My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree2Thanks
  • 2 Post By SDK
Reply
 
LinkBack Thread Tools Display Modes
April 29th, 2018, 08:53 AM   #1
Newbie
 
Joined: Apr 2018
From: Earth

Posts: 1
Thanks: 0

Proof of Beal conjecture

If
$$a^x+b^y=c^z$$
for positive integers $a$, $b$, $c$, $x$, $y$ and $z$ such that $x$, $y$, $z>2$ then $a$, $b$ and $c$ have a common prime factor.

Proof
Let $a=ue$, $b=uf$ and $c=ug$ such that $u$, $e$, $f$ and $g$ are positive integers.
\begin{align*}
(ue)^x+(uf)^y&=(ug)^z\\
u^xe^x+u^yf^y&=u^zg^z
\end{align*}
Let $y=x\pm m$ and $z=y\pm n$ implying $z=x\pm m\pm n$ or $z=x\pm m\mp n$ such that $m$ and $n$ are integers.
If $x\neq y\neq z$ then $m$ and $n$ are positive integers and if $z=x\pm m\mp n$, $\pm m\mp n\neq0$, implying $m\neq n$.
If $x=y\neq z$ then $m=0$, if $x\neq y=z$ then $n=0$ and if $x=z\neq y$, then either if $z=x\pm m\pm n$, then $m$, $n=0$ or if $z=x\pm m\mp n$, then $\pm m\mp n=0$, implying $m=n$.
\begin{align*}
u^xe^x+u^{x\pm m}f^{x\pm m}&=u^{x\pm m\pm n}g^{x\pm m\pm n}\tag{1.1}
\end{align*}
or
\begin{align*}
u^xe^x+u^{x\pm m}f^{x\pm m}&=u^{x\pm m\mp n}g^{x\pm m\mp n}\tag{1.2}
\end{align*}
$y=x\pm m$ therefore $x=y\pm m$. Therefore
\begin{align*}
u^{y\pm m}e^{y\pm m}+u^yf^y&=u^{y\pm n}g^{y\pm n}\tag{2}
\end{align*}
$z=y\pm n$ therefore $y=z\pm n$ and $x=z\pm m\pm n$ or $z=x\pm m\mp n$ therefore $x=z\pm m\pm n$ and $x=z\pm m\mp n$. Therefore
\begin{align*}
u^{z\pm m\pm n}e^{z\pm m\pm n}+u^{z\pm n}f^{z\pm n}&=u^zg^z\tag{3.1}
\end{align*}
or
\begin{align*}
u^{z\pm m\mp n}e^{z\pm m\mp n}+u^{z\pm n}f^{z\pm n}&=u^zg^z\tag{3.2}
\end{align*}
If $u>1$ and from equations 1.1 and 1.2,
\begin{equation*}
\begin{aligned}
u^x=u^{x\pm m}=u^{x\pm m\pm n}>1
\end{aligned}
\qquad\text{or}\qquad
\begin{aligned}
u^x=u^{x\pm m}=u^{x\pm m\mp n}>1
\end{aligned}
\end{equation*}
from equation 2,
$$u^{y\pm m}=u^y=u^{y\pm n}>1$$
from equations 3.1 and 3.2,
\begin{equation*}
\begin{aligned}
u^{z\pm m\pm n}=u^{z\pm n}=u^z>1
\end{aligned}
\qquad\text{or}\qquad
\begin{aligned}
u^{z\pm m\mp n}=u^{z\pm n}=u^z>1
\end{aligned}
\end{equation*}
then $a$, $b$ and $c$ have a common prime factor. Therefore, from equations 1.1 and 1.2,
\begin{equation*}
\begin{aligned}
u^x&>1\\
u^x&>u^0\\
x&>0
\end{aligned}
\qquad\text{and}\qquad
\begin{aligned}
u^{x\pm m}&>1\\
u^{x\pm m}&>u^0\\
x\pm m&>0\\
x&>\pm m
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
u^{x\pm m\pm n}&>1\\
u^{x\pm m\pm n}&>u^0\\
x\pm m\pm n&>0\\
x&>\pm m\pm n
\end{aligned}
\qquad\text{or}\qquad
\begin{aligned}
u^{x\pm m\mp n}&>1\\
u^{x\pm m\mp n}&>u^0\\
x\pm m\mp n&>0\\
x&>\pm m\mp n
\end{aligned}
\end{equation*}
To get all positive integers which $x$ is greater than, we eliminate the options $x>0$, $x>-m$ and $x>-m-n$. Therefore $x>m$ and $x>m+n$ or $x>\pm m\mp n$.
From equation 2,
\begin{equation*}
\begin{aligned}
u^y&>1\\
u^y&>u^0\\
y&>0
\end{aligned}
\qquad\text{and}\qquad
\begin{aligned}
u^{y\pm m}&>1\\
u^{y\pm m}&>u^0\\
y\pm m&>0\\
y&>\pm m
\end{aligned}
\qquad\text{and}\qquad
\begin{aligned}
u^{y\pm n}&>1\\
u^{y\pm n}&>u^0\\
y\pm n&>0\\
y&>\pm n
\end{aligned}
\end{equation*}
To get all positive integers which $y$ is greater than, we eliminate the options $y>0$, $y>-m$ and $y>-n$. Therefore $y>m$ and $y>n$.
From equations 3.1 and 3.2,
\begin{equation*}
\begin{aligned}
u^z&>1\\
u^z&>u^0\\
z&>0
\end{aligned}
\qquad\text{and}\qquad
\begin{aligned}
u^{z\pm n}&>1\\
u^{z\pm n}&>u^0\\
z\pm n&>0\\
z&>\pm n
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
u^{z\pm m\pm n}&>1\\
u^{z\pm m\pm n}&>u^0\\
z\pm m\pm n&>0\\
z&>\pm m\pm n
\end{aligned}
\qquad\text{or}\qquad
\begin{aligned}
u^{z\pm m\mp n}&>1\\
u^{z\pm m\mp n}&>u^0\\
z\pm m\mp n&>0\\
z&>\pm m\mp n
\end{aligned}
\end{equation*}
To get all positive integers which $z$ is greater than, we eliminate the options $z>0$, $z>-n$ and $z>-m-n$. Therefore $z>n$ and $z>m+n$ or $z>\pm m\mp n$.
Recall if $x\neq y\neq z$ then $m$ and $n$ are positive integers and if $z=x\pm m\mp n$, $\pm m\mp n\neq0$, implying $m\neq n$.
Recall $x>m$ and $x>m+n$ or $x>\pm m\mp n$. Therefore in this case the least positive integers that $x$ is greater than are 1 if $x>m$ and $x>\pm m\mp n$ such that $m=1$ and $n=2$ or $x>\pm m\mp n$ such that $m=2$ and $n=1$ and 2 if $x>m$ such that $m=2$ or if $x>m+n$ such that $m$, $n=1$.
$y>m$ and $y>n$. Therefore the least positive integer that $y$ is greater than is 1 if $m$, $n=1$ and 2 if $m\neq n$ and either $n=2$ or $m=2$.
$z>n$ and $z>m+n$ or $z>\pm m\mp n$. The least positive integers that $z$ is greater than are 1 if $z>n$ and $z>\pm m\mp n$ such that $m=2$ and $n=1$ or if $z>\pm m\mp n$ such that $m=1$ and $n=2$ and 2 if $z>n$ if $m=2$ or if $z>m+n$ such that $m$, $n=1$.
If $x=y\neq z$ then $m=0$. Hence $x>n$, $y>n$ and $z>n$. In this case the least positive integer that $x$, $y$ and $z$ is greater than is 1 such that $n=1$.
If $x\neq y=z$ then $n=0$. Therefore $x>m$, $y>m$ and $z>m$. The least positive integer that $x$, $y$ and $z$ is greater than is 1 such that $m=1$.
If $x=z\neq y$, then either if $z=x\pm m\pm n$, then $m$, $n=0$ or if $z=x\pm m\mp n$, then $\pm m\mp n=0$, implying $m=n$. We eliminate the options $m$, $n=0$, $x>\pm m\mp n$ and $z>\pm m\mp n$ to get the least positive integer that $x$, $y$ and $z$ are greater than. $x>m$ and $x>m+n$ therefore $x>1$ and $x>2$ respectively if $m$, $n=1$. $y>m$ and $y>n$ therefore $y>1$ if $m$, $n=1$. $z>n$ and $z>m+n$ therefore $z>1$ and $z>2$ respectively if $m$, $n=1$.
Therefore, from all the cases above, if $x$, $y$, $z>2$ then $a$, $b$ and $c$ share a common prime factor.
iMan is offline  
 
April 29th, 2018, 10:55 PM   #2
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 502
Thanks: 280

Math Focus: Dynamical systems, analytic function theory, numerics
Damn straightforward algebraic manipulation was the key the whole time. Why didn't anyone else ever think of that?
Thanks from Maschke and topsquark
SDK is online now  
April 30th, 2018, 12:25 AM   #3
Senior Member
 
Joined: Oct 2009

Posts: 608
Thanks: 186

Quote:
Originally Posted by iMan View Post
If
$$a^x+b^y=c^z$$
for positive integers $a$, $b$, $c$, $x$, $y$ and $z$ such that $x$, $y$, $z>2$ then $a$, $b$ and $c$ have a common prime factor.

Proof
Let $a=ue$, $b=uf$ and $c=ug$ such that $u$, $e$, $f$ and $g$ are positive integers
I can choose any such integers? So can I just choose $u=1$, $e=a$, $f=b$ and $g=c$? Note that the conjecture says exactly that we must prove that $u>1$ can occur, right?

Quote:
\begin{align*}
(ue)^x+(uf)^y&=(ug)^z\\
u^xe^x+u^yf^y&=u^zg^z
\end{align*}
Let $y=x\pm m$ and $z=y\pm n$ implying $z=x\pm m\pm n$ or $z=x\pm m\mp n$ such that $m$ and $n$ are integers.
If $x\neq y\neq z$ then $m$ and $n$ are positive integers and if $z=x\pm m\mp n$, $\pm m\mp n\neq0$, implying $m\neq n$.
If $x=y\neq z$ then $m=0$, if $x\neq y=z$ then $n=0$ and if $x=z\neq y$, then either if $z=x\pm m\pm n$, then $m$, $n=0$ or if $z=x\pm m\mp n$, then $\pm m\mp n=0$, implying $m=n$.
\begin{align*}
u^xe^x+u^{x\pm m}f^{x\pm m}&=u^{x\pm m\pm n}g^{x\pm m\pm n}\tag{1.1}
\end{align*}
or
\begin{align*}
u^xe^x+u^{x\pm m}f^{x\pm m}&=u^{x\pm m\mp n}g^{x\pm m\mp n}\tag{1.2}
\end{align*}
$y=x\pm m$ therefore $x=y\pm m$. Therefore
\begin{align*}
u^{y\pm m}e^{y\pm m}+u^yf^y&=u^{y\pm n}g^{y\pm n}\tag{2}
\end{align*}
$z=y\pm n$ therefore $y=z\pm n$ and $x=z\pm m\pm n$ or $z=x\pm m\mp n$ therefore $x=z\pm m\pm n$ and $x=z\pm m\mp n$. Therefore
\begin{align*}
u^{z\pm m\pm n}e^{z\pm m\pm n}+u^{z\pm n}f^{z\pm n}&=u^zg^z\tag{3.1}
\end{align*}
or
\begin{align*}
u^{z\pm m\mp n}e^{z\pm m\mp n}+u^{z\pm n}f^{z\pm n}&=u^zg^z\tag{3.2}
\end{align*}
If $u>1$ and from equations 1.1 and 1.2,
Hmm, "If $u>1$"? If $u>1$, then the entire thing is obvious, no? why waste more time on this. I'm more interested in the case that $u=1$. Did you cover that case, I don't see it?
Micrm@ss is online now  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
beal, conjecture, proof



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Simple proof for Beal's Conjecture mohakanari Number Theory 6 November 20th, 2015 07:45 AM
The proof of the beal's conjecture lwgula Number Theory 18 November 4th, 2014 11:26 AM
A simple proof to Beal's Conjecture DrRaj Number Theory 7 September 4th, 2014 10:00 AM
Proof Beal's conjecture HuyThang1981 Number Theory 1 January 7th, 2014 09:40 AM
Proof of bealís conjecture akhil verma Number Theory 29 July 21st, 2013 05:51 AM





Copyright © 2018 My Math Forum. All rights reserved.