Elegant proof of Fermats Last theorem?

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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 lets 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 lets multiply the original Fermats equation by factor k, so ------> a^n*k + b^n*k = c^n*k, lets 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 lets now put that into 2a^nb^n = c^2n - b^2n - a^2n -----------> lets divide now the whole eqation 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.
 

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D H
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Dividing a2n by an yields an, not a2.

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