3. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{100}-y^{100}=2^{99}(x-y), \\
x^{200}-y^{200}=2^{199}(x-y)
\end{array}(x \neq y) .\right.
$$
Let $x=2a, y=2b$. Then
$$
\left\{\begin{array}{l}
a^{100}-b^{100}=a-b, \\
a^{200}-b^{200}=a-b
\end{array}(a \neq b) .\right.
$$

Thus, $a^{100}-b^{100}=a^{200}-b^{200}$
$$
\begin{array}{l}
=\left(a^{100}-b^{100}\right)\left(a^{100}+b^{100}\right) \\
\Rightarrow a^{100}+b^{100}=1 .
\end{array}
$$

Combining with
$$
\begin{array}{l}
a-b=a^{100}-b^{100} \\
=(a-b)\left(a^{99}+a^{98} b+\cdots+a b^{98}+b^{99}\right),
\end{array}
$$

we can prove that if $ab \neq 0$, the above equation does not hold.
Therefore, $ab=0$.
Hence, $(x, y)=(2,0)$ or $(0,2)$ are the solutions to the system of equations.