10.26 Find the largest perfect square such that when the last two digits of this perfect square are subtracted from it, the result is still a perfect square (assuming the digits subtracted are not all $0$).
(4th All-Russian Mathematical Olympiad, 1964)
[Solution] Let this number be $x^{2}$, where $x$ is a positive integer. The last two digits of the number are $y$, where $y$ is a positive integer, and $041$, then from $z \leqslant 4$ we know
$$
x^{2}-100 z^{2} \geqslant 42^{2}-1600>100 .
$$

Thus, the largest perfect square number sought is $41^{2}=1681$.