Let a be any positive integer. Let q be the quotient and r be remainder. Then a = bq + r where q and r are also positive integers and 0 ≤ r < b

Taking b = 3, we get

a = 3q + r; where 0 ≤ r < 3

When, r = 0 = ⇒ a = 3q

When, r = 1 = ⇒ a = 3q + 1

When, r = 2 = ⇒ a = 3q + 2

Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m.

⇒ Squares of 3q = (3q)^{2}

= 9q^{2} = 3(3q)^{2} = 3 m where m is some integer.

Square of 3q + 1 = (3q + 1)^{2}

= 9q^{2} + 6q + 1 = 3(3q^{2} + 2q) + 1

= 3m +1, where m is some integer

Square of 3q + 2 = (3q + 2)^{2}

= (3q + 2)^{2}

= 9q^{2} + 12q + 4

= 9q^{2} + 12q + 3 + 1

= 3(3q^{2} + 4q + 1) + 1

= 3m + 1 for some integer m.

∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.