Problem:
Let m be a positive integer, and let a0​,a1​,…,am​ be a sequence of real numbers such that a0​=37,a1​=72,am​=0, and
ak+1​=ak−1​−ak​3​
for k=1,2,…,m−1. Find m.
Solution:
For 1≤k≤m−1, we have ak+1​ak​=ak​ak−1​−3. Let bk​=ak​ak−1​ for 1≤k≤m. Then b1​=a1​a0​=72⋅37=3⋅8⋅3⋅37=3⋅888 and bk+1​=bk​−3. Hence b889​=0 and bk​>0 for 1≤k≤888. Thus a889​=0 and m=889​.
The problems on this page are the property of the MAA's American Mathematics Competitions