anemone said:
My solution:
Given $b_1=3$, $b_2=3$, and for $n \ge 2$, $b_{n+1}b_{n-1}=b_n^2+2007$.
Build a table for a few terms for $n>2$ to check if there is a pattern to be observed:
[TABLE="class: grid, width: 800"]
[TR]
[TD]n[/TD]
[TD]$b_{n+1}=\dfrac{b_n^2+2007}{b_{n-1}}$[/TD]
[TD]$\dfrac{b_{n+1}}{b_n}$[/TD]
[/TR]
[TR]
[TD]2[/TD]
[TD]$b_3=\dfrac{b_2^2+2007}{b_1}=\dfrac{3^2+2007}{3}=672$[/TD]
[TD]-[/TD]
[/TR]
[TR]
[TD]3[/TD]
[TD]$b_4=\dfrac{b_3^2+2007}{b_2}=\dfrac{672^2+2007}{3}=151197$[/TD]
[TD]$\dfrac{b_4}{b_3}=\dfrac{151197}{672} \approx 224.9955357 \approx 225$[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]$b_5=\dfrac{b_4^2+2007}{b_3}=\dfrac{151197^2+2007}{672}=34018653$[/TD]
[TD]$\dfrac{b_5}{b_4} \approx 224.99555 \approx 225$[/TD]
[/TR]
[TR]
[TD]5[/TD]
[TD]$b_6=\dfrac{b_5^2+2007}{b_4}=\dfrac{34018653^2+2007}{151197}=7654045728$[/TD]
[TD]$\dfrac{b_6}{b_5} \approx 224.99556 \approx 225$[/TD]
[/TR]
[TR]
[TD]6[/TD]
[TD]$b_7=\dfrac{b_6^2+2007}{b_5}=\dfrac{7654045728^2+2007}{34018653}=1.72212627\times10^{12}$[/TD]
[TD]$\dfrac{b_7}{b_6} \approx 225$[/TD]
[/TR]
[TR]
[TD]7[/TD]
[TD]$b_8=\dfrac{b_7^2+2007}{b_6}=\dfrac{(1.72212627 \times10^{12})^2+2007}{7654045728}=3.8747 \times10^{14}$[/TD]
[TD]$\dfrac{b_8}{b_7} \approx 225$[/TD]
[/TR]
[TR]
[TD]$\vdots$[/TD]
[TD]$\vdots$[/TD]
[TD]$\vdots$[/TD]
[/TR]
[/TABLE]
That is, we have generated a "geometric-like" sequence with the common ratio that takes the approximate value of 225 (in fact, the so-called common ratio is increasing as $n$ increases).
Another thing that is worth mentioning here is as $n$ increases, the value of $b_n$ increases monumentally that the addition of the value 2007 in the numerator brings no significant increment to the numerator and we can thus treat the whole equation of $b_{n+1}b_{n-1}=b_n^2+2007$ as $b_{n+1}b_{n-1}=b_n^2$ or $\dfrac{b_{n+1}}{b_n}=\dfrac{b_n}{b_{n-1}}$, i.e. the terms of $b_n$ generate a geometric sequence.
$\therefore \dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}= \dfrac{b_{2007}}{b_{2006}}+\dfrac{b_{2006}}{b_{2007}}=\dfrac{b_{2007}}{b_{2006}}+\dfrac{1}{\dfrac{b_{2007}}{b_{2006}}}<225+\dfrac{1}{225} <225.00444$
Hence the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$ is 225.
You have shown that the ratio $x_n = \dfrac{b_{n+1}}{b_n}$ is very close to 225. But suppose it always stays just sufficiently
less than 225 to ensure that $x_n + x_n^{-1} = \dfrac{b_{n+1}^2+b_{n}^2}{b_{n+1}b_{n}}$ is also less than 225. In that case, "the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$" would be 224, not 225. As far as I can see, that appears to be what actually happens.
The condition $x_n + x_n^{-1} < 225$ is equivalent to the condition that $x_n$ should lie between the two roots of the equation $x + x^{-1} = 225$, or $x^2 - 225x + 1 = 0$. The larger of the two roots is $\frac12\bigl(225 + \sqrt{50621}\bigl) \approx 224.99556.$ So we have to ask whether $x_n$ stays below that value.
Divide both sides of the equation $b_{n+1}b_{n-1}=b_n^2+2007$ by $b_nb_{n-1}$ to get $x_{n+1} = \dfrac{2007}{b_{n-1}b_n} + x_n$. Apply that relation inductively to see that $$x_n = 2007\left(\frac1{b_{n-1}b_n} + \frac1{b_{n-2}b_{n-1}} + \ldots + \frac1{b_4b_3}\right) + x_3.$$ Next, the sequence $(x_n)$ is increasing, and $x_3 = b_4/b_3 \approx 224.9955357 >224.9$. Therefore $x_n = b_{n+1}/b_n >224.9$ whenever $n\geqslant3.$ Thus $b_n > 224.9b_{n-1}$ and (again by an inductive argument) $b_n > 224.9^{n-3}b_3 = 672\cdot 224.9^{n-3}.$ It follows that $$x_n < \frac{2007}{672^2}\left(\frac1{224.9^{2n-7}} + \frac1{224.9^{2n-9}} + \ldots + \frac1{224.9}\right) + x_3.$$ If we replace the geometric series in the large brackets by the corresponding infinite geometric series, which has sum $\dfrac1{224.9(1-224.9^{-1})}$ then we get the estimate $$x_n < \frac{2007}{672^2\cdot 224.9(1-224.9^{-1})} + x_3 \approx 0.0000198 + 224.9955357 = 224.9955555 .$$ This is less than $224.99556$, so it appears that $x_n+x_n^{-1}$ always stays below 225.
I hate to rely on such delicate numerical evidence, especially since my little calculator only works to 8 significant figures. I would much prefer to have an analytic argument showing that the answer to the problem is $224$, but I don't see how to do that.