$30 \cdot 7$ Let $M^{2}=a^{2}+b^{2}+c^{2}$, where $a, b$ are consecutive integers, and $c=ab$, then $M$
(A) is always even.
(B) is sometimes odd.
(C) is always odd.
(D) is sometimes rational.
(E) is always irrational.
(18th American High School Mathematics Examination, 1967)
[Solution 1] Let $a=n$, then $b=n+1, \quad c=n(n+1)$.
From the problem, we have
$$
\begin{aligned}
M^{2} & =n^{2}+(n+1)^{2}+n^{2}(n+1)^{2} \\
& =n^{4}+2 n^{3}+3 n^{2}+2 n+1 \\
& =\left(n^{2}\right)^{2}+n^{2}+1^{2}+2 n^{2} \cdot n+2 n^{2}+2 \cdot n \\
& =\left(n^{2}+n+1\right)^{2},
\end{aligned}
$$

Therefore, $M=n^{2}+n+1$.
Since $n^{2}$ and $n+1$ must be one odd and one even, their sum must be odd.
Thus, the answer is $(C)$.
[Solution 2] $M=\sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{(a-b)^{2}+2 a b+(a b)^{2}}$
Since $a, b$ are consecutive integers, then $(a-b)^{2}=1$, and $a b$ is even. Therefore,
$$
M=\sqrt{1+2 a b+(a b)^{2}}=a b+1 \text {, }
$$

Thus, $M$ is odd.
Therefore, the answer is $(C)$.