New Year Problem (2)

Consider a number $A_n = \underbrace{11111\cdots11111}_{\text{Number of 1's}=n} = \displaystyle \sum_{k=0}^{n-1} 10^k = \frac {10^n-1}9$ . Then

$\begin{aligned} \underbrace{333\cdots3}_n4^2 & = (30A_n + 4)^2 \\ & = 900A_n^2 + 240A_n + 16 \\ & = 900 \left(\frac {10^n-1}9\right)A_n + 240A_n + 16 \\ & = 100 \left(10^n-1 \right)A_n + 240A_n + 16 \\ & = 10^{n+2}A_n + 100A_n + 40A_n+ 16 \\ & = \underbrace{111\cdots1}_n\underbrace{000\cdots0}_{n+2}+ \underbrace{111\cdots1}_n00+\underbrace{444 \cdots4}_n0 + 16 \\ & = \underbrace{111\cdots 1}_{n+1}\underbrace{555\cdots5}_n6 \end{aligned}$

Put $n=2019$ and it is "Yes" .

The inspiration to this problem, it came to my mind that $34^2=1156, 334^2= 111556, 3334^2= 11115556$ and so on. I still need to find a logical solution that works for a general case.

$\underbrace{1111111 ... 1111111}_\text{Number of 1's=2020} \underbrace{5555555 ... 5555555}_\text{Number of 5's=2019} 6$

$= \underbrace{11111 ... 11111}_\text{2 * 2020} + \underbrace{44444 ... 44444}_\text{2020} + 1$

$= \frac{10^{2 \cdot 2020} - 1}{9} + 4 \cdot \frac{10^{2020} - 1}{9} + 1$

$= \frac{10^{2 \cdot 2020} - 1}{9} - 2 \cdot \frac{10^{2020} - 1}{9} + 2 \cdot \frac{10^{2020} - 1}{3} + 1$

$= \frac{10^{2 \cdot 2020} - 1 - 2(10^{2020} - 1)}{9} + 2 \cdot \frac{10^{2020} - 1}{3} + 1$

$= \frac{10^{2 \cdot 2020} - 2 \cdot 10^{2020} + 1}{9} + 2 \cdot \frac{10^{2020} - 1}{3} + 1$

$= (\frac{10^{2020} - 1}{3})^2 + 2 \cdot \frac{10^{2020} - 1}{3} + 1$

$= (\frac{10^{2020} - 1}{3} + 1)^2$