30-39 In decimal, a six-digit number is called a "special square number" if it satisfies the following conditions:
(i) None of its digits are zero;
(ii) It is a perfect square;
(iii) The first two digits, the middle two digits, and the last two digits are all perfect squares when considered as two-digit numbers.
How many special square numbers are there?
(A) 0.
(B) 2.
(C) 3.
(D) 8.
(E) 9.
(31st American High School Mathematics Examination, 1980)
[Solution] If $N$ is a special square number, by the problem's condition, there exist six positive one-digit integers $A$, $B$, $C$, $a$, $b$, $c$ satisfying
$$
N=10^{4} A^{2}+10^{2} B^{2}+C^{2}=\left(10^{2} a+10 b+c\right)^{2} .
$$

Let $d=10 b+c$, then $d \leqslant 99$. This necessarily implies $A=a$,
otherwise, if $a<\mathrm{A}$,
$$
\begin{aligned}
N & =\left(10^{2} a+d\right)^{2} \leqslant 10^{4} a^{2}+2(100) 99 a+(99)^{2} \\
& <10^{4} a^{2}+2(10)^{4} a+10^{4}=10^{4}(a+1)^{2} \leqslant N,
\end{aligned}
$$

which is clearly a contradiction!
Since the middle two digits and the right two digits of $N$ are both perfect squares, and the largest of these is 8181, we have

then
$$
\begin{array}{c}
8181 \geqslant 10^{2}, \\
B^{2}+C^{2}=\left(10^{2} a+10 b+c\right)^{2}-10^{4} a^{2} \\
=10^{3}(2 a b)+10^{2}\left(b^{2}+2 a c\right)+10(2 b c)+c^{2} .
\end{array}
$$

Thus, we can see that $2 a b \leqslant 8$. Since $N$ has no zero digits, then $a \geqslant 4$.
This means either $b=0$ or $a=4, b=1$. In the latter case,
$$
N=(410+c)^{2}=168100+820 c+c^{2} .
$$

Since $N$ has no zero digits, we know $c \geqslant 1$. But in this case, either the middle two digits or the leftmost two digits are not perfect squares.
Therefore, the latter case is impossible, so $b=0$.
Thus, $N=\left(10^{2} a+c\right)^{2}=10^{4} a^{2}+10^{2}(2 a c)+c^{2}$.
Similarly, $a \geqslant 4, c \geqslant 4$ and $2 a c$ is an even two-digit perfect square.
It is easy to verify that $a=8, c=4, N=646416$; or $a=4, c=8$, $N=166464$, satisfy the problem's conditions.
Therefore, the answer is $(B)$.