Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$. Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.
So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod 1000$. I  don't know how to proceed since 1000 seems to be a bit high.

In a practical way, the number $N$ is, $\forall p>=0$ $$p\times1000+(100a+10b+c)$$
We can ignore the prefix $p\times1000$ as it doesn't affect the $3$ last digits of $N^2$
We want $c$ and the last digit of $c^2$ to be equal
$$c\equiv c^2\pmod{10}$$
That only works for 
$$\bbox[7px,border:1px solid gray]{c\in\{0,1,5,6\}}$$
For $b$, we want the second digit of $(10b+c)^2$ to be equal to $b$ (note: $\lfloor x\rfloor$ denotes the integer part of $x$)
$$b\equiv \lfloor\frac{(10b+c)^2}{10}\rfloor\pmod{10}$$
Plugging our $c$ values, $b$ can only be $0$ if $c=0$, also $0$ if $c=1$, $2$ if $c=5$ and $7$ if $c=6$
$$\bbox[7px,border:1px solid gray]{c=0\implies b = 0}\\
\bbox[7px,border:1px solid gray]{c=1\implies b = 0}\\
\bbox[7px,border:1px solid gray]{c=5\implies b = 2}\\
\bbox[7px,border:1px solid gray]{c=6\implies b = 7}\\$$
Not so many valid candidates after all.
Finally, for $a$, we want the third digit of $(100a+10b+c)^2$ to be equal to $a$
$$a\equiv \lfloor\frac{(100a+10b+c)^2}{100}\rfloor\pmod{10}$$
We have only $4$ cases to check since $b$ and $c$ values are strongly correlated. Injecting our $(b,c)$ values,we get
$$\bbox[7px,border:1px solid gray]{c=b=0\implies a = 0}\\
\bbox[7px,border:1px solid gray]{c=1,b=0\implies a = 0}\\
\bbox[7px,border:1px solid gray]{c=5,b=2\implies a = 6}\\
\bbox[7px,border:1px solid gray]{c=6,b=7\implies a = 3}\\$$
Meaning the last $3$ digits of $N$ can be any of
$$000, 001, 625, 376$$
Since we want the smaller valid number $>1$, the answer is
$$\bbox[8px,border:1px solid blue]{376}\\$$
