Croatian Mathematical Olympiad

Let $\varphi = \frac{1+\sqrt{5}}{2}$ be the positive root of the quadratic polynomial $t^2 - t - 1$. The first player can win if the ratio of the starting numbers is in the set
$$\langle 0, \frac{1}{\varphi} \rangle \cup \{1\} \cup \langle \varphi, +\infty \rangle.$$
Let $M$ and $m$ be positive integers such that $M \geq m$. We claim that the player who is next on the turn when $M$ and $m$ are on the board wins if and only if $m = M$ or $M > \varphi m$. The claim is clear for $m = M$, so let us assume that $m \neq M$.
If $M < \varphi m$, then the player who plays next must pass on the pair of numbers $(m, M-m) \neq (m, 0)$ and $m > \varphi(M-m)$. Hence it is enough to show that the player who plays with a pair such that $M > \varphi m$ can either win or pass on a pair $(m', M')$ with $m' < M' < \varphi m'$.
The player wins if he plays with the numbers such that $m \mid M$. Let us assume that $M > \varphi m$ and $M = qm + r, 0 < r < m$. If $q \geq 2$, then player may pass on $(m, r)$, as well as on $(m, m+r)$. He will pass on $(m, r)$ if $r < m < \varphi r$. Otherwise, if $m > \varphi r$, he will pass on $(m, m+r)$, so the other player will pass on $(m, r)$ and we know that with these numbers the player on the turn wins. If $q = 1$, the player passes on $(m, r)$. We claim that $r \neq m < \varphi r$. Indeed, since $m + r > \varphi m$, we have $(\varphi - 1)m < r$, i.e. $m < \varphi r$.
Hence, the first player can in each move either win or pass on a pair $(m', M')$ such that $m' < M' < \varphi m'$.