4. If $a_{0}=b_{0}=1, a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}(n=1,2, \cdots)$, then $a_{1993}^{2}-2 b_{1993}^{2}=(\quad)$.
A. 1
B. 1993
C. 2
D. 2000
4. A.

Given $a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$, thus $a_{n}+\sqrt{2} b_{n}=(\sqrt{2}+1)\left(a_{n-1}+\sqrt{2} b_{n-1}\right)$, and $a_{n}-$ $\sqrt{2} b_{n}=(1-\sqrt{2})\left[a_{n-1}-\sqrt{2} b_{n-1}\right]$. Therefore, $a_{n}^{2}-2 b_{n}^{2}=-\left(a_{n-1}^{2}-2 b_{n-1}^{2}\right)(n \geqslant 1)$, so $a_{n}^{2}-2 b_{n}^{2}=$ $\left(a_{0}^{2}-2 b_{0}^{2}\right)(-1)^{n}=(-1)^{n+1}$. Hence, $a_{1993}^{2}-2 b_{1}^{2}{ }_{993}=(-1)^{1994}=1$.