67th Czech and Slovak Mathematical Olympiad

The set $\{-1,3,5,-2,6,10\}$ attests that $M$ can have $6$ elements: The sum of any two numbers from the triplet $(-1,3,5)$ is a power of two and the same is true for triplet $(-2,6,10)$. For the sake of contradiction, assume that some set $M$ has more than $6$ elements.

Clearly, $M$ can't contain three (or more) non-positive numbers. Hence it contains at least five positive numbers. Denote by $x$ the largest positive number in $M$ and by $a,b,c,d$ some four other positive numbers in $M$. Consider pairs $x+a,x+b,x+c,x+d$. They are all larger than $x$ and less than $2x$. The open interval $(x,2x)$ contains at most one power of two, hence at least three of the four sums are not a power of two. Without loss of generality, assume those are $x+a,x+b,x+c$. Considering the triplets $(a,b,x),(a,c,x),(b,c,x)$ we infer that all $a+b,a+c,b+c$ are powers of two. However, this is impossible. Without loss of generality, let $a=\max \{a,b,c\}$. Then $a+b$ and $a+c$ both lie in $(a,2a)$, hence at least one of them is not a power of two, a contradiction.