`A=\underbrace{111...1}_{2n}-\overbrace{222...2}^{n}`
`=1\underbrace{000...0}_{n-1}1.\underbrace{111...1}_{n}-2.\overbrace{111...1}^{n}`
`=(1\underbrace{000...0}_{n-1}1-2).\underbrace{111...1}_{n}`
`=\underbrace{999...9}_{n}.\underbrace{111...1}_{n}`
`=\underbrace{333...3}_{n}.3.\underbrace{111...1}_{n}`
`=\underbrace{333...3}_{n}.\underbrace{333...3}_{n}`
`=(\underbrace{333...3}_{n})^2`
Vì `\underbrace{333...3}_{n}∈\mathbbZ`
`⇒A` là `SCP` `(đpcm)`
