Question

prove that the square of any positive integers is of the form of 4q, 4q+1 for some integer q

Answer 1

Step-by-step explanation:

a=bq+r

where b=4

r=0,1,2,3,

a=4q+0

a=4q+1

a=4q+2

a=4q+3

4q+0

(4q+0)^2

(a+b)^2=a^2+2ab=b^2

(4q+0)^2=4q^2+2(4q)(0)+0^2

=16q^2+0

taking 4 common

4(4q^2+0)

let( 4q+0)=q

=4q

4q +1

(4q+1)^2

applying the formula of (a+b)^2

(4q+1)^2=4q^2+2(4q)(1)+1^2

=16q^2+8q+1

taking 4 common

4(4q^2+2q)+1

let(4q^2+2q)=q

=4q+1

4q +2

(4q+2)^2

applying the formula of (a+b)^2

(4q+2)^2=4q^2+2(4q)(2)+2^2

=16q^2+16q+4

taking 4 common

4(4q^2+4q+1)

let(4q^2+4q+1)=q

=4q

4q +3

(4q+3)^2

applying the formula of (a+b)^2

(4q+3)^2=4q^2+2(4q)(3)+3^2

=16q^2+24q+9

taking 4 common

4(4q^2+6q+2)+1

let(4q^2+6q+2)=q

=4q+1

By this we can conclude that the square of any positive integers is of the form of 4q, 4q+1 for some integer q

hope u r satisfied....