Let odd integer = 2x + 1

(2x + 1)^2 / 8

= (4x^2 + 4x + 1) / 8

= [4x(x + 1) + 1] / 8

= 4x(x + 1) / 8 + 1/8

= x(x + 1) / 2 + 1/8

Since x(x+1) is always even, being the product of 2 consecutive integers,

then x(x+1) / 2 is always an integer, so 1/8 is always remainder: that's the .125 you were getting.