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−ak3
for k=1,2,…,m−1. Find m.
Solution:
For 1≤k≤m−1, we have ak+1ak=akak−1−3. Let bk=akak−1 for 1≤k≤m. Then b1=a1a0=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