Fermat's Last theorem short proof
in [Math]
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From: Jeroen on 26 Feb 2007 07:31 bassam king karzeddin schreef: >> I'm not a mathematician, but maybe I can help: > > I'm also not a mathematician and this proof is ment mainly to nonmathematicians There is no proof, because the first line is wrong given the counterexample of Randy Poe. You stated that: (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p Randy replied: "Are you claiming this is true in general? Counterexample: x=3, y=4, z=5, p=5. (x+y+z)^p = 248832 (x^p + y^p + z^p) = 4392 (x+y)(x+z)(y+z) = 560 => p*N(x,y,z) = (248832 - 4392)/560 = 436.5 " You didn't reply to this counterexample yet.... Jeroen
From: Hisanobu Shinya on 25 Feb 2007 22:49 > 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. > > 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: bassam king karzeddin on 26 Feb 2007 00:38 > bassam king karzeddin > >> I'm not a mathematician, but maybe I can help: > > > > I'm also not a mathematician and this proof is meant > mainly to non mathematicians and was invented by a non mathematician but great genius few centuries back > > There is no proof, because the first line is wrong > given the > counterexample of Randy Poe. You stated that: > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + > ) + x^p+y^p+z^p > > Randy replied: > > "Are you claiming this is true in general? > > Counterexample: > x=3, y=4, z=5, p=5. > (x+y+z)^p = 248832 > (x^p + y^p + z^p) = 4392 > (x+y)(x+z)(y+z) = 560 > > => p*N(x,y,z) = (248832 - 4392)/560 = 436.5 > " > > You didn't reply to this counterexample yet.... > > Jeroen You may ask Randy himself or check replies or calculations Regards B.Karzeddin
From: Randy Poe on 26 Feb 2007 10:51 On Feb 23, 2:13 pm, Dan Cass <d...(a)sjfc.edu> wrote: > > On Feb 21, 9:51 am, bassam king karzeddin > > <bas...(a)ahu.edu.jo> wrote: > > > 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) > > > Are you claiming this is true in general? > > > Counterexample: > > x=3, y=4, z=5, p=5. > > (x+y+z)^p = 248832 > > (x^p + y^p + z^p) = 4392 > > (x+y)(x+z)(y+z) = 560 > > Correction: (x+y)(x+z)(y+z) = 504, > and then (248832 - 4392)/504 is 485 = 5*97 > > 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.) OK, we'll take that as valid. Now let's look at the rest of the argument. >> 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) OK. So if a FLT counter-example exists, then there will be x, y >0, z<0 such that this identity holds. >> CASE-1 >> If (p=3) implies N (x, y, z) = 1, Why does that follow? Perhaps there is something about N(x,y,z) we are not being told. I will ask OP for the third time to provide his proof of this identity (which I will now accept as true) so I understand what N(x,y,z) is. It would be nice if he could answer the question I just asked as well. If for the third time he refuses, I'll abandon this thread. - Randy
From: Randy Poe on 26 Feb 2007 11:04
On Feb 21, 9:51 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > 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 > > (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. In that case, yes this is true with x,y>0, z<0 if there exists a FLT counterexample with p=3, > > 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, 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) OK, if xyz is not divisible by 3, then none of x, y or z are divisible by 3. But (x+y+z) could still be divisible by 3. If it is, then (x+y+z)^3 must be divisible by 27 so the left hand side, as you say, is divisible by 9. Why isn't the right hand side divisible by 3? Suppose (x+y+z) = 3k for some integer k. Then (x+y) = 3k-z, and 3 does not divide z, therefore 3 does not divide (x+y). Similarly for (x+z), (y+z). All right, your contradiction argument seems OK to me unless I'm missing something. I wish you'd actually *provided* your argument instead of leaving it for readers to guess. So it follows that xyz must be 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) Well, 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. - Randy |