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.
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.
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.
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