$31 \cdot 53$ satisfies the system of equations
$$
\left\{\begin{array}{l}
x^{2}-3 x y+2 y^{2}-z^{2}=31, \\
-x^{2}+6 y z+2 z^{2}=44, \\
x^{2}+x y+8 z^{2}=100 .
\end{array}\right.
$$

The number of ordered triples $(x, y, z)$ is
(A) 0.
(B) 1.
(C) 2.
(D) A finite number greater than 2.
(E) Infinitely many.
(31st American High School Mathematics Examination, 1980)
［Solution］Adding and reorganizing the three known equations, we get
$$
\begin{array}{c}
\left(x^{2}-2 x y+y^{2}\right)+\left(y^{2}+6 y z+9 z^{2}\right)=175, \\
(x-y)^{2}+(y+3 z)^{2}=175 .
\end{array}
$$

or
Since the remainder of a perfect square when divided by 4 can only be 0 or 1, the sum of two perfect squares when divided by 4 can only have a remainder of 0, 1, or 2.
However, the right side of the equation above is $175 \div 4=43 \cdots \cdots 3$, so the equation has no solution. Therefore, the answer is $(A)$.