From ec7d0edb71b5a07a5ed168b04d47820107b633ff Mon Sep 17 00:00:00 2001
From: "Robin P. Clark" <robin48gx@gmail.com>
Date: Thu, 16 Jul 2015 14:23:23 +0100
Subject: [PATCH] ok ok found a flaw in it....

---
 papers/fermat/fermat.tex | 45 ++++++++++++++++++++++++++++++++++++++--
 1 file changed, 43 insertions(+), 2 deletions(-)

diff --git a/papers/fermat/fermat.tex b/papers/fermat/fermat.tex
index a3e6f0d..a62eb06 100644
--- a/papers/fermat/fermat.tex
+++ b/papers/fermat/fermat.tex
@@ -89,10 +89,14 @@ this can be re-written as:
 
 \begin{equation}
 \label{eqn:primesexpanded1}
-    \prod{cbpf(a,b)}^n   \big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}   \big)  = c^n \; .
+    \prod{cbpf(a,b)}^n   \big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}   \big)  = c^n \; ,
 \end{equation}
 
-
+That is to say the inner bracket on the left hand side is an addition of co-prime numbers,
+\begin{equation}
+\label{eqn:primesexpanded2}
+    \prod{ubpf(a)^n}  \perp   \prod{ubpf(b)^n}    \; .
+\end{equation}
 
 \section{Properties of numbers viewed as products of bags of prime factors}
 
@@ -371,6 +375,43 @@ that is larger than any found in $a$ or $b$.
 If this occurs, the new larger prime will not be present in $c^n$.
 Thus for whole numbers, where $\prod bpf(c^n)$ contains multiple primes  $ a^n + b^n \neq c^n  \; where \; n < 2$.
 
+% \subsection{Another way to look at it}
+% 
+% Take the highest prime factor in the result $c^n$.
+% Try to reproduce it using integers with $a^n$ and $b^n$.
+% 
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% IS it true that if there are any common factors a^n +b^n = c^n cannot exist
+% even for n >= 2 ????
+%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\subsection{Hole in the theory..... pythagorean squares.... and bollocks}
+
+Pythagorean squares all contain uncommon prime factors in their addition and create
+a large prime to the power of 2 as a result.
+Fermat's last theorem is for powers, the $n$ factor, of greater than 2.
+Consider where all the cubed prime factors in $a$ and $b$ are uncommon (the is $a$ and $b$ are co-prime).
+The result of the addition will create a number that is the product of prime
+numbers; why could some combination not produce a number with $n$ times the power of
+all its primes.
+Consider that although the numbers $a$ and $b$ are co-prime, there will
+be prime numbers `missing' from both $a$ and $b$. If both numbers
+added are odd, the prime factor 2 has to pop-up.
+If 2 is a common prime factor but 3 is not that will pop-up and so on.
+
+But I am stumped here. If all the numbers in $a$ and $b$ are co-prime, i.e $a \perp b$,  
+$a^n + b^n$ could, magically, produce a result where the result $n$ number of the same prime, 
+or even several prime factors, but all to the power of $n$, i.e. $c^n$ where $c$ 
+is the prime addition of $a^n + b^n$. I just cannot see why not, so all this has proved nothing. Arrrrghhhhh.
+Probably good for the soul to look at prime numbers a-lot.....
+
+Maybe you can take prime triplets say, and try to work back in some general way and show no $a^3 + b^3$ integer solution exists?
+Or a general way of showing $a^n + b^n \; where \; n > 2$ cannot produce a simple $n$ number of all prime factors shown in the result....
+That is the last thing........ that is the only way flt  can have an exception, where $a \perp b$, AND 
+it produces a $c^n$ where c is an integer.....
 
 \section{Further work}