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? 

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 $(0,0,1),(1,1,2),(2,2,3),(3,3,4),(4,4,5),(5,5,6),(6,6,7),(7,7,8),(8,8,9)$
