由 lskuo 於 星期五 十月 18, 2013 11:18 amsbs 寫到:證明11,111,1111,......均非完全平方數
For any odd number x,
x = 2n + 1,
x^2 = 4n^2 + 4n + 1 = 4n(n+1) + 1
Now for number s_m
s_m = 1 + 10 + 10^2 + ... + 10^m = 11 + 100*(1 + 10 + ... + 10^(m-2)) = 3 + 4k
where k = 2 + 25*(1+10 + ... + 10^(m-2))
Since mod(s_m,4) =3, where mod(x,y) means the reminder of x/y, one can conclude that s_m is not the square of a number.