Re: Karzeddin's evolving conjectures ...

On Sun, 18 Nov 2007 "quasi" wrote:
On Sun, 18 Nov 2007 17:17:46 EST, tommy1729
<tommy1729@xxxxxxxxx>
wrote:

quasi wrote:

------ Counterexample [quasi, 11/18/07]:

p = 3, x = 13, y = 67, z = 30296, n = 526893

wow

this is amazing.

how did you find that ?

brute force or theory ?

Brute force together with some theoretical bounds and
congruence
relations to make the search tractable.

Once I figured out how to search more efficiently, it
took Maple less
than a second to find the above counterexample.

There are lots of other counterexamples ...

Here are a few more, all found (by Maple) within a
few seconds:

p = 5, x = 1, y = 23, z = 5084, n =
14523267475345

p = 7, x = 71, y = 129, z = 203593268,
n = 3887786672352973512576873404989164620970442043

p = 11, x = 115, y = 157, z = 8207715608141992,
n =
3842337520612857177688598507988032012089
4268395124748492950018265246505853082241
9492064308604791214474112030720863001333
31544215457657636860941268330090287

p = 13, x = 3, y = 53, z = 322383253196,
n =
3963168721969862174284675399763335812027
9405696205039257879767501575534539805521
0807682614533368106554877111728517731296
5302193944214105

quasi

What a nice and hard work for counterexamples:
I can see, that some of Karzeddin's conjectures
have to strong or too week relation to FLT.
When beginning with n=3 it was mentioned by
Mr. Bassam : k^3 = m^3 +n^3 +2mnk
as impossible for rationals and it does:
(See my final proof for this and how it could
be more elementary and correct ?)
Also this conjecture has too strong relation to
FLT and so on is more easy to proof it.
Why later Mr. Bassam writes:
( z^3 = x^3 +y^3 +2nxyz
as impossible for rationals )
I don't know but this sentence has much week
relation to proper FLT case and so on
rational solutions according to Your
and others counterexamples...
I'd like to notice, that for true relation to
FLT it should be written in general:
(when case for x;y;z not divided by exponent n )
t^n = a^n + b^n + 2*n^u abtp
where p value appears from:
T^n = nAB(2T+A+B)Ext where for perfect n power:
it was taken A = a^n ;
B = b^n ;
2T+A+B = t^n ;
n*Ext = n^nu p^n ;
then T = n^u abtp ;
once x;y;z were substituted:
z=T+A+B; x=T+B; y=T+A;
for n=3:
T^3 = 3ABT(2T+A+B)
and or x or y or z should be divided by 3
and so on here Ext=1 and p=1
for n=5:
T^5 = 5ABT(2T+A+B)[2T^2 +2(A+B)T +A^2 +AB +B^2]
also Ext = 2T^2 +2(A+B)T +A^2 +AB +B^2
then as we take:
5*Ext = 5^5u p^5
so it will be the fall of x;y;z; not divided by 5
Now if it could be also stated, that u value
at least is 2: u>=2 :
So the interesting will be, if for such
developed sentences:
t^n = a^n +b^n +2*n^u abtp
where natural n;t;a;b;p are of gcd =1
and u>=2 You can find counterexamples ?
(it is necessary to work only with n as prime values
as to follow composite n = p1*p2 comes etc. like for FLT)
I think not, because this equation gives the size
of ideal number p. It was only taken in account as if
we have such n*Ext = n^nu p^n.
(trying to solve from another side is to take A;B;
and 2T+A+B as ideal powers:
and then looking for:
T^n = n*(abt)^n Ext
also for n=5:
T^5 = 5*(abt)^5 (2T^2 +2(a^5 +b^5)T +|
|+a^10 +a^5 b^5 +b^10 | then taken:
T = abtP
P^5 = 5[2(abtP)^2 + 2(a^5 +b^5)abtP +|...|]
then for sure 5|P and taken P =5^u p
5^(5u-1) p^5 =
= 2(5^u abtp)^2 + 2(a^5 +b^5)5^u abtp +|
|+ a^10 +a^5 b^5 +b^10
Now for optional and natural a;b;t; and
of gcd=1 we'll not find some natural p.
There is to do much more as with the first
way attempt. For the 1-st way attempt it is
to consider p bigger than t and as I already
used to mention: 5;a;b;t;p of gcd=1.
(here it is to see absolute term:
Abst = a^10 +a^5b^5 +b^10 and as
5^u and p divides this term so it is
prime to a or b or to Abst = a^5 +b^5
where its divisors composing t so also to t )
What is Your opinion ?

With the Best Regards
and Season Greetings

Ro-Bin
.

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