$8 \cdot 11$ In the equations $z^{x}=y^{2 x}$ (1), $2^{z}=2 \cdot 4^{x}$ (2), $x+y+z=$
16 (3), the integer roots for $x, y, z$ are respectively
(A) $3,4,9$.
(B) $9,-5,12$.
(C) $12,-5,9$.
(D) $4,3,9$.
(E) $4,9,3$.
(3rd American High School Mathematics Examination, 1952)
[Solution] From (1) we get $z=y^{2}$.
From (2) we have $2^{z}=2 \cdot 2^{2 x}=2^{2 x+1}$, which gives $z=2 x+1$.
$\therefore x=\frac{z-1}{2}=\frac{y^{2}-1}{2}$. Substituting into (3) gives $\frac{y^{2}-1}{2}+y+y^{2}=16$, which simplifies to $3 y^{2}+2 y-33=0$.

Solving for $y$ yields the only integer root $3$, and $x=4$, $z=9$. Therefore, the answer is $(D)$.