2005 AIME II 第 11 题

Multiply the recurrence by $a_k$ to see that the products $a_k a_{k-1}$ drop by $3$ at each step

$a_1 a_0 = 2664 = 3 \cdot 888,$ and $a_m = 0$ exactly when the product $a_m a_{m-1}$ reaches $0$

Multiplying the recurrence by $a_k$ gives $a_{k+1} a_k = a_k a_{k-1} - 3,$ so the products $b_k = a_k a_{k-1}$ form an arithmetic sequence with common difference $-3.$ Since $b_1 = 72 \cdot 37 = 2664 = 3 \cdot 888,$ we get $$\begin{aligned} b_k &= 2664 - 3(k - 1) \\ &= 3(889 - k). \end{aligned}$$

Thus $b_k \gt 0$ for $k \le 888,$ so no term before $a_{889}$ can vanish (and the recurrence never divides by zero), while $b_{889} = a_{889} a_{888} = 0$ with $a_{888} \ne 0.$ Hence $a_{889} = 0,$ and $m = 889.$