More explanation in the examples

after comments from Ian Dixon.

papers/fermat/fermat.tex

@@ -319,16 +319,57 @@ again this means for the highest prime factor, $p^t$, $a+b$ must equal 1. In oth
 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}
+\subsection{trivial case: single prime number in $c^n$}
 
 Take the trivial case where $n=3$ and  $c$ has the prime number 7 as one of its prime~factors:
 %
 $$ a^n + b^n = 7^n = 343 \; . $$
 %
-In order to get the prime factor $7^3$ in the result both a and b must have the prime number 7 in them.
-That is the numbers $a$ and $b$ must both have the number 7 as a common prime factor
-to get seven as a prime factor in the result.
-Any other number will not give a 7 in the bag of prime numbers representation of the result.
+In order to get the prime factor $7^3$ in the result both $a$ and $b$ must have some proportion of $7^3$ in them.
+That is the numbers $a$ and $b$ must both have the number $7^3$ as a common prime factor
+to get $7^3$ as a prime factor in the result.
+Any other number will not give a $7^3$ in the bag of prime numbers representation of the result.
+Thus to make $ a^n + b^n = 343 $ both a and b could contain fractional quantities of  $7^3$
+but not both.
+Thus for whole numbers, where $\prod bpf(c^n)$ contains a single prime  $ a^n + b^n \neq c^n  \; where \; n < 2$.
+
+\subsection{trivial case: multiple prime numbers in the bags}
+
+Take the trivial case where $n=3$ and  $c$ has the prime numbers 13 and 11 as its prime~factors:
+%
+say $ c = \prod \{ 13,13,13,11 \} = 24167$ cubing this gives $ \prod \{13,13,13,13,13,13,13,13,13,11,11,11\} $ or $ \prod \{13^9,11^3\} $ .
+A strategy of trying to preserve the prime factors under addition can now be attempted.
+To preserve the primes both $13^3$ and $11$ must be present in both a and b.
+So trying $a = \{13^3,11\}$ and  $b = \{13^3,11\}$ taking cubes gives $a^3 + b^3 = \prod \{13^9,11^3,2\}$
+Here both primes $13^3$ and $11$ have been preserved in the addition but there is an extra factor in the result, i.e. the $2$.
+Adding any other prime factors to either $a$ or $b$ makes the result too
+large. Adding the minimum quantity to both in order to preserve the prime factors
+gives a result with the prime factor $2$ in it as well.
+%
+% 
+\paragraph{Looking at just the highest prime factor}
+For numbers to be equal their highest prime factors must have the same index (or power).
+To get the result $c^n$ from the addition $13^3$ must be present in both $a$ and $b$, 
+if it is present singularly in $a$ and $b$, it will 
+be present twice in the result (i.e. adding the prime $2$) to the result product.
+\paragraph{thinking about preserving $13^3$ in the result $c^n$}
+So to preserve $13^3$ in the result; consider $a = \prod \{ 13,13,13\}$ and $b = \{ 13,13,13\}$.
+Adding them cubed; $a^3+b^3 = \prod \{ 13^9\} + \prod \{ 13^9\}$ which can be re-written as
+ $ \prod \{13^9\} (1 + 1) \}$ which gives $\prod \{ 13^9,2 \}$
+; the extra prime factor of 2 means that while $13^3$ was preserved a new prime factor popped up in the result.
+%; as $c$ is to the power of n
+%it should be $2^3$.
+In general this means $a$ and $b$ being whole numbers
+cannot  make the equation $a^n+b^n=c^n$ true.
+%
+Or in  other words it comes back to the addition $ a^n + b^n = c^n $ preserving the common prime factors in the result $c^n$,
+but not the uncommon factors. 
+%
+\paragraph{Case where $a^n$ and $b^n$ may have a large number of uncommon factors}
+A prime number may be produced by the addition
+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$.
 
 
 \section{Further work}