July 28th, 2014, 10:58 AM  #1 
Senior Member Joined: Sep 2010 Posts: 218 Thanks: 20  Fermat's Last again
Since the proof of FLT posted by MrAwojobi is based on algebraic approach to the problem I think itâ€™s reasonable to outline main steps of another way to prove it by means of algebra. The following has been deduced. 1. To satisfy equation $\displaystyle a^n+b^n=c^n$ it is required $\displaystyle a=uwv +v^n; b=uwv+w^n;c=uwv+v^n+w^n$ In the case of $\displaystyle n$ dividing $\displaystyle b$ these expressions are modified. 2. The $\displaystyle a^n+b^n$ is a product of $\displaystyle a+b=(u_p)^n$ and $\displaystyle a^{n1}a^{n2}b+â€¦+b^{n1}=(c_p)^n.$ Then $\displaystyle c=(u_p)(c_p); u=(u_p)(u_s)$. 3. $\displaystyle a^n+b^n =2(uwv)^n+n(uwv)^{n1}(v^n+w^n)+â€¦+nuwv[v^{n(n1)}+w^{n(n1)}]+v^{n*n}+w^{n*n}$ is a sum of polynomials divided by c $\displaystyle (uvw)^n+n(uwv)^{n1}v^n+â€¦+n(uwv)v^{n(n1)}$; (i) $\displaystyle (uwv)^n+n(uwv)^{n1}w^n+â€¦+n(uwv)w^{n(n1)}$; (ii) $\displaystyle v^{n*n}+w^{n*n}$. (iii) 4. The long division of (iii) by either (i) or (ii) results in remainder not divisible by $\displaystyle u$ i.e by factor $\displaystyle u_p$ of $\displaystyle c$. This contradicts previous conclusion and proves the FLT. 

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