Question

What is the solution of this statement

Answer 1

let a be any positive odd integer,here b=4 there exist unique integer q and r
such that a= 4q+r , 0 <r <4
possible remainders are 0,1,2,3.
case1. if r=0
a=4q+r
a=4q+0 (a is not equal to 4q)
it is an even integer.
case2. if r= 1
a=4q+r
a=4q+1 (it is an odd integer)
case3. if r=2
a=4q+r
a=4q+2 (it is an even integer)
case4. if r=3
a=4q+r
a=4q+3 (it is an odd integer)
from the above cases we conclude that every positive odd integer is of the form 4q+1, 4q+3, where q is some integer.