On Thursday, December 20, 2007 at 12:07:02 AM UTC+2, Roman B. Binder wrote:
> Bassam King Karzeddin on Dec 16, 2007 wrote:
> > I'm afraid, quasi may not had heard you and your
> > proof is not a proof
> I'm afraid , that he likes to complete such proof too:
> Anyway for Your k^3 -m^3 -n^3 = 2mnk conjecture
> I have just found much more simple developments:
> I'll write it in brief with my symbols:
> (there will be more comfortable to compare it
> with other and true FLT developments)
> also for questionable t^3 -a^3 -b^3 = 2abt........(1)
> for a;b;t natural values and of gcd=1..........(B.C.)
> we'd like to find some proper t = a+b-R
> where R also some natural value and for sure
> once t>a and t>b so a>R and b>R .
> Now You should retrieve following forms:
> 1): (3R-a)bt = R[R^2 +3(t-b)a]
> where "b" or "t" could be divided by R but not "a"
> and even with no factor of R = r1*r2*...
> once r1|a so r1|Ls but r1^2|Rs and once (B.C.):
> a;b;t should be of gcd=1
> 2): (3R-b)at = R{R^2 +3(t-a)b]
> now "a" or "t" could be divided by R but not "b"
> and with no factor of R = r1*r2*...
> once r1|b so r1|Ls but r1^2|Rs and once (B.C.):
> 3): (3R+t)ab = R[R^2 +3(a+b)t]
> now "a" or "b" could be divided by R but not "t"
> and with no factor of R = r1*r2*...
> once r1|t so r1|Ls but r1^2|Rs and once (B.C.):
> Together 1),2) and 3): We can not find any factor
> of R for to divide a;b or t except of R=1.
> but once R = a+b-t and where acc. to eq.(1)
> two of a;b;t values should be odd and one even
> (all even values is excluded with (B.C.)
> so R should be even number: Rmin =2
> Inconsistency for retrieved R=1.
> Q. E. D.
> >
> > About mine, I'm not interested in particular cases
> > but the general one where I hope I will make it
> > available to school students
> You are dreaming at least very hard...
> Case t^3 - a^3 -b^3 = 2abt is not a real FLT
> fall. We can call it pseudoFLT.
> True cases for FLT with n=3:
> 1) t^3 = 3^(3u-1) a^3 +b^3 +2*3^u abt
> 2) 3^(3u-1) t^3 = a^3 +b^3 +2*3^u abt
> >
> > You should realize by now that your famous form:
> >
> > T^n = nAB( T+A+B)ExtExtn
> > doesn't work as was damaged by quasi with some little
> > conversion
> You already used to rewrite developed by me forms
> with certain fault:
> T^n = nAB(2T+A+B)Ext
> where Ext varies according to n prime value:
> for n=3 Ext = 1
> for n=5 Ext = 2T^2 +2(A+B)T +A^2 +AB +B^2
> etc. etc.
> Once "quasi" used to find his counterexamples
> so lets say not for proper general form like:
> t^n = a^n +b^n +2*n^u abtp
> (such forms covers only the first fall of FLT:
> no one of X;Y;Z is divided by exponent n )
> Where should be taken at least u=2
> then p value will be limited from for ex. for n=5:
> 5^(5u-1) p^5 = Ext
> > Also my form was damaged by quasi, but I could find
> > presently some way to escape completely-I hope
> >
> > keep trying with good luck
> >
> How much luck have we already ?
> Developments acc. the k^3 -m^3 -n^3 = 2mnk
> also pseudoFLT conjecture could be hardly
> adapted into proper FLT 1-st falls for
> bigger exponents...
> Anyway life is very interesting...
> > My Regards
> > Bassam Karzeddin
> My Regards too
> and Happy New Year
> Ro-Bin
This guy seems to be lost & got fixed minded on an issue that he only understands

BKK