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

Solution:
Let a be the positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5.
So, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.
(6q)² = 36q² = 6(6q²)
= 6m, where m is any integer.
(6q + 1)² = 36q² + 12q + 1
= 6(6q² + 2q) + 1
= 6m + 1, where m is any integer.
(6q + 2)² = 36q² + 24q + 4
= 6(6q² + 4q) + 4
= 6m + 4, where m is any integer.
(6q + 3)² = 36q² + 36q + 9
= 6(6q² + 6q + 1) + 3
= 6m + 3, where m is any integer.
(6q + 4)² = 36q² + 48q + 16
= 6(6q² + 7q + 2) + 4
= 6m + 4, where m is any integer.
(6q + 5)² = 36q² + 60q + 25
= 6(6q² + 10q + 4) + 1
= 6m + 1, where m is any integer.
Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m.

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.

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