Is it perfect square?

$\boxed{\text{Yes}}$ : $\large \left(\underbrace{33\dots33}_{2017}4\right)^2 = \underbrace{11\dots11}_{2018}\underbrace{55\dots55}_{2017}6.$

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A useful fact is: $\underbrace{11\dots11}_n = \sum_{i=0}^{n-1} 10^i = \frac{10^n - 1}9.$

The given number may be written as $\underbrace{11\dots11}_{4036} + 4\cdot \underbrace{11\dots11}_{2018} + 1 \\ = \frac{10^{4036} - 1} 9 + 4\cdot \frac{10^{2018} - 1} 9 + 1 \\ = \frac{10^{4036}} 9 + \frac{4 \cdot 10^{2018}} 9 + \frac 4 9 \\ = \left(\frac{10^{2018}} 3 + \frac 2 3\right)^2 \\ = \left(\frac{1 + 9\cdot \overbrace{11\dots11}^{2018}}3 + \frac 2 3\right)^2 \\ = \left(1 + \underbrace{33\dots33}_{2018}\right)^2 \\ = \left(\underbrace{33\dots33}_{2017}4\right)^2.$

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Another approach, less thorough but more intuitive, is to observe: $\begin{array}{rl} 34^2 & = 1156 \\ 334^2 & = 111556 \\ 3334^2 & = 11115556 \\ 33334^2 & = 1111155556 \\ & \vdots \end{array}$ After going through a handful of examples, most people are willing to believe the pattern continues indefinitely!

We can prove by induction that it continues indeed. If $\underbrace{33\dots33}_{n}4^2 = \underbrace{11\dots11}_{n+1}\underbrace{55\dots55}_n6$ , then $\underbrace{33\dots33}_{n+1}4^2 = (3\cdot 10^{n+1} + \underbrace{33\dots33}_n4)^2 \\ = (3\cdot 10^{n+1})^2 + 2\cdot 3\cdot 10^{n+1}\cdot \underbrace{33\dots33}_n4 + (\underbrace{33\dots33}_n4)^2 \\ = 9\cdot 10^{2n+2} + 6\cdot \underbrace{33\dots33}_n4\cdot 10^{n+1} + \underbrace{11\dots11}_{n+1}\underbrace{55\dots55}_n6 \\ = 9\underbrace{00\dots00}_{2n+2} + 2\underbrace{00\dots00}_n4\underbrace{00\dots00}_{n+1} + \underbrace{11\dots11}_{n+1}\underbrace{55\dots55}_n6 \\ = 11\underbrace{11\dots11}_n5\underbrace{55\dots55}_n6 \\ = \underbrace{11\dots11}_{n+2}5\underbrace{55\dots55}_{n+1}6.$