8.29 Find all integers not exceeding 1000 such that the last two digits of their square are the same but not zero.
(Kyiv Mathematical Olympiad, 1962)
[Solution] Note that the last digit of a square number is not zero, so the last digit of a square number can only be
$$
1,4,5,6,9 \text {. }
$$

First, the last two digits of a perfect square cannot both be odd.
This is because the tens digit of the square of an odd number is even, while the units digit is odd.
Therefore, the last two digits of a perfect square cannot be $11,55,99$.
In addition, if the units digit of a perfect square is 6, then its tens digit is odd.
Therefore, the last two digits of a perfect square cannot be 66.
Thus, we only need to find perfect squares whose last two digits are 44.
Since
$$
(50 k+m)^{2}=2500 k^{2}+100 k m+m^{2}
$$

then $50 k+m$ and $m$ have the same last two digits in their squares. Therefore, we only need to find those in $\{1,2$, $\cdots, 50\}$ whose squares have the last two digits as 44.
We calculate $12^{2}, 18^{2}, 22^{2}, 28^{2}, 32^{2}, 38^{2}, 42^{2}, 48^{2}$ and find that only
$$
12^{2}=144,38^{2}=1444 \text {. }
$$

Thus, the numbers with the property in question are
$$
50 k+12, \quad 50 k+38
$$

where $k \in\{0,1,2, \cdots, 19\}$.
Therefore, there are 40 perfect squares whose last two digits are the same among the numbers not exceeding 1000.