By induction: If we have $x$ such that $x^2-x\equiv 0\pmod{10^n}$, then we need to find a digit $k$ with $(10^nk+x)^2-(10^nk+x)\equiv 0\pmod{10^{n+1}}$. This can be done, since the congruence reduces to $10^n(2xk-k+\frac{x^2-x}{10^n})\equiv 0\pmod{10^{n+1}}$, ie $k(2x-1)\equiv -\frac{x^2-x}{10^n}\pmod{10}$, and $2x-1\equiv 1\pmod{10}$.

$x_n=10^n-(5^{2^{n-2}}-1),(mod \ 10^n),n>1$

Rust wrote: $x_n=10^n-(5^{2^{n-2}}-1),(mod \ 10^n),n>1$ That doesn't seem to work for n=2, or any n in fact..

It is work for any n>3. We have $x^2-x=x(x-1)=(-x)(-x+1)=(5^{2^{n-2}}-1)5^{2^{n-2}}=0(mod \ 10^n)$, because $5^{2^{n-2}}-1=0(mod 2^n),n>2,5^{2^{n-2}}=0(mod 5^n),n>3$. $x=10^n-(5^{2^{n-1}}-1)$ work for any n.