robert80
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Elegant proof of Fermats Last theorem?
Hello to all. I have found an elegant would be solution of Fermats člast theorem and I would like to kindly ask you where is the mistake, since I am not skilled in Math...
Proof: Let us suppose that a,b,c are coprimes, so if we construct the from a,b,c the smallest triangle for solution of the Fermats Last Theorem. so let's suppose that the sollution exist, a^n + b^n = c^n so if we sqare the equation it will hold true that (a^n + b^n)^2 = (c^n)^2 so -----> a^2n + b^2n + 2a^nb^n = c^2n ----------> 2a^nb^n = c^2n - b^2n - a^2n, from the number theory we know that it follows that c^2n - b^2n is devidable by a^n, so c^2n - b^2n = a^n*k where k is the element of natural numbers. so let's multiply the original Fermats equation by factor k, so ------> a^n*k + b^n*k = c^n*k, let's now substitute the term a^n*k by c^2n - b^2n so:-------> c^2n - b^2n + b^n*k = c^n*k -------->b^n*(k - b^n) = c^n*(k - c^n), since b and c are coprimes b^n = (k - c^n)*m and c^n = (k - b^n)*m where m again is the element of Natural numbers. so--------> c^n + b^n*m = b^n + c^n*m, we see that m is 1, so -----> k = c^n + b^n let's now put that into 2a^nb^n = c^2n - b^2n - a^2n -----------> let's divide now the whole equation by a^n ----------------> 2b^n = c^n + b^n - a^2 -------------> b^n = c^n - a^2 and since a^n + b^n = c^n ---------> b^n = a^n + b^n - a^2 ------------> a^n = a^2 -------------> n = 2 if the solution of the fermats last theorem exists.
I believe n is the elemnt of odd natural numbers, and not natural, but for the proof itself this is not vital I suppose...
Thank you guys in advance.
Hello to all. I have found an elegant would be solution of Fermats člast theorem and I would like to kindly ask you where is the mistake, since I am not skilled in Math...
Proof: Let us suppose that a,b,c are coprimes, so if we construct the from a,b,c the smallest triangle for solution of the Fermats Last Theorem. so let's suppose that the sollution exist, a^n + b^n = c^n so if we sqare the equation it will hold true that (a^n + b^n)^2 = (c^n)^2 so -----> a^2n + b^2n + 2a^nb^n = c^2n ----------> 2a^nb^n = c^2n - b^2n - a^2n, from the number theory we know that it follows that c^2n - b^2n is devidable by a^n, so c^2n - b^2n = a^n*k where k is the element of natural numbers. so let's multiply the original Fermats equation by factor k, so ------> a^n*k + b^n*k = c^n*k, let's now substitute the term a^n*k by c^2n - b^2n so:-------> c^2n - b^2n + b^n*k = c^n*k -------->b^n*(k - b^n) = c^n*(k - c^n), since b and c are coprimes b^n = (k - c^n)*m and c^n = (k - b^n)*m where m again is the element of Natural numbers. so--------> c^n + b^n*m = b^n + c^n*m, we see that m is 1, so -----> k = c^n + b^n let's now put that into 2a^nb^n = c^2n - b^2n - a^2n -----------> let's divide now the whole equation by a^n ----------------> 2b^n = c^n + b^n - a^2 -------------> b^n = c^n - a^2 and since a^n + b^n = c^n ---------> b^n = a^n + b^n - a^2 ------------> a^n = a^2 -------------> n = 2 if the solution of the fermats last theorem exists.
I believe n is the elemnt of odd natural numbers, and not natural, but for the proof itself this is not vital I suppose...
Thank you guys in advance.