From 117312cae72e78eb80ae45542e10aa1000ca8953 Mon Sep 17 00:00:00 2001 From: "Robin P. Clark" Date: Mon, 6 Jul 2015 16:04:02 +0100 Subject: [PATCH] comments from Ian Dixon --- papers/fermat/fermat.tex | 12 +++++++----- 1 file changed, 7 insertions(+), 5 deletions(-) diff --git a/papers/fermat/fermat.tex b/papers/fermat/fermat.tex index 3815ba7..96ce5fc 100644 --- a/papers/fermat/fermat.tex +++ b/papers/fermat/fermat.tex @@ -272,21 +272,22 @@ That is to say the highest prime number in the bag $bpf(a^n + b^n)$ must be the same as the highest prime factor in the bag $bpf(c^n)$. -\subsection{Case where the highest prime factor in $pbf(c)$ is a single instance} +\subsection{Case where the highest prime factor in $bpf(c)$ is a single instance} Due to the destruction of non-common prime factors under addition both $a$ and $b$ must contain the highest prime in $c$. If $a$ and $b$ are whole numbers they either create a result with the highest prime more than once, or it is destroyed by addition. -For $a^n + b^n = c^n$, for the highest prime, this means $a+b=1$. +For $a^n + b^n = c^n$, for the highest prime, a and b must contain a proportion +of the highest prime factor in $c$; this means a and b {\em for that prime factor} would have to add up to one. This means that where $a$ and $b$ are $ > 1$; $a^n + b^n \neq c^n$ for whole numbers. This concept can be extended to numbers where there are duplicate highest primes. %% Simple case where only one of highest prime factor in c^n % describe contradiction for simple case: -\subsection{Case where the highest prime factor in $pbf(c)$ is a multiple instance} +\subsection{Case where the highest prime factor in $bpf(c)$ is a multiple instance} %% case where highest prime factor in c^n may be duplicated. The highest prime factor in the bag may be duplicated. @@ -296,7 +297,7 @@ a largest prime in $c$ (to a power $t$ which is one or more), i.e. $p^t$. When $c$ is taken to the power $n$, $c^n$, that means this prime factor becomes $p^{tn}$. % -Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{(tn)}$, for that prime in the result. +Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{tn}$, for that prime in the result. % Because prime numbers are by definition indivisible by other whole numbers, the only way to get a prime number taken to @@ -306,7 +307,8 @@ This means both a and b must contain this prime factor {\em in some proportion} so that $a p^{tn} + b p^{tn} = p^{tn} $ satisfy the highest prime in $c$. % In order for this to be true $a$ and $b$ must both be fractions of a whole number: -again this means $a+b$ must equal 1. +again this means for the highest prime factor, $p^t$, $a+b$ must equal 1. In other words a proportion of the +highest prime factor in $c$ must exist in $a$ and $b$ such that they add up to one. Thus where $a$ and $b$ are $ > 1$; $a^n + b^n \neq c^n$ for whole numbers. \subsection{trivial case}