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Math Help - Solutions to this diophantine equation.

  1. #1
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    Solutions to this diophantine equation.

    Here is a problem I enjoyed so I decided to share it.

     \mbox{Find all solutions to }3^n = 2^m-1\mbox{ for all }n,m \in \mathbb{N}.

    Also how do i do the natural numbers '|N' in latex? (I'm new to it) EDIT: Thanks
    Last edited by whipflip15; October 25th 2008 at 12:44 AM.
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  2. #2
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    Quote Originally Posted by whipflip15 View Post
    Here is a problem I enjoyed so I decided to share it.

     \mbox{Find all solutions to }3^n = 2^m-1\mbox{ for all }n,m \in N.

    Also how do i do the natural numbers '|N' in latex? (I'm new to it)
    \mathbb{N} gives \mathbb{N}

    CB
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  3. #3
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    Quote Originally Posted by whipflip15 View Post
    Here is a problem I enjoyed so I decided to share it.

     \mbox{Find all solutions to }3^n = 2^m-1\mbox{ for all }n,m \in \mathbb{N}.
    Obviously m=2, n=1 is a solution. (So is m=1, n=0 if you believe that 0\in\mathbb{N}.) If m>2 then 2^m-1\equiv-1\!\!\pmod8. But powers of 3 are all congruent to 1 or 3 (mod 8). So there are no more solutions.
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