Lời giải:
Đặt $\underbrace{111...1}_{n-1}=a$ thì $9a+1=10^{n-1}$
\(C=\underbrace{444....4}_{n}\underbrace{88...8}_{n-1}9\)
\(=\underbrace{444...4}_{n}.10^n+\underbrace{88..888}_{n-1}.10+9\)
\(=4(\underbrace{11...1}_{n-1}.10+1).10^n+8.\underbrace{111....1}_{n-1}.10+9\)
\(=4(a.10+1).(9a+1).10+8.a.10+9=3600a^2+840a+49=(60a+7)^2\) là bình phương của stn
Ta có đpcm.