Fermat's Last theorem short proof
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From: Hisanobu Shinya on 27 Feb 2007 03:27 > > > 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. > > Hence, your 9-times-something argument does not seem > to work. > > > > > > > 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: bassam king karzeddin on 27 Feb 2007 03:38 > > > > 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 My Regards Bassam Karzeddin AL Hussein bin Talal University JORDAN
From: Hisanobu Shinya on 27 Feb 2007 03:50 > > > > > 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
From: bassam king karzeddin on 27 Feb 2007 07:59 > 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 > 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 08:27
> 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 If you do the steps you will simply arrive to the following equation (x+y)^(p-1) = p*x*y*N(x,y) + (x^p + y^p) / (x+y) Now, assuming (p) is prime factor of (x+y), then accordingly you get (1 Mod p) on the left hand side of the equation and (0 Mod p) on the right hand side of the same above equation because it will be a multiple of (p) A contradiction implies that is not possible with all integer numbers, and hence no counter example exists. Similarly you may apply assuming either (x=o), or (y=0) My Regards Bassam Karzeddin AL Hussein bin Talal University JORDAN |