II. (16 points) Find all three-digit numbers such that the last three digits of their square are the original two-digit number.



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Let the three-digit number $m$ satisfy the condition, then $m^{2}=k \times 1000+$ $m$, where $k$ is a natural number. Therefore,
$$
m(m-1)=k \times 1000=k \times 2^{3} \times 5^{3} .
$$
$\because m, m-1$ are two coprime natural numbers, and one is odd while the other is even, hence the odd number among $m, m-1$ is a multiple of $5^{3}$, and the even number is a multiple of $2^{3}$.

If $m=125,375,625,875$, then $m-1=124,374$, 624,874 , among which only 624 is a multiple of $2^{3}$;

If $m-1=125,375,625,875$, then $m=126,376$, 626,876 , among which only 376 is a multiple of $2^{3}$.
In summary, the three-digit numbers sought are 625 or 376.