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If $A^n+B^n=C^n$ is Fermat's last theorem and $A^x+B^y=C^z$ is Beal's conjecture , then what is $A^x+A^y=A^z$ ? Is there any conjecture like this? Just curious to know.

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"Fermont"?? That should be almost-almost an heresy, a blasphemy for anyone trying to be a mathematician. Google it and, please, write it correctly. –  DonAntonio Jun 11 '13 at 6:06
    
Divide by $A^{\min\{x,y,z\}}$ and mod out $A$, then you see that this is only possible for $A= 2$ and $x=y$, $z = x+1$. –  martini Jun 11 '13 at 6:17
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@DonAntonio Calm down and just catechize him by pressing the edit button. May the spirit of Oiler be with you, brother. –  Vÿska Jun 11 '13 at 6:21
    
Nop. I just point out the sins, I don't cleans them. Luckily the mathematical Inquisition is on recess, otherwise... –  DonAntonio Jun 11 '13 at 6:25
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Don't worry, @Rajasankar : I was just goofing around. Now go and sin no more, my son. –  DonAntonio Jun 11 '13 at 6:36

2 Answers 2

up vote 3 down vote accepted

Fermat's last theorem is not $A^n+B^n=C^n$, but rather

There are no positive integers $A,B,C,n$ with $n>2$ such that $A^n+B^n=C^n$.

Beal's conjecture is not $A^x+B^y=C^z$, but rather

If $A,B,C,x,y,z$ are positive integers and $x,y,z>2$ such that $A^x+B^y=C^z$, then $A,B,C$ have a common prime factor.

The treatment of integer solutions of $A^x+A^y=A^z$ with positive integers $A,x,y,z$ is trivial in comparison: If $A=1$, we get $1+1=1$, so no solution. If $A>1$, then clearly $z>y$ and $z>x$ and wlog. $y\ge x$ so that after dividing by $A^x$ we get $1+A^{y-x}=A^{z-x}$. The right hand side is a multiple of $A$ because $z>x$. Hence $A^{y-x}$ must noit be a multiple of $A$, which implies $x=y$. But then we get $2=A^{z-x}$, i.e. $A=2$, $z=x+1(=y+1)$ as only solutions.

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There are the trivial solutions $2^x+2^x=2^{x+1}$ and $0^x+0^y=0^z$. If $x < y$ or $A > 2$, then $A^x+A^y < A^{y+1}$, so no other integer solutions to that equation exist.

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