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

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.

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how do you do >= in Tex? – fosho Sep 29 '12 at 20:24
Please try to avoid double  in titles. \geq yields $\geq$, and \leq yields $\leq$. – yunone Sep 29 '12 at 20:24
Oh ok thanks. Sorry about the title. – fosho Sep 29 '12 at 20:25
Use a \geq for greater than/equal to and \leq for less than/equal to. – Michael Dyrud Sep 29 '12 at 20:25
What are the trivial solutions? $y=0$; $x=1,2|y$; $x=2, 6|y$; and in general $2\cdot 3^{x-1}|y$? – Hagen von Eitzen Sep 29 '12 at 20:29

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.

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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

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 I do not see a question. This seems a bit chatty or such. Seems like a case of ' not a real question '. – mick Oct 11 '12 at 15:54