The 3rd problem of the mathematical class competition:

Find the six-digit square numbers such that, when the first three digits are erased, the square root of the number is obtained.
Let's denote the square root of the number by $x$, $x^{2}$ is a six-digit number, so $x$ is a three-digit number, and the first three digits of $x^{2}$ are removed, leaving the number formed by the last three digits of $x^{2}$. Therefore, the last three digits of $x^{2}-x=x(x-1)$ are 0, meaning it is divisible by 1000. Our task is to find three-digit numbers $x$ such that $x(x-1)$ is divisible by $1000=125 \cdot 8$.

Let's examine divisibility by 125. Since $x(x-1)$ is divisible by 125, it is also divisible by 5. The greatest common divisor of $x$ and $x-1$ is 1, so only one of $x$ and $x-1$ can be divisible by 5. Therefore, the one divisible by 5 must also be divisible by 125, meaning that one of $x$ and $x-1$ is divisible by 125.

Similarly, it can be shown that one of the two factors is divisible by 8. However, it is impossible for the same factor to be divisible by both 125 and 8, as then it would be divisible by $125 \cdot 8=1000$, making it at least a four-digit number. Therefore, one of $x$ and $x-1$ is divisible by 125, and the other by 8. We need to find a multiple of 125 whose neighbor, either 1 less or 1 more, is divisible by 8. This neighbor must also be even and divisible by 4, so it is sufficient to try the odd multiples of 125: 125, 375, 625, and 875. Among their neighbors, those ending in 26 and 74 are not divisible by 4, and 124 and 876 are not divisible by 8, leaving only two pairs: $x=376$ and $x-1=375$, as well as $x=625$ and $x-1=624$.

Indeed: $376^{2}=141376$ and $625^{2}=390625$.