Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 27 Feb 2007 00:05 > Dear ALL > > Yes, there are few things I kept them hidden, they may look like conjectures, but they don't have a life time, may be for few hours or days I wanted you to know the truth, because will not be accepted from me > > First- something very interesting about N (x, y, z), > I did not mention, since that wasn't needed to prove > the first case- when (p) is not a factor of (x*y*z)- > and that is too simple to prove > > p, (x+y), (x+z), (y+z), N (x, y, z) are all coprime > pair wise And (x+y+z) will always be divisible by (p) for assumed counter example Gcd(p, N(x, y, z)) = 1, for both cases, that is to say if p divides xyz or p doesn't divides xyz What does that mean? all the prime factors exponents of N(x,y,z) must be a multiples of p in both cases similarley for all the prime factors of (x+y)*(x+z)*(y+z) when p is not a factor of x*y*z but will be p^(kp-1)*a^p, when p is a prime factor of x*y*z , so then Rad(x+y+z) = p*Rad{(x+y)*(x+z)*(y+z)*N(x,y,z)} where, Rad n = product of all prime numbers of n then what is left? when p is a prime factor xyz implies p is a prime factor of one and only one of (x,y,z) Shall I contiue.... B.Karzeddin Regards > > Second- and according to "ONT" or "odd number > theorem" > > Gcd {(x+y), (x^p+y^p) / (x+y)} = p^k, where (k=0), if > p is not a factor of (x+y), and (k=1), if p is a > prime factor of (x+y) > > And that was due to a very valuable note from a > member in sci.math QUASI whom we do miss and I do > appreciate. > > On a comment to a previous thread of mine Fermat > Last Secret > Proof > Let p be a prime factor of (x+y), assume (x=a Mod > p), then (y= -a Mod p), implies > (x^p+y^p) / (x+y) = p*a^(p-1) Mod p, which is a > multiple of (p^1), the rest follows directly > > Third- if you whish to see the prime factorization of > (x^p+y^p), then assume (z=0), in the general equation > I provided in the beginning of this thread > > Now, still the proof is too short, but can you get > the complete picture?? > > IF YOU CAN'T I WILL... > > My Regards > 27,TH FEB.2007 > Bassam Karzeddin > AL Hussein Bin Talal University > JORDAN
From: Dik T. Winter on 27 Feb 2007 10:32 In article <29745573.1172583841740.JavaMail.jakarta(a)nitrogen.mathforum.org> Raymond Burhoe <nospamtorcb(a)powerpuff.com> writes: .... > > 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. Indeed, x^2 + x.y + x.z + y^2 + y.z = 0 mod (z - y), so z = y yields a zero. Substutiing y for z we get (x + y)^2 + 2.y^2 = 0, which is the contradiction. Trying the same for p = 7, I get: (x + y)^4 + 5.y^2.(x + z)^2 + 2.y^4 = 0, also a contradiction. For 11 I find: (x + y)^8 + 15.y^2.(x + y)^6 + 42.y^4.(x + y)^4 + 30.y^6.(x + y)^2 + 5.y^8 = 0 again a contradiction. And for 13, similar with coefficients 1, 22, 99, 132, 55 and 6. > If the > contradiction holds for all p > 4, then he may indeed have proved case 1. Upto now, it looks pretty convincing. So, can we always write N(x, y, y) as a sum of squares, with positive coefficients? Under condition of course that your statement "For case 1 of FLT ..." is correct. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Hisanobu Shinya on 27 Feb 2007 01:08 > 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? > > (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. > 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: Richard Henry on 27 Feb 2007 11:37 On Feb 26, 12:29 pm, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > Yes, and from the beginning I have shown you only a proof when > p doesn't divide xyz, and in general the complete proof is so simple that an average mathematician can 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 whome 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.
From: bassam king karzeddin on 27 Feb 2007 02:32
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, 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 |