Să\;se\; arate \;că\; numărul\; a = \underbrace{111…1}_\text{n} \underbrace{555…5}_\text{n-1}6\; este \;pătrat \;perfect.Rezolvare:
\underbrace{111…1}_\text{n} \underbrace{555…5}_\text{n-1}6=\\ \;\\ =\underbrace{111…1}_\text{n} \underbrace{555…5}_\text{n} + 1 = \\ \;\\ =\underbrace{111…1}_\text{n} \cdot10^n + 5\cdot\underbrace{111…1}_\text{n} + 1=\\ \;\\ =\underbrace{111…1}_\text{n}\cdot(10^n+5)+1 =\\ \;\\=\underbrace{111…1}_\text{n}\cdot(\underbrace{999..9}_\text{n}+1 +5)+1 = \\ \;\\ =\underbrace{111…1}_\text{n}\cdot(\underbrace{999..9}_\text{n} +6)+1 = \\ \; \\=\underbrace{111…1}_\text{n}\cdot\underbrace{999..9}_\text{n} + \underbrace{111…1}_\text{n}\cdot6 + 1 = \\ \;\\ =\underbrace{111…1}_\text{n}\cdot 9 \cdot \underbrace{111…1}_\text{n} + \underbrace{111…1}_\text{n}\cdot 2 \cdot 3 + 1 =\\ \;\\ = (3\cdot \underbrace{111…1}_\text{n})^2 + 2 \cdot(3\cdot \underbrace{111…1}_\text{n}) + 1 =\\ \;\\ =(3\cdot \underbrace{111…1}_\text{n} + 1)^2
