Solving $2^{x-z} + 2^{y-z} = 1$ Suppose we have an equation $2^{x-z} + 2^{y-z}  = 1$,where $x, y, z$ are integers from $0$  to $9$.  $x, y, z$ can have same values. I guess $x$ and $y$ should have $9$ possible values like $x=y=0$ and $z=1$ and so on . Or should it have more values? 
 A: The only possible case is $x-z = -1 = y-z\implies x = y = z-1$. Thus $z = 1,2,3,4,5,6,7,8,9$, and $x = y = 0,1,2,3,4,5,6,7,8$
A: Without loss of generality let $x \leq y$ ad write $1+2^{y-x}=2^{z-x}$. We can show that $x=y$ is necessary. Assume that opposite false, i.e. $y>x$, then left side is odd integer, therefore right side must be odd integer as well, in other words $z-x=0$. But that would mean $1+2^{y-x}=1$, or $2^{y-x}=0$, impossible. Therefore $x=y$ and so $2=2^{z-x}$, i.e. $z-x=1$. Thus we have found all solutions $x=y=z-1$ (subjected to the set constrains).
Another way to look this (perhaps more intuitive way) is to look at binary representations of $2^x+2^y=2^z$. Since all three numbers $2^x,2^y,2^z$ have just one $1$ in its binary representation, the sum on the left must make one $1$. The only way for this is if $x=y$ again.
A: WLOG, let's assume $x\leq y$ and since $2^{x-z} + 2^{y-z} = 1 \iff 2^x+ 2^y=2^z \implies z \gt x, z\gt y$.
Since, $y \geq x \implies y-x \geq 0 \implies 2^{y-x} \geq 1$ and $2^{y-x} \in \mathbb{Z}$ and
similarily, $z\gt x \implies 2^{z-x}>1$ and $2^{z-x} \in \mathbb{Z}$
Now, $2^x+2^y=2^z \iff 2^{z-x} - 2^{y-x} =1$
Since, the difference of two even numbers can't be odd, so, $2^{y-x}$ has to be $1$ which $\implies y-x=0\implies x=y$ and
thus, $2^{z-x} = 1+1 = 2 \implies z-x = 1 \implies z = x+1$.
Thus solutions for $(x,y,z)$ are $(0,0,1),(1,1,2),(2,2,3),(3,3,4),(4,4,5),(5,5,6),(6,6,7),(7,7,8),(8,8,9)$
A: Another attempt:
$2^x+2^y=2^z;$
$2^z-2^y=2^x;$
Note: $z>y$;
$z=y+a$, $a$, positive integer.
$2^y(2^a-1)=2^x;$
Implies $a=1$(Why?).
Hence $x=y$; and  $z=y+1$.
The possible cases are?
Check: $2^x+2^x=2^{x+1}=2(2^x)$.
