IMC 2025 · Day 2 · P7

Note that $M$ is closed under addition since $x + y = \frac{2x + 2y}{2}$. Therefore it is closed under multiplication by an arbitrary natural number. Also, $M$ contains some odd numbers since $x \in M \Longrightarrow \frac{x + 2x}{2} = \frac{3x}{2} \in M$, and we can repeat this until all factors of 2 are removed from a number.

Let $d$ be the greatest common divisor of all members from $M$. Then $M \subseteq d \mathbb{Z}_{>0}$ and $d$ is odd. Since $d$ can be represented in the form $$d = v_1 a_1 + v_2 a_2 + \dots + v_k a_k - v_{k+1} a_{k+1} - v_{k+2} a_{k+2} - \dots - v_n a_n$$ for some $a_i \in M$ and $v_i \in \mathbb{Z}_{>0}$, there exist two members of $M$ with difference $d$.

Let $c$ be the minimal element of $M$, $c < a$, and $a, a + d \in M$. We choose the largest $x \in M$ such that $x < a$. Then the only elements of $M$ in the interval $[x, a + d]$ are $x$, $a$, and $a + d$, since $M \subseteq d\mathbb{Z}$. Meanwhile, $x < \frac{x + a}{2} < a$, so $x + a$ must be odd, hence $x + a + d$ is even, and then $x < \frac{x + a + d}{2} < a + d$ means $x = a - d$. Thus, the following implication holds: $$(a, a + d \in M \text{ and } c < a) \ \Longrightarrow \ a - d \in M.$$ Similarly, by setting $x \in M$ to be the smallest such that $x > a$ (we know that such an element exists, since $2a \in M$), we obtain $$a - d, a \in M \ \Longrightarrow \ a + d \in M.$$ We thus get that $M$ is obtained as the set of elements of the arithmetic progression $c + kd$ ($k \in \mathbb{Z}_{\ge 0}$). Obviously, this set satisfies both of the properties (a) and (b). Hence, we have proved that $M$ satisfies these properties if and only if $M = \{ nd \mid m \leqslant n \in \mathbb{N} \}$ for some $m \in \mathbb{N}$ and odd $d$.