Question

Square of any positive integer is of 3rd or 3k+1 for some positive integer k

Answer 1

Let the positive integer be a divided by 3 gives r as remainder nd q as quotient
a=bq+r
r=0, 1, 2
a=3q
a=3q+1
a=3q+2
case... 1
(a)^2=(3q)^2
a^2=9q^2
=3(3q^2)
=3k(where k=3q^2)
case... 2
a^2=(3q+1)^2
a^2=9q^2+6q+1
=3(3q^2+2q)+1
=3k+1(where k=3q^2+2q)
case.... 3
a^2=(3q+2)^2
=9q^2+12q+4
=9q^2+12q+3+1
=3(3q^2+4q+1) +1
=3k+1(where k=3q^2+4q+1)
Since, in all three cases either we get a^2=3k
or
a^2=3k+1
hence, the square of any positive integer is in the form 3k or 3k+1