Problem:
The terms of the sequence (ai​) defined by an+2​=1+an+1​an​+2009​ for n≥1 are positive integers. Find the minimum possible value of a1​+a2​.
Solution:
The definition gives
a3​(a2​+1)=a1​+2009,a4​(a3​+1)=a2​+2009,a5​(a4​+1)=a3​+2009.
Subtracting adjacent equations yields a3​−a1​=(a3​+1)(a4​−a2​) and a4​−a2​=(a4​+1)(a5​−a3​). Suppose that a3​−a1â€‹î€ =0. Then a4​−a2â€‹î€ =0, a5​−a3â€‹î€ =0, and so on. Because ∣an+2​+1∣≥2, it follows that 0< ∣an+3​−an+1​∣=∣an+2​+1∣∣an+2​−an​∣​<∣an+2​−an​∣, that is, ∣a3​−a1​∣>∣a4​−a2​∣> ∣a5​−a3​∣>⋯, which is a contradiction. Therefore an+2​−an​=0 for all n≥1, which implies that all terms with an odd index are equal, and all terms with an even index are equal. Thus as long as a1​ and a2​ are integers, all the terms are integers. The definition of the sequence then implies that a1​=a3​=a2​+1a1​+2009​, giving a1​a2​=2009=72â‹…41. The minimum value of a1​+a2​ occurs when {a1​,a2​}={41,49}, which has a sum of 90.
The problems on this page are the property of the MAA's American Mathematics Competitions