Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 26 Apr 2007 02:25 > On 25 abr, 23:14, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > > > > ****************************************************** > > > * > > > ...***Sigh***....you obviously have, again, > stopped > > > taking your meds, > > > but you are right in one little, tiny thing: FLT > is > > > driving you > > > utterly crazy. Of course, you are NOT a > > > mathematician... > > > Regards > > > Tonio > > > > Yes I'm not a mathematician, but certainly I'm the > MATHEMATICS > > > > Regards > > B.Karzeddin- > > ****************************************************** > ***** > Sure dude...hehe: of course you are! > Regards > Tonio > Sorry Tonio, I was joking What did I mean EXACTLY IS If you-only-YOU are a MATHEMATICIAN then I'M the MATHEMATICS Of course you may deduce the BIG DIFFERENCE. I shall give you an additional HINT, where you may be able to deduce it alone BY YOUR SELF I shall make it ALSO available even to the FOOLS, with my respect and apologies to ALL "If (M) is positive square free number>=1, (C) is positive integer, then, ANY positive integer (E), can be expressed as the following well known equation E = M*C^2 Good Luck Tonio B.Karzeddin Al Hussein bin Talal University JORDAN
From: bassam king karzeddin on 26 Apr 2007 03:01 > In article > <6okq23p8fd8lj9rchb311b9pj6e997jquh(a)4ax.com>, > quasi <quasi(a)null.set> wrote: > > > On Mon, 23 Apr 2007 23:27:25 GMT, V <V(a)telenet.be> > wrote: > > > > >gcd (x+y, (x^n+y^n)/(x+y)) = gcd (x+y, n) if x and > y are coprime and n > > >is odd and > 1. > > > > > >x^n+y^n = gcd(x+y,n)^2 * (x+y)/gcd(x+y,n) * > ((x^n+y^n)/(x+y))/gcd(x+y,n) > > > > > >x^n+y^n is divisible by (x+y) and > gcd(x+y,((x^n+y^n)/(x+y))/gcd(x+y,n))=1 > > > > > >For c^n to be a solution, c needs to be a multiple > of (x+y), > > > > The error is right here (the above line). > > There's already an error above that. > > x = 1, y = 2, n = 27. > > gcd( x + y, (x^n + y^n) / ((x + y) gcd( x + y, n )) = > 3. > > -- > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for > email) In fact Gerry Myerson is a brave fighter, his weapons only numbers that no one can beat Let me make the trial befor the last-till I find those notes, that are hopefully are not thrown by my wife. Given, two distinct, coprime non zero integers (x & y), where (x & y) aren't square numbers, and m = gcd ((x+y), n), where (n) is square free odd positive integer then this implies the following theorem: Gcd ((x+y), (x^n+y^n)/(x+y)) = Rad (m), Where Rad (m) equals the product of all the prime factors of (m), that is to say Rad (m) is square free number that divides (x^n+y^n), Regards ???? ???????? Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: bassam king karzeddin on 26 Apr 2007 03:37 In article > <13246142.1172051785962.JavaMail.jakarta(a)nitrogen.math > forum.org>, > bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: > > > I would like to know a counter example to the > following integer equation > > where I think there is no one at all, hoping to be > mistaken > > > > PUZZLE > > > > IF (n, m, k) are three positive distinct coprime > integers ,where (m, k) are > > odd integers, > > > > then the following integer equation doesn't have > any solution in the whole > > number system as defind above > > > > n^3 = m^3 + k^3 + 2*n*m*k > Gerry Myerson wrote > There is some discussion of this equation in Mordell, > Diophantine > Equations, especially starting on page 78 and > starting on page 130. > > -- > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for > email) Did any body find any feed back available freely on the internet about this issue, and before I add another one, that will not be mine again. THANKS TO ALL. My Regards B.Karzeddin
From: bassam king karzeddin on 26 Apr 2007 04:17 Dear All "If (M) is positive square free number>=1, (C) is positive integer, then, ANY positive integer (E), can be expressed as the following well known equation E = M*C^2 Regards B.Karzeddin Al Hussein bin Talal University JORDAN
From: bassam king karzeddin on 26 Apr 2007 11:48
> Dear All > > "If (M) is positive square free number>=1, (C) is > positive integer, then, ANY positive integer (E), can > be expressed as the following well known equation > > E = M*C^2 So, you may associate it with the most famous-Einstein's Equation, that balance energy (E), with mass (M), and speed of light (C), and you might wonder what is the relation of FLT with Einstein's Equation, is it a MAGIC, AND you will NEVER know, but later they will write books about it..., and FERMAT, EINSTEIN ... ARE REALLY LAUGHING > Regards > > B.Karzeddin > Al Hussein bin Talal University > JORDAN |