Four. (20 points) The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=$ $b_{0}=1, a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$, $(n=1,2 \ldots)$. Find the value of $a_{2001}^{2}-2 b_{2001}^{2}$.
$$
\begin{array}{l}
a_{0}^{2}-2 b_{0}^{2}=-1, \\
a_{1}^{2}-2 b_{1}^{2}=1, \\
\ldots \ldots
\end{array}
$$

Conjecture $a_{n}^{2}-2 b_{n}^{2}=(-1)^{n+1}(n=0,1 \ldots)$.
We will prove this by mathematical induction.
Assume the proposition holds for $n=k$, that is,
$$
a_{k}^{2}-2 b_{k}^{2}=(-1)^{k+1} \text {. }
$$

Then for $n=k+1$,
$$
\begin{array}{l}
a_{k+1}^{2}-2 b_{k+1}^{2}=\left(a_{k}+2 b_{k}\right)^{2}-2\left(a_{k}+b_{k}\right)^{2} \\
=-\left(a_{k}^{2}-2 b_{k}^{2}\right)=(-1)^{k+2} .
\end{array}
$$

Therefore, by mathematical induction, the conjecture is true.
Thus, $a_{2001}^{2}-2 b_{2001}^{2}=1$.