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
= 9q2 = 3(3q)2 = 3 m where m is some integer.
Square of 3q + 1 = (3q + 1)2
= 9q2 + 6q + 1 = 3(3q2 + 2 q) + 1
= 3m +1, where m is some integer
Square of 3q + 2 = (3q + 2)2
= (3q + 2)2
= 9q2 + 12q + 4
= 9q2 + 12q + 3 + 1
= 3(3q2 + 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.