# Are there any solutions to $\frac{3^n - 2^n}{2^k-3^n} = N$

Are there any solutions to $\frac{3^n - 2^n}{2^k-3^n} = N$ for $n$, $k$, $N$ $\in\mathbb{N}$, greater than 2.

This is related to a previous answered question: Are there any solutions to $2^n-3^m=1$

There are no appropriate solutions to $2^k-3^n = 1$ which would make this trivial.

There is an equivalent question which is $\frac{2^{k-n} - 1}{2^k-3^n} = M$.

Things I've tried:

Several variants on diophantine equations. I've not yet found anything that involves quotients, etc. any pointers here would be welcome.

Several things to do with modulus/cyclic groups. The best outcome of this has been the above alternate form which doesn't seem any closer really.

Experimentally, there are no solutions with $n, k < 1000$. There are some pretty tight limits on what k can be as you need $2^k-3^n < 3^n - 2^n$ but also $2^k > 3^n$ which is only one or two $k$'s for every $n$. Also so far I've not seen any $N > 16$ which is a bit puzzling given the size of the values in the fraction. This makes me think there may be a limit based argument. Any pointers to where to start here would also be welcome.

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It is not too hard to show that there are no solutions. The assumption that $2^k-3^n$ divides $3^n-2^n$ implies that the former quantity divides $2^{k-n}-1$ and, further, that $2^k < 3^{n+1}$, so that $k < (n+1) \log (3)/\log(2)$. We thus have $$\left| 2^k - 3^n \right| < 2^{(n+1) \log (3)/\log(2)-n}.$$ On the other hand, standard lower bounds for linear forms in $2$ logarithms show that $$\left| 2^k - 3^n \right| \geq \min \left\{ 2^k, 3^n \right\}^{0.9},$$ say, with precisely $23$ exceptions (this is a result from de Weger's thesis; the largest exception is with $(k,n)=(84,53)$). Putting these inequalities together gives you what you want.