Question

Show that the square of any positive integer is of the form 3p or 3p+1

Answer 1

For any positive integer a, there exists positive integer b and r such that

a = 3b + r where 0<r<3 i.e. r = 0,1,2

now

a^2 = (3b+r)^2

= (3b)^2 + 2*(3b)*r + r^2

= 9b^2 + 6br + r^2

= 3(3b^2 + 2br) + r^2

= 3 p + r^2 , where p is an integer

case 1, r=0

a^2 = 3p

case 2: r= 1

a^2 = 3p + 1

case 3: r = 2

a^2 = 3p + 2^2

= 3p + 4

= 3p +. 3 + 1

= 3(p+1) + 1

thus any square number of positive integer can be expressed in the form of either 3p or 3p+1