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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
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=== Three-Cube Theorem ===
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=== Leibniz' differentiation rule ===
  
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By Fermat’s little theorem, <math>k \in {}^{\omega }{\mathbb{Z}}</math> is sum of three cubes if and only if
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For <math>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>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\curvearrowright a(x) = a(\curvearrowright x)</math> and <math>\curvearrowright b(x) = b(\curvearrowright x)</math>, it holds that<div style="text-align:center;"><math>\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).</math></div>
  
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<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>
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==== Proof: ====
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<div style="text-align:center;"><math>\begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right) &={\left( {\uparrow}_{a(\curvearrowright x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright x,t)-f(x,t)){\downarrow}t}+{\uparrow}_{b(x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{a(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).\square\end{aligned}</math></div>
  
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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>
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=== Beal's theorem ===
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Equation <math>a^m + b^n = c^k</math> for <math>a, b, c \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies gcd<math>(a, b, c) > 1.</math>
  
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=== Fickett's Theorem ===
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==== Proof: ====
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For <math>b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr}</math>, the function <math>f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0</math> is continuous in <math>q, r \in {}^{\omega} \mathbb{R}_{>0}</math> and <math>(q_0, r_0) = \left(\check{1}, \check{1}\right)</math> solves the equation. Every further solution in fractions yields after exponentiation gcd<math>(a, c) > 1</math> and thus proves the claim.<math>\square</math>
  
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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>m := 2n - 1</math>, it can be stated for the exact standard measure <math>\mu</math>, where <math>\mu</math> for <math>n = 2</math> needs to be replaced by the Euclidean path length <math>L</math>, that:
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=== Conclusion: ===
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The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd<math>(a, b, c) > 1</math> that no <math>n \in {}^{\omega}\mathbb{N}_{\ge 3}</math> satisfies <math>a^n + b^n = c^n</math> for arbitrary <math>a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square</math>
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<div style="text-align:center;"><math>\hat{m} &lt; r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) &lt; m.</math></div>
 
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==== Proof: ====
 
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Since the underlying extremal problem has its maximum for rectangles with the side lengths <math>s</math> and <math>s + 2d0</math>, min <math>r = s/(3s - 2d0) \le r \le</math> max <math>r = (3s - 2d0)/s</math> holds. The proof for <math>n &gt; 2</math> is analogous.<math>\square</math>
 
  
 
== Recommended reading ==
 
== Recommended reading ==

Revision as of 08:15, 4 March 2024

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

[math]\displaystyle{ \tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)). }[/math]

Proof:

[math]\displaystyle{ \begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right) &={\left( {\uparrow}_{a(\curvearrowright x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright x,t)-f(x,t)){\downarrow}t}+{\uparrow}_{b(x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{a(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).\square\end{aligned} }[/math]

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:

For [math]\displaystyle{ b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr} }[/math], the function [math]\displaystyle{ f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0 }[/math] is continuous in [math]\displaystyle{ q, r \in {}^{\omega} \mathbb{R}_{\gt 0} }[/math] and [math]\displaystyle{ (q_0, r_0) = \left(\check{1}, \check{1}\right) }[/math] solves the equation. Every further solution in fractions yields after exponentiation gcd[math]\displaystyle{ (a, c) \gt 1 }[/math] and thus proves the claim.[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]

Recommended reading

Nonstandard Mathematics

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