Question

show that every positive even integer is of the form 4q or 4q+2​

Answer 1

Let a is a positive even integer. We apply the division algorithm with a and b=4

Since 0≤r<b, the possible remainders are 0,1,2 and 3

That is, a can be 4q ,or 4q + 1, or 4q + 2,or 4q + 3 where q is the quotient .However, since a is even, a cannot be 4q + 1 or 4q + 3. But 4q and 4q+2they are both divisible by 2 . Therefore any even integer is of the form 4q or 4q+ 2

Answer 2

let a be any positive integer

and 3m , 3m+1

case - 1

a=3

a^2 =(4m)^2

a^2=16m^2

a^2=4m(4m). (where q=m(4m))

a^2=4q

case -2

a=4m+1

a^2=(4m+1)^2

a^2=(4m^2)+(1^2)+2(4m+1)

a^2= 16m^2+1+8m

a^2=16m^2+8m+1

a^2=4m(4m+2)+1. (where q=m(4m+2))

a^2= 4q+1