My Math Forum Proof of Beal conjecture

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 April 29th, 2018, 07:53 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.
 April 29th, 2018, 09:55 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 378 Thanks: 205 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
April 29th, 2018, 11:25 PM   #3
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Quote:
 Originally Posted by iMan 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?

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