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Robin P. Clark 2015-06-30 09:09:40 +01:00
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10
papers/fermat/fermat.tex
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@ -125,10 +125,12 @@ Thus only common prime factors in $a$ and $b$ are preserved
as a result of equation~\ref{eqn:primesexpanded1}.
This is simply because in addition
the common prime factors can be extracted, $a+b \equiv \prod bfp(a) + \prod bfp(b)$
re-writing $\prod cbpf(a,b) \big( \prod ubpf(a) + \prod ubpf(b) \big)$:
This means the uncommon prime factors of $\big( \prod ubpf(a) + \prod ubpf(b) \big)$
extracting the common prime factors this becomes $\prod cbpf(a,b) \big( \prod ubpf(a) + \prod ubpf(b) \big)$:
this means the uncommon prime factors of $\big( \prod ubpf(a) + \prod ubpf(b) \big)$
are lost and the $\prod cbpf(a,b)$ preserved.
%
Because of this property of addition of numbers in relation to preserved
prime factors, it can be used to make inferences on the equation $a^n+b^n = c^n$.
%
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This means for $a+b$ and $a^n+b^n$ the only prime factors preserved (i.e. in $c^n$)
@ -218,7 +220,7 @@ 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 $a+b=1$.
This means that where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers.
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.
@ -245,7 +247,7 @@ so that $a p^{tn} + b p^{tn} = p^{tn} $ satisfy the highest prime in $c$.
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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.
Thus where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers.
Thus where $a$ and $b$ are $ > 1$; $a^n + b^n \neq c^n$ for whole numbers.
\subsection{trivial case}