What a beautiful problem! The winning positions (for the player who just moved) are those where the stacks are within a factor of the golden ratio $\phi$, i.e. the $(a,b)$ with $$1/\phi < b/a < \phi.$$

Say that $a\leq b$. The legal moves result in positions of the form $(a,x)$. This is in the band of winning positions whenever $x \in (a/\phi,a \phi)$ an interval whose length is exactly $a$.

Since the possible $x$'s are spaced $a$ apart, exactly one such $x$ in the band. Then, $(a,x)$ is a winning move unless it is $(a,b)$ itself in the band, since we must have $x<b$. We've shown that there's a move to a winning position exactly when we're not a winning position, which inductively proves that these are the winning positions.

I found this solution by writing code to find the winning positions, and noticing the band pattern and suspecting the golden ratio.

What a beautiful problem! The winning positions (for Bob, the player who just moved) are those where the stack sizes are within a factor of the golden ratio $\phi$, i.e. the $(p,q)$ with $$1/\phi < q/p < \phi.$$ So, the winning strategy is to make the stacks as even as possible to keep this invariant true after your move.

Say that $p\leq q$. The legal moves result in positions of the form $(p,q')$. This is in the band of winning positions whenever $q' \in (p/\phi,p \phi)$ an interval whose length is exactly $p$.

Since the possible $q'$ are spaced $p$ apart, exactly one such $q'$ lies in the band. Then, $(p,q')$ is a winning move unless it is $(p,q)$ itself in the band, since we must have $q'<q$. We've shown that there's a move to a winning position exactly when we're not a winning position, which inductively proves that these are the winning positions.

I found this solution by writing code to find the winning positions, and noticing the band pattern and suspecting the golden ratio.