I2.4 If $111111111111-222222=(R+S)^{2}$ and $S>0$, find the value of $S$.
$\begin{array}{l}111111111111-222222=(1+S)^{2} \\ 111111(1000001-2)=(1+S)^{2} \\ 111111 \times 999999=(1+S)^{2} \\ 3^{2} \times 111111^{2}=(1+S)^{2} \\ 1+S=333333 \\ S=333332\end{array}$