Fermat's Last theorem short proof
in [Math]
|
From: bassam king karzeddin on 26 Feb 2007 20:35 >> I am laughing out loud. Just when Dr. Poe points out > that you have proved only a very weak statement (that > xyz is not divisible by 3) you let us all know that > you have not posted the complete proof of FLT. > > So why did you not point out before that you did not > post a complete proof? > > Still giggling, MO I really did, I wish to have time to present it in your languge keep giggling please My Regards B.Karzeddin
From: bassam king karzeddin on 26 Feb 2007 20:49 > In article > <11886980.1172521796572.JavaMail.jakarta(a)nitrogen.math > forum.org> bassam king karzeddin <bassam(a)ahu.edu.jo> > writes: > > Randy Poe: > > > I agree that if p=3 and (x,y,z) is a FLT counter > > > example, then assuming xyz <> 0 mod 3 leads to > > > a contradiction. Hence (x,y,z) FLT counter > r example > > > implies that xyz = 0 mod 3. > > > Yes, and from the beginning I have shown you only > y a proof when > > p doesn't divide xyz, and in general the complete > e proof is so simple > > that an average mathematician can do > > For a general p? I do not know as I do not follow it > very closely, but > Sophie Germain proved that if p and 2p+1 are both > prime that > x^p + y^p + z^p = 0 > implies that one of x, y or z is divisible by p (p = > 3 and p = 5 both > fall into this special case, and since that time such > primes are called > Sophie Germain primes). From that time on FLT is > split in two cases: > (1): None of x, y and z is divisible by p (the easy > sy case) > (2): One of x, y and z is divisible by p (the > he difficult case). > Sophie Germain further proved Case 1 for all p less > than 100 and Legendre > extended it to all numbers less than 197. I didn't get that Do you main sir that proof is based on numerical test we know that primes are infinit, if so then it is not a proof unless I'm mistaken > > I do not know what proof you actually did find, > perhaps it is Sophie Germain's > proof, perhaps not. But the first case you have to > check thoroughly is > p=7, which is not a Sophie Germain prime. I will gladly check, but give me some time to provide you with a detailed report about p=7, despite my general check My Regards B.Karzeddin AL Hussein bin Talal University > -- > 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 26 Feb 2007 20:56 > 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. > > > 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. Sorry Sir, he didn't because he was assuming a counter example B.Karzeddin > -- > 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 26 Feb 2007 21:29 > > 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? > > Maybe you are using the division algorithm: > > Given integers a and b, there exist q and r < |a| > such that > > b = aq + r. > > However, you have no information on q and r. However, you are too far from here!! Regards > > > > > > > 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 > > > > (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), > > 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
From: Raymond Burhoe on 26 Feb 2007 22:43
> 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. > > > 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/ |