On Saturday, March 17, 2007 at 4:12:46 AM UTC+2, bassam king karzeddin wrote:
> 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
Rarely a matheematicker acadimic or a historiean understand anything!
This actually was a type of common math I used to befor I did realize the many fallacies about real numbers !

BKK

On Wednesday, August 16, 2023 at 12:50:42 AM UTC+3, bassam karzeddin wrote:
> On Saturday, March 17, 2007 at 4:12:46 AM UTC+2, bassam king karzeddin wrote:
> > 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
> Rarely a matheematicker acadimic or a historiean understand anything!
> This actually was a type of common math I used to befor I did realize the many fallacies about real numbers !
>
> BKK
..you see, the Whole problem is about factoring!

BKK