Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 21 Apr 2007 13:59 > 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 don¹t 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) But, Still Rad(m) divides (x^2+y^2), doesn't it? Regards B.Karzeddin
From: bassam king karzeddin on 21 Apr 2007 14:05 Dear All With out finding my old notes yet, I think I meant the following: Given, two distinct, coprime non zero integers (x & y), where (x & y) aren't square numbers, and m = gcd ((x+y), n), where (n) is odd integer 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), Regards ???? ???????? Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: bassam king karzeddin on 21 Apr 2007 14:22 > Conjectures are easy, proofs are hard. > quasi If you make conjectures that lasts for centuries, then you didn't solve the problem If you make conjectures that lasts for few years, then you solved almost half the problem If you make conjectures that lasts for a year or few months, then you have completely solved the problem That is the importance of conjectures as far as I can see Regards B.Karzeddin
From: bassam king karzeddin on 22 Apr 2007 05:06 Dear All I have already given you the short proof scattered here and there in this thread, but can you gather the picture. Some already have got it, HAVEN'T YOU?? Some are still waiting, AREN'T YOU?? But the majority are careless, never mind!! This thread has already found it's way to the outside world I think, and I shall also make it find it's way to other planets. For those who are waiting I shall tell them to keep waiting, because I shall give it to you DROP BY DROP until you get HEALED REGARDS B.Karzeddin
From: essam abd allah on 23 Apr 2007 02:38
as you said " 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 " can you tell me where can i send that proof to record it and to discover the error if it is found . and thanks to all |