On Thursday, March 5, 2020 at 4:24:48 PM UTC+2, bassam karzeddin wrote:
> On Wednesday, February 21, 2007 at 5:51:51 PM UTC+3, karzeddin wrote:
> > Fermat's Last theorem short proof
> >
> > We have the following general equation (using the general binomial theorem)
> >
> > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
> >
> > Where
> > N (x, y, z) is integer function in terms of (x, y, z)
> > P is odd prime number
> > (x, y, z) are three (none zero) co prime integers?
> >
> > Assuming a counter example (x, y, z) exists such that (x^p+y^p+z^p=0)
> >
> >
> > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
> >
> > CASE-1
> > If (p=3) implies N (x, y, z) = 1, so we have
> >
> > (x+y+z)^3 =3* (x+y)*(x+z)*(y+z)
> >
> > Assuming (3) does not divide (x*y*z), then it does not divide (x+y)*(x+z)*(y+z),
> > So the above equation does not have solution
> > (That is by dividing both sides by 3, you get 9 times an integer equal to an integer which is not divisible by 3, which of course is impossible
> > I think proof is completed for (p=3, and 3 is not a factor of (x*y*z)
> >
> > My question to the specialist, is my proof a new one, more over I will not feel strange if this was known few centuries back
> >
> > Thanking you a lot
> >
> > Bassam King Karzeddin
> > Al-Hussein Bin Talal University
> > JORDAN
> You (academic mainstreams) don't deserve at all to know more about it, FOR SURE
> bkk