$x,y,z\in \{1,2,3,4,5,6,7,8,9\}$ such that $\sqrt{\underset{"2000"times}{\underbrace{xxxx\mathrm{.....}xx}}-\underset{"1000"times}{\underbrace{yyyy\mathrm{.....}y}}}=\underset{"1000"times}{\underbrace{zzzz\mathrm{.....}zz}}$ then

$$\begin{gathered} x.\frac{{10}^{2000}-1}{9}-y.\frac{{10}^{1000}-1}{9}=\frac{z^2.{({10}^{1000}-1)}^2}{81} \\ {10}^{1000}=k\Rightarrow x.\frac{(k^2-1)}{9}-y.\frac{(k-1)}{9}=\frac{z^2{(k-1)}^2}{81} \\ x(k+1)-y=\frac{z^2.(k-1)}{9};9x(k+1)-9y=z^2.k-z^2 \\ k(z^2-9x)=z^2+9x-9y\left\{z^2\ne 9x\Rightarrow k(z^2-x)>z^2+9x-9y\right\} \\ {z}^2=9x\Rightarrow x=1,z=3,y=2;x=4,z=6,y=8 \end{gathered}$$