Ta có `\underbrace{99...9}_{n}\overbrace{00...0}^{n}25`
`=\underbrace{99...9}_{n}\overbrace{00...0}^{n+2}+25`
`=(10^n-1).10^{n+2}+25`
`=10^{2n+2}-10^{n+2}+25`
`=(10^{n+1})^2-2.5.10^{n+1}+5^2`
`=(10^{n+1}-5)^2`
`⇒\underbrace{99...9}_{n}\overbrace{00...0}^{n}25` là bình phương của :
$\begin{cases}10^{n+1}-5\\5-10^{n+1}\end{cases}$