I've encountered the following question in my research:


Let $A$ be a subset of
$\mathbb{Z}/n\mathbb{Z}$. Let me call $A$ "sum-free" if there is no solution to
$x+y=z$ for $x,y,z \in A$ with **distinct** $x$ and $y$.
The addition is of course mod $n$.


**Question:** For which $n$ is the set $A = \{2^x \!\!\mod n: x>0\}$ (the set of powers of $2$)
sum-free? I'm assuming that $n$ is odd.


Does anyone have an idea how to approach this question? 



When are the powers of 2 sum-free mod n?
This question is very similar to the one [here](https://mathoverflow.net/questions/172706/primes-dividing-2a2b-1/172714), and the heuristic should apply equally well. In particular, $A$ is sum-free if and only if there does not exist a $k$ with $k \ne \frac{n+1}{2}$ so that $k$ and $1-k$ are both in $A$, and the chances of this occurring are approximately $e^{-r^{2}/\phi(n)}$, where $r = |A|$. Hence, one should only expect $A$ to be sum-free if $|A| \approx \sqrt{\phi(n)}$. This is a very infrequent situation, but it is certainly not impossible. (In particular, it is easy to see that if $n = 2^{k} - 1$, then $A$ is sum-free. Also, one could look for divisors $n$ of $2^{k} - 1$ with $n \gg k^{2}$ to find other examples where $A$ is sum-free.)


