i suspect some of the confusion here is arising because of the term "remainder". i believe the "remainder" obtained by the OP is the remainder of (x+b)^{n}uponpolynomialdivision by 2x+a.

this is *not* the same remainder obtained byintegerdivision. for example, if x = 1, y = 3, z = 4, and n = 1 (so that a = 2, and b = 3), then the integer remainder is 0, because:

2x+a = 2+2 = 4

x+b = 1+3 = 4

and 4 divides 4.

it is in fact true that:

x+b = (1/2)(2x+a) + (b-(a/2)), but the quantity b-(a/2) is, in this case, 2, which is not 0 (nor even equivalent to 0 mod 2x+a = 4).