An Elementary Proof of Fermat's 'Last Theorem'
in [Math]
|
Prev: Zugswang in my proof of No Odd Perfect Numbers #675 Correcting Math
Next: If Economics is one of the "Soft Sciences" how can it be trusted determine so much in society?
From: A. N. Emus on 14 Jul 2010 11:59 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} where m >= 1 must be some integer. Rewrite {2} as 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. Rearrange {3} as 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. 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. *******
From: master1729 on 14 Jul 2010 13:06 > 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
From: Tim Little on 15 Jul 2010 00:46 On 2010-07-14, A. N. Emus <lainiech(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
From: Bobby Simione on 15 Jul 2010 03:23 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
From: curious george on 15 Jul 2010 08:19
"Tim Little" <tim(a)little-possums.net> wrote in message news:slrni3t4js.jrj.tim(a)soprano.little-possums.net... > On 2010-07-14, A. N. Emus <lainiech(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 ). |