My Math Forum A faint light on FLT?

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 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 $\displaystyle N^m|X+Y-Z$ 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
 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

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