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

If we let $1_n= \underbrace{1111....1}_{n}$
$a^2 + b = c$ 
$\frac {a^2}{1_n} + \frac {b}{1_n} = \frac {c}{1_n}$
$x^2*1_n + y = z*(10^n + 1)$
Let's think about what that means:
If $x^2 = 10j + k$ we have
$10j + k + y \equiv z$.  And we carry $j$ or $j+1$
Then we have $j +k (+1)\equiv 0$.  Which means $j+k(+1) = 10$.
This can occur if:
$x=1$ so $k=1$ and $j=0$, $j+k(+1) \equiv 1,2 \not \equiv 0$.
$x = 2$ so $k=4$ and $j=0$, $j+k(+1) \equiv 4,5\not equiv 0$.
$x=3$ so $k=9$ and $j=0$ and $j+k(+1)\equiv 0$ if we carry.
$x=4$ so $k=6$ and $j=1$ and $j+k(+1)\not \equiv 0$.
$x=5$ so $k=5$ and $j=2$ and $j+k(+1)\not \equiv 0$.
$x=6$ so $k=6$ and $j=3$ and $j+k+1\equiv 0$.
$x=7$ so $k=9$ and $j=4$ and $j+k(+1) \not \equiv 0$.
$x=8$ so $k =4$ and $j=6$ and $j+k\equiv 0$.
$x=9$ so $k =1$ and $j = 8$ and $j+k(+1)\equiv 0$.
Now we repeat this and carry again to get the third digit.  Assume $n > 2$  As $j+k(+1) = 10$ we must carry $1$ and get $j+k+1$ so this can only happen if $j+k+1 = 10$; not $j+k=10$.
So this can occur if:
$x=3; k=9;j=0$ and $z= j+1=1$ and $9+y \equiv 1$ so $y = 2$.
$x=6; k=9;j=3$ and $z = j+1 = 4$ and $36+y\equiv 4$ so $y=8$.
$x=9; k=1; j=8$ and $z = j+1=9$ and $81+y\equiv 9$ so $y=8$ but.. then we don't carry the $1$.  we must have $y=18$ but that's not a single digit. 
But if $n = 2$ we can have:
$x^2 = 10j + k$ and $x^2*11 = 100j + 10(k+j) + k$ and $x^2*11 + y= 100j + 10(k+j) + k+y$ where $k+y = z$ and $k+j = 10$ and $j+1 = z$
Then we can have $x=8;k=4;j=6;z=7$ and $y = 3$. (i.e  $8^2*11 + 3= 707$ or $(88)^2 + 33 = 7777$
