Fermat's Last theorem short proof
in [Math]
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From: Hisanobu Shinya on 27 Feb 2007 02:44 > > Fermat's Last theorem short proof > > > > We have the following general equation (using the > > general binomial theorem) > > > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + > > x^p+y^p+z^p > > > > Where > > N (x, y, z) is integer function in terms of (x, y, > z) > > > > P is odd prime number > > (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 > > Why? All right. I got this, and I think this is clever. > > > > > (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), > > Why? > > If x = 4, y = 5, z = 7, for instance, then > > x + y is divisible by 3, which is relatively prime to > xyz. Besides, 3 is also a divisor of another factor (y + z). Hence, your 9-times-something argument does not seem to work. > > > So > > Not 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) > > > > My question to the specialist, is my proof a new > one, > > more over I will not feel strange if this was > known > > few centuries back > > > > Thanking you a lot > > > > Bassam King Karzeddin > > Al-Hussein Bin Talal University > > JORDAN > > Dah~~~~! > > Why are there so many replies to this simple argument!
From: bassam king karzeddin on 27 Feb 2007 02:57 > do > > > > Some have already got it as you see from the > replies > > > > I wonder also why simply you deny me all my efforts > to settle up this issue for ever. > > > > Is it because I give my knowledge for Free??? > > > > I can't write papers because mainly I don't have > time, and I don't know how to write introduction and > references, I also can't waite or tolerate refrees > opinions especially when I'm a teacher and they are > the students who are going to learn > > > > Here, many outstanding mathematicians used to have > the final word in this topic but not any more, where > they deliberately disappear now and may be working in > secret, (the main code for success}- ask their first > teacher home I do respect to assure this fact, but > the story it self didn't end.! and the final laugh > will be for FERMAT > > > > if my proofs (which I will post completely whenever > I get free time ) are stolen then > > > > I swear I will add and triple your WORLD THAT IS > FULL OF buzzls for another few centuries, and the > final word will be for the new born that is not a > mind kind. > > Good start. But you have a long way to go to be as > funny as James > Harris. > > I don't get what do you mean , though I'm not an actor Do you main James Harris is the fake brain storm that proceed the real?? But really speaking, he does have many points that I have proved regiously Regards Bassam Karzeddin
From: Hisanobu Shinya on 27 Feb 2007 03:00 > Hello TO ALL > > The only remaining case -PROOF > > Now, you may easily derive the following identity for > a counter example when (p) is a prime factor OF > (x*y*z) > > Let x^p + y^p = z^p, where (x,y,z) are three positive > coprime integers, then the following identity must > hold > > (x*y*z)^p = (x+y)*(z-x)*(z-y)*N(x,y,z) > where > > (x+y),(z-x),(z-y), N(x,y,z) are all coprime pair > wise > > Very Simple to prove, (no need to waste time on > that) > > Now, (p^k*p) divides the left hand side of the above > identity, where as p^(k*p-1) only divides the right > hand side of the above Identity, p^(k*p - 1) divides also the left side. I do not know why the number k appeared, but it is clear that given an integer A divisible by p^m, it is also divisible by p^a for a = 0, 1,..., m. > where k is positive > integer number,and as I had shown you today from > other post. > > Therefore, the identity can't hold true with integer > numbers defined above AND THERE IS NO COUNTER > EXAMPLE > > Hence THE PROOF IS INDEED COMPLETED > > I was hoping some one would share me this vectory, > but For those who love mathematics AND TRUTH > DEFINDERS- DON'T LET IT GO, there are many things to > come > > You may add any suitable references, and then write > it in any suitable languge, beside let the JOURNALS > KNOW ABOUT IT > > KEEP IT FOR GENARATIONS > > And here, I WILL NOT STOP > > MY BEST REGARDS > > Bassam Karzeddin > > AL Hussein bin Talal University > > JORDAN
From: bassam king karzeddin on 27 Feb 2007 03:03 > > In article > > > <1172505870.803300.256810(a)a75g2000cwd.googlegroups.com > > > > "Randy Poe" <poespam-trap(a)yahoo.com> writes: > > > On Feb 21, 9:51 am, bassam king karzeddin > > n <bas...(a)ahu.edu.jo> wrote: > > > > We have the following general equation (using > > g the general binomial theorem) > > > > > > > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + > > + x^p+y^p+z^p > > > > > > > > Where > > > > N (x, y, z) is integer function in terms of > (x, > > , y, z) > > > > P is odd prime number > > > > (x, y, z) are three (none zero) co prime > > e integers? > > > > > > > > Assuming a counter example (x, y, z) exists > such > > h 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) > > > > > > All right, you didn't answer but somebody else > > e did. I have > > > now seen a validation of this identity including > > g the > > > statement that N(x,y,z) = 1 for p = 3. > > > > See my proof of it in my previous article. For p > = > > 1, N(x,y,z) = 0 and > > for p = 3, N(x, y, z) = 1. That is also easily > > proven: (x + y + z)^p > > - x^p - y^p - x^p is a homogenous polynomial of > > degree 3, and so it is > > 3.(x + y)(x + z)(y + z). For p= 5 it is: > > x^2 + x.y + x.z + y^2 + y.z + z^2. > > > For case 1 of FLT, if (z - y) = b^p, then N(x.y.z) = > z^(p - 3) (mod b). If your statement about N(x,y,z) > when p = 5 is correct, then a simple contradiction is > reached that proves case 1 for p = 5. If the > contradiction holds for all p > 4, then he may indeed > have proved case 1. I'm very happy, that truth is know becoming to be seen by many I shall be happer to be mistaken Regards B.Karzeddin > > > > > > > You are still leaving a lot out. Dividing both > > h sides by 3, > > > I get (x+y+z)^3/3 = (x+y)*(x+z)*(y+z) > > > > You forget on the right hand side (x^3 + y^3 + > > z^3)/3. > > -- > > dik t. winter, cwi, kruislaan 413, 1098 sj > > amsterdam, nederland, +31205924131 > > home: bovenover 215, 1025 jn amsterdam, > nederland; > > http://www.cwi.nl/~dik/
From: bassam king karzeddin on 27 Feb 2007 03:12
> > Hello TO ALL > > > > The only remaining case -PROOF > > > > Now, you may easily derive the following identity > for > > a counter example when (p) is a prime factor OF > > (x*y*z) > > > > Let x^p + y^p = z^p, where (x,y,z) are three > positive > > coprime integers, then the following identity must > > hold > > > > (x*y*z)^p = (x+y)*(z-x)*(z-y)*N(x,y,z) > > > where > > > > (x+y),(z-x),(z-y), N(x,y,z) are all coprime pair > > wise > > > > Very Simple to prove, (no need to waste time on > > that) > > > > Now, (p^k*p) divides the left hand side of the > above > > identity, where as p^(k*p-1) only divides the > right > > hand side of the above Identity, > > p^(k*p - 1) divides also the left side. I do not know > why the number k appeared, but it is clear that given > an integer A divisible by p^m, it is also divisible > by p^a for a = 0, 1,..., m. Obvious, but my aim is make it visible to School Children beside non-mathematicians So, you see one side of the identity has p as a prime factor where as the other side doesn't, then you know the rest... Regards B.Karzeddin > > > > where k is positive > > integer number,and as I had shown you today from > > other post. > > > > Therefore, the identity can't hold true with > integer > > numbers defined above AND THERE IS NO COUNTER > > EXAMPLE > > > > Hence THE PROOF IS INDEED COMPLETED > > > > I was hoping some one would share me this vectory, > > but For those who love mathematics AND TRUTH > > DEFINDERS- DON'T LET IT GO, there are many things > to > > come > > > > You may add any suitable references, and then > write > > it in any suitable languge, beside let the > JOURNALS > > KNOW ABOUT IT > > > > KEEP IT FOR GENARATIONS > > > > And here, I WILL NOT STOP > > > > MY BEST REGARDS > > > > Bassam Karzeddin > > > > AL Hussein bin Talal University > > > > JORDAN |