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
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