［Solution］The last digit of a perfect square can only be $1,4,5,6,9,0$.
This allows us to eliminate $(B),(C)$.
Furthermore, for a perfect square ending in 5, the last two digits should be 25.
By examining the last two digits of the squares in groups $(A),(D)$, we find that $(A)$ is the correct choice. Therefore, the answer is $(A)$.
$33 \cdot 17$ Given that among the following four sets of values for $x$ and $y$, there is exactly one set for which $\sqrt{x^{2}+y^{2}}$ is an integer. This set of $x$ and $y$ values is
(A) $x=88209, y=90288$.
(B) $x=82098, y=89028$.
(C) $x=28098, y=89082$.
(D) $x=90882, y=28809$.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1991)