On Thursday, April 6, 2017 at 7:01:38 PM UTC+3, bassam king karzeddin wrote:
> > On Wed, 19 Apr 2006 20:48:47 +0200, "Ulrich Diez"
> > <eu_an...@web.de.invalid> wrote:
> >
> > >bassam king karzeddin wrote:
> > >
> > >> "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 (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 "
> > >
> > >I'm not all too experienced in maths.
> > >I think not all points of the proof were outlined in
> > detail.
> >
> > That's because it's not a full proof. Rather it's a
> > proof based on a
> > conjecture.
> >
> > >So not everything is clear to me yet.
> > >Please let me summarize the "bassam king
> > karzeddin"-proof in my words:
> > >
> > >Let v = -m , you get:
> > > x^ n + y^n - v*(p^(2^k)) = 0
> > > x^ n + y^n = v*(p^(2^k))
> > >
> > >m and p are prime to each other.
> > >-> v and p are prime to each other also/p is not a
> > prime-factor of v.
> > >->
> > >When looking at the exponents of the prime-factors
> > of
> > >(x^n+y^n), you will for n >3 in any case find for at
> > least one
> > >prime-factor (which is called p in the "bassam king
> > karzeddin"-
> > >equation), that it's exponent is a non-negative
> > integer-power of 2
> > >(, which means that the whole expression cannot be
> > an
> > >odd positive integer-power > 1 of another
> > integer-number).
> > >
> > >Did I get this right?
> >
> > Yes, that's the argument. The unproved part is the
> > equation:
> >
> > x^ n + y^n + m*(p^(2^k)) = 0
> >
> > There's no proof that for any distinct coprime
> > integers x,y, you can
> > always find m,p,k satisfying the equation together
> > with the
> > requirements that p is prime, k is a nonnegative
> > integer, and m is an
> > integer relatively prime to p.
> >
> > >If not: Please forgive my ignorance and tell me
> > about the error in
> > >my understanding.
> > >If so: Please forgive my ignorance --I'm not a
> > mathematician at all-- but
> > >how to prove this?
> >
> > Good question. As far as I can see, no proof or even
> > partial proof of
> > the conjectured equation has been provided by
> > Karzeddin. His only
> > claim is that "no counterexample exists", which is
> > obviously not a
> > proof.
> >
> > >(Maybe there is an example where all prime-factor-
> > >exponents are not non-negative integer-powers of 2
> > although the
> > >expression does not have the form x^n+y^n=z^n. I
> > didn't find one
> > >yet and I hope that no one else will,
> >
> > Right -- maybe there is a counterexample. but so far,
> > the conjecture
> > has survived brute force attack.
> >
> > >but perhaps a formal proof might be a remedy for all
> > scepticalness.)
> >
> > A formal proof would be spectacular, since FLT would
> > then be just a
> > simple corollary.
> >
> > I don't have any feeling for why the conjecture
> > should be true, hence,
> > a counterexample wouldn't surprise me. In fact, I'm
> > surprised that the
> > conjecture didn't fall immediately to a simple brute
> > force test with
> > relatively small values of x,y,n.
> >
> > quasi
>
> And, there will be no counter example for sure
>
> Regards
> Bassam King Karzeddin
> 6th, April, 2017

Aren't the number theorists still clueless about it?

Bkk