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Robin P. Clark 2015-07-06 16:04:02 +01:00
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12
papers/fermat/fermat.tex
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@ -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}