Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.
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.