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\begin{document}
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\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
\maketitle 
 
\today 

\paragraph{Keywords:} fermat; prime;
%\small

\abstract{ % \em 
} % abstract



\section{Introduction}
Fermat's Last Theorem
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 these positive integers into constituent primes}

Any positive integer can be represented as a collection (or bag) of prime numbers multiple 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 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, $cbpf(a,b)$.

\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
\end{equation}

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
\end{equation}




Adding two prime numbers at any power greater than 1
and then taking a root means getting an irrational number.



% 
% \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}

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 $.
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.

That means that ${ubpf(a)^n}$ and ${ubpf(b)^n}$ multiply the prime numbers in $cbpf(a,b)$
$n$ times each.
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) }
\end{equation}

% 
% \begin{equation}
% \label{eqn:primesexpanded1}
%         \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}    = \frac{c^n}{ \prod{cbpf(a,b)}^n }
% \end{equation}
% 
% 
% $c^n$ must contain $ \prod{cbpf(a,b)}^n $
% %Try to find a and b such that a^2 + b^2 = 144;
% 
% 
% \begin{equation}
% \label{eqn:primesexpanded1}
%         \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}    = \frac{c^n}{ \prod{cbpf(a,b)}^n }
% \end{equation}
% 

%It should be even because its multiplied by 2.
% It must have all the common factors of $a$ and $b$ twice but the uncommon factors only once.
% This seems to be an apparent contradiction.
% It means the $2 \prod{cpf(a,b)}^n $ term is  multiplied by at least one other prime number. % and therefore cannot have an nth root.
% A number must consist of n times of all its prime number can give an integer nth root.
% Because a and b are different they must consist of at least one difference in prime numbers.
% 
% Taking equation~\ref{eqn:primesexpanded}
% and re-writing:
% \begin{equation}
% \label{eqn:primesexpanded2}
% 	\sqrt[n]{2}^n \prod{cbpf(a,b)}^n  \prod{ubpf(a)^n}  \prod{ubpf(b)^n} = c^n
% \end{equation}
% 
% 
% Taking the nth root of both sides of equation~\ref{qn:primesexpanded2} gives
% 
% \begin{equation}
% \label{eqn:primesexpanded2}
% 	\sqrt[n]{2}  \prod{cbpf(a,b)} (\prod{ubpf(a)}   \prod{ubpf(b)}) = c 
% \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.

%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.

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\today
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