3. If $x-y=2, x^{2}+y^{2}=4$, then the value of $x^{1998}+y^{1996}$ is ( ).
(A) 4
(B) $1996^{2}$
(C) $2^{1996}$
(D) $4^{1996}$
3. (C).

From $x-y=2$
squaring gives $x^{2}-2 x y+y^{2}=4$.
Also, $x^{2}+y^{2}=4$.
(3) - (2) gives $2 x y=0 \Rightarrow x y=0$.
Therefore, at least one of $x, y$ is 0, but since $x^{2}+y^{2}=4$, only one of $x, y$ can be 0, the other being 2 or -2. In either case,
$$
x^{1996}+y^{1996}=2^{1996} .
$$