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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.

Thanks in advance.

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  • $\begingroup$ 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. $\endgroup$
    – Tob Ernack
    Jun 8, 2018 at 4:01

2 Answers 2

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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.

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  • $\begingroup$ I do like big hammer solutions, but given my limited background this answer doesn't do very much to aid understanding. I'll tentatively mark this as answering the question fully. If you or someone else finds a modulus that makes this solution more obvious I might change the accepted answer. For now, I'll try to read up on how these bounds are derived. $\endgroup$
    – Brett Berger
    Jun 8, 2018 at 1:55
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One of the solutions is $(k,m,n)=(3,3,5)$

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  • $\begingroup$ Oh thank you! I was slightly mistaken, this is a necessary but not sufficient condition for an exception. It turns out F(251) = 0 even though there is a trio with k = 3. So it's likely I missed quite a few potential solutions. $\endgroup$
    – Brett Berger
    Jun 8, 2018 at 0:17

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