Fermat's Last theorem short proof
in [Math]
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From: Joseph Parranto on 7 Mar 2007 11:41 Trust me when I say that I really have little to offer in terms of verifying your math. I will take the formulas and convert them into a readable fashion for interest, but from the outset one of your terms immediately causes problems - and that is that there is "an infinite sum" attached and that alone makes it not a proof. It must be self-contained and computable across the entire set of integers, not just the few you have here. So every time you say that it is too much work for you, imagine how little interest there is from anyone else. Beal's Conjecture is on the Internet and you should really take a shot at it - your method may work but I cannot see that this is the correct f(x) Joe
From: Joseph Parranto on 7 Mar 2007 12:25 b^(n-m) / (abs (a))^n < (m^m)*(n-m)^(n-m) / n^n, There is some extraordinary carelessness here. Is this b^(n-m) / (abs (a))^n < [ (m^m)*(n-m)^(n-m) / n^n ] or b^(n-m) / (abs (a))^n < (m^m)*(n-m)^[ (n-m) / n^n ] or b^(n-m) / (abs (a))^n < (m^m)*[ (n-m)^(n-m) / n^n ]
From: bassam king karzeddin on 7 Mar 2007 23:40 Dear All this is a reply to Joseph Parranto and may be others who would like to apply my formula Subject: obtain one real root for the following ODD degree equation, that is rational function of x f (x) = x^n + a*x^m + b = 0 Where (a & b) are ANY two none-zero rational numbers n is ANY odd positive integer m is any positive integer less than n solution: case-1) Let, m is odd positive integer less than n let, (k = n - m) let, r = (b^k / a^n)^(1/m), where, r is the arithmetical mth root of (b^k/a^n) Let, s = (a^n/b^k)^(1/n), where, s is the arithmetical nth root of (a^n/b^k) if, b^k / {abs (a)}^n < (m^m)*(k^k) / (n^n), then x = - ((b/a)^(1/m))*[1 - r/m + (2*n-m+1)*r^2 /(2!*m^2) - (3*n-2*m+1)*(3*n-m+1)*r^3 / (3!*m^3) + (4*n-3*m+1)*(4*n-2*m+1)*(4*n-m+1)*r^4 / (4!*m^4) -(5*n - 4*m+1)*(5n-3m+1)*(5n-2m+1)*(5n-m+1)*r^5/(5!m^5) +...] If, b^k / {abs (a)}^n > (m^m)*(k^k) /(n^n) x = - ( b^(1/n))*[1 - s/n + (2*m-n+1)*s^2 / (2!*n^2) - (3*m-2*n+1)*(3*m-n+1)*s^3 / (3!*n^3) + (4*m - 3*n+1)*(4*m - 2*n+1)*(4*m -n+1)*s^4 / (4!*n^4) - (5m-4n+1)*(5*m-3*n+1)*(5m-2n+1)*(5m-n+1)*s^5/(5!*n^5) +...] Case - 2 Let, m be even positive integer less than n , then, let (x=1/y), in the above equation, and substitute & get equation of the previous solved form for one real root. In fact, this is nothing but a VERY little introduction to obtain all roots to any polynomials in terms of their coefficients, It would be very nice from you to make it readable by others, so they may check it to any degree of accuracy they require, and report back honestly their comments Thanking you Bassam Karzeddin Al Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 8 Mar 2007 00:19 > b^(n-m) / (abs (a))^n < (m^m)*(n-m)^(n-m) / n^n, > > There is some extraordinary carelessness here. Is > this > > b^(n-m) / (abs (a))^n < [ (m^m)*(n-m)^(n-m) / > n^n ] > or > > b^(n-m) / (abs (a))^n < (m^m)*(n-m)^[ (n-m) / n^n > ] > or The following IS correct b^(n-m) / (abs (a))^n < (m^m)*[(n-m)^(n-m)] / n^n, where (abs) means the absolute value > > b^(n-m) / (abs (a))^n < (m^m)*[ (n-m)^(n-m) / n^n > ] Regards Joseph B.Karzeddin
From: bassam king karzeddin on 8 Mar 2007 00:40
Dear Mathematicians This is about angle division or angle multiplication or angle section in a triangle I would like to state the following theorem, hopping this would be INTERESTING, as, I might show you later. If (Alpha and Beta) are two angles in the same triangle and (n) is positive integer,(m) is positive integer, less than n ,such that: n*Alpha = Beta Mod (PI) OR: n*Alpha - m*(PI) = Beta Where, Angle (PI) = 180 degree OR: (PI) = 3.14159265... Then, the sides of (Alpha-Beta) Triangle are of the ratio: 1 :: absolute value of f(x) ::absolute value of g(x) Where, f (x) and g (x) are two polynomial equations of degrees (n) and (n-1) successively in real variable (x). And f (x) is defined as the following: f (x) = Sum {from i=0,to, i=[n/2]} (-1)^i*[(n-i)! *x^(n-2*i) / {(n-2*i)! *i!}] Where, ! denotes the factorial of a number,and, [..] denotes the least integer floor function g (x) is a polynomial of the same kind of f (x), but, with (n-1) degree. My Regards Bassam Karzeddin Al Hussein bin Talal University JORDAN |