Question

Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Answer 1

Let a be an arbitrary positive integer.

Then, by Euclid’s division algorithm, corresponding to the positive integers a and 5, there exist non-negative integers m and r such that

a = 5m + r, where 0 ≤ r < 5

`\implies` a² = (5m + r)² = 25m² + r² + 10mr ......[∵ (a + b)² = a² + 2ab + b²]

`\implies` a² = 5(5m² + 2mr) + r², where, 0 < r < 5 …(i)

Case I: When r = 0,

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

a² = 5(5m²) = 5q

Where, q = 5m² is an integer.

Case II: When r = 1,

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

a² = 5(5m² + 2m) + 1

`\implies` a² = 5q + 1

Where, q = (5m² + 2m) is an integer.

Case III: When r = 2,

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

a² = 5(5m² + 4m) + 4

= 5q + 4

where, q = (5m² + 4m) is an integer.

Case IV: When r = 3,

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

a² = 5(5m² + 6m) + 9

= 5(5m² + 6m) + 5 + 4

= 5(5m² + 6m + 1) + 4

= 5q + 4

Where, q = (5m² + 6m + 1) is an integer.

Case V: When r = 4,

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

a² = 5(5m² + 8m) + 16

= 5(5m² + 8m) + 15 + 1

`\implies` a² = 5(5m² + 8m + 3) + 1

= 5q + 1

Where, q = (5m² + 8m + 3) is an integer.

Hence, the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.