
6. Find the largest possible size of a set $\mathbb{M}$ of integers with the following property: Among any three distinct numbers from $\mathbb{M}$, there exist two numbers whose sum is a power of 2 with non-negative integer exponent.

(Ján Mazák)


Solution. The set $\{-1,3,5,-2,6,10\}$ attests that $\mathbb{M}$ 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 $\mathbb{M}$ has more than 6 elements.

Clearly, $\mathbb{M}$ 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 $\mathbb{M}$ and by $a, b, c, d$ some four other positive numbers in $\mathbb{M}$. Consider pairs $x+a, x+b, x+c$, $x+d$. They are all larger than $x$ and less than $2 x$. The open interval $(x, 2 x)$ 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 $(a, 2 a)$, hence at least one of them is not a power of two, a contradiction.

## First Round of the 67th Czech and Slovak Mathematical Olympiad (December 12th, 2017) $\mathbb{N} / /(10)$
