Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 19 Apr 2007 15:12 > 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, Regards B.Karzeddin I have already shown you –in this thread- a valid proof- when (n) is odd prime number
From: Gerry Myerson on 19 Apr 2007 19:23 In article <wpSdnWHR17REQbrbnZ2dnUVZ_qWvnZ2d(a)hamilton.edu>, Rick Decker <rdecker(a)hamilton.edu> 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. If all you're trying to do is show that (x^n + y^n) / (x + y) need not be an integer, there are much smaller examples, e.g., (2^2 + 1^2) / (2 + 1). -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: bassam king karzeddin on 19 Apr 2007 15:37 > On Thu, 19 Apr 2007 14:37:09 EDT, essam abd allah > <fmgret12(a)yahoo.com> > wrote: > > >i have really found a simple proof of Fermat's Last > Theorem > >how can i announce this solution ? > > and what is the amount of prizes on this proof > ??? and where can i send that proof ??can you help > p me ?? > > There is no longer any prize -- the problem has > already been solved by > Wiles & Taylor. > > As far as what you should do with a simple proof, I > think you should > post it here in sci.math, but include your real name > in the message, > so if it's correct, you will still get credit. If > it's not correct, > and you should _assume_ it's probably not correct > based on history and > common sense, the responses may help you see where > you went wrong. > > quasi Welcome back Quasi In fact once you made a very intelligent note to the whole issue quasi,(only one line) and I think I have got it in the back of my head, like a pigeon it flayed again, but it is there – in my mind, I will search for it. My Regards B.Karzeddin My Name is REAL
From: quasi on 19 Apr 2007 22:17 On Thu, 19 Apr 2007 19:37:09 EDT, bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: >> On Thu, 19 Apr 2007 14:37:09 EDT, essam abd allah >> <fmgret12(a)yahoo.com> >> wrote: >> >> >i have really found a simple proof of Fermat's Last >> Theorem >> >how can i announce this solution ? >> > and what is the amount of prizes on this proof >> ??? and where can i send that proof ??can you help >> p me ?? >> >> There is no longer any prize -- the problem has >> already been solved by >> Wiles & Taylor. >> >> As far as what you should do with a simple proof, I >> think you should >> post it here in sci.math, but include your real name >> in the message, >> so if it's correct, you will still get credit. If >> it's not correct, >> and you should _assume_ it's probably not correct >> based on history and >> common sense, the responses may help you see where >> you went wrong. >> >> quasi > >Welcome back Quasi > >In fact once you made a very intelligent note to the whole issue quasi,(only one line) > > and I think I have got it in the back of my head, like a pigeon it flayed again, but it is there � in my mind, I will search for it. > >My Regards >B.Karzeddin >My Name is REAL Hi REAL (just kidding). Ok, so now the question is whether your claim of having a short proof of FLT is also REAL. I'm not trying to discourage you, but realistically, the likelihood that your proof is valid is very, very low. As far as I remember, you stated some appealingly simple conjectures which, if true, would give an instant proof of FLT. Moreover, your conjectures appear to be strictly stronger than FLT in the sense that the truth of your conjectures easily proves FLT but it's not obvious that the truth of FLT implies the truth of your conjectures. Also, your conjectures survived several brute force attempts to find a counterexample. However, as far as I remember, there was no proof, and from what I can see, that's still the case. Conjectures are easy, proofs are hard. Admittedly, conjectures are fun to make -- it's one of the things I enjoy doing. Moreover, I feel it's a worthwhile endeavor, even if it turns out that the conjectures are too hard to be resolved. However, if I recall correctly, your earlier claim of a proof of FLT was based on assuming that, since no one seemed to be able to disprove your conjectures, that therefore they were true, and hence you felt entitled to use that "truth" to prove FLT. That logic is clearly flawed -- all you had was some nice conjectures, but no proof. Contrary to what a few posters have repeatedly claimed, there is no conspiracy in sci.math to suppress alternative proofs. That's the beauty of usenet -- anyone can play. People with all different math backgrounds can read and respond to your argument, not just professional mathematicians. Thus, if your proof is correct and if the reasoning is not too muddled, you are sure to get some positive responses. On the other hand, unless the proof is expressed clearly and at a certain minimum level of mathematical rigor, many potential responders won't even bother to look at it. quasi
From: Gerry Myerson on 19 Apr 2007 21:21
In article <8917073.1177024391960.JavaMail.jakarta(a)nitrogen.mathforum.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) |