now for the finale, can't catch up ^3 and above
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flt.tex
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\documentclass[a4paper,10pt]{article}
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\usepackage[utf8]{inputenc}
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%opening
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\title{Fermat Primes reasoning}
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\author{R.P.Clark}
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\begin{document}
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\maketitle
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\begin{abstract}
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Fermat reasoning for proving $a^N+b^N = c^N$ does not exist as an integer solution for values of $N \ge 3$.
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\end{abstract}
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\section{Definitions}
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\subsection{Numbers as collections or bags of prime numbers}
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For the reasoning that follows it is helpful to think of
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integers as bags\footnote{Bags are like axiomatic sets except bags can hold duplicate values.} of prime numbers that can be multiplied together, or bag of Prime Numbers (BPN).
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For instance the number, say 12 can be factored as $2\times2\times3$
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or $2^2 \times 3$.
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\paragraph{Additional as a destructor of co-prime factors}
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Consider addition using numbers as bags of prime numbers.
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Consider adding 12 and 21 i.e. $12+21=33$.
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$$ (2\times2\times3 + 3\times7) $$
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Clearly the common factor of three can be moved outside the brackets
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$$ 3 \times (2\times2 + 7) $$
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This means that the co-prime elements within the brackets
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will not be preserved, but destroyed / removed from the resultant BPN.
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The BPN for 33 is $3\times11$, the 2 and 7 prime numbers
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in the brackets have been removed.
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\paragraph{Power Balanced Prime Numbers}
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A number is prime balanaced for a given integer $N$, if all its prime factors exist $N*k$
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times where $k$ is a positive integer. as the PBPN is relevant for a given N value
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PBPN is defined as a boolean function PBPN(N) where it is true or false
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according to its BPN content.
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For instance, the number 25 i.e. $5^2$ is prime balanced for $N=2$: 100 is also a PBPN(2)
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$5^2\times2^2 = 100$.
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The number 200 i.e. $5^2\times2^3$ is not a PBPN(2) because the power of 2 ($2^3$) is not cleanly divisible by 2.
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\subsection{Roots: integer or trancendental results}
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In order for a root to have an integer result it must have a `Prime~Balanced'
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PBPN. If it does not the root will contain a trancedental fractional power.
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Consider 200 in the case above. Its square root is $5^1\times2^{3/2}$.
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Any number greater than 1 with a fractional root will be transendental
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because it cannot resolve back to a BPN.
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\subsection{Relatively Co-Prime}
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If two numbers have at least some co-prime factors they are `Relatively co-prime' (RCP).
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Thus numbers that can share some prime factors and have co-prime as well
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are RCP.
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\section{Back to the Fermat conjecture}
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\subsection{Discussion with $N=2$}
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$$ a^N + b^N = c^N$$
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For the $c^N$ part to have an integer in $c$ the result of the addition must be
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a PBPN.
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Also both $a$ and $b$ must be less than $c$.
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However they must add up to a PBPN.
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This means a and b must be relatively co prime.
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Consider the Pythagoras example for a set square 3,4,5.
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$$4^2 + 3^2 = 5^2$$
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Viewing this as bags of prime numbers we have $2^3 + 3^2 = 5^2$.
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Here the numbers being added are RCP (actually co-prime as well).
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The addition has removed the co-prime factors of $2$ and $3$ and resulted
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in a PBPN(2) number with 5 as its only prime number.
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Now consider multiplying the numbers in the addition
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by a PBPN(2) number. The simplest is $2^2$. So
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$$ 2^2\times3^2 + 2^2\times4^2 = 100$$
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or
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$$ 2^2(3^2 + 4^2) = 5^2\times2^2 $$
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The number 100 is true for PBPN(2) and therfore has an integer square root.
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\subsection{Discussion with $N=3$}
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For the equation below
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$$ a^3 + b^3 = c^3$$
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to be true for some integers a,b and c
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the following conditions must be met.
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\begin{itemize}
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\item a and b must be less than c
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\item a and b must be RCP
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\end{itemize}
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\end{document}
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