Art and Ben play a game while sharing an $m \times n$ chocolate bar. They take turns breaking the bar into two rectangular pieces along one of the lines and eating the smaller piece. (If the two pieces are equal, they can choose which piece to eat.) Whoever is left with the last $1 \times 1$ square of chocolate loses. If Art moves first, describe all pairs $(m, n)$ for which Ben has a winning strategy.
The answer is all pairs $(m, n)$ such that the ratio

$$
R=\frac{m+1}{n+1}
$$

is an integer power of two, that is, the pairs $\left(a-1,2^{k} a-1\right)$ and $\left(2^{k} a-1, a-1\right)$ for $a \geq 2$ and $k \geq 0$. We will show that

(a) If $R$ is a power of two, then it will not remain so after one move;

(b) If $R$ is not a power of two, then some move will make it one.

To prove (a), simply note that at each move, one of the dimensions is decreased at most by half, so either $m+1$ or $n+1$ is diminished by a factor strictly between 1 and 2 . Thus the new value of $R$ is sandwiched between two powers of two: $R / 2$ and $R$ in the first case, $R$ and $2 R$ in the second.

As for (b), assume $m \geq n$ (the other case is symmetric) so $2^{k}<R<2^{k+1}$ for some $k \geq 0$. We have

$$
2^{k}(n+1)<m+1<2^{k+1}(n+1) \text {. }
$$

These inequalities imply that one can move to a bar that is $\left[2^{k}(n+1)-1\right] \times(n+1)$, with $R$-value $2^{k}$.

If $R$ is initially a power of two, Ben wins by making it that way at each move; otherwise Art wins by the same strategy (in either case ending at the $1 \times 1$ square, where $R=1$ ).