Thread: Solutions to this diophantine equation.

1. Solutions to this diophantine equation.

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

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

Also how do i do the natural numbers '|N' in latex? (I'm new to it) EDIT: Thanks

2. Originally Posted by whipflip15
Here is a problem I enjoyed so I decided to share it.

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

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

CB

3. Originally Posted by whipflip15
Here is a problem I enjoyed so I decided to share it.

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