An Elementary Proof of Fermat's 'Last Theorem'
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From: Chip Eastham on 15 Jul 2010 19:32 On Jul 15, 8:19 am, "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 ). It is completely incorrect at the point where it is claimed that x^n + y^n = z^n for n > 2 necessarily implies x^2 + y^2 = z^2: > >> Choices [1] and [2] are not possible, > >> otherwise the left side of {1} > >> would be either less than or greater > >> than the right side. As Tim points out, this is untrue for real arguments x,y,z > 0, and the OP gives no explanation for how integrality of x,y,z might be used to fill the gap. regards, chip
From: smallfrey on 15 Jul 2010 19:13
Here's some FLT trivia related to inequalities. 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) where n is a natural number and x and y are positive real numbers. (I had a proof for n=3, but the dog ate 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. (All the prime factors of R, S, and T are of the form pk+1.) Then r divides t^p-s^p, s divides t^p-r^p, and t divides r^p+s^p. By the above inequality, r+s>t>r,s and hence each of r, s, and t has a prime factor of the form pk+1. (Since t divides (r+s)*[(r^p+s^p)/(r+s)]and t is greater than r+s, t and (r^p+s^p)/(r+s) must have a common factor.) This is of little consequence unless all the prime factors of r, s, and t are of the form pk+1. |