Question

Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Answer 1

Let a be an arbitrary positive integer.

Then by Euclid’s division algorithm, corresponding to the positive integers a and 4, there exists non-negative integers m and r, such that

a = 4m + r, where 0 ≤ r < 4

`\implies` a² = 16m² + r² + 8mr

Where, 0 ≤ r < 4  ......(i)  [∵ (a + b)² = a² + 2ab + b²]

Case I: When r = 0,

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

a² = 16m² 

= 4(4m²)

= 4q

Where, q = 4m² is an integer.

Case II: When r = 1,

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

a² = 16m² + 1 + 8m

= 4(4m² + 2 in) + 1

= 4q + 1

Where, q = (4m² + 2m) is an integer.

Case III: When r = 2,

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

a² = 16m² + 4 + 16m

= 4(4m² + 4m + 1)

= 4q

Where, q = (4m² + 4m + 1) is an integer.

Case IV: When r = 3,

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

a² = 16m² + 9 + 24m

= 16m² + 24m + 8 + 1

= 4(4m² + 6m + 2) + 1

= 4q + 1

Where, q = (4m² + 6m + 2) is an integer.

Hence, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.