The natural numbers $a, b, c$, formed by the same $n$ digits $x$, $n$ digits $y$, and $2n$ digits $z$ satisfy $a^2 + b = c$ 
Given that the natural numbers $a, b, c$ are formed by the same $n$
  digits $x$, $n$ digits $y$, and $2n$ digits $z$ respectively. For any
  $n \geq 2$, find the digits $x, y, z$ such that $a^2 + b = c$

Greetings, I was doing this question above and I couldn't figure out how to do.
Here's my progress so far: 

The relation $a^2 + b = c$ means $$(xxx...x)^2 + (yy..yy) = (zzz...zz)$$ where the $x\text{ and }y$ are $n$ in number and the $z$ is in $2n$ in number. This simplifies to $$x^2(11···1)^2 +y(11···1) = z(11···1)$$
  and I couldn't figure out further. :(

By some trial and error, I found $(333)^2 + 222 = 111,111$. 
Any help would be appreciated. I tried but couldn't find out even a similar question somewhere else. Please inform if this question has been asked before.
Thank You

Let
$$f(n) = \sum_{i=0}^{n-1} 10^i \tag{1}\label{eq1}$$
Thus, as you've shown,
$$x^2 f^2(n) + yf(n) = zf(2n) \implies f(n)\left(x^2 f(n) + y\right) = zf(2n) \tag{2}\label{eq2}$$
Also, note that
$$f(2n) = 10^n f(n) + f(n) = f(n)\left(10^n + 1\right) \tag{3}\label{eq3}$$
Thus, substituting that into \eqref{eq2} and dividing by the common factor of $f(n)$ gives
$$x^2f(n) + y = z(10^n + 1) \tag{4}\label{eq4}$$
First, there's the trivial case of $x = y = z = 0$. The rest of this solution will assume that $z \gt 0$. Next, note that
$$f(n) = \frac{10^n - 1}{9} \tag{5}\label{eq5}$$
Thus, substituting \eqref{eq5} into \eqref{eq4} gives
$$\begin{equation}\begin{aligned}
x^2\left(\frac{10^n - 1}{9}\right) + y & = z(10^n + 1) \\
x^2(10^n - 1) + 9y & = 9z(10^n + 1) \\
x^2(10^n) - x^2 + 9y & = 9z(10^n) + 9z \\
(x^2 - 9z)(10^n) & = x^2 - 9y + 9z
\end{aligned}\end{equation}\tag{6}\label{eq6}$$
The RHS is less than $100$ (update: actually, it's less than $1000$, but it can be $100$; see the end for more details) and greater than $-10$, so for $n \ge 2$, this means that $x^2 - 9z = x^2 - 9y + 9z = 0$. Note $x^2 - 9z = 0$ occurs only for $x = 3, z = 1$ and $x = 6, z = 4$. This gives, from $x^2 - 9y + 9z = 0$, that $y = 2$ for the first part and $y = 8$ for the second part.
Another way to see this is that the RHS of \eqref{eq4}, as $z$ is a digit, in base $10$, would be $z$, followed by $n - 1$ zeros, and then $z$ again. On the LHS, $y$ is just a single digit. This means for $n \ge 2$, the digits in $x^2f(n)$ must basically all "disappear" when $y$ is added, i.e., $x^2f(n)$ must be just slightly less (i.e., $\lt 9$) than a power of $10$. Checking the squares of $x$ multiplied by $11$, you can see this only occurs if $x = 3$, so $x^2 = 9$ (with it being $1$ less than a power of $10$), or $x = 6$ so $x^2 = 36$ (with it being $4$ less than a power of $10$). For $x = 3$, this means $z = 1$, which requires that $y = 2$, while for $x = 6$, this means $z = 4$ and $y = 8$.
Note the second solution comes from the first one by multiplying by $4$ in \eqref{eq4}. You can see this changes $x = 3$ to $x = 6$, $y = 2$ to $y = 8$ and $z = 1$ to $z = 4$.
In conclusion, there are $3$ solutions, with the sets of digits $(x,y,z)$ being $(0,0,0)$, $(3,2,1)$ and $(6,8,4)$.
Update: As I saw from the answer by fleablood, I made a mistake in my handling of \eqref{eq6}. The RHS of it is definitely less than $1000$, so what I wrote above is true for $n \ge 3$. However, the RHS can be equal to $100$, which means that for $n = 2$, there's also the solution $(8,3,7)$.
