Fermat's Last theorem short proof
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From: bassam king karzeddin on 27 Feb 2007 08:37 > > > > > > 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). > > > > > > My couterexample above is not sufficient to > > disprove > > > your argument, since it can be easily shown by > > > Fermat's Little Theorem and the counterexample > to > > > Fermat's Last Theorem that > > > > > > x + y = z mod 3; > > > > > > hence, if x + y is divisible by 3, then so is z, > > > contradictory to the first case assumption. > > > > > > However, I would like to present the following > > > example: > > > > > > x = 7, y = 10, z = 47. > > > > > > x + y = 17 is not divisible by 3; > > > > > > x + z = 54 = 3*18; > > > > > > y + z = 57 = 3*19. > > > > > > I hope I did the arithmetic right. > > > > > > > It is well established that a counter example MUST > > form a Triangle > > > > In my previous posts, I stated that a counter > example > > must form an integer triangle (with one side only > > even integer), and the largest angle between > (PI/2, > > PI/3) > > > > So, you see it follows directly > > > > Yours are not a counter example > > > > > Where is the previous post? > > > > My Regards > > > > Bassam Karzeddin > > AL Hussein bin Talal University > > JORDAN You may see at the following link, even I may have posted it befor http://mathforum.org/kb/message.jspa?messageID=4695333&tstart=0 My Regards Bassam Karzeddin
From: Hisanobu Shinya on 27 Feb 2007 08:55 > > 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 > > > > > Gcd((x+y)*(z-x)*(z-y), N(x,y,z))= p, > > > 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 > > SORRY, a flaw was found by me few hours after posting > proof of this case and it is no more valid, but I > have another one I will be happy to read another argument. > > > You may add any suitable references, and then > write > > it in any suitable languge, beside let the > JOURNALS > > KNOW ABOUT IT > > > > > > > > > > MY BEST REGARDS > > > > Bassam Karzeddin > > > > AL Hussein bin Talal University > > > > JORDAN
From: bassam king karzeddin on 27 Feb 2007 08:56 > 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. Hello Sir Where is your previous article please My Regards Bassam 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 09:00 > > > 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 > > > > > > > > Gcd((x+y)*(z-x)*(z-y), N(x,y,z))= p, > > > > > 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 > > > > SORRY, a flaw was found by me few hours after > posting > > proof of this case and it is no more valid, but I > > have another one > > I will be happy to read another argument. > > > > > > You may add any suitable references, and then > > write > > > it in any suitable languge, beside let the > > JOURNALS > > > KNOW ABOUT IT > > > > > > > > > > > > > > > > MY BEST REGARDS > > > > > > Bassam Karzeddin > > > > > > AL Hussein bin Talal University > > > > > > JORDAN I have already corrected that argument B.Karzeddin
From: Dik T. Winter on 27 Feb 2007 19:52
In article <JE4oIF.KpL(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > 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, Sorry, this was wrong. It should be: (x + y)^4 + 5.y^2.(x + y)^2 + 3.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. I find now (but can not yet prove it) that p*N(x,y,y) is: sum{k = 0 -> (p-3)/2} C(p, 2k + 1) (x + y)^(p-3-2k).y^(2k). valid for all odd p in the range 3..17. > 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. And, apparently, if that statement is true, it gets more convincing, the more I try. But a proof is needed for the expansion above. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |