# How would I approach this diophantine equation with powers of two? $(6k-1)2^{2(n-1)} = 9l-1$

A friend of mine recently gave me the follwing diophantine equation ($k, l, etc. \in \Bbb N$) and asked me whether I could solve it, i.e. "find all solutions" (I'm still a high school student, so please bear with me and please suggest improvements regarding this question in case parts of the question should remain unclear):

$$(6k-1)2^{2n-1} = 18l-2$$

The only thing I managed to do though was reforming it to:

$$(6k-1)2^{2(n-1)} = 9l-1$$

Solving linear diophantine equation is no problem and if $n$ is set to a certain number, I would even say it is very obious to find all solutions to the resulting equation, e.g. if $n=1$ then there is a solution at every $k=3m$ and $l=2m$. But if $n$ isn't set I simply can't find an approach to this problem, at least currently. Therefore, my friend and I would really appreciate if someone could help us out on this. Thank you!

HINT $$\frac{9l-1}{6k-1} = 4^{n-1}$$ so $${9l-1} = 4^{n-1}r$$ with some integer $r$.

For $n-1$ = 2, there are only two numbers, mod $16$, that satisfy $9l\equiv 1$

Any number that satisfies $9l=1 \pmod {64}$ must be one of eight possibilities (four cases for each of the two numbers that work mod $16$). Only one of these, $l\equiv 57$, works mod $64$.

Now try the four cases stemming from $57$, in mod $256$. You will find there is just one solution. Now move to mod $1024$ and there is again one solution.

If you keep going you may find some power $4^m$ for which there is no solution; each new power involves checking only a few numbers. Once you have a power for which there are no solutions, you can stop --- no higher power of 4 will work either.

That reduces the problem to the cases where $n-1\in\{0,1,2,3,4,\dots, m-1\}$ and you know how to handle each of those.

If you don't come to a "stopper," (and I believe you won't) you will see a pattern emerge. That may let you solve all the cases without considering individual powers past $4^{10}$ or so.