6. If the prime $p \equiv 3(\bmod 4)$, and $a^{2}+b^{2} \equiv 0(\bmod p)$, then $(\quad)$
A. $p|a, p| b$
B. $p \times a, p \times b$
C. $p \mid a$ and $p \nmid b$
D. $p \nmid a, p \mid b$
6. A Prompt: If $p \mid a$, then $p \mid b$,

If $p$ divides $a$, then $a^{p-1}=\left(a^{2}\right)^{\frac{p-1}{2}}$
$$
\begin{array}{l}
\equiv\left(-b^{2}\right)^{\frac{-1}{2}} \\
\equiv-b^{p-1}(\bmod p)
\end{array}
$$

Also, $a^{p-1} \equiv b^{p-1} \equiv 1(\bmod p)$ (Fermat's Little Theorem), contradiction!