1. Let $n$ be a three-digit number, satisfying $100 \leqslant n \leqslant 999$. Find all $n$ such that the last three digits of $n^{2}$ are equal to $n$.
1. The original proposition is equivalent to $1000 \mid\left(n^{2}-n\right)=n(n-1)$. Since $(n, n-1)=1$ and only one of them is even, and $1000=$ $2^{3} \times 5^{3}$, there are the following 4 cases:
a) $n$ is a multiple of 1000, then $n$ is not a three-digit number;
b) $n-1$ is a multiple of 1000, then $n$ is not a three-digit number;
c) $n$ is a multiple of $2^{3}$, and $n-1$ is a multiple of $5^{3}$. Let $n=2^{3} a, n-1=5^{3} b$, then $2^{3} a-5^{3} b=1$.
By the Euclidean algorithm, it is easy to get: $a=47, b=3$. Thus, $n=376$.
d) $n$ is a multiple of $5^{3}$, and $n-1$ is a multiple of $2^{3}$. Similarly, let $n=5^{3} c, n-1=2^{3} d$, from $5^{3} c-2^{3} d=1$, we get: $c=5$, $d=78$. Thus, $n=625$.
In summary: $n=376$ or $n=625$.