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= Welcome to MWiki = | = Welcome to MWiki = | ||
== Theorem of the month == | == Theorem of the month == | ||
− | === | + | === Fermat's Last Theorem === |
− | + | For all <math>p \in {}^{\omega }{\mathbb{P}_{\ge 3}}</math> and <math>x, y, z \in {}^{\omega }{\mathbb{N}^{*}}</math>, always <math>x^p + y^p \ne z^p</math> holds and thus for all <math>m \in {}^{\omega }{\mathbb{N}_{\ge 3}}</math> instead of <math>p</math>. | |
==== Proof: ==== | ==== Proof: ==== | ||
− | + | Because of [[w:Fermat's little theorem|<span class="wikipedia">Fermat's little theorem</span>]], rewritten, <math>f_{akp}(n) := (2n + a - kp)^p - n^p - (n + a)^p \ne 0</math> is to show for <math>a, k, n \in {}^{\omega }{\mathbb{N}^{*}}</math> where <math>kp < n</math>. | |
+ | <div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;"> | ||
+ | <div style="font-weight:bold;line-height:1.6;">Proof details</div> | ||
+ | <div class="mw-collapsible-content">From <math>x := n, y:= n + a</math> and <math>z := 2n + a + d</math> where <math>d \in {}^{\omega }{\mathbb{N}^{*}}</math>, it follows due to <math>z^p \equiv y, y^p \equiv y</math> and <math>z^p \equiv z</math> first <math>d \equiv 0 \mod p</math>, then <math>d = \pm kp</math>. Since <math>x + y = 2n + a > z</math> is required, <math>f_{akp}(n)</math> is chosen properly.</div></div> | ||
+ | |||
+ | [[w:Mathematical induction|<span class="wikipedia">Induction</span>]] for <math>n</math> implies the claim due to the case <math>m = 4</math><ref name="Ribenboim">[[w:Paulo Ribenboim|<span class="wikipedia">Ribenboim, Paulo</span>]]: ''Thirteen Lectures on Fermat's Last Theorem'' : 1979; Springer; New York; ISBN 9780387904320, p. 35 - 38.</ref> and <math>y > x > p</math><ref name="loccit">loc. cit., p. 226.</ref>: | ||
+ | |||
+ | '''Induction basis''' <math>(n \le p): f_{akp}(n) \ne 0</math> for all <math>a, k</math> and <math>p</math>. Let <math>r \in {}^{\omega }{\mathbb{N}_{< p}}</math>. | ||
+ | |||
+ | '''Induction step''' <math>\,(n = q + r \; \rightarrow \; n^{*} = n + p):</math> Let <math>f_{akp}(n^{*}) \ge 0</math>, but <math>f_{akp}(n) < 0</math>, since <math>f_{akp}(n)</math> is [[w:Monotonic function|<span class="wikipedia">strictly monotonically increasing</span>]] and otherwise nothing to prove. | ||
+ | <div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;"> | ||
+ | <div style="font-weight:bold;line-height:1.6;">Proof details</div> | ||
+ | <div class="mw-collapsible-content">The strict monotonicity follows from (continuously) differentiating by <math>n</math> such that <math>f_{akp}(n)' = p(2(2n + a - kp)^{p - 1} - n^{p - 1} - (n + a)^{p - 1}) > 0</math>.</div></div> | ||
+ | |||
+ | It holds <math>f_{akp}(n^{*}) = (\int_0^{n^{*}}{f_{akp}(v)}dv)' \ne 0</math>, since <math>(n^{*})^{p + 1} + (n^{*} + a)^{p + 1}</math> does not divide <math>((n^{*})^p + (n^{*} + a)^p)^2</math> after separating the positive factor as [[w:Polynomial long division|<span class="wikipedia">polynomial division</span>]] shows.<math>\square</math> | ||
+ | <div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;"> | ||
+ | <div style="font-weight:bold;line-height:1.6;">Proof details</div> | ||
+ | <div class="mw-collapsible-content"><math>\int_0^{n^{*}}{f_{akp}(v)}dv = ((2n^{*} + a - kp)^{p + 1} / 2 - (n^{*})^{p + 1} - (n^{*} + a)^{p + 1})/(p + 1) + t = ((2n^{*} + a - kp)^{(p + 1)/2} \pm \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})^2/(2p + 2) + t</math> for <math>t \in {}^{\omega}{\mathbb{Q}}</math> where the third binomial formula <math>r^2 - s^2 = (r \pm s)^2 := (r + s)(r - s)</math> was used. After separating the negligible factor <math>\hat{2}((2n^{*} + a - kp)^{(p + 1)/2} + \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})/(p + 1)</math>, the derivative is just <math>(\hat{2}(2n^{*} + a - kp)^{(p - 1)/2} - \hat{2}((n^{*})^p + (n^{*} + a)^p)/\sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})</math>. After squaring the terms, the polynomial division gives <math>(n^{*})^{p - 1} + (n^{*} + a)^{p - 1} + a^2(n^{*})^{p - 1}(n^{*} + a)^{p - 1}/((n^{*})^{p + 1} + (n^{*} + a)^{p + 1})</math> as recalculating by multiplication confirms.</div></div> | ||
== Recommended reading == | == Recommended reading == | ||
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics] | [https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics] | ||
+ | |||
+ | == References == | ||
+ | <references /> | ||
[[de:Hauptseite]] | [[de:Hauptseite]] |
Revision as of 07:06, 6 May 2020
Welcome to MWiki
Theorem of the month
Fermat's Last Theorem
For all [math]\displaystyle{ p \in {}^{\omega }{\mathbb{P}_{\ge 3}} }[/math] and [math]\displaystyle{ x, y, z \in {}^{\omega }{\mathbb{N}^{*}} }[/math], always [math]\displaystyle{ x^p + y^p \ne z^p }[/math] holds and thus for all [math]\displaystyle{ m \in {}^{\omega }{\mathbb{N}_{\ge 3}} }[/math] instead of [math]\displaystyle{ p }[/math].
Proof:
Because of Fermat's little theorem, rewritten, [math]\displaystyle{ f_{akp}(n) := (2n + a - kp)^p - n^p - (n + a)^p \ne 0 }[/math] is to show for [math]\displaystyle{ a, k, n \in {}^{\omega }{\mathbb{N}^{*}} }[/math] where [math]\displaystyle{ kp < n }[/math].
Induction for [math]\displaystyle{ n }[/math] implies the claim due to the case [math]\displaystyle{ m = 4 }[/math][1] and [math]\displaystyle{ y > x > p }[/math][2]:
Induction basis [math]\displaystyle{ (n \le p): f_{akp}(n) \ne 0 }[/math] for all [math]\displaystyle{ a, k }[/math] and [math]\displaystyle{ p }[/math]. Let [math]\displaystyle{ r \in {}^{\omega }{\mathbb{N}_{< p}} }[/math].
Induction step [math]\displaystyle{ \,(n = q + r \; \rightarrow \; n^{*} = n + p): }[/math] Let [math]\displaystyle{ f_{akp}(n^{*}) \ge 0 }[/math], but [math]\displaystyle{ f_{akp}(n) < 0 }[/math], since [math]\displaystyle{ f_{akp}(n) }[/math] is strictly monotonically increasing and otherwise nothing to prove.
It holds [math]\displaystyle{ f_{akp}(n^{*}) = (\int_0^{n^{*}}{f_{akp}(v)}dv)' \ne 0 }[/math], since [math]\displaystyle{ (n^{*})^{p + 1} + (n^{*} + a)^{p + 1} }[/math] does not divide [math]\displaystyle{ ((n^{*})^p + (n^{*} + a)^p)^2 }[/math] after separating the positive factor as polynomial division shows.[math]\displaystyle{ \square }[/math]
Recommended reading
References
- ↑ Ribenboim, Paulo: Thirteen Lectures on Fermat's Last Theorem : 1979; Springer; New York; ISBN 9780387904320, p. 35 - 38.
- ↑ loc. cit., p. 226.