**Authors:** Phil Aaron Bloom

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null set $\{(s,t,u)|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null set $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

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[v1] 2018-12-08 21:32:29

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