Let a be any positive integer
Let b = 4
By dividing algorithm rule for a and b : a = 4q + r, for q>0,and 0 ≤ r <
The possible remainders are 0, 1, and 3
So, a = 4q , 4q+1, 4q+2, 4q+3, where qis the
However, since a is
a cannot be 4q = 2(2q) or 4q+2 = 2(2q+1), since they were divisible by
Therefore, any odd integer is of the form 4q + or 4q +