diff --git a/papers/fermat/fermat.tex b/papers/fermat/fermat.tex
index 2d1c033..3815ba7 100644
--- a/papers/fermat/fermat.tex
+++ b/papers/fermat/fermat.tex
@@ -77,7 +77,7 @@ The numbers $a$ and $b$ may have common and  uncommon prime factors; these can b
 three 'bags', those only in $a$; $ubpf(a)$, those only in $b$; $ubpf(b)$ and those common to both; $cbpf(a,b)$.
 A `Set' in mathematics is a collection of objects that may have only one of each type of element. 
 A `bag' is similar to a Set, except that it may have duplicates.
-The number $32$ is represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$.
+The number $32$ for instance, can be represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$.
 
 Viewing the addition of $a^n +b^n$ as products of bags of common and uncommon~{\pfs}:
 \begin{equation}
@@ -124,7 +124,7 @@ are the common ones, i.e. cbpf(a,b).
 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, $a+b \equiv \prod bfp(a) + \prod bfp(b)$
+the common prime factors can be extracted, $a+b \equiv \prod bpf(a) + \prod bpf(b)$
 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.
@@ -132,6 +132,66 @@ 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 $a^n+b^n = c^n$.
 %
+\subsubsection{trivial example, single prime factor preserved}
+%
+Consider $bpf(182)=\{2,7,13\}$ and $bpf(2365)=\{5,11,43\}$ these have no common prime factors
+so adding them should result in a number which when factored contains none of the primes in $182$ and $2365$.
+Adding $182+2365=2574$; so taking prime factors $bpf(2574)=\{3,3,283\}$.
+The loss of uncommon prime factors for this case of addition is shown to be true.
+
+%
+Now consider two numbers with one common factor $bpf(49665)=\{3,7,5,11,43\}$ and 
+$bpf(322) = \{2,7,23\}$. These have one common factor, $7$.
+%
+The addition of these numbers is: $ 322 + 49665 \equiv \prod \{2,7,23\} + \prod \{3,7,5,11,43\} $; as the 7 is added twice
+it can be taken outside of the addition and multiplied by what remains of it
+$7(\prod \{2,23\} + \prod \{3,5,11,43\})$. 
+%
+This means the unique properties
+of the uncommon prime factors (those within the bracket) will be destroyed,
+and thus they will not appear in the result
+of the addition. The number 7 will be multiplied by the number in brackets, 
+and thus that prime factor will survive in the result.
+%
+So, $ 322 + 49665 = 49987$: $bpf(49987) = \{7,37,193\}$.
+As expected the common prime factor,7, exists in the result of the addition
+but the uncommon prime factors have disappeared.
+
+\subsubsection{trivial example, multiple prime factor preserved}
+
+Consider a repeated prime factor (i.e. a prime $p$ to the power $t$ $p^t$). The same rules apply.
+%
+For this prime factor to be preserved in the result
+of an addition, it must be in both summed quantities at an equal or greater power (or
+number of duplicates of that prime in the bag).
+%
+Consider two numbers with $11^2$ in one number and $11$ in the other: $bpf(110)=\{11*2*5\}$
+and $bpf(67639)=\{13,11,11,43\}$. The only common prime factor is 11 once.
+%
+%
+%
+
+Adding $67639+110 = 67749$, $bpf(67749) = \{3,11,2053\}$. As expected the prime factor
+11 only appears once in the result, because only one 11 can be taken as a common factor; $11(\prod\{3,2053\}+\prod \{13,11,43\})$
+ addition has been to the lone 11 within the brackets and thus `dissolved' it from the result.
+
+To get $11^2$ preserved as a prime factor is must appear twice, i.e. on both sides of the 
+addition. 
+%
+%
+Taking a new number with two prime factors of 11, say 58685,
+$bpf(58685)=\{5,11,11,97\}$ and adding this 
+to $bpf(67639)=\{13,11,11,43\}$:
+$58685+67739 = 126324 \equiv \prod\{5,11,11,97\}+\prod\{13,11,11,43\} $;
+because $11^2$ can be taken out of the bracket it can be re-written thus:
+$11\times 11(\prod\{5,97\} + \prod \{13,43\}) .$
+%
+%
+This means the result $58685+67739 = 126324$, should contain $11$ twice as a prime factor
+but all the uncommon prime factors should not be 
+present in the result, i.e. $bpf(126324)=\{2,2,3,3,11,11,29\}$.
+%
+%
 %
 This means for $a+b$ and $a^n+b^n$ the only prime factors  preserved (i.e. in $c^n$)
 are those common to $a$ and $b$.