Find all solutions to the equation $3^x - 3^y = 3$ where $x, y$ are positive integers (or show there are none).
I know there are no solutions to this, so I know I can set up the proof for 2 cases: $x \le y$ and $y\le x$, but not too sure how to go from here.
If we have $3^x-3^y=3$, then dividing by $3$ we get: $$\frac{3^x}{3}-\frac{3^y}{3}=1$$
Which we can simplify to: $$3^{x-1}-3^{y-1}=1$$
To do your proof by cases we can look at $x\ge y$.
If we list powers of three out:
$3^0=1, 3^1=3, 3^2=9 ....$ They all differ by at least 3 never by one so no solutions! You can do the otherwise and argue the same way !