$30 \cdot 2$ If positive integers $a, b, c$ satisfy $a^{2}+b^{2}=c^{2}$, and $a$ is a prime number, then the two numbers $b, c$
(A) are both odd.
(B) are both even.
(C) one is odd and one is even.
(D) are both composite.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution] Since $a^{2}=c^{2}-b^{2}=(c+b)(c-b)$, and $a$ is a prime number, and $c+b>c-b$, then
$$
c-b=1, \quad \text { and } c+b=a^{2} \text {. }
$$

Since $c-b$ is odd, then $c$ and $b$ must be one odd and one even. Therefore, the answer is $(C)$.