Fermat's Last theorem short proof
in [Math]
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From: V on 23 Apr 2007 19:02 essam abd allah wrote: > Hello bassam > do you think the best place is " Annals of mathematics " to post my proof ? > and what is the period of Annals of mathematics to response to me ? > and what is the easy way to force the world to accept my proof , can you tell me ? > thanks to you Note that you may need to hurry to be first because I am going to post here a "simple" seventeenth century math proof of FLT (only for odd powers of n), which will hopefully not contain an fallacy (but I'm confident it does not). -- Vik
From: Aatu Koskensilta on 23 Apr 2007 19:08 On 2007-04-23, essam abd allah wrote: > and what is the easy way to force the world to accept my proof , can you tell me ? Threaten to cry if they refuse to accept the proof. If that doesn't work, lock yourself up in your room, play HIM as loud as possible and draw pictures of gothic fairies in your diary. -- Aatu Koskensilta (aatu.koskensilta(a)xortec.fi) "Wovon man nicht sprechen kann, daruber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: V on 23 Apr 2007 19:27 bassam king karzeddin wrote: > 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), This very much looks like parts of my proof :), but I don't need Rad fn. gcd (x+y, (x^n+y^n)/(x+y)) = gcd (x+y, n) if x and y are coprime and n is odd and > 1. x^n+y^n = gcd(x+y,n)^2 * (x+y)/gcd(x+y,n) * ((x^n+y^n)/(x+y))/gcd(x+y,n) x^n+y^n is divisible by (x+y) and gcd(x+y,((x^n+y^n)/(x+y))/gcd(x+y,n))=1 For c^n to be a solution, c needs to be a multiple of (x+y), but (x+y)^n is already larger than x^n+y^n, so there cannot be a solution for c^n. It is quite easy to prove that if x and y are not coprime, there cannot be a solution either, because the existance of a solution for the FLT equation with x and y not coprime implies the existance of a solution with x and y coprime and vice versa. -- Vik
From: quasi on 23 Apr 2007 20:57 On Mon, 23 Apr 2007 23:27:25 GMT, V <V(a)telenet.be> wrote: >gcd (x+y, (x^n+y^n)/(x+y)) = gcd (x+y, n) if x and y are coprime and n >is odd and > 1. > >x^n+y^n = gcd(x+y,n)^2 * (x+y)/gcd(x+y,n) * ((x^n+y^n)/(x+y))/gcd(x+y,n) > >x^n+y^n is divisible by (x+y) and gcd(x+y,((x^n+y^n)/(x+y))/gcd(x+y,n))=1 > >For c^n to be a solution, c needs to be a multiple of (x+y), The error is right here (the above line). None of your above analysis concerning common factors justifies your claim that c is a multiple of x+y. All you know is that c^n is a multiple of x+y. Let's take the easy case where x+y and (x^n+y^n)/(x+y) have no common factor. Letting a=x+y and b=(x^n+y^n)/(x+y), you have c^n=ab where a,b are coprime. That does not imply that c is a multiple of a (unless a is prime). So your proof fails. quasi
From: Gerry Myerson on 23 Apr 2007 21:58
In article <6okq23p8fd8lj9rchb311b9pj6e997jquh(a)4ax.com>, quasi <quasi(a)null.set> wrote: > On Mon, 23 Apr 2007 23:27:25 GMT, V <V(a)telenet.be> wrote: > > >gcd (x+y, (x^n+y^n)/(x+y)) = gcd (x+y, n) if x and y are coprime and n > >is odd and > 1. > > > >x^n+y^n = gcd(x+y,n)^2 * (x+y)/gcd(x+y,n) * ((x^n+y^n)/(x+y))/gcd(x+y,n) > > > >x^n+y^n is divisible by (x+y) and gcd(x+y,((x^n+y^n)/(x+y))/gcd(x+y,n))=1 > > > >For c^n to be a solution, c needs to be a multiple of (x+y), > > The error is right here (the above line). There's already an error above that. x = 1, y = 2, n = 27. gcd( x + y, (x^n + y^n) / ((x + y) gcd( x + y, n )) = 3. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |