Find n such that $2^n equiv xmod 3^k$

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It looks like for every $k ge 1$ and x is not a multiple of 3, $2^n equiv xmod 3^k$ as a unique solution (modulo $2 times 3^k-1$).



How to prove it?
How to find such n given k and x?



Thanks.




modular-arithmetic




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    That is called $2$ being a primitive root $mod 3^k$ en.wikipedia.org/wiki/Primitive_root_modulo_n
    – fleablood
    Aug 3 at 11:52














up vote
2
down vote

favorite
1












It looks like for every $k ge 1$ and x is not a multiple of 3, $2^n equiv xmod 3^k$ as a unique solution (modulo $2 times 3^k-1$).



How to prove it?
How to find such n given k and x?



Thanks.




modular-arithmetic




share|cite|improve this question

















  • 3




    That is called $2$ being a primitive root $mod 3^k$ en.wikipedia.org/wiki/Primitive_root_modulo_n
    – fleablood
    Aug 3 at 11:52












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





It looks like for every $k ge 1$ and x is not a multiple of 3, $2^n equiv xmod 3^k$ as a unique solution (modulo $2 times 3^k-1$).



How to prove it?
How to find such n given k and x?



Thanks.




modular-arithmetic




share|cite|improve this question













It looks like for every $k ge 1$ and x is not a multiple of 3, $2^n equiv xmod 3^k$ as a unique solution (modulo $2 times 3^k-1$).



How to prove it?
How to find such n given k and x?



Thanks.





modular-arithmetic





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edited Aug 3 at 11:24









pointguard0

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asked Aug 3 at 11:11









Damien

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  • 3




    That is called $2$ being a primitive root $mod 3^k$ en.wikipedia.org/wiki/Primitive_root_modulo_n
    – fleablood
    Aug 3 at 11:52












  • 3




    That is called $2$ being a primitive root $mod 3^k$ en.wikipedia.org/wiki/Primitive_root_modulo_n
    – fleablood
    Aug 3 at 11:52







3




3




That is called $2$ being a primitive root $mod 3^k$ en.wikipedia.org/wiki/Primitive_root_modulo_n
– fleablood
Aug 3 at 11:52




That is called $2$ being a primitive root $mod 3^k$ en.wikipedia.org/wiki/Primitive_root_modulo_n
– fleablood
Aug 3 at 11:52










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You need to show that $2^2*3^k-1ne1pmod3^k+1$. That follows by induction by cubing $1+3^k-1+m*3^kpmod3^k+1$.

To find $n$, write $x$ in base $3$. First, $n=0$ or $1$ depending on $xpmod3$. Then add $0,2$ or $4$ to $n$ so that $2^n=xpmod 9$; add $0,6$ or $12$ to get it correct mod $27$, and so on.






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    You need to show that $2^2*3^k-1ne1pmod3^k+1$. That follows by induction by cubing $1+3^k-1+m*3^kpmod3^k+1$.

    To find $n$, write $x$ in base $3$. First, $n=0$ or $1$ depending on $xpmod3$. Then add $0,2$ or $4$ to $n$ so that $2^n=xpmod 9$; add $0,6$ or $12$ to get it correct mod $27$, and so on.






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      up vote
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      accepted










      You need to show that $2^2*3^k-1ne1pmod3^k+1$. That follows by induction by cubing $1+3^k-1+m*3^kpmod3^k+1$.

      To find $n$, write $x$ in base $3$. First, $n=0$ or $1$ depending on $xpmod3$. Then add $0,2$ or $4$ to $n$ so that $2^n=xpmod 9$; add $0,6$ or $12$ to get it correct mod $27$, and so on.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        You need to show that $2^2*3^k-1ne1pmod3^k+1$. That follows by induction by cubing $1+3^k-1+m*3^kpmod3^k+1$.

        To find $n$, write $x$ in base $3$. First, $n=0$ or $1$ depending on $xpmod3$. Then add $0,2$ or $4$ to $n$ so that $2^n=xpmod 9$; add $0,6$ or $12$ to get it correct mod $27$, and so on.






        share|cite|improve this answer













        You need to show that $2^2*3^k-1ne1pmod3^k+1$. That follows by induction by cubing $1+3^k-1+m*3^kpmod3^k+1$.

        To find $n$, write $x$ in base $3$. First, $n=0$ or $1$ depending on $xpmod3$. Then add $0,2$ or $4$ to $n$ so that $2^n=xpmod 9$; add $0,6$ or $12$ to get it correct mod $27$, and so on.







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        share|cite|improve this answer











        answered Aug 3 at 12:06









        Empy2

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