Fermat's Last theorem short proof
in [Math]
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From: Gerry Myerson on 26 Apr 2007 22:34 In article <24935012.1177585315116.JavaMail.jakarta(a)nitrogen.mathforum.org>, bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: > > 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), Note that as you are assuming n is square-free it follows that m = gcd ( x + y, n ) is also square-free so Rad(m) is just m. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: sonnenrain17 on 27 Apr 2007 06:02 > 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 That is true indeed! You can even remove the assumption that M is a square free number and that C is positive: take C=1 or C=-1.
From: Tonico on 27 Apr 2007 11:46 On Apr 26, 1:25 pm, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > 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- Hide quoted text - > > - Show quoted text - ******************************************************************* I never claimed, nor anyone that I can think of, that I am the only mathematician, or that only I am a mathematician. You're claim that any positive integer (what in maths is sometimes called a natural number) E can be expressed as m*c^2, for m square free and C positive integer is awesome! I wonder whether you are the one that also discovered that doors serve the same to go out AND to go in, or that one can go up stairs and ON THE SAME stair also go down...! If you were, I wonder how come you haven't yet received the Nobel Prize Regards Tonio
From: bassam king karzeddin on 27 Apr 2007 10:59 > On Apr 26, 1:25 pm, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > > 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- Hide quoted text - > > > > - Show quoted text - > ****************************************************** > ************* > I never claimed, nor anyone that I can think of, that > I am the only > mathematician, or that only I am a mathematician. > You're claim that any positive integer (what in maths > is sometimes > called a natural number) E can be expressed as > m*c^2, for m square > free and C positive integer is awesome! I wonder > whether you are the > one that also discovered that doors serve the same to > go out AND to go > in, or that one can go up stairs and ON THE SAME > stair also go > down...! > If you were, I wonder how come you haven't yet > received the Nobel > Prize > Regards > Tonio > I told you Tonio, it is a Joke only and still you can't play with, let us see something of yours other than non-mathematics About Nobel Prize, Do you think it is opened for any one? Fermat himself and many many didn't, so what? Regards B.Karzeddin
From: bassam king karzeddin on 27 Apr 2007 11:06
> In article > <24935012.1177585315116.JavaMail.jakarta(a)nitrogen.math > forum.org>, > bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: > > > > 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), > > Note that as you are assuming n is square-free it > follows > that m = gcd ( x + y, n ) is also square-free > so Rad(m) is just m. > > -- > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for > email) Yes Sir I thought I have changed my (n) to be not a square number-odd positive integer Thanking your valuable notes B.Karzeddin |