Join 1 more 6 , find 1 more 4 and 5.

Call the number $A= 666\cdots 666$ having $n$ numbers of sixes. Now $A^2 = (\underbrace{666\cdots 666}_{\text{n 6's}} )^2=36\left(\dfrac{10^n-1}{9}\right)^2=\dfrac{4}{9}\,(10^{2n} -2\cdot 10^n +1)$ Since $\begin{aligned} 10^{2n} -2 \cdot 10^n & = 10^n(10^n -2) \\&= 10^n(\underbrace{999\cdots 9}_{\text{ n-1 9's}}8)\\& =\underbrace{999\cdots 9}_{\text{ n-1 9's}}8\underbrace{000\cdots 0}_{\text{n 0's}}\end{aligned}$ Therefore We have the following $\begin{aligned} B= 10^{2n} -2\cdot 10^n +1 & = \underbrace{999\cdots 9}_{\text{ n-1 9's}}8\underbrace{000\cdots 0}_{\text{n -1 0's}}1\\& = \underbrace{999\cdots 9}_{\text{ n-1 9's}}0 + 8\underbrace{000\cdots 0}_{\text{n -1 0's}}1 \end{aligned}$ Let $a_0=81$ and observe that $a_1 = 801 = 720+81=720+a_0, \ a_2= 8001 = 7200+ 801 =7200+a_1$ and $\forall n\geq 2$ we have following recursive formula (by induction can be proved) $a_{n-1} = 72\times 10^{n-1} + a_{n-2}$ which when divided by $9$ gives $R = \sum_{k=1}^{n-1}8\cdot 10^{k} + 9 = \underbrace{88\cdots 8}_{\text{n-1 8's}} 9$ Thus we prove $\dfrac{B}{9} = \underbrace{11\cdots 11}_{\text{n-1 1's }} 0+ \underbrace{88\cdots 8}_{\text{n-1 8's}} 9$ and further multiplying gives $\begin{aligned}\dfrac{4B}{9} & = \underbrace{44\cdots 44}_{\text{n-1 4's}} 0+3\underbrace{55\cdots 55}_{\text{n-1's}} 6\\ A^2 & = (\underbrace{666\cdots 666}_{\text{n 6's}} )^2=\underbrace{44\cdots 44}_{\text{n-1 4's}}3\underbrace{55\cdots 55}_{\text{n-1 5's}} 6\end{aligned}$ Proof for $4\times \underbrace{88\cdots 8}_{\text{n-1 8's}} 9 = 3\underbrace{55\cdots 55}_{\text{n-1 5's}} 6$ If $a_0= 89$ then $a_1 =889 = 800+a_0 , \ a_2 = 8889 = 8000+a_1$ Giving us the recursive formula $\forall n\geq 2$ ( by induction can be proved ) $a_{n-1} = 8\cdot 10^{n-1} +a_{n-2}$ and hence $4a_{n-1} = 32\cdot 10^{n-1} +4a_{n-2} = \sum_{k=1}^{n-1} 32\cdot 10^k + 36 = 3\underbrace{55\cdots 55}_{\text{n-1 5's}}$