Problem 915

We claim this is possible if and only $m+n$ is a power of 2 . Let $q=m / n$, so the numbers on the board are $q$ and $1 / q$.
Impossibility. The main idea is the following.
Claim - Suppose $p$ is an odd prime. Then if the initial numbers on the board are $-1(\bmod p)$, then all numbers on the board are $-1(\bmod p)$.

Proof. Let $a \equiv b \equiv-1(\bmod p)$. Note that $2 \not \equiv 0(\bmod p)$ and $a+b \equiv-2 \not \equiv 0(\bmod p)$. Thus $\frac{a+b}{2}$ and $\frac{2 a b}{a+b}$ both make sense modulo $p$ and are equal to $-1(\bmod p)$.
Thus if there exists any odd prime divisor $p$ of $m+n$ (implying $p \nmid m n$ ), then
$$q \equiv \frac{1}{q} \equiv-1 \quad(\bmod p) .$$
and hence all numbers will be $-1(\bmod p)$ forever. This implies that it's impossible to write 1 , whenever $m+n$ is divisible by some odd prime.
₪ Construction. Conversely, suppose $m+n$ is a power of 2 . We will actually construct 1 without even using the harmonic mean.

Note that
$$\frac{n}{m+n} \cdot q+\frac{m}{m+n} \cdot \frac{1}{q}=1$$
and obviously by taking appropriate midpoints (in a binary fashion) we can achieve this using arithmetic mean alone.