Question

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

Answer 1

Any positive integer x can be written as 3q or 3q + 1 or 3q + 2.

Case I When x = 3q: In this case, we obtain

x² = (3q)²

= 9q²

= 3(3q²)

= 3m, where m = 3q²

Case II When x = 3q + 1: In this case, we obtain

x² = (3q + 1)²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1, where m = 3q² + 2q

Case III When x = 3q + 2: In this case, we obtain

x² = (3q + 2)²

= 9q² + 12q + 4

= (9q² + 12q + 3) + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1, where m = 3q² + 4q + 1

Hence, x² is of the form 3m or 3m + 1 but not of the form 3m + 2.