Problem:
Let a sequence be defined as follows: a1​=3,a2​=3, and for n≥2, an+1​an−1​=an2​+2007. Find the largest integer less than or equal to a2007​a2006​a20072​+a20062​​.
Solution:
The fact that the equation an+1​an−1​=an2​+2007 holds for n≥2 implies that an​an−2​=an−12​+2007 for n≥3. Subtracting the second equation from the first one yields an+1​an−1​−an​an−2​=an2​−an−12​, or an+1​an−1​+an−12​=an​an−2​+an2​. Dividing the last equation by an−1​an​ and simplifying produces an​an+1​+an−1​​= an−1​an​+an−2​​. This equation shows that an​an+1​+an−1​​ is constant for n≥2. Because a3​a1​=a22​+2007,a3​=2016/3=672. Thus an​an+1​+an−1​​=3672+3​=225, and an+1​=225an​−an−1​ for n≥2. Note that a3​=672an​>3=a2​. Furthermore, if an​>an−1​, then an+1​an−1​=an2​+2007 implies that
an+1​=an−1​an2​​+an−1​2007​=an​(an−1​an​​)+an−1​2007​>an​+an−1​2007​>an​.
Thus by mathematical induction, an​>an−1​ for all n≥3. Therefore the recurrence an+1​=225an​−an−1​ implies that an+1​>225an​−an​=224an​ and therefore an​≥2007 for n≥4. Finding an+1​ from an+1​an−1​=an2​+2007 and substituting into 225=an​an+1​+an−1​​ shows that an​an−1​an2​+an−12​​=225−an​an−1​2007​. Thus the largest integer less than or equal to the original fraction is 224​.
The problems on this page are the property of the MAA's American Mathematics Competitions