Fermat's Last theorem short proof
in [Math]
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From: Randy Poe on 26 Feb 2007 16:11 On Feb 26, 3:29 pm, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > "what you've shown as far as I can tell is that > > if x^3 + y^3 + z^3 = 0 has a solution, then one of > > x, y or z must be divisible by 3. " > > > 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 example > > implies that xyz = 0 mod 3. > > 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. I'm not denying anything. I agree that as far as I can tell, you have a valid proof that if (x,y,z) is a FLT counter-example for p=3, then xyz = 0 mod 3. - Randy
From: Dik T. Winter on 26 Feb 2007 19:22 In article <1172505080.079287.259690(a)z35g2000cwz.googlegroups.com> "Randy Poe" <poespam-trap(a)yahoo.com> writes: > On Feb 23, 2:13 pm, Dan Cass <d...(a)sjfc.edu> wrote: .... > > Based on a few Maple calcs, I think the identity > > (x+y+z)^p > > =p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p > > gives an integer coefficient polynomial for N(x,y,z) > > whenever p is odd and at least 3. > > Ah. OK. Glad somebody answered. OP refused to. > (Not sure how I got that arithmetic wrong.) It is easy to prove that N(x,y,z) is an integer function when p and odd prime. Look at (x+y+z)^p - (x^p+y^p+z^p), which is a polynomial in x, y and z. Set x+y = 0, and see (when p is odd) that it vanishes, so the resulting polynomial is divisible by x+y. Similar for x+z and y+z. Let's now see whether it is divisible by p. In (x+y+z)^p we look at the coefficient of x^k.y^l.z^m with k+l+m = p. It is pretty easy to see that it is C(p, k).C(p-k, l). We can exclude those cases where one of k, l or m is equal to p (or two of them equal to 0). And when p is prime and k != 0, C(p, k) is divisible by p, when k = 0 and l != 0 and C(p-k, l) is divisible by p. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 26 Feb 2007 19:54 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 <bas...(a)ahu.edu.jo> wrote: > > 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 > > > > (x+y+z)^3 =3* (x+y)*(x+z)*(y+z) > > All right, you didn't answer but somebody else did. I have > now seen a validation of this identity including 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 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: Dik T. Winter on 26 Feb 2007 20:43 In article <11886980.1172521796572.JavaMail.jakarta(a)nitrogen.mathforum.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 example > > implies that xyz = 0 mod 3. > 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 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 case) (2): One of x, y and z is divisible by p (the 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 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. -- 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 20:13
Dear ALL Yes, there are few things I kept them hidden, 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 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 |