Fermat's Last theorem short proof
in [Math]
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From: quasi on 20 Apr 2007 09:04 On Fri, 20 Apr 2007 07:49:01 EDT, essam abd allah <fmgret12(a)yahoo.com> wrote: >thanks to all >but ..... >I want to know is there any awards on that proof ? No money awards but certainly you would get fame. quasi
From: Johann Wiesenbauer on 20 Apr 2007 07:30 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? I haven't read all the answers here, but the simplest way to see that this is true indeed is based on the following easy observations as to f(x,y,z) = (x+y+z)^p-x^p-y^p-z^p. 1) The coefficient p!/(i!j!k!) of x^i y^j z^k in the expansion of (x+y+z)^p is certainly divisible by p, if p is a prime and ijk /= 0, showing that p|f(x,y,z) 2. f(x,y,z)= 0, if x=-y or x=-z or y=-z, showing that this expression is divisible by x+y, x+z, y+z. 3. p, x+y,x+z,y+z are irreducible elements in the factorial ring Z[x,y,z], hence f(x,y,z) is also divisible by their product. Cheers, Johann
From: bassam king karzeddin on 20 Apr 2007 15:49 > thanks to all > but ..... > I want to know is there any awards on that proof ? > because my proof is correct and i show it to other > professional in mathematics and did not discovered > any error on it . > can you help me ? > is there any awards on that proof ? > despite of my proof is very easy and depends on truth > that is known to all people . > please help me ......... Assuming your proof is rigorous, and quite short, then it is really a tragedy, let us try to learn something from the story of Wile's proof –and how success in mathematics could be made despite all the efforts of scientists and engineers who invented the Internet and soon biologists will join them to rearrange their genetic order and invent the brain aid and finally –hopefully soon- a new born NOT of their kind- will take the task of the mathematicians for ever •He had to work very secretly for so many years to obtain a very heavy proof that is based on the shoulders of hundreds of giants and needs few thousands to carry and convey •He had a very powerful position and reputation Despite all that, he didn't trust SCI..MATH COMMUNITY, because he knew in advance that –THEY ARE HELPLESS, and no body will read them even after the proof was been announced •Not only that, but although he didn't trust the JOURNALS OPINIONS, because he knew in advance that most of them are intelligent thieves, or stupid when the issue is so important •So, using his power, he arranged a conference with out telling the main aim, and people used to sleep while he was lecturing on the third day before he announced- that only proves FLT, then he sat down for a final stop • But latter, the proof was flawed •After nearly a year the gap in his proof was filled up and their proof was • accepted by the brain masters gangue, which of course all of you accept with out understanding But still the hope is there and lies on the supports of the true mathematicians If your proof is rigorous, understandable, and not so long So, if you want to relax, announce it here for example and see where do you stand And if it is been confirmed, I will tell you a way to force the whole world to accept it, because mathematical proofs is not a matter of democracy. About the awards I don't know and quasi promised a fame Good Luck Bassam Karzeddin
From: bassam king karzeddin on 20 Apr 2007 17:09 > In article > <21287047.1177033447144.JavaMail.jakarta(a)nitrogen.math > forum.org>, > bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: > > > > In article > > > > <8917073.1177024391960.JavaMail.jakarta(a)nitrogen.mathf > > > orum.org>, > > > bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: > > > > > > > > bassam king karzeddin wrote: > > > > > > Dear All > > > > > > > > > > > > As a generalization to one of my posts in > this > > > > > thread > > > > > > > > > > > > > > > > > > Given, two distinct, coprime non zero > integers > > > > > > > > > (x & y), > > > > > > > > > > > > Theorem- (new or old, I don©t care), > precisely > > > I > > > > > don't know > > > > > > > > > > > > If, (n & m) are two positive integers, > where > > > > > > > > > > > > m = gcd ((x+y), n), > > > > > > > > > > > > then this implies the following theorem: > > > > > > > > > > > > Gcd ((x+y), (x^n+y^n)/(x+y)) = Rad (m), > > > > > > > > > > > > Where Rad (m) equals the product of all the > > > prime > > > > > factors of (m), that is to say > > > > > > Rad (m) is square free number that divides > > > > > (x^n+y^n), > > > > > > > > > > Oh? Perhaps you need another condition, since > > > > > > > > > > x = 15 and y = 49 are coprime and if we pick > n = > > > 8 > > > > > then > > > > > > > > > > m = gcd(x+y, n) = gcd(64, 8) = 8 > > > > > > > > > > but 15^8 + 49^8 = (16617746730113)(2) which > isn't > > > > > even > > > > > divisible by x+y. > > > > > > > > > > > > > > > Regards, > > > > > > > > > > Rick > > > > > > > > Yes Rick, > > > > and thank you very much for the note > > > > > > > > In fact, and for the purpose of FLT, you may > assume > > > either (n) is odd > > > > positive integer > > > > > > > > OR (x & y), are both odd-distinct-coprime- > > > integers, > > > > > > If x = 3 and y = 1 then (x^2 + y^2) / (x + y) is > not > > > an integer. > > > > > > -- > > > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for > > > email) > > > > Yes, I always do those silly mistakes, un > intentionally, but any way, they > > actually help to drag others for discussion, and > therefore I will stick to > > the condition where (n) is odd positive integer, > since the issue is FLT, and > > more over the even case is a few lines proof only > > > > I should like to thank you sincerely for the note > > x = 8, y = 1, n = 9. m = gcd( x + y, n ) = 9. > (x^n + y^n) / (x + y) = 14913801. > gcd(x + y, (x^n + y^n) / (x + y)) = 9. > Rad(m) = 3. > > -- > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for > email) yes, in fact one can't do any thing in the presence of a single counter example I have to go back to my old notes and see if I did it the right way from memory but certainly there was something absolutely correct there. I should also apologize for saying - drag for discussion-and-many thanks for noting that counter example My Keen Regards B.Karzeddin
From: bassam king karzeddin on 21 Apr 2007 13:53
> bassam king karzeddin wrote: > > Dear All > > > > As a generalization to one of my posts in this > thread > > > > > > Given, two distinct, coprime non zero integers > > (x & y), > > > > Theorem- (new or old, I don't care), precisely I > don't know > > > > If, (n & m) are two positive integers, where > > > > m = gcd ((x+y), n), > > > > then this implies the following theorem: > > > > Gcd ((x+y), (x^n+y^n)/(x+y)) = Rad (m), > > > > Where Rad (m) equals the product of all the prime > factors of (m), that is to say > > Rad (m) is square free number that divides > (x^n+y^n), > > Oh? Perhaps you need another condition, since > > x = 15 and y = 49 are coprime and if we pick n = 8 > then > > m = gcd(x+y, n) = gcd(64, 8) = 8 > > but 15^8 + 49^8 = (16617746730113)(2) which isn't > even > divisible by x+y. > > > Regards, > > Rick But, still Rad(m) divides (x^n +y^n), Doesn't it? Regards B.Karzeddin |