Remainder of $2^y$ and $3^y$

Question

When is $$2^y\mod 3^x = 1$$

where $x,y\geq0$ and $x,y$ are integers. I know the trivial solutions but can anyone please provide non-trivial solutions. Thanks.

Answer 1

Let $x\ge 2$. Then since $2$ is a primitive root of $3^2$, it is a primitive root of $3^x$. It follows that $2$ has order $\varphi(3^x)=2\cdot 3^{x-1}$ modulo $3^x$. (One can prove the order result with less machinery.)

Thus the "trivial" Euler's Theorem solutions are the only ones.

Answer 2

Do not know if you consider this not trivial, but by Euler's theorem, and the fact that $\phi(3^x)=2\cdot3^{x-1}$, you get the set of solutions $(x,k\cdot2\cdot3^{x-1})$ for any $k,x$ positive integers