$2 \cdot 115$ Find the sum of the squares of all real numbers satisfying $x^{256}-256^{32}=0$.
(A) 8 .
(B) 128 .
(C) 512 .
(D) 65536 .
(E) $2\left(256^{32}\right)$.
(30th American High School Mathematics Examination, 1979)
［Solution］Notice that
$$
\begin{aligned}
x^{256}-256^{32} & =x^{2^{8}}-\left(2^{8}\right)^{32}=x^{2^{8}}-2^{2^{8}} \\
& =\left(x^{2^{7}}+2^{2^{7}}\right)\left(x^{2^{6}}+2^{2^{6}}\right) \cdots(x+2)(x-2),
\end{aligned}
$$

Thus, the real roots of the given equation are $\pm 2$, and the sum of the squares of the real roots is 8. Therefore, the answer is $(A)$.