Question

show that the square of any positive integers is of the form 3m or 3m+1,for sime integer m,

Answer 1

let a be positive integer when divided by 3 gives quotient q and remainder r.
then a=3q+r. (by Euclid's lemma)
where, 0=<r<3 so,r =0,1,2
when r=0
a = 3q+0
a^2 = (3q)^2. (squaring both side)
=9q^2
= 3(3q^2)
=3m where m=3q^2

when r=1
a =3q +1
a^2= (3q+1)^2. (squaring both side)
= 9q^2 + 6q +1
=3(3q^2 + 2q) +1
=3m + 1 where m =(3q^2 +2q)

when r = 2
a= 3q+2
a^2= (3q+2)^2. (squaring both side)
= 9q^2 + 12q+4
= (9q^2 + 12q + 3)+1
= 3(3q^2 + 4q + 1) +1
=3m + 1 where m=(3q^2+4q+1)

Hence square of every positive integer is in form 3m,3m+1 .
-H.P

Answer 2

$\textbf{Answer is in Attachment !}$