Too simple and beautiful, to be true?

Starting from the left hand side of the equality and considering the decimal expansion of the numbers, we get:

$\underbrace{11\ldots1}_{2n}-\underbrace{22\ldots2}_{n}=\left(10^{2n-1}+10^{2n-2}+\ldots+10+1\right)-2\times\left(10^{n-1}+10^{n-2}+\ldots+10+1\right)$

$=\frac{10^{2n}-1}{10-1}-2\times\frac{10^{n}-1}{10-1}=\frac{1}{9}\times\left(10^{2n}-2\times10^{n}+1\right)$

$=\frac{1}{9} \times(10^n-1)^2=9\times\left(\frac{10^{n}-1}{10-1}\right)^2$

$=9\times{(\underbrace{11\ldots1}_{n})}^2$

$=(\underbrace{33\ldots3}_{n})^2$

Hence, the statement is both beautiful and true .

$\begin{aligned} LHS & = \underbrace{111\cdots 1}_{2n} - \underbrace{222\cdots 2}_n \\ & = {\color{#3D99F6}\frac {10^{2n}-1}9} - \color{#D61F06} 2 \left(\frac {10^n-1}9\right) \\ & = {\color{#3D99F6} \left(\frac {10^n}3\right)^2-\frac 19} - \color{#D61F06} 2 \left(\frac {10^n}3\right)\left(\frac 13\right) + \frac 29 \\ & = \left(\frac {10^n}3\right)^2 - 2 \left(\frac {10^n}3\right)\left(\frac 13\right) + \frac 19 \\ & = \left(\frac {10^n}3-\frac 13\right)^2 \\ & = \bigg(\frac {\overbrace{999\cdots 9}^n}3\bigg)^2 \\ & = \big(\underbrace{333\cdots 3}_n\big)^2 \end{aligned}$

True , $LHS = RHS$ .