Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$
For $0 < k < m$, we have


$a_{k}a_{k+1} = a_{k-1}a_{k} - 3$.


Thus the product $a_{k}a_{k+1}$ is a [monovariant](https://artofproblemsolving.com/wiki/index.php/Monovariant): it decreases by 3 each time $k$ increases by 1.  For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$, so when $k = \frac{37 \cdot 72}{3} = 888$, $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$, our answer.

Note: In order for $a_{m} = 0$ we need $a_{m-2}a_{m-1}=3$ simply by the recursion definition.