Number Theory


A Simple, Direct Proof of Fermat's Last Theorem

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 $r\in\mathbb{R},q\in\mathbb{Q},r>2q^n,n,q,r>0$. 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 $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$ that we subsequently prove null : We show, 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 $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

Comments: 3 Pages.

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Submission history

[v1] 2018-12-08 21:32:29
[v2] 2018-12-11 10:03:58
[v3] 2019-01-09 21:44:36
[v4] 2019-01-14 21:02:12
[v5] 2019-01-19 20:14:58
[v6] 2019-01-22 19:55:05
[v7] 2019-01-29 20:38:53
[v8] 2019-01-31 20:27:12
[v9] 2019-02-13 19:33:18
[vA] 2019-02-18 09:09:13
[vB] 2019-02-23 19:59:18
[vC] 2019-03-02 20:18:33
[vD] 2019-03-07 18:34:05
[vE] 2019-03-14 09:19:34
[vF] 2019-03-24 17:43:58
[vG] 2019-04-15 21:48:03

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