Fermat's Last theorem short proof
in [Math]
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From: neilist on 24 May 2007 10:15 On May 24, 7:52 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: <snip more Harris-ian spewing> You prove yourself more and more the James Harris everyone loves to detest. Except for Quinn Tyler Jackson, who LOVES you so, so dearly.
From: neilist on 24 May 2007 10:47 On May 24, 9:19 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: <snip> Ah, James, you admit to drinking while mathematic-ing (Is that a word? Well, now it is.). No wonder your posts are full of drunken tirades that you vomit from your distended belly. I'm sure you just drink to wash down the "solution" of Quinn's "mathematics" from your mouth?
From: bassam king karzeddin on 24 May 2007 11:57 > > On Thu, 24 May 2007 06:57:20 -0500, quasi > > <quasi(a)null.set> wrote: > > > > >On Thu, 24 May 2007 06:00:36 EDT, bassam king > > karzeddin > > ><bassam(a)ahu.edu.jo> wrote: > > > > > >>> In article > > >>> > > > <24975251.1176279446561.JavaMail.jakarta(a)nitrogen.math > > > >>> forum.org>, > > >>> bassam king karzeddin <bassam(a)ahu.edu.jo> > writes: > > >>> >... > > >>> > > >(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) > > >>> >... > > >>> > > >>> The gap in the argument is in the marked > lines. > > For > > >>> instance, > > >>> 3 does not divide 1*2*7, but it does divide > > >>> (1+2)*(1+7)*(2+7). > > >>> > > >>> > > >>> William C. Waterhouse > > >>> Penn State > > >>> > > >>I noticed your reply today only, and really > > speaking I didnt get the wisdom of your reply > > separately > > >>I wonder also if you realize that the issue is > > about FLT, where I have made it very clear in my > > original thread that for assumed counter example > (x, > > y, z) must form a triangle, where also mentioned > for > > a counter example to be considered the following > > inequality must hold true always: > > >> > > >>Sqrt (x^2+y^2) > z > Sqrt (x^2+y^2-x*y) > > >> > > >>Which also means that the angle opposing the > larger > > side (z) must be between (60, 90) degrees > > > > > >3 does not divide 4*7*8, but it does divide > > (4+7)*(4+8)*(7+8). > > > > I withdraw the above example. > > > > It's not applicable in the context of the problem. > > > > Assuming the equation > > > > (x+y+z)^3 =3*(x+y)*(x+z)*(y+z) > > > > you are definitely correct in your statement that > if > > 3 does not divide > > x*y*z then 3 does not divide (x+y)*(x+z)*(y+z). > > > > So yes, you can prove FLT for p=3, case 1, but > that's > > the absolute > > easiest case. What about p=3, case 2? > > > > quasi > > Think about this simple equation, and then every > thing may come so easily quasi: > > If (x, y, z) are assumed counter example to the FLT > with already given definitions previously, > > Then the following integer equation doesn't have any > y solution in the whole positive integer numbers > > (x+y-z)^n / (z-x)*(z-y) > > Where (n) is positive integer >2, > And the inequality > > z>(x+y-z)>0 > > What I mean to say quasi is the following IF m = (x+y-z)^n / (z-x)*(z-y) where (x,y,z) are coprime positive integers (pair wise) such that (x<y<z), and (n) is positive integer>2 where also the inequality Sqrt (x^2+y^2) > z > Sqrt (x^2+y^2-x*y) then (m) can't be an integer number > My Regards > Bassam Karzeddin > Al Hussein bin Talal University > JORDAN > .
From: bassam king karzeddin on 24 May 2007 12:42 > On May 24, 7:52 am, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > <snip more Harris-ian spewing> > > You prove yourself more and more the James Harris > everyone loves to > detest. > > Except for Quinn Tyler Jackson, who LOVES you so, so > dearly. > I wish to understand what are you talking about, and what are your true intentions, and who is that Quinn Tyler Jackson ? a poet I suppose! B.Karzeddin
From: neilist on 24 May 2007 17:53
On May 24, 4:42 pm, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > On May 24, 7:52 am, bassam king karzeddin > > <bas...(a)ahu.edu.jo> wrote: > > > <snip more Harris-ian spewing> > > > You prove yourself more and more the James Harris > > everyone loves to > > detest. > > > Except for Quinn Tyler Jackson, who LOVES you so, so > > dearly. > > I wish to understand what are you talking about, and what are your true intentions, and who is that Quinn Tyler Jackson ? > a poet I suppose! > > B.Karzeddin Of course Harris, you are well aware of Quinn's poetry, in wooing you into the night with sonnets of love. He professed his love to you, saying "James, James" so many times in public. |