一般項為2k位
前k位1,接著(k-1)位5,個位6
11.........155........56
=111.........1(2k位)+44........4(k位)+1
=(1/9)*999.........9+(4/9)*999.........9+1
=(1/9)*(10^2k-1)+(4/9)*(10^k-1)+1
=(1/9)*10^2k+(4/9)*10^k+4/9
=[(10^k+2)/3]^2............(1)
10^k=1(mod3),10^k+2=0(mod3)
所以(1)為整數平方,得證