$33 \cdot 18$ If $\sqrt{x^{2}+y^{2}}$ is a positive integer. Then among the following four sets of values, $x$ and $y$ can only be
(A) $x=25530, y=29464$.
(B) $x=37615, y=26855$.
(C) $x=15123, y=32477$.
(D) $x=28326, y=28614$.
(China Beijing Junior High School Mathematics Competition, 1990)
[Solution] Given that $x^{2}+y^{2}$ should be a perfect square, its unit digit can only be one of 0, 1, 4, 5, 6, 9.
Thus, we can first eliminate (C) and (D).
Furthermore, if (B) were true, since the last two digits of $x^{2}+y^{2}$ are 50, it cannot be a perfect square. This is also not valid.
Therefore, the answer is (A).