Question

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

Answer 1

Let a be an arbitrary positive integer.

Then, by Euclid’s division algorithm, 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

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

`\implies` a² = 6(6q² + 2qr) + r² … (i)

Where, 0 ≤ r < 6

Case I: When r = 0,

Then putting r = 0 in equation (i), we get

a² = 6(6q²)

= 6m

Where, m = 6q² is an integer.

Case II: When r = 1,

Then putting r = 1 in equation (i), we get

a² = 6(6q² + 2q) + 1

= 6m + 1

Where, m = (6q² + 2q) is an integer.

Case III: When r = 2,

Then putting r = 2 in equation (i), we get

a² = 6(6q² + 4q) + 4

= 6m + 4

Where, m = (6q² + 4q) is an integer.

Case IV: When r = 3,

Then putting r = 3 in equation (i), we get

a² = 6(6q² + 6q) + 9

= 6(6q² + 6q) + 6 + 3

`\implies` a² = 6(6q² + 6q + 1) + 3

= 6m + 3

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

Case V: When r = 4,

Then putting r = 4 in equation (i), we get

a² = 6(6q² + 8q) + 16

= 6(6q² + 8q) + 12 + 4

`\implies` a² = 6(6q² + 8q + 2) + 4

= 6m + 4

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

Case VI: When r = 5,

Then putting r = 5 in equation (i), we get

a² = 6(6q² + 10q) + 25

= 6(6q² + 10q) + 24 + 1

`\implies` a² = 6(6q² + 10q + 4) + 1

= 6m + 1
Where, m = (6q² + 10q + 4) is an integer.

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