By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Let a be an arbitrary positive integer, then corresponding to the positive integers a and 6, there exist, non-negative integers q and r such that

a = 6q + r, where 0 ≤ r < 6

Squaring both the sides using (a + b)² = a² + 2ab + b²

⇒ a² = (6q + r)² = 36q² + r² + 12qr

⇒ a² = 6(6q² + 2qr) + r²

where,0 ≤ r < 6

Case I When r = 0 we get

⇒ a² = 6(6q²) = 6m

where, m = 6q² is an integer.

Case II when r = 1 we get

⇒ a² = 6(6q² + 2q) + 1 = 6m + 1

where, m = (6q² + 2q) is an integer.

Case III When r = 2 we get

⇒ a² = 6(6q² + 4q) + 4 = 6m + 4

where, m = (6q² + 4q) is an integer.

Case IV When r = 3 we get

⇒ a² = 6(6q² + 6q) + 9

⇒ a² = 6(6q² + 6a) + 6 + 3

⇒ a² = 6(6q² + 6q + 1) + 3 = 6m + 3

where, m = (6q + 6q + 1) is an integer.

Case V when r = 4 we get

⇒ a² = 6(6q² + 8q) + 16

⇒ a² = 6(6q² + 10q) + 24 + 1

⇒ a² = 6(6q² + 8q + 2) + 4 = 6m + 4

where, m = (6q² + 8q + 2) is an integer

Case VI When r = 5 we get

⇒ a² = 6(6q² + 10q) + 25

⇒ a² = 6(6q² + 10q) + 24 + 1

⇒ a² = 6(6q² + 10q + 4) + 1 = 6m + 1

where, m = (6q² + 10q + 1) + 1 is an integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.