## Elementary Set Theory Used To Prove FLT

**Authors:** Phil A. 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 $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ for which $n$ is any given positive natural number, $r$ is positive real and $q$ is positive rational such that the set of triples $\{((r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},2^\frac{2}{n}q)\}$ is not empty with $(r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},(2^\frac{2}{n}q)\in\mathbb{N}$. We relate this set of triples to $\{z,y,x|z,y,x\in\mathbb{N}\}$ for which the transposed \emph{Fermat equation} $z^n-y^n=x^n$ holds. We demonstrate, for any given value of $n$, that $2^\frac{2}{n}q=x$. Clearly, for $n>2$, the term $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational. Consequently, for values of $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Comments:** 2 Pages.

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### Submission history

[v1] 2018-11-01 11:11:48

[v2] 2018-11-05 17:07:33

[v3] 2018-11-06 14:00:16

[v4] 2018-11-09 17:10:10

[v5] 2018-11-25 17:58:44

[v6] 2018-12-09 22:58:42

[v7] 2019-01-10 11:46:28

[v8] 2019-01-15 22:07:17

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