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. 
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]

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 
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 
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 $\displaystyle N^mX+YZ$ Sentences used refer to OP Theorem D:$\displaystyle N^5X+YZ$ Proof: $\displaystyle T+SR+L = T+SR+EX+FYDZ = T+SR+E^2T+F^2SD^2R = T+SR+(N^2e+1)^2T+(N^2f+1)^2S(N^2d+1)^2R = $ $\displaystyle 2(T+SR)+N^4(eT+fSdR)+2N^2eT+2N^2fS2N^2dR\rightarrow N^4T+SR+L2(T+SR)2N^2eT2N^2fS+2N^2dR =$ $\displaystyle T+SR+L2(T+SR)2(E1)T2(F1)S+2(D1)R = T+SR+L2(ET+FSDR) = T+SR+L2(X+YZ)$ and as $\displaystyle N^3X+YZ$, we conclude $\displaystyle N^3T+SR+L$ (1) So,if $\displaystyle T+SR=N^2W$ and $\displaystyle X+YZ=N^3U$, then $\displaystyle N^3T+SR+L = N^2W+EX+FYDZ\rightarrow$ (sentence II) $\displaystyle N^4(N^2W+EX+FY)^ND^NZ^N\rightarrow$ (sentence III) $\displaystyle N^4(EX+FY)^N+N^3W(EX+FY)^{N1}Z^{2N}/R^N=$ (Fermat’s little one) $\displaystyle [N^2(eX+fY)+X+Y]N+N^3WZ^{2N}/(X+Y)\rightarrow$ (sentence III) $\displaystyle N^4N^3(eX+fY)(X+Y)^{N1}+(X+Y)^N+N^3WZ^{2N}/(X+Y)\rightarrow$ (Fermat’s little one) $\displaystyle N^4N^3(eX+fY)(X+Y)+(X+Y)(N^3U+Z)^N+N^3W(X+Y)Z^{2N}\rightarrow$ (sentence III) $\displaystyle N^4N^3(eX+fY)(N^3U+Z)+(N^3U+Z)Z^N+N^3W(N^3U+Z)Z^{2N}\rightarrow N^4N^3(eX+fY)Z+N^3UZ^N+Z^{N+1}+N^3WZZ^{2N}\rightarrow$ (sentence I) $\displaystyle N^4N^3(eX+fY)+N^3UZ^{N1}+Z^N+N^3WZ^{2N1}\rightarrow $ (Fermat’s little one) $\displaystyle N^4N[(E1)X+(F1)Y]+N^3U+Z^N+N^3WZ^{2N1} = N(EX+FY)+N(X+Y)+N^3U+Z^N+N^3WZ^{2N1}\rightarrow$ (as $\displaystyle N^3X+YZ$) $\displaystyle N^4N(EX+FY)+NZ+N^3U+Z^N+N^3WZ^{2N1}$ . Likewise: $\displaystyle N^4N(EXDZ)NY+N^3UY^N+N^3W+Y^{2N1}$ and $\displaystyle N^4N(FYDZ)NX+N^3UX^N+N^3W+X^{2N1}$ Calling $\displaystyle B=X^{2N1}+Y^{2N1}Z^{2N1}$,if we add the last three rels and given that $\displaystyle X^N+Y^N=Z^N$, we conclude: $\displaystyle N^42NLN(X+YZ)+3N^3U+3N^3W+B\rightarrow N^42NL+3(X+YZ)+3N(T+SR)+B $ (2) But $\displaystyle T^{N^N} = (TT^{N1})^N = [T(N^2t+1)]^N$ (3) and by sentence V (post #13) with k=2,b=1 and a=t we can write (3) as: $\displaystyle T^{N^N} = T^N[1+N^3t+(N1)N^5t^2/2+N^7A]$(4) $\displaystyle \rightarrow N^7 2T^{N^N}2T^N2N^3tT^NN^6t^2T^N+N^5t^2T^N = 2T^{N^N}2T^N2N(T^{N1}1)T^NN^2(T^{N1}1)^2T^N +N(T^{N1}1)^2T^N$ and as $\displaystyle N^2T^N–T$ we can claim: $\displaystyle N^7 2T^{N^N}2T^N2N(T^{N1}1)T^NN^2(T^{N1}1)^2T+N(T^{N1}1)^2T = 2T^{N^N}+2N(Ν^21)T^N(N^2N+2)T^{2N1}N(N1)T$. Likewise $\displaystyle N^7 2S^{N^N}+2(N^21)S^NN(N+1)S^{2N1}N(N1)S$ and $\displaystyle N^7 2R^{N^N}2(N^21)R^N+N(N+1)R^{2N1}+N(N1)R$. 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^72P+2(Ν^21)(T^N+S^NR^N)N(N1)(T+SR)N(N+1)B = $(by lemma A,post #12) $\displaystyle 2P4(Ν^21)(X+YZ)N(N1)(T+SR)N(N+1)B$ . Following the same reasoning and starting from (4) we come to: $\displaystyle N^62P+4Ν(X+YZ)+N(T+SR)NB$. Substracting the two last rels: $\displaystyle N^64N^2(X+YZ)+N^2B+N^2(T+SR)\rightarrow N^44(X+YZ)+B+(T+SR)$ Substracting the last rel.from (2) we deduce: $\displaystyle N^42NL(X+YZ)+(3N1)(T+SR)$ (5) and as $\displaystyle N^2L $(why?) we imply: $\displaystyle N^3(3N1)(T+SR)\rightarrow$(sentence I) $\displaystyle N^3T+SR$ and by e.g.theorem A (post #12) $\displaystyle N^4(N2)(X+YZ)\rightarrow$ (sentence I) $\displaystyle N^4X+YZ $ and by e.g.(1) $\displaystyle N^3L$ and then (5) commands: $\displaystyle N^4(3N1)T+SR\rightarrow$ (sentence I) $\displaystyle N^4T+SR$ and back to theorem A again gently $\displaystyle N^5X+YZ$ Fortuna Audaces Iuvat To be continued 
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 

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 