An Elementary Proof of Fermat's 'Last Theorem'
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From: alainverghote on 15 Jul 2010 12:19 On 15 juil, 14:19, "curious george" <bu...(a)bunch.net> wrote: > "Tim Little" <t...(a)little-possums.net> wrote in message > > news:slrni3t4js.jrj.tim(a)soprano.little-possums.net... > > > > > > > On 2010-07-14, A. N. Emus <laini...(a)msn.com> wrote: > > [ To prove: x^n + y^n = z^n {1} has no integer solutions for n > 2 ...] > >> By symmetry we can suppose x < y. Clearly, y < z. > > [...] > >> x^m(x^2) + y^m(y^2) = z^m(z^2), {2} > > [...] > >> x^m(x^2) + (x^m + a)y^2 = (x^m + b)z^2, {3} > >> where a and b must be some positive integers. Clearly, a < b. > > [...] > >> x^m(x^2 + y^2 - z^2) = bz^2 - ay^2. {4} > > >> There are three choices: > > >> [1] x^2 + y^2 - z^2 < 0. > >> [2] x^2 + y^2 - z^2 > 0. > >> [3] x^2 + y^2 - z^2 = 0. > > >> Choices [1] and [2] are not possible, otherwise the left side of {1} > >> would be either less than or greater than the right side. > > > Why? Your reasoning is certainly incorrect for non-integer solutions > > to {1}, and you have given no reason to suppose that it holds for > > integers. > > > - Tim > > Tim, how about telling the rest of us why you think his "reasoning is > certainly incorrect". How incorrect is it? 51% or 99.99999%. an > unsubstantiated answer to an unsubstantiated claim ( according to you ).- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Good evening, Three integer numbers out four can be written as difference of two squares. uneven integer numbers and multiple of 4 . For instance let us begin: 7^2 + ... = ... 49 = 2*24+1 = (24+1)^2-24^2 so 7^2+24^2 = 25^2 10^2+... = ... , 100=4*(25)*(1) or 4*A*B ?!! so 4*(25)*(1)= 26^2-24^2 We can complete the equality: 10^2+24^2 = 26^2 In fact with power two ,lots of numbers will verify a pythagorician equation. For higher powers we unfortunately don't know so simple relations between kind/class of integers and sum or difference of powers, Alain
From: master1729 on 15 Jul 2010 10:46 bobbysim wrote : > On Jul 14, 10:06 pm, master1729 <tommy1...(a)gmail.com> > wrote: > > > An Elementary Proof of Fermat's 'Last Theorem' > > > by A. N. Emus > > > > > The equation x^n + y^n = z^n > > > {1} > > > has no solution in positive integers x, y, z, n > if > > > n > 2. > > > > > Proof: > > > > > By symmetry we can suppose x < y. Clearly, y < > z. > > > > > Rewrite {1} as > > > > > x^m(x^2) + y^m(y^2) = z^m(z^2), > > > 2) = z^m(z^2), {2} > > > > > where m >= 1 must be some integer. > > > > > Rewrite {2} as > > > > > x^m(x^2) + (x^m + a)y^2 = (x^m + b)z^2, > > > + b)z^2, {3} > > > > > where a and b must be some positive integers. > > > Clearly, a < b. > > > > > Rearrange {3} as > > > > > x^m(x^2 + y^2 - z^2) = bz^2 - ay^2. > > > z^2 - ay^2. {4} > > > > > There are three choices: > > > > > [1] x^2 + y^2 - z^2 < 0. > > > > > [2] x^2 + y^2 - z^2 > 0. > > > > > [3] x^2 + y^2 - z^2 = 0. > > > > > Choices [1] and [2] are not possible, otherwise > the > > > left side of {1} would be either less than or > > > greater > > > than the right side. This leaves [3] as the only > > > choice. Then, from {4}, bz^2 - ay^2 = 0. This > is > > > not > > > possible considering a < b and y < z. > > > > > ******* > > > > not bad , [1] and [3] are indeed impossible. > > > > but you didnt prove [2] to be impossible. > > > > x^2 + y^2 - z^2 > 0 > > > > x^3 + y^3 - z^3 < 0 > > > > e.g. x = 5 y = 6 z = 7. > > > > tommy1729- Hide quoted text - > > > > - Show quoted text - > > Agreed, case [2] can be proved true if you consider > that x < y, and > notice that you can use the triangle inequality to a > suitable norm. > (l^n norm for the correct n ;) ). > -Robert i am both happy and amazed that you alone noticed that the problem lies with [2] and not with [1] as the other people think. the happy is a compliment to you. the amazed is a compliment to the other posters 'past posts' but a dissapointment to the recent. also it would be nice if the original poster would comment ... but often that is not the case in flt posts unless to whine ( musatov inverse 19 harris style ) to say something potentially meaningfull ; x^n + y^n - z^n > 0 is required for x(n+1) + y^(n+1) = z^(n+1) which might be the final generalization of the OP idea(s). not enough for a short proof of FLT of course. regards the master tommy1729
From: master1729 on 15 Jul 2010 10:50 x^n + y^n - z^n > 0 is required for x^(n+1) + y^(n+1) = z^(n+1) * corrected typo *
From: smallfrey on 15 Jul 2010 11:53 Here's a bit of FLT trivia. If n>2, [(x^n + y^n)^(1/n) - x]^(1/n) + [(x^n + y^n)^(1/n) - y]^(1/n) > (x+y)^1/n. (I used to have a proof for this when n=3, but I lost it.) Barlow's formulae implied by a first-case solution of Fermat's equation a^p+b^p=c^p are (c^p-b^p)/(c-b)=R^p, (c^p-a^p)/(c-a)=S^p, (a^p+b^p)/(a+b)=T^p, c-b=r^p, c-a=s^p, and a+b=t^p where rR=a, sS=b, tT=c, and gcd(r,R)=gcd(s,S)=gcd(t,T)=1. The above inequality implies that r+s>t>r,s and hence that each of r, s, and t has a prime factor of the form pk+1. (All the prime factors of R, S, and T are of the form pk+1.) This isn't of much significance unless all of the factors of r, s, and t are of the form pk+1.
From: smallfrey on 15 Jul 2010 11:58
Oops. I forgot to say that x and y are positive real numbers. |