A [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) is defined over [non-negative](https://artofproblemsolving.com/wiki/index.php/Non-negative) integral indexes in the following way: $a_{0}=a_{1}=3$, $a_{n+1}a_{n-1}=a_{n}^{2}+2007$.
Find the [greatest integer](https://artofproblemsolving.com/wiki/index.php/Floor_function) that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$
Solution 1
Define a function $f(n)$ on the non-negative integers, as \[f(n) = \frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_{n}}\]
We want $\left\lfloor f(2006) \right\rfloor$. 
Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$. Dividing through by $a_{n}a_{n-1}$, we get 
\[.\phantom{------------} \frac{a_{n+1}}{a_{n}} = \frac{a_{n}}{a_{n-1}} + \frac{2007}{a_{n}a_{n-1}} \phantom{------------} (1)\]
and dividing through by $a_{n}a_{n+1}$, we get 
\[.\phantom{------------} \frac{a_{n-1}}{a_{n}} = \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} \phantom{------------} (2)\]
Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get
\[\frac{a_{n+1}}{a_{n}} + \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} = \frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}} + \frac{2007}{a_{n}a_{n-1}}\]
i.e.
\[f(n)+ \frac{2007}{a_{n}a_{n+1}} = f(n-1) + \frac{2007}{a_{n}a_{n-1}}\]
Summing over $n=1$ to $n=2006$, we notice that most of the terms on each side cancel against the corresponding term on the other side. We are left with
\[f(2006) + \frac{2007}{a_{2006}a_{2007}} = f(0) + \frac{2007}{a_{1}a_{0}}\]
We have $f(0) = 2$, and $2007/a_0a_1 = 2007/9 = 223$. So
\[f(2006) = 2 + 223 - \frac{2007}{a_{2006}a_{2007}} = 224 + \left( 1 - \frac{2007}{a_{2006}a_{2007}}\right)\]
Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$, this means that the sequence $(a_n)$ is increasing. Since $a_3=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$. It follows that $\left\lfloor f(2006) \right\rfloor = \boxed{224}$.

Solution 2
We are given that
$a_{n+1}a_{n-1}= a_{n}^{2}+2007$,
$a_{n-1}^{2}+2007 = a_{n}a_{n-2}$.
Add these two equations to get

$a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})$
$\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}$.
This is an [invariant](https://artofproblemsolving.com/wiki/index.php/Invariant). Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$, the above equation means
$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$.
We can thus calculate that $b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225$. Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$, notice that $b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}$. This means that
$b_{2007}+\frac{1}{b_{2007}} a_{t+1}.\]
This completes the induction.
Lemma 3 : For $n \geq 1,$
\[a_n a_{n+1} > 9k\]
Using (1) and Lemma 2, for $n \geq 1,$
\[a_{n+1}a_n \geq a_{n+1}a_{n-1}  = a_n^2 + 9k  > 9k\]
Finally, using (3), for $n \geq 1,$
\[\frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.\]
Using lemma 3, the largest integer less than or equal to this value would be $k + 1$.

Solution 5 (pure algebra)
We will try to manipulate $\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\frac{a_1^2+a_2^2}{a_1a_2}$. 
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}$
Using the recurrence relation, 
$= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1}  = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}$
Applying the relation to $a_0a_2$, 
$= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}$
We can keep on using this method to get that 
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}$
This telescopes to 
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}$
or 
$\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}$
Finding the first few values, we notice that they increase rapidly, so $\frac{2007}{a_{2006}a_{2007}} < 1$. Calculating the other values, 
$\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}$.
The greatest number that does not exceed this is $\boxed{224}$

Solution 6 (using limits)
Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$, and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$, respectively. It will be clear why I decided to factor these expressions as I did momentarily. 
Next, let's see what the expression $\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$. For $n = 1$, we get $\frac{1 + 224^2}{224}$, the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}$. It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$, while the second-largest term in the denominator is smaller than $224^2$. Thus, the floor of this expression will come out to be $224$ as well.
Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$, the degrees of both the numerator and denominator increase by $2$, because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2$). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\approx224:1$ ratio between the two. 
For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$, they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\geq2$, whose $\lim_{k\to\infty}=$ $\boxed{224}$.
~anellipticcurveoverq