diff --git a/papers/fermat/fermat.tex b/papers/fermat/fermat.tex
index 71f3728..9e09706 100644
--- a/papers/fermat/fermat.tex
+++ b/papers/fermat/fermat.tex
@@ -18,20 +18,20 @@
 \usepackage{ifthen}
 \usepackage{lastpage}
 
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{corollary}[theorem]{Corollary}
+
 \def\layersep{1.8cm}
 
 \linespread{1.0}
-
+\title{flt primes}
 \begin{document}
                                    % numbers at outer edges
 \pagenumbering{arabic}                        % Arabic page numbers hereafter
-\author{R.Clark$^\star$,  \\     
-         $^\star${\em Energy Technology Control, UK. r.clark@energytechnologycontrol.com} \and $^\dagger${\em University of Brighton, UK} 
-}
-
-%\title{Developing a rigorous bottom-up modular static failure mode modelling methodology}
-\title{fermat}
-%\nodate
+\author{R.P. Clark}
+ 
 \maketitle 
  
 \today 
@@ -50,86 +50,104 @@ states that no three positive integers a, b, and c can satisfy the
 equation $a^n + b^n = c^n$ for any integer value of n greater than two.
 
 
+\section{Breaking positive integers into constituent products of bags of primes}
 
-\section{Breaking these positive integers into constituent primes}
-
-Any positive integer can be represented as a collection (or bag) of prime numbers multiple together.
+Any positive integer can be represented as a collection (or bag) of prime numbers multiplied together.
 A function $bpf()$ or `bag of prime factors' is defined to represent this.
 \begin{equation}
 	\prod{bpf(a)}^n + \prod{bpf(b)}^n = c^n
 \end{equation}
 
-The function $bpf()$ will always contain 1.
+%The function $bpf()$ will always contain 1.
 
-The numbers $a$ and $b$ may have common and will have uncommon prime factors; these can be collected into
-three bags, those only in a $ubpf(a)$, those only in b, $ubpf(b)$ and those common, $cbpf(a,b)$.
+The numbers $a$ and $b$ may have common and  uncommon prime factors; these can be collected into
+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.
+Thus the number $32$ is represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$.
 
+Viewing the addition of $a^n +b^n$ as products of bag of  prime factors:
 \begin{equation}
 \label{eqn:primesexpanded0}
-    2 \prod{cbpf(a,b)}^n  \prod{ubpf(a)^n}  +  \prod{cbpf(a,b)}^n \prod{ubpf(b)^n} = c^n
+    \prod{cbpf(a,b)}^n  \prod{ubpf(a)^n}  +  \prod{cbpf(a,b)}^n \prod{ubpf(b)^n} = c^n \; ,
 \end{equation}
 
-this can be re-written as
+this can be re-written as:
 
 \begin{equation}
 \label{eqn:primesexpanded1}
-    2 \prod{cbpf(a,b)}^n   \big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}   \big)  = c^n
-\end{equation}
-
-These are all prime numbers, and although some may be repeated within their bags
-a prime number can only exist in one of the bags.
-
-Also all these prime numbers are greater than two and therefore odd.
-
-So this becomes a product of a list of prime numbers in ${cbpf(a,b)}$.
-The common prime factors between a and b multiplied
-by the uncommon prime numbers.
-Let $\prod{ubpf(a)^n}  +    \prod{ubpf(b)^n = k$.
-
-\begin{equation}
-\label{eqn:primesexpanded2}
-    2 \prod{cbpf(a,b)}^n   k  = c^n
+    \prod{cbpf(a,b)}^n   \big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}   \big)  = c^n \; .
 \end{equation}
 
 
 
+\section{Properties of numbers viewed as products of bags of prime factors}
 
-Adding two prime numbers at any power greater than 1
-and then taking a root means getting an irrational number.
+\subsection{ Primes are guaranteed not preserved in addition }
+
+Only common prime factors are guaranteed preserved as a result of equation~\ref{eqn:primesexpanded1}.
+%
+This means for $a^n+b^n$ the only prime factors guaranteed to be in $c^n$
+are are those common in $a$ and $b$.
+%
+Adding numbers creates a dissolving of the prime factors in the result.
+Addition of primes causes the highest prime factors to become
+lost, but increases the number of smaller prime factors.
+%
+Consider $43 +21 = 64$. These primes add up to a result with 
+a bag of six twos i.e. $bpf(64) = \{2,2,2,2,2,2\}$ or more conventionally $64=2^6$.
+%
+If a prime is added to another prime number the result
+cannot be a prime number, simply because all prime numbers above two are odd; 
+the result of the addition must even and therefore have at least a prime factor of two.
 
 
-
-% 
-% \begin{equation}
-% \label{eqn:primesexpanded}
-%      \prod{cbpf(a,b)}^n  \big( \prod{ubpf(a)^n}  +   \prod{ubpf(b)^n} \big) = c^n
-% \end{equation}
-% 
-% \begin{equation}
-% 	a^n + b^n = \prod{bpf(c)^n}
-% \end{equation}
-% 
-% 
-% %assuming its true $c^n$ must be $ 2 \prod{cpf(a,b)}^n \prod{upf(a)^n} \prod{upf(b)^n} $
-% 
-% 
-% % 
-% % \begin{equation}
-% % 	2 \prod{cpf(a,b)}^n  =  \frac{\prod{bpf(c)^n}}{\prod{upf(a)^n} \prod{upf(b)^n}}
-% % \end{equation}
-
-\section{conditions for having a integer root}
+\subsection{conditions for having a integer root}
 
 To have an integer root $n$ all prime numbers that comprise the number to be rooted must be at least
 to the power of $n$.
 Consider the square root of 144.
 This can be written as
-$12 \times 12$ or breaking it down into prime numbers
-$2  \times 2  \times 3 \times 2  \times \times 2 \times 3$ or $ 2^4 \times 3^2 $.
+$12 \times 12$ or representing it as a bag of  prime numbers
+$\{2,2,2,2,3,3\}$ or conventionally as $ 2^4 \times 3^2 $.
 Taking the square root means halving the powers $ \sqrt{2^4 \times 3^2} = 2^2 \times 3$.
 
-To get an nth root you need all the prime numbers that comprise
-that number to be at the power of n or greater.
+To get a whole number $n^{th}$ root   all the prime numbers that comprise
+that number must be at the power of n or greater.
+
+% So this becomes a product of a list of prime numbers in ${cbpf(a,b)}$.
+% The common prime factors between a and b multiplied
+% by the uncommon prime numbers.
+% Let $\prod{ubpf(a)^n}  +    \prod{ubpf(b)^n} = k$.
+% 
+% \begin{equation}
+% \label{eqn:primesexpanded2}
+%     \prod{cbpf(a,b)}^n   k  = c^n
+% \end{equation}
+
+% Adding two prime numbers at any power greater than 1
+% and then taking a root means getting an irrational number.
+
+\subsection{}
+
+\begin{equation}
+\label{eqn:primesexpanded21}
+     \prod{cbpf(a,b)}^n  \big( \prod{ubpf(a)^n}  +   \prod{ubpf(b)^n} \big) = c^n
+\end{equation}
+
+\begin{equation}
+	a^n + b^n = \prod{bpf(c)^n}
+\end{equation}
+
+
+%assuming its true $c^n$ must be $ 2 \prod{cpf(a,b)}^n \prod{upf(a)^n} \prod{upf(b)^n} $
+
+
+% 
+% \begin{equation}
+% 	2 \prod{cpf(a,b)}^n  =  \frac{\prod{bpf(c)^n}}{\prod{upf(a)^n} \prod{upf(b)^n}}
+% \end{equation}
+
 
 That means that ${ubpf(a)^n}$ and ${ubpf(b)^n}$ multiply the prime numbers in $cbpf(a,b)$
 $n$ times each.
@@ -137,10 +155,65 @@ These are a component of $c^n$.
 
 
 \begin{equation}
-\label{eqn:primesexpanded1}
-    2 \prod{cbpf(a,b)}^n      = \frac{c^n}{\big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}  \big) }
+\label{eqn:primesexpanded22}
+    \prod{cbpf(a,b)}^n      = \frac{c^n}{\big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}  \big) }
 \end{equation}
 
+
+\section{contradiction}
+%
+For $a^n + b^n = c^n$ to be true for whole numbers $ > 2$, the highest prime factors on both sides of the equation must be equal.
+%
+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}
+
+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.
+
+
+%% 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}
+%% case where highest prime factor in c^n may be duplicated.
+
+The highest prime factor in the bag may be duplicated.
+Taking the value $c$ as the product of a bag of prime numbers, it must have
+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 $p^{t+n}$.
+%
+Therefore, for that highest prime in $c$,   $a^n + b^n$ must add up to $p^{(t+n)}$, 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
+a power $p^t$ by addition is to add proportions that add up to one $p^t$.
+%
+This means both a and b must contain this prime factor {\em in some proportion}
+so that $a p^{t+n} + b p^{t+n} = p^{t+n} $ satisfy the highest prime in $c$.
+
+In order for this to be true $a$ and $b$ must both fractions of a whole number.
+
+\subsection{trivial case}
+
+
+Take the trivial case where $c$ has the prime number 7:
+%
+$$ a^n + b^n = 7^n = 49 $$
+%
+In order to get the prime factor 7 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 result.
+
+
+
+
 % 
 % \begin{equation}
 % \label{eqn:primesexpanded1}
@@ -181,15 +254,15 @@ These are a component of $c^n$.
 % \end{equation}
 % 
 
-
-
-Which means that a product of $c$ is a root of 2, it is therefore irrational
-and not a whole number.
-
-
-If $c$ is even 2 can be divided from each side until only
-both $c$ and $ \prod{cbpf(a,b)}  \prod{ubpf(a)}   \prod{ubpf(b)} $
-are odd. The $\sqrt[n]{2}$ term remains. The result $c$ is therefore irrational.
+% 
+% 
+% Which means that a product of $c$ is a root of 2, it is therefore irrational
+% and not a whole number.
+% 
+% 
+% If $c$ is even 2 can be divided from each side until only
+% both $c$ and $ \prod{cbpf(a,b)}  \prod{ubpf(a)}   \prod{ubpf(b)} $
+% are odd. The $\sqrt[n]{2}$ term remains. The result $c$ is therefore irrational.
 
 %Adding $a^n$ and $b^n$ where a and b are different means adding primes to th power of N
 %which means they have no integer nth root.
diff --git a/papers/fermat/to_primes.c b/papers/fermat/to_primes.c
new file mode 100644
index 0000000..7eb2967
--- /dev/null
+++ b/papers/fermat/to_primes.c
@@ -0,0 +1,25 @@
+#include <stdio.h>
+
+int main()
+{
+
+    int number, div;
+    printf("Enter a number to know its prime factor: ");
+    scanf("%d", &number);
+
+    printf("\nThe prime factors of %d are: \n\n", number);
+
+    div = 2;
+
+    while (number != 0) {
+        if (number % div != 0)
+            div = div + 1;
+        else {
+            number = number / div;
+            printf("%d ", div);
+            if (number == 1)
+                break;
+        }
+    }
+    return 0;
+}