17. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1} a_{n}-1=a_{n}^{2} \text {. }
$$
(1) Prove: $\sqrt{2 n-1} \leqslant a_{n} \leqslant \sqrt{3 n-2}$;
(2) Find the integer $m$, such that $\left|a_{2 \cos }-m\right|$ is minimized.
17. (1) From the given information, it is easy to see that $\{a_n\}$ is an increasing sequence, and all its terms are positive. Since $a_{n+1} a_{n} - 1 = a_{n}^2$, we have, for $k \geq 2$,
$$
a_{k} = a_{k-1} + \frac{1}{a_{k-1}}.
$$

From this, we get
$$
a_{k}^2 = \left(a_{k-1} + \frac{1}{a_{k-1}}\right)^2 = a_{k-1}^2 + \frac{1}{a_{k-1}^2} + 2 > a_{k-1}^2 + 2,
$$

which means $a_{k}^2 - a_{k-1}^2 > 2$.
Thus, $a_{n}^2 - a_{1}^2 = \sum_{k=2}^{n} (a_{k}^2 - a_{k-1}^2) > 2(n-1)$, i.e.,
$$
a_{n}^2 > a_{1}^2 + 2(n-1) = 2n - 1.
$$

Therefore, $a_{n} > \sqrt{2n-1}$ for $n > 1$.
Also, $a_{n}^2 - a_{1}^2 = \sum_{k=2}^{n} (a_{k}^2 - a_{k-1}^2)$
$$
\begin{array}{l}
= 2(n-1) + \sum_{k=2}^{n} \frac{1}{a_{k-1}^2} \\
\leq 2(n-1) + (n-1) \times 1 = 3n - 3,
\end{array}
$$

which means $a_{n}^2 \leq 3n - 3 + a_{1}^2 = 3n - 2$.
Therefore, $a_{n} \leq \sqrt{3n-2}$.
Combining these results, we have $\sqrt{2n-1} \leq a_{n} \leq \sqrt{3n-2}$.
(2) From the results in (1), we know $\sqrt{4009} \leq a_{2000} \leq \sqrt{6013}$, i.e., $631$ when $b_{n} > 0$.
Thus, the equation $a_{n+1}^2 = a_{n}^2 + 2 + \frac{1}{a_{n}^2}$ can be rewritten as
$$
2n + 1 + b_{n+1} = 2n - 1 + b_{n} + 2 + \frac{1}{2n - 1 + b_{n}}.
$$

From this, we get
$$
b_{n+1} = b_{n} + \frac{1}{2n - 1 + b_{n}} \leq b_{n} + \frac{1}{2n - 1}.
$$

By induction, we have
$$
b_{n+1} \leq b_{1} + 1 + \frac{1}{3} + \cdots + \frac{1}{2n-3} + \frac{1}{2n-1}.
$$

Since $b_{1} = 0$, we have
$$
b_{2008} \leq 1 + \frac{1}{3} + \cdots + \frac{1}{4005} + \frac{1}{4007}.
$$

To estimate the value of the right-hand side, we segment it as follows:
$$
\begin{aligned}
b_{2008} \leq & + \left(\frac{1}{3} + \frac{1}{5} + \frac{1}{7}\right) + \left(\frac{1}{9} + \frac{1}{11} + \cdots + \frac{1}{20}\right) + \\
& \cdots + \left(\frac{1}{243} + \cdots + \frac{1}{721}\right) + \left(\frac{1}{729} + \cdots + \frac{1}{2185}\right) + \\
& \left(\frac{1}{2187} + \cdots + \frac{1}{4007}\right).
\end{aligned}
$$

The first parenthesis on the right-hand side has 3 terms, with the largest term being $\frac{1}{3}$; the second parenthesis has 9 terms, with the largest term being $\frac{1}{9}$; ...; the sixth parenthesis has 729 terms, with the largest term being $\frac{1}{729}$; finally, the seventh parenthesis has 911 terms, with the largest term being $\frac{1}{2187}$. Therefore, $b_{2008} < 8$. Combining this with $a_{n}^2 = 2n - 1 + b_{n}$, we get
$$
a_{2008}^2 < 4010 - 1 + 8 < 4032.25 = 63.5^2.
$$

Thus, $63 < a_{2008} < 63.5$. Therefore, $m = 63$.
(Yang Chunhong provided)