Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 16 Mar 2007 15:35 > Phishing are we? > Casting doubt where none exists? > Feiging misunderstanding? > Shrouded in the veil of poor English skills? > In search of mailing addresses? > Wanting to seed dissent? > Deflecting from the truth? > > We have seen you before...and we will see you again... SO NICE POEM SURE YOU WILL..... Regards B.Karzeddin
From: Joseph Parranto on 16 Mar 2007 16:53 Not a poem - just a mirror for you to see your reflection, and others to see your transparency.
From: bassam king karzeddin on 16 Mar 2007 17:42 Dear News Group If AB is a diameter of a sphere, you can always find two points (C and D ) on the surface of the same sphere such that the distances: (AB)^3 = (AC)^3 + (CD)^3 + (DB)^3 Regards Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: bassam king karzeddin on 16 Mar 2007 17:59 Dear All 1) Any four (non-coplaner) points in space-in a stationary position or on a random motion- lie exactly at a unique surface of a sphere-at any moment of time- having finite radius 2)Any number of straight lines fitting inside a sphere and intersecting at any point on a surface of a smaller sphere having the same center such that each line consist of two segments where the product of each two segments for the same line is constant for all lines 3)Any four (non-coplaner) points in space contain a unique sphere with finite radius that is touching the plans formed by the four points each at one unique point only. Thanking Your patience Bassam Karzeddin AL-Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 16 Mar 2007 18:12
Dear All Fermat's Last SECRET or Fermat's last theorem for non zero integers (x, y, z)belong to Z*, where (n) is positive integer and (n>2) x^n + y^n + z^n = 0 Doesn't have any solution in the whole number system Can simply be reduced to the case where (n) is odd positive integer, and (n > 1) In fact it can be reduced further to the case where (n) is odd prime number, and (x, y, z) are coprime in pairs, but I will consider (n) as odd positive integer, where (n > 1) What I think that Fermat realized that the following equation holds true always, then his last theorem becomes so obvious to conclude. The equation is the following, where a counter example simply doesn?t exist .. "If (x) and (y) are two distinct integers belong to Z*, where (x) and (y) are prime to each other, and (n) is odd positive integer grater than one, then there exist three integers numbers (m, k, p), where (m) belongs to Z*, (k) belongs to N, (p) is prime number, where (m) and (p) are prime to each other, such that: x^ n + y^n + m*(p^(2^k)) = 0 " where N represents the natural numbers N = {0,1,2,3,4,...} and Z* represents the whole non zero integer numbers Z* = {...,-4,-3,-2,-1,1,2,3,4,...} Of course, one simply can prove that the later term m*(p^(2^k)) can?t be equal to (z^n), where (z) belongs to (Z*), other wise assume z^n = m*(p^(2^k)) Then, (p) must be a prime factor of (z), Let, z = t*(p^s), where (t) is integer number belongs to (Z*), (s) is positive integer, where also (t) and (p) are prime to each other. This implies: z^n = m*(p^(2^k)) = (t^n)*(p^(n*s)) From which, we conclude that: n*s = 2^k Since, the left hand side of the above equation has at least one odd prime factor belongs to (n), and the right hand side of the same equation doesn?t have any odd prime factor, we conclude that (n*s) is not equal to (2^k), and similarly z^n is not equal to m*(p^(2^k)), a contradiction to our assumption of equality, Hence, Fermat equation as defined earlier (x^n + y^n +z^n = 0) is impossible with non zero integer numbers belong to Z* I hope the proof is completed which was based on a Conjecture, and later Quasi suggested that the value of (k) is either equal to zero or one, then ....will continue Best Regards "Copyright (c) 2006, Bassam Karzeddin. All rights reserved" Bassam Karzeddin Al-Hussein Bin Talal University JORDAN |