$7 \cdot 23$ If $M=\left\{(x, y)|| \operatorname{tg} \pi y \mid+\sin ^{2} \pi x=0\right\}, N=\{(x, y) \mid$ $\left.x^{2}+y^{2} \leqslant 2\right\}$, then the number of elements in $M \cap N$ is
(A) 4
(B) 5 .
(C) 8 .
(D) 9 .
(China High School Mathematics League, 1993)
［Solution］From the condition that the sum of non-negative integers is 0, for $M$ we have $\left\{\begin{array}{l}\operatorname{tg} \pi y=0, \\ \sin ^{2} \pi x=0,\end{array}\right.$ which yields $\left\{\begin{array}{ll}x=k & (k \in \mathbb{Z}), \\ y=m & (m \in \mathbb{Z}) .\end{array}\right.$
Thus, the set $M$ consists of all integer points (lattice points) on the coordinate plane. For $N$, from $\quad x^{2}+y^{2} \leqslant 2=(\sqrt{2})^{2}$, then $N$ is the set of all points within the closed circle centered at the origin with radius $\sqrt{2}$.

It is not difficult to calculate that $k=-1,0,1, \quad m=-1,0,1$, as shown in the figure, there are 9 integer points within the closed circle. Therefore, the answer is $(D)$.