# Solutions to $2^{2^k}-5=2^m+3^n$

I'm quite close to proving a particular function has infinitely many zeros of the form $2^{2^k}-5$, $k\ge 3$. The exceptional cases happen when $2^{2^k}-5$ can be written as $2^m+3^n$. Working mod 3, it's easy to show that $m$ must be odd.

Are there solutions to $2^{2^k}-5=2^m+3^n$ over the positive integers with $k\ge 3$? How would one go about finding them, or proving them impossible?

I know the trio $(2, 3, 1)$ is a solution, hence the $k\ge 3$ restriction. It would be sufficient for my purpose to prove there are infinitely many fixed $k$ where there is no solution, but seeing as how I have found no solutions at all with higher $k$, I am curious about ways to prove anything conclusive.

• Just a thought, it would be nice to find upper bounds on the Hamming weight of the binary expansion of $3^n$, since $2^t = 2^m + 2^2 + 1 + 3^n$ implies that $3^n = 2^t - 2^m - 2^2 - 1$ has fairly many $1$'s for large $t, m$. I don't know if this approach is feasible. Jun 8, 2018 at 4:01
The only solutions to the slightly more general ($S$-unit) equation $$2^t-5=2^m+3^n$$ are with $$(m,n) \in \left\{ (1,0), (1,2), (3,1), (3,5), (5,3) \right\}.$$ There is probably a way to prove this locally (i.e. by considering the equation modulo a well-chosen modulus), but if one is willing to use a big hammer, one gets this from using linear forms in $p$-adic and complex logarithms. Specifically, you can use bounds for linear forms in $2$-adic logarithms (applied to $3^n+5$) to show that $m$ is "small". One then applies lower bounds for linear forms in complex logarithms to $3^n-2^t$ (so to $n \log 3 - t \log 2$) to get a contradiction for large enough $n$ and $t$. A reasonably short calculation finishes the proof.
One of the solutions is $(k,m,n)=(3,3,5)$