<s>[INST] How prove this congruence equation has four zeros solution Question:
let congruence equation
$$\begin{cases}
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{m}}(\mod 10)\\
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{m-1}a_{m}}(\mod 10^2)\\
\cdots\cdots\cdots\cdots\\
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{1}a_{2}\cdots a_{m}}(\mod 10^m)
\end{cases}$$
show that: this congruence equation has $four$ zeros solution.
where $m\ge 2,a_{i}\in\{0,1,2,\cdots,9\}$
My try: 
for example 
when $m=2$ then 
$25$and $76$ ,$0,1$this four solution  such
first $0$ and $1$ is such it
and 
$$25^2=625\equiv 5(\mod 10),76^2=5776\equiv 6(\mod 10)$$
$$25^2=625\equiv25(\mod 10^2),76^2=5776\equiv 76(\mod 10^2)$$
$$25^2=625\equiv 625(\mod 10^3),76^2=5776\equiv 776(\mod 10^3)$$
But My problem I can't prove it,Thank you very much!
 [/INST] As @Hurkyl pointed out, if we let $x=\overline{a_1a_2\ldots a_m}$, it suffices to prove that $x^2 \equiv x \pmod{10^m}$ has exactly four solutions $\pmod{10^m}$.
This is straightforward; we get $2^m(5^m)=10^m \mid x(x-1)$. Note that $\gcd(x, x-1)=1$, so we have four cases:
\begin{align}
x \equiv 0 \pmod{2^m}, x \equiv 0 \pmod{5^m} \\
x \equiv 0 \pmod{2^m}, x \equiv 1 \pmod{5^m} \\
x \equiv 1 \pmod{2^m}, x \equiv 0 \pmod{5^m} \\
x \equiv 1 \pmod{2^m}, x \equiv 1 \pmod{5^m}
\end{align}
For each case, Chinese Remainder Theorem ensures the existence of a unique solution for $x \pmod{10^m}$. We thus get exactly four solutions.
</s>