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July 28th, 2014, 10:58 AM   #1
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Fermat's Last again

Since the proof of FLT posted by MrAwojobi is based on algebraic approach to the problem I think it’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
$\displaystyle a^{n-1}-a^{n-2}b+…+b^{n-1}=(c_p)^n.$
Then $\displaystyle c=(u_p)(c_p); u=(u_p)(u_s)$.
3. $\displaystyle a^n+b^n =2(uwv)^n+n(uwv)^{n-1}(v^n+w^n)+…+nuwv[v^{n(n-1)}+w^{n(n-1)}]+v^{n*n}+w^{n*n}$
is a sum of polynomials divided by c
$\displaystyle (uvw)^n+n(uwv)^{n-1}v^n+…+n(uwv)v^{n(n-1)}$; (i)

$\displaystyle (uwv)^n+n(uwv)^{n-1}w^n+…+n(uwv)w^{n(n-1)}$; (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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