For how many numbers $X^2 \equiv X \mod 10^n$? I've been looking through this post:

Square of four digit number $a$

And wanted to see if there's a way to generalize the idea, in order to answer the following question:

For how many numbers $X^2 \equiv X \mod 10^n$?

Also, is $9376$ the only four digit number such that $9376^2 \equiv 9376 \mod 10^4$
Thanks

The number of distinct numbers modulo $10^n$ solving
$$x^2\equiv x\pmod{10^n}\tag1$$
is four.
To see this by the Chinese remainder theorem $x$ solves (1) iff both
$$x^2\equiv x\pmod{2^n}\tag2$$
and
$$x^2\equiv x\pmod{5^n}\tag3.$$
These congruences state that $p^n\mid x(x-1)$ for $p=2$ or $5$.
As $x$ and $x-1$ have no common factor then $p^n\mid x$ or $p^n
\mid(x-1)$. So the solutions are $x\equiv 0$ or $1\pmod{p^n}$.
By the Chinese remainder theorem again these can be assembled into
four solutions modulo $10^n$. Two are obvious: $0$ and $1$. The other
two are $x_n$ and $1-x_n$ where $x_n\equiv 0\pmod{2^n}$ and
$x_n\equiv 1\pmod{5^n}$. These can be found by applying the extended
Euclidean algorithm to $2^n$ and $5^n$.
For $n=4$ you have $x_4=9376$, as you say, so $1-9376\equiv 625\pmod{10^4}$. So in this case $x_4$ is the only solution
between $10^3$ and $10^4-1$ inclusive, that is "with four digits".
