Which version of this question is right? 
Find digits $x,y,z$ such that the equality $$\sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}}} = \overline{\underbrace{zz\cdots z}_{n}}$$ holds for at least two values of $n \in \mathbb{N}$, and in that case find all $n$ for which this equality is true.

Is my method below correct?
Let $A_n=\overline{\underbrace{11\ldots 1}_{n}}$.  Then we have $$\sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}}} =  \sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$n$}}\cdot 10^n+xA_n-yA_n}$$ and thus equals $\sqrt{\smash[b]{A_n(x10^n+x-y)}}$.
Lemma. A number of the form $A_n(x10^n+x-y)$ can't be a perfect square.
Proof. We know that $A_n = 4k+3$, which is never a perfect square, except possibly when $n = 1$. Therefore in the case $\gcd(A_n, x10^n+x-y) = 1$, we know that the only time a solution will occur is when $n=1$. So we can disregard this case. Now, if $\gcd(A_n, x10^n+x-y) \neq 1$, that must mean $x=0$, an impossibility. $\square$
 A: No, unfortunately your solution is not correct. 
Your proof breaks down when you make the claim: 

Now, if $\gcd(A_n, x10^n+x-y) \neq 1$, that must mean $x=0$, an impossibility.

If say $n = 2$, $x = 1$, $y = 2$, and $z = 3$, then we have $\gcd(A_n, x10^n+x-y) = \gcd(11,99) = 11 \neq 1$, but $x = 1 \neq 0$.
So your claim is false in this case.

Hint for solving the problem: Try writing the given equation as: 
$$\sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}}} = \overline{\underbrace{zz\cdots z}_{n}}$$
$$\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}} = (\overline{\underbrace{zz\cdots z}_{n}})^2$$
$$\dfrac{x(10^{2n}-1)}{9} - \dfrac{y(10^n-1)}{9} = \dfrac{z^2(10^n-1)^2}{9^2}$$
$$9x(10^n+1) - 9y = z^2(10^n-1)$$
$$9x(10^n-1) + (18x-9y) = z^2(10^n-1)$$
$$(9x-z^2)(10^n-1) = 9y-18x$$
Can you figure out values of $n,x,y,z$ that work from this equation? Remember that $x,y,z$ are digits which must be between $0$ and $9$.
