Let the positive integer = a

According to Euclid’s division algorithm,

a = 6q + r, where 0 ≤ r < 6

a² = (6q + r)² = 36q² + r² + 12qr [∵(a+b)² = a² + 2ab + b²]

a² = 6(6q² + 2qr) + r²   …(i), where,0 ≤ r < 6

When r = 0, substituting r = 0 in Eq.(i), we get

a² = 6 (6q²) = 6m, where, m = 6q² is an integer.

When r = 1, substituting r = 1 in Eq.(i), we get

a² + 6 (6q² + 2q) + 1 = 6m + 1, where, m = (6q² + 2q) is an integer.

When r = 2, substituting r = 2 in Eq(i), we get

a² = 6(6q² + 4q) + 4 = 6m + 4, where, m = (6q² + 4q) is an integer.

When r = 3, substituting r = 3 in Eq.(i), we get

a² = 6(6q² + 6q) + 9 = 6(6q² + 6a) + 6 + 3

a² = 6(6q² + 6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer.

When r = 4, substituting r = 4 in Eq.(i) we get

a² = 6(6q² + 8q) + 16

= 6(6q² + 8q) + 12 + 4

⇒ a² = 6(6q² + 8q + 2) + 4 = 6m + 4, where, m = (6q² + 8q + 2) is integer.

When r = 5, substituting r = 5 in Eq.(i), we get

a² = 6 (6q² + 10q) + 25 = 6(6q² + 10q) + 24 + 1

a² = 6(6q² + 10q + 4) + 1 = 6m + 1, where, m = (6q² + 10q + 1) is integer.

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

Hence Proved