My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Reply
 
LinkBack Thread Tools Display Modes
August 26th, 2015, 12:54 PM   #31
Member
 
Joined: Aug 2015
From: Chiddingfold, Surrey

Posts: 57
Thanks: 3

Math Focus: Number theory, Applied maths
Many thamks. Yes please help. Page 4 has not appeared nor an updated page 5. I have been using Tiny Pic uk to produce the URLs It would certainly be better as one document/image.

At age 87 I need help with these things.
magicterry is offline  
 
August 26th, 2015, 02:14 PM   #32
Member
 
Joined: Aug 2015
From: Chiddingfold, Surrey

Posts: 57
Thanks: 3

Math Focus: Number theory, Applied maths
[[IMG]http://[/IMG]
magicterry is offline  
August 26th, 2015, 02:22 PM   #33
Member
 
Joined: Aug 2015
From: Chiddingfold, Surrey

Posts: 57
Thanks: 3

Math Focus: Number theory, Applied maths
[IMG]htIMG]http://i60.tinypic.com/28ssoys.jpg[/IMG]tp://[/IMG]

Updates page 5
magicterry is offline  
August 27th, 2015, 10:19 AM   #34
Member
 
Joined: Aug 2015
From: Chiddingfold, Surrey

Posts: 57
Thanks: 3

Math Focus: Number theory, Applied maths
apologies

The link in my last message has not inserted the revised page 5, but it does take you to it. (very slowly on my computer)

My apologies to Bruno59 for interrupting your thread. I think I should have started a new one. My excuse being a newcomer.

Proving FLT in a simple way is probably impossible - it took Andrew Wiles seven years to reach his proof. If Fermat had not claimed to have a proof, I doubt that anyone would have attempted it. The best we can do is throw some light on the subject
magicterry is offline  
September 5th, 2015, 04:54 AM   #35
Senior Member
 
Joined: May 2013

Posts: 118
Thanks: 10

This is another way of proving theorem D (post #19),which we believe is simpler (!).And besides,it provides a perspective for higher exponent m in $\displaystyle N^m|X+Y-Z$
Sentences used refer to OP
Theorem D:$\displaystyle N^5|X+Y-Z$
Proof:
$\displaystyle T+S-R+L = T+S-R+EX+FY-DZ = T+S-R+E^2T+F^2S-D^2R = T+S-R+(N^2e+1)^2T+(N^2f+1)^2S-(N^2d+1)^2R =
$
$\displaystyle 2(T+S-R)+N^4(eT+fS-dR)+2N^2eT+2N^2fS-2N^2dR\rightarrow N^4|T+S-R+L-2(T+S-R)-2N^2eT-2N^2fS+2N^2dR =$
$\displaystyle T+S-R+L-2(T+S-R)-2(E-1)T-2(F-1)S+2(D-1)R = T+S-R+L-2(ET+FS-DR) = T+S-R+L-2(X+Y-Z)$
and as $\displaystyle N^3|X+Y-Z$, we conclude $\displaystyle N^3|T+S-R+L$ (1)
So,if $\displaystyle T+S-R=N^2W$ and $\displaystyle X+Y-Z=N^3U$, then
$\displaystyle N^3|T+S-R+L = N^2W+EX+FY-DZ\rightarrow$ (sentence II) $\displaystyle N^4|(N^2W+EX+FY)^N-D^NZ^N\rightarrow$
(sentence III) $\displaystyle N^4|(EX+FY)^N+N^3W(EX+FY)^{N-1}-Z^{2N}/R^N=$ (Fermat’s little one) $\displaystyle [N^2(eX+fY)+X+Y]N+N^3W-Z^{2N}/(X+Y)\rightarrow$
(sentence III) $\displaystyle N^4|N^3(eX+fY)(X+Y)^{N-1}+(X+Y)^N+N^3W-Z^{2N}/(X+Y)\rightarrow$ (Fermat’s little one)
$\displaystyle N^4|N^3(eX+fY)(X+Y)+(X+Y)(N^3U+Z)^N+N^3W(X+Y)-Z^{2N}\rightarrow$ (sentence III)
$\displaystyle N^4|N^3(eX+fY)(N^3U+Z)+(N^3U+Z)Z^N+N^3W(N^3U+Z)-Z^{2N}\rightarrow N^4|N^3(eX+fY)Z+N^3UZ^N+Z^{N+1}+N^3WZ-Z^{2N}\rightarrow$ (sentence I)
$\displaystyle N^4|N^3(eX+fY)+N^3UZ^{N-1}+Z^N+N^3W-Z^{2N-1}\rightarrow $ (Fermat’s little one) $\displaystyle N^4|N[(E-1)X+(F-1)Y]+N^3U+Z^N+N^3W-Z^{2N-1} = N(EX+FY)+N(X+Y)+N^3U+Z^N+N^3W-Z^{2N-1}\rightarrow$
(as $\displaystyle N^3|X+Y-Z$) $\displaystyle N^4|N(EX+FY)+NZ+N^3U+Z^N+N^3W-Z^{2N-1}$ .
Likewise: $\displaystyle N^4|N(EX-DZ)-NY+N^3U-Y^N+N^3W+Y^{2N-1}$ and $\displaystyle N^4|N(FY-DZ)-NX+N^3U-X^N+N^3W+X^{2N-1}$
Calling $\displaystyle B=X^{2N-1}+Y^{2N-1}-Z^{2N-1}$,if we add the last three rels and given that $\displaystyle X^N+Y^N=Z^N$, we conclude:
$\displaystyle N^4|2NL-N(X+Y-Z)+3N^3U+3N^3W+B\rightarrow N^4|2NL+3(X+Y-Z)+3N(T+S-R)+B $ (2)
But $\displaystyle T^{N^N} = (TT^{N-1})^N = [T(N^2t+1)]^N$ (3)
and by sentence V (post #13) with k=2,b=1 and a=t we can write (3) as:
$\displaystyle T^{N^N} = T^N[1+N^3t+(N-1)N^5t^2/2+N^7A]$(4)
$\displaystyle \rightarrow N^7| 2T^{N^N}-2T^N-2N^3tT^N-N^6t^2T^N+N^5t^2T^N = 2T^{N^N}-2T^N-2N(T^{N-1}-1)T^N-N^2(T^{N-1}-1)^2T^N +N(T^{N-1}-1)^2T^N$
and as $\displaystyle N^2|T^N–T$ we can claim:
$\displaystyle N^7| 2T^{N^N}-2T^N-2N(T^{N-1}-1)T^N-N^2(T^{N-1}-1)^2T+N(T^{N-1}-1)^2T = 2T^{N^N}+2N(Ν^2-1)T^N-(N^2-N+2)T^{2N-1}-N(N-1)T$.
Likewise $\displaystyle N^7| 2S^{N^N}+2(N^2-1)S^N-N(N+1)S^{2N-1}-N(N-1)S$ and $\displaystyle N^7| -2R^{N^N}-2(N^2-1)R^N+N(N+1)R^{2N-1}+N(N-1)R$.
Calling $\displaystyle P=T^{N^N}+S^{N^N}-R^{N^N}$ (see also post #14) & adding the last three rels we end to:
$\displaystyle N^7|2P+2(Ν^2-1)(T^N+S^N-R^N)-N(N-1)(T+S-R)-N(N+1)B = $(by lemma A,post #12)
$\displaystyle 2P-4(Ν^2-1)(X+Y-Z)-N(N-1)(T+S-R)-N(N+1)B$ .
Following the same reasoning and starting from (4) we come to:
$\displaystyle N^6|2P+4Ν(X+Y-Z)+N(T+S-R)-NB$.
Substracting the two last rels:
$\displaystyle N^6|4N^2(X+Y-Z)+N^2B+N^2(T+S-R)\rightarrow N^4|4(X+Y-Z)+B+(T+S-R)$
Substracting the last rel.from (2) we deduce:
$\displaystyle N^4|2NL-(X+Y-Z)+(3N-1)(T+S-R)$ (5)
and as $\displaystyle N^2|L $(why?) we imply:
$\displaystyle N^3|(3N-1)(T+S-R)\rightarrow$(sentence I) $\displaystyle N^3|T+S-R$
and by e.g.theorem A (post #12)
$\displaystyle N^4|(N-2)(X+Y-Z)\rightarrow$ (sentence I) $\displaystyle N^4|X+Y-Z $
and by e.g.(1) $\displaystyle N^3|L$ and then (5) commands:
$\displaystyle N^4|(3N-1)T+S-R\rightarrow$ (sentence I) $\displaystyle N^4|T+S-R$
and back to theorem A again gently $\displaystyle N^5|X+Y-Z$

Fortuna Audaces Iuvat

To be continued
bruno59 is offline  
September 6th, 2015, 09:54 AM   #36
Member
 
Joined: Aug 2015
From: Chiddingfold, Surrey

Posts: 57
Thanks: 3

Math Focus: Number theory, Applied maths
Greetings

Good to see you continuing with your proof. Hope no flaws will be found. My effort is to provide a visual aid to understanding the theorem (not proving it) and it is applicable to all powers from 2 on. Gets a bit intricate to show a diagram with power higher than 3, but I have one for power 5 to show if interested
magicterry is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
faint, flt, light



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
if light cannot bend then how refraction & reflection of light works? Ganesh Ujwal Physics 4 December 14th, 2014 06:25 PM
Can someone shed light on this? sivela Calculus 2 March 18th, 2012 03:46 PM
How does the inverse square law of light effect light intens moore778899 Elementary Math 0 January 16th, 2011 07:20 AM
Integral *not for the faint of heart* shaese Calculus 2 May 1st, 2009 12:25 PM
Light Switches mahuirong Algebra 9 January 1st, 2008 12:30 PM





Copyright © 2019 My Math Forum. All rights reserved.