$31 \cdot 28$ satisfies the equation $x^{4} y^{4}-10 x^{2} y^{2}+9=0$ for the number of different pairs $(x, y)$ (where $x, y$ are positive integers) is
(A) 0.
(B) 3.
(C) 4.
(D) 12.
(E) infinite.
(20th American High School Mathematics Examination, 1969)
［Solution］The given equation is equivalent to
$$
\left(x^{2} y^{2}-1\right)\left(x^{2} y^{2}-9\right)=0 \text {, }
$$

Therefore,
$$
x y= \pm 1 \text { or } \pm 3 \text {. }
$$

Thus, the different pairs of positive integers that satisfy the equation are $(1,1),(1,3)$ or $(3,1)$. Hence, the answer is $(B)$.