$33 \cdot 19$ Given $z=\sqrt{x^{2}+y^{2}}$, the set of integer pairs $(x, y)$ that can make $z$ a positive integer is
(A) $(15213,32477)$.
(B) $(37615,26855)$.
(C) $(28326,28614)$.
(D) $(25530,29464)$.
(China Beijing High School Mathematics Competition, 1990)
[Solution] Consider the last digits of the squares of the four sets of numbers $(A),(B),(C),(D)$:
$$
(9,9),(5,5),(6,6),(0,6) \text {, }
$$

Then the last digits of their sums are $8,0,2,6$, respectively. Since the last digit of a perfect square can only be $0,1,4,5,6,9$, we can eliminate $(A),(C)$.
Next, consider the sums of the digits of the two numbers in set $(B)$:
$$
3+7+6+1+5=21,2+6+8+5+5=26,
$$

These are numbers of the form $3k+1$ and $3k+2$, respectively. When squared, they become numbers of the form $3k+1$ and $3k+1$. Thus, their sum is a number of the form $3k+2$.
Since a number of the form $3k+2$ cannot be a perfect square, we can eliminate $(B)$.
Therefore, the answer is $(D)$.