Is this number a perfect square?

For any integer $N$ consider the number $S_N \; = \; \underbrace{111\cdots1}_{\times N}\underbrace{222\cdots2}_{\times (N+1)}5$ for then $\begin{aligned} S_N & = \; \frac{10^N-1}{9} \times 10^{N+2} + 2\frac{10^{N+1}-1}{9}\times10 + 5 \\ & = \; \tfrac1910^{2N+2} - \tfrac1910^{N+2} + \tfrac2910^{N+2} - \tfrac{20}{9} + 5 \\ & = \; \tfrac19\left(10^{2N+2} + 10^{N+2} + 25\right) \; = \; \tfrac19\left(10^{N+1} + 5\right)^2 \end{aligned}$ and hence $S_N$ is the square of $\tfrac13\big(10^{N+1} + 5\big)$ .

$\textbf{First Solution:}$

Similar to @Mark Hennings solution :

Let $N= \underbrace{11\dots11}_{2018}\times10^{2020}+\underbrace{22\dots22}_{2019}\times10+5$

$N= \frac{1}{9}(10^{2018}-1)\times10^{2020}+\frac{2}{9}(10^{2019}-1)\times10+5$

$N=\frac{1}{9}(10^{4038}+2\times5\times10^{2019}+25)=[\frac{1}{3}(10^{2019}+5)^2]$

$N=\Big(\large\frac{1\overbrace{00\dots00}^{2018}5}{3}\Big)^2=(\underbrace{33\dots33}_{2018}5)^2$ , hence $N$ is a perfect square.

$\textbf{Second Solution:}$

Note that $9N= 1\underbrace{00\dots00}_{2017}1\underbrace{00\dots00}_{2018}25=10^{4038}+10^{2020}+25=(10^{2019}+5)^2\implies N$ is a perfect square.

$\textbf{Third Solution}:$

There is a lemma that states: (and it was the inspiration to the problem)

Let $b$ be an integer greater than $5$ . For each positive integer $n$ , consider the number:

$x_n=\underbrace{11\dots1}_{n-1}\underbrace{22\dots2}_{n}5$

written in base $b$ . Then the following holds true if and only if $b=10$ ; that there exists a positive integer $M$ such that for every integer $n$ greater than $M$ , the number $x_n$ is a perfect square.

My logic : nobody just comes with a 4038 digit number and asks whether it is a perfect square, so it must be a perfect square :p