## problem Let a sequence be defined as follows: $a_{1}=3$, $a_{2}=3$, and for $n \geq 2$ $ a_{n+1} a_{n-1}={a_{n}}^{2}+2007 $ Find the largest integer less than or equal to $\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007} a_{2006}}$ ## solution The most important part of this solution will be establishing an expression relating neighbors in the sequence. We can take the original definition and rewrite it to be true for all $n\geq1$ $\begin{aligned} a_{n+1} a_{n-1}={a_{n}}^{2}+2007;\quad n\geq 2 \\ a_{n+2} a_{n}={a_{n+1}}^{2}+2007;\quad n\geq 1 \end{aligned}$ Subtracting the two equations will result in something easier on the eyes $a_{n+1} a_{n-1}-a_{n+2} a_{n}={a_{n}}^{2}-{a_{n+1}}^{2}$ now the trick is to show that this leads to an equation with both sides looking similar. $a_{n+1} a_{n-1}+{a_{n+1}}^{2}={a_{n}}^{2}+a_{n+2} a_{n}$ $ \frac{a_{n-1}+a_{n+1}}{a_{n}}=\frac{a_{n}+a_{n+2}}{a_{n+1}}$ The two sides of this equation perform the same operations on a set of neighbors in the sequence. This means, by induction, all expressions of this form will have the same value. Since $a_1$ and $a_2$ are already known, we only need $a_3$ to be able to evaluate. $\frac{a_{n-1}+a_{n+1}}{a_{n}}=\frac{a_1+a_3}{a_{2}}$ To find $a_3$, we use the original definition. $\begin{aligned} a_{3} a_{1}&={a_{2}}^{2}+2007\\ a_3\cdot3&={3}^{2}+2007\\ a_3&=672\end{aligned}$ And we use $a_1,a_2,a_3$ in to evaluate as planned $\frac{a_1+a_3}{a_{2}}=\frac{3+672}{3}=225$ This leads to a more linear recursive definition of the sequence. $\begin{aligned}\frac{a_{n-1}+a_{n+1}}{a_{n}}&=225\\ {a_{n-1}+a_{n+1}}&=225a_n\\ a_{n+1}&=225a_n-a_{n-1} \end{aligned}$ Note now that $a_{n+1}>a_n$ must be true (for $n\geq2$) since the negation $a_{n+1}\leq a_n$ leads to contradiction $\begin{aligned}a_{n}&\geq 225a_n-a_{n-1}\\a_{n- 1}&\geq224a_n \quad \text{false}\end{aligned}$ One step further, this also tells us that $a_n>a_{n-1}$