When the square of any odd number, greater than 1, is divided by 8, it always leaves remainder :

Solution (By Examveda Team)

Let the number be N = 2x + 1
N² = (2x + 1)²
    = 4x² + 1 + 4x
    = 4x (x + 1) + 1
Clearly, 4x (x + 1) is always divisible by 8 since one of x and (x + 1) is even which when multiplied by 4 is always divisible by 8.
Hence, required remainder = 1