From d466bb8bdde7eaa6d9050a7795341adaa0bbd2ed Mon Sep 17 00:00:00 2001 From: "Robin P. Clark" Date: Tue, 14 Jul 2015 15:21:57 +0100 Subject: [PATCH] More explanation in the examples after comments from Ian Dixon. --- papers/fermat/fermat.tex | 51 ++++++++++++++++++++++++++++++++++++---- 1 file changed, 46 insertions(+), 5 deletions(-) diff --git a/papers/fermat/fermat.tex b/papers/fermat/fermat.tex index 5fc5c13..a3e6f0d 100644 --- a/papers/fermat/fermat.tex +++ b/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}