False

By Euclid’s lemma, b = a × q + r, 0 ≤ r ≤ a Here, b is any positive integer.

According to the question, a = 3 and b = 3q + r for 0 ≤ r ≤ 2

So any positive integer is of the form 3k, 3k + 1 or 3k + 2.

Squaring each of these terms,

(3k)² = 9k² = 3m [where, m = 3k²]

and (3k + 1)² = 9k² + 6k + 1 [Using (a + b)² = a² + 2ab + b²]

⇒ (3k + 1)² = 3(3k² + 2k) + 1 = 3m + 1 [where, m = 3k² + 2 k]

Also (3k + 2)² = 9k² + 12 k + 4 [Using (a + b)² = a² + 2ab + b²]

⇒ (3k + 1)² = 9k² + 12k + 3 + 1

⇒ (3k + 1)² = 3(3k² + 4k + 1) + 1

⇒ (3k + 1)² = 3m + 1 [where m = 3k² + 4k + 1]

Hence, the square of any positive integer cannot be of the form 3m + 2, where m is natural number.