ||
− | The | + | The underlying extremal problem has its maximum for rectangles with side lengths <math>s</math> and <math>s + \hat{\iota}</math>. Putting <math>q := 3 - \hat{\iota}\tilde{s}</math> implies min <math>r = \tilde{q} \le r \le</math> max <math>r = q</math>. The proof for <math>n > 2</math> works analogously.<math>\square</math> |
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== Recommended reading == | == Recommended reading == |
Revision as of 04:35, 1 November 2022
Welcome to MWiki
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
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{ \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{ L }[/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]