Problem:
The terms of the sequence (aiβ) defined by an+2β=1+an+1βanβ+2009β for nβ₯1 are positive integers. Find the minimum possible value of a1β+a2β.
Solution:
The definition gives
a3β(a2β+1)=a1β+2009,a4β(a3β+1)=a2β+2009,a5β(a4β+1)=a3β+2009.
Subtracting adjacent equations yields a3ββa1β=(a3β+1)(a4ββa2β) and a4ββa2β=(a4β+1)(a5ββa3β). Suppose that a3ββa1βξ =0. Then a4ββa2βξ =0, a5ββa3βξ =0, and so on. Because β£an+2β+1β£β₯2, it follows that 0< β£an+3ββan+1ββ£=β£an+2β+1β£β£an+2ββanββ£β<β£an+2ββanββ£, that is, β£a3ββa1ββ£>β£a4ββa2ββ£> β£a5ββa3ββ£>β―, which is a contradiction. Therefore an+2ββanβ=0 for all nβ₯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 a1β and a2β are integers, all the terms are integers. The definition of the sequence then implies that a1β=a3β=a2β+1a1β+2009β, giving a1βa2β=2009=72β
41. The minimum value of a1β+a2β occurs when {a1β,a2β}={41,49}, which has a sum of 90.
The problems on this page are the property of the MAA's American Mathematics Competitions